LECTURE NOTES ON
APPLIED THERMODYNAMICS BY Dr. T.R.Seethram (Mech.Engg. Department, PESIT, Bangalore)
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CHAPTER 3 GAS POWER CYCLES 3.1. Introduction:- Two important applications of thermodynamics are power generation and refrigeration. Both are usually accomplished by systems that operate on thermodynamic cycles. Hence thermodynamic cycles are usually divided into two general categories, viz., “power cycles” and “refrigeration cycles”. Power or refrigeration cycles are further classified as “gas cycles” and “vapour cycles”. In the case of gas cycles, the working substance will be in gaseous phase throughout the cycle, where as in vapour cycles, the working substance will be in liquid phase in one part of the cyclic process and will be in vapour phase in some other part of the cycle. Thermodynamic cycles are also classified as “closed cycles” and “open cycles”. In closed cycles, the working fluid is returned to its original state at the end of each cycle of operation and is recirculated. In an open cycle, the working substance is renewed at the end of each cycle instead of being re-circulated. In automobile engines, the combustion gases are exhausted and replaced by fresh air-fuel mixture at the end of each cycle. Though the engine operates in a mechanical cycle, the working substance does not go through a complete thermodynamic cycle. 3.2. Basic Considerations in the Analysis of Power Cycles:- The cycles encountered in actual devices are difficult to analyse because of the presence of friction, and the absence of sufficient time for establishment of equilibrium conditions during the cycle. In order to make an analytical study of a cycle feasible, we have to make some idealizations by neglecting internal irreversibilities and complexities. Such cycles resemble the actual cycles closely but are made up totally of internal reversible processes. These cycles are called ideal cycles. 3.3. Carnot Power Cycle:- The T-s and p-v diagrams for a Carnot power cycle are shown in Fig.3.1. The cycle consists of two reversible adiabatic and two reversible isothermal processes, The working of the cycle is as follows: Process 1-2:Reversible isothermal heating of the working substance from state1to state 2. Process 2-3 :- Isentropic expansion of the working substance from state 2 to state 3. During this process work is done by the working substance on the surroundings. Process 3-4:- Reversible isothermal cooling of the working substance from state 3 to state 4. Process 4-1:- Isentropic compression of the working substance so that it comes back to its initial state. During this process work is done on the working substance by the surroundings.
2
p
T
1 Th
Tc
1
2
2
4
4
3
3 s
v
Fig. 3.1: T-s and p-v diagrams for a Carnot power cycle
Expression for Thermal Efficiency of the Cycle Net work output from the cycle = Wn = ∫ dW . 1-2-3-4-1
∫dW
By first law for a cyclic process we have
1-2-3-4-1 2
2
3
4
1
= ∫ dQ = ∫dQ + ∫dQ + ∫dQ + ∫dQ 1-2-3-4-1
1
2
3
4
4
Or Wn = ∫TdS + 0 + ∫TdS + 0 = Th [S2 – S1] + Tc [S4 – S3] 1
3
Since S4 = S1 and S3 = S2, we have Wn = (S2 – S1)[Th – Tc] ………………………(3.1) Assuming that the working substance behaves as a perfect gas and since process 1 – 2 is isothermal we have S2 – S1 = m R ln(p1 / p2). Substituting this expression in Eq. (3.1) we have Wn = m R ln(p1 / p2) [Th – Tc] ……………….(3.2)
3
2
External heat supplied per cycle = Qs = Q1 – 2 = ∫TdS = Th[S2 – S1] 1
(S2 – S1)[Th – Tc] Thermal Efficiency = ηCarnot = Wn / Qs = --------------------------------Th[S2 – S1] Or
ηCarnot = [Th – Tc] / Tc = 1 – Th / Tc .....................................(3.3)
Carnot cycle can be executed in a closed system (a piston and cylinder device or in a steady flow device. It can be seen from Eq. (3.3) that the thermal efficiency depends only on two temperatures Th and Tc and is independent of working substance. The Carnot cycle is the most efficient cycle that can be executed between a heat source at temperature Th and a heat sink at temperature Tc. But reversible isothermal heat transfer process is difficult to achieve in practice, because, it would require very large heat exchangers and it would take a very long time (a power cycle in a typical engine has to be completed in a fraction of a second). Therefore it is not practical to build an engine that would operate on a cycle that closely approximates a Carnot cycle. The real value of the Carnot cycle comes from the fact that it is used as a standard against which the actual or other ideal power cycles are compared. It can be seen from Eq. (3.3) that the thermal efficiency of the Carnot power cycle increases with increase in Th and with decrease in Tc. Hence in actual or other ideal cycles attempts are made in increasing the average temperature at which heat is supplied or by decreasing the average temperature at which heat is rejected. It should also be noted that the source and sink temperatures that can be used in practice have their limitations. The highest temperature in the cycle is limited by the maximum temperature the components of the engine can withstand and the lowest temperature is limited by the temperature of the cooling medium used in the cycle such as the atmospheric air, ocean, lake or a river. 3.4. Illustrative examples on Carnot cycle Example 3.1:- A Carnot cycle using air as the working substance works between temperature limits of 900 K and 300 K. The pressure limits are 60 bar and 1 bar. Determine (i) pressure at salient points of the cycle, (ii) the heat supplied per unit mass of air, (iii) net work output per unit mass of air, (iv)mean effective pressure and (v) thermal efficiency of the cycle Given:- Refer to T – s diagram shown in Fig. E3.1. Tmax = T1 = T2 = 900 K ; Tmin = T3 = T4 = 300 K ; pmax = p1 = 60 bar ; pmin = p3 = 1 bar. For air the following property values are assumed: Cp = 1.005 kJ/kg – K; γ = 1.4 ;
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Hence Cv = Cp / γ = 1.005 / 1.4 = 0.718 kJ/kg-K ; R = 1.005 – 0.718 = 0.287.kJ/(kg-K) To find:- (i) p2 ; p4 ; (ii) q 1-2 ; (iii) wn (iv) mep ;(v) η Carnot.
Solution:T q 1-2 T1
T3 q 3-4 s Fig. E3.1: T – s diagram for example 3.1
(i) Process 2 – 3 is isentropic. For a perfect gas undergoing isentropic process we have p2 / p3 = (T2 / T3)γ / (γ – 1) Or
p2 = p3 (T2 / T3)γ / (γ – 1) = 1 x [ 900 / 300 ] 1.4 / 0.4 = 46.77 bar.
Similarly for process 4 – 1 we have p4 = p1 (T4 / T1)γ / (γ – 1) = 60 x [ 300 / 900 ] 1.4 / 0.4 = 1.283 bar (ii) Applying first law for process 1 – 2 we have q 1-2 − w1-2 = (u2 – u1) = Cv(T2 – T1) = 0 because T2 = T1.
5
2
Hence
q1-2 = w1-2 = ∫pdv = p1v1 ln(p1/p2), as process 1-2 is isothermal. 1
For a perfect gas p1v1 = RT1. Hence q1-2 = w1-2 = RT1 ln(p1/p2) = 0.287 x 900 x ln (60 / 46.77) = 64.34 kJ/kg. (iii) For process 3 -4 we have q3-4 = w3-4 = RT3 ln (p3 / p4) = 0.287 x 300 x ln (1 / 1.283) = − 21.46 kJ/kg. ( negative sign indicates that during this process heat is rejected by air to the surroundings). Net work output = wn = ∑w = ∑q = q1-2 + q3-4 = 64.34 – 21.46 = 42.88 kJ/kg (iv) mean effective pressure is given by Net work output wn mep = ------------------------------------- = ---------Stroke volume of the piston (v2 – v1) Now (v2 – v1) = [(RT2 / p2) – (RT1/p1)] = RT1 [ 1 / p2 – 1/p1] = 287 x 900 x [ (1/ 46.77) – (1 / 60)] x (1 / 10 5) = 0.0122 m3/kg.
Hence
42.88 x 1000 mep = ----------------- = 35.15 x 10 5 N / m2 = 35.15 bar 0.0122
Example 3.2:- The maximum pressure and temperature in a Carnot gas power cycle are limited to 20 bar and 400 C. The volumetric ratio of isentropic compression is 6 and volumetric ratio of isothermal expansion is 1.5. Assuming that air is the working substance and the volume of air at the beginning of isothermal expansion is 0.1 m3, determine (i) the minimum temperature in the cycle, (ii) change in entropy during isothermal expansion process, (iii) thermal efficiency of the cycle, (iv) power output from the cycle if there are 200 cycles per minute and (v) mean effective pressure. Solution: The T-s and p-v diagram for the cycle are shown in figure below.
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Given:- p1 = 20 bar ; T1 = 400 + 273 = 673 K = T2 ; V4 / V1 = 6 ; V2 / V1 = 1.5 ; V1 = 0.1 m3; For air Cp = 1.005 kJ/(kg-K); γ = 1.4 ; R = 0.287 kJ/(kg-K); Cv = 0.718 kJ/(kg-K). .
To find :- (i) T3 ; (ii) S2 – S1 ; (iii) η Carnot ; (iv) Wn ; (v) MEP. p
1
T
Isothermal Process 2 Isentropic process
1
2
T1
4
T3 4
3
3
V
s (i) Since process 1 – 2 is isothermal and air is assumed to behave as a perfect gas, it follows that p1V1 = p2V2 .Hence p2 = (V1 / V2) p1 = (1/1.5) x 20 = 13.33 bar. Process 4-1 is isentropic. Hence T4= (V1 / V4) (γ – 1) T1 = (1/6) 0.4 x 673 = 328.7 K. Also T3 = T4 = 328.7 K. (ii) mass of air = m = (p1V1) / (RT1) 20 x 10 5 x 0.1 = ------------------- = 1.035 kg 287 x 673 Change in entropy for process 1-2 for a perfect gas is given by S2 – S1 = m [ Cv ln (T2 / T1) + R ln (V2 / V1) ] = 1.035 x [ 0 + 0.287 x ln (1.5) ] = 0.1204 kJ/kg-K. (iii) η Carnot = (T1 – T3) / T1 = [673 – 328.7] / 673 = 0.5116 = 51.16 %.
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(iv) Heat supplied per cycle = Qs = Q1-2 = W1-2 = mRT1 ln(V2 / V1) = 1.035 x 0.287 x 673 x ln (1.5) = 81.06 kJ/cycle. Net work output per cycle = Wn = η Carnot Qs = 0.5116 x 81.06 = 41.47 kJ/cycle. Power output = P = Wc Nc, where Nc = Number of cycles executed per unit time. Hence
P = 41.47 x 200 /60 = 138 kJ/s (kW).
Net work output per cycle in J/cycle Wn (v) MEP = -------------------------------------------- = --------------------Swept volume per cycle in m3 / cycle (V3 – V1) Now for process 2-3 we have V3 = V2 (T2 / T3) 1 / (γ – 1) = 1.5 x 0.1 x (673 / 328.7) 2.5 = 0.9 m3. 41.47 x 1000 Hence MEP = ---------------- = 51837.5 N / m2 (0.9 – 0.1) Example 3.3:- In an air-standard Carnot cycle, 110 kJ/kg of heat is transferred to the working fluid at 1110 K. Heat is rejected at 273 k. The minimum pressure in the cycle is 1 bar. Find (i) thermal efficiency, (ii) mean effective pressure. Solution: The T – s diagram for the cycle is shown in Fig. E3.3. Given :- T1 = T2 = 1110 K ; T3 = T4 = 273 K; q 1-2 = 110 kJ/kg ; p3 = 1 bar. To find:- (i) η Carnot ; (ii) MEP (T1 – T3) (1110 – 273) (i)η Carnot = ------------- = ----------------- = 0.754 = 75.4 %. T1 1110 (ii) MEP = wn / (v3 – v1)
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wn = η Carnot q1-2 = 0.754 x 110 = 82.94 kJ/kg Applying I law for process 1 – 2 we have q 1-2 = w1-2 = RT1 ln (v2 / v1) Hence
v2 / v1 = exp [q1-2 / RT1] = exp [ 110 / (0.287 x 1110)] = 1.4124.
Also
v3 = RT3 / p3 = 287 x 273 / (1 x 10 5) = 0.7835 m3 / kg.
Process 2-3 is isentropic. Hence T2 v2 (γ – 1) = T3 v3 (γ – 1) Or
v2 = (T3 / T2) 1 / (γ – 1) v3 = (273 / 1110) 2.5 x 0.7835 = 0.0235 m3 / kg
Therefore
Therefore
v1 = 0.0235 / 1.4124 = 0.0166 m3 / kg. 82.94 x 1000 MEP = ----------------------(0.7835 – 0.0166) = 1.082 x 10 5 N/m2 = 1.082 bar.
3.5 Air Standard Cycles: In gas power cycles, the working fluid will be in gaseous phase throughout the cycle. Petrol engines (gasoline engines), diesel engines and gas turbines are familiar examples of devices that operate on gas cycles. All these devices are called “Internal combustion engines” as the fuel is burnt within the boundaries of the system. Because of the combustion of the fuel, the composition of the working fluid changes from a mixture of air and fuel to products of combustion during the course of the cycle. However, considering that air is predominantly nitrogen which hardly undergoes any chemical reaction during combustion, the working fluid closely resembles air at all times. The actual gas power cycles are complex. Hence actual gas cycles are approximated by ideal cycles by making the following assumptions called “air standard assumptions”. Air standard assumptions:- (i) The working fluid is air which continuously circulates in a closed loop. (ii). Air behaves as a perfect gas.
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(iii) All the processes that make up the cycle are internally reversible. (iv) The combustion process is replaced by a heat addition process from an external source. (v) The exhaust process is replaced by a heat rejection process that restores the working substance to its original state. (vi) Changes in kinetic and potential energies of the working substance is very small and hence negligible. A cycle which is analyzed making use of these assumptions is called an “air standard cycle”. The air standard assumptions make the thermodynamic analysis very simple without significantly deviating from the actual cycle. This simplified model will help to study qualitatively the influence of major parameters on the performance of the cycle. 3.5. Air standard Otto cycle: Otto cycle is the ideal cycle for spark ignition engines. The cycle is named after Nikolaus A Otto, a German who built a four – stroke engine in 1876 in Germany using the cycle proposed by Frenchman Beau de Rochas in 1862.The p – V and T – s diagrams for an Otto cycle are shown in Fig. 3.2.The cycle consists of the
p
T
3
2
Constant volume process
Isentropic Process
3
4
4
2
1
1 V
s
Fig. 3.2: p-V and T-s diagrams for Otto cycle. following processes. Process 1 – 2: Isentropic compression of air from state 1 to state 2. During this process work is done on air by the surroundings. Process 2 -3 : Constant volume of heating of air from state 2 till the maximum permissible temperature is reached.
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Process 3 – 4: Isentropic expansion of air from state 3 to state 4. During this process work is done by air on the surroundings. Process 4 – 1: Constant volume cooling of air till the air comes back to its original state. 3.5.1. Expressions for Net work output and thermal efficiency Net work output per unit mass of air = wn = w1-2 + w2-3 + w3-4 + w4-1 By I law for the cycle we have w1-2 + w2-3 + w3-4 + w4-1 = q1-2 + q 2-3 + q 3-4 + q 4-1. Also q 1-2 = q 3-4 = 0 as processes 1-2 and 3-4 are isentropic. Hence
wn = q 2-3 + q 4-1. ……………………………..(3.4)
Since both the processes 2-3 and 3-4 are at constant volume, applying I law for these two processes we have q2-3 = (u3 – u2) = Cv[T3 – T2] ……………………………(3.5) and
q4-1 = (u1 – u4) = Cv[T1 – T4] ……………………………(3.6)
It should be noted that q4-1 will be negative (T1 < T4) as heat is transferred from the working substance to the surroundings. Hence the net work output in terms of temperatures at the four salient points of the cycle is given by wn = Cv[T3 – T2] + Cv[T1 – T4] ………………………….(3.7) Thermal efficiency is given by Net work output wn η Otto = ---------------------- = -------Heat Supplied q2-3 Cv[T3 – T2] + Cv[T1 – T4] [T4 – T1] η Otto = ------------------------- -------- = 1 − ----------------Cv[T3 – T2] [T3 – T2] …………………………(3.8) Eq. (3.8) gives the expression for thermal efficiency of the Otto cycle in terms of the temperatures at the salient points of the cycles. It is possible to express the net work output and thermal efficiency of the Otto cycle in terms two parameters namely (i) the
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comprssion ratio, Rc and the maximum cycle temperature ratio, t. The compression ratio is defined as the ratio of volume of air befor compression to the volume of air after compression; i.e., Rc = V1 / V2 and the maximum cycle temperature ratio is defined as the ratio of the maximum temperature in the cycle to the minimum temperature in the cycle; i.e., t = T3 / T1. Now Process 1 – 2 is isentropic. Hence T1V1(γ – 1) = T2V2(γ – 1) T2 / T1 = (V1/V2) (γ – 1) = Rc(γ – 1) ……………………………………….(3.9)
Or
Similarly we have for ptocess 3-4, T3 / T4 = (V4/V3) (γ – 1) = (V1/V2) (γ – 1) = Rc(γ – 1) ……………………... (3.10) Therefore
T2 T3 [T3 – T2] ---- = ------ = ------------ = Rc(γ – 1) T1 T4 [ T4 – T1]
Suibstituting this in Eq.(3.8) we have 1 ηOtto = 1 − ------------------- ………………(3.11) Rc(γ – 1) It can be seen from Eq. (3.11) that the thermal efficiency of the Otto cycle depends only on the compression ratio Rc. The efficiency increases with increase in Rc. The increase is steep at low values of Rc , but becomes flatter as Rc exceeds 8 as shown in Fig. 3.3.
ηOtto 0.6 0.4
0.2
2
4
6
8
10
12 14 Rc Fig.3.3: Effect of Rc on thermal efficiency of Otto cycle
12
16
For very high compression ratios (Rc >10), the temperature of the air fuel mixture in an actual petrol engine will be so high as to cause pre-ignition of the fuel leading to “knocking” of the engine and hence should not be used. Therefore the compression ratio cannot be increased arbitrarily to have higher efficiency. Between the same temperature limits the thermal efficiency of the Otto cycle is less than that for a Carnot cycle.In order to know how far the Otto cycle deviates from Carnot cycle, a parameter called “relative efficiency” is defined as the ratio of the thermal efficiency of the Otto cycle to the thermal efficiency of a Carnot cycle working between the same temperature limits. i.e., Air standard efficiency of Otto cycle Relative efficiency of Otto cycle = ----------------------------------------------- ……..(3.12) Efficiency of Carnot cycle working between same temperature limits. 3.5.2. Condition for optimum work output from an Otto cycle Net work output per unit mass of air is given by Eq. (3.7) : wn = Cv[T3 – T2] + Cv[T1 – T4] ………………………….(3.7) = Cv T1 [ T3 / T1 – T2 / T1 – T4 / T1 + 1 ] T3 / T1 is the ratio of maximum temperature in the cycle to the minimum temperature in the cycle and is called “maximum cycle temperature ratio” and is denoted by ‘t’. t Now T4 / T1 = (T4 / T3) x (T3 / T1) = ---------- ……………………………………(3.13) Rc(γ – 1) Hence
wn = CvT1 [ t – Rc(γ – 1) – t / Rc(γ – 1) + 1] ………………...(3.14)
For given values of ‘t’ and T1, wn depends only on Rc. Hence for optimum output, dwn / dRc = 0. i.e.,
dwn / dR = Cv T1 [−(γ – 1) Rc(γ – 2) − t (1 – γ)Rc− γ ] = 0
or
Rc 2(γ – 1) = t
or
Rc = t 1 / 2(γ – 1) = Rc*…………………………………..(3.15)
If this value of Rc is substituted in Eq. (3.14) we get the expression for maximum work output as (wn)maximum = CvT1[ t − t ½ - t / t1/2 + 1]
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3.13 (wn)Maximum = Cv T1 [ √ t – 1 ] 2 …………………………….(3.16)
Or
Thermal efficiency corresponding to maximum work output is therefore given by (ηOtto) * = 1 − 1 / Rc* (γ – 1) = 1 – [1 / √ t ] ………………(3.17) 3.5.3. Illustrative examples on Otto cycle Example 3.4:- An ideal Otto cycle has a compression ratio of 8. The conditions at the beginning of compression stroke are 100 kPa and 17 C. If the heat added during the cycle is 800 kJ/kg find (i) temperatures and pressures at salient points of the cycle, (ii) net work output per unit mass of air, (iii)thermal efficiency of the cycle, (iv) mean effective pressure, (v) compression ratio corresponding to maximum work output, (vi) maximum work output and (vii) thermal efficiency corresponding to maximum work output. Solution: The p – V diagram for the cycle is shown below. Vs = Stroke Volume = V1 – V2.
p
Vc = ClearanceVolume =V2=V3
3
Given:- Rc = V1 / V2 = 8. T1 = 17 + 273 = 290 K.
2
p1 = 100 kPa. 4 Heat supplied = q2-3 = 800 kJ/kg.
1
Vc
V
Vs = V1 – V2
To find:- (i) T2, p2, T3,p3, T4, p4 (ii) wn; (iii) ηOtto; (iv) MEP;
(v) Rc* ; (vi) (wn) maximum (i) Or
Process 1-2 is isentropic. Hence T1V1(γ – 1) = T2V2(γ – 1) T2 = T1 (V1/V2) (γ – 1) = 290 x 8 0.4 = 666 K.
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3.14 Also Or
p1V1 / T1 = p2V2 / T2 . p2 = p1 (V1/V2) (T2 / T1) = 100 x 8 x (666 / 290) = 1837.24 kPa = 18.3724 bar.
Now q2-3 = Cv(T3 – T2) or T3 = T2 + q2-3 / Cv Hence Also Hence
T3 = 666 + 800 / 0.718 = 1780.2 K. p3V3 / T3 = p2V2 / T2 and V3 = V2. p3 = p2 (T3 / T2) = 18.3724 x (1780.2 / 666) = 49.11 bar.
Process 3-4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1) = T3 (V2 / V1) (γ – 1) = 1780.2 x (1/8) 0.4 = 775 K. Also Hence
p4V4 / T4 = p1V1 / T1 and V4 = V1. p4 = p1 (T4 / T1) = 100 x (775 / 290) = 267.24 kPa.
(ii)Heat rejected per unit mass of air = q4-1 = Cv (T1 – T4) = 0.718 x (290 – 775) = − 348.23 kJ/kg (Negative sign for q4-1 indicates that heat is transferred from air to the surroundings during this process). Hence Net work output per unit mass of air = wn = 800 – 348.23 = 451.77 kJ/kg. (iii)Thermal efficiency = η Otto = wn / q2-3 = 451.77 / 800 = 0.565 = 56.5 %. Check for thermal efficiency: η Otto = 1 – 1 / Rc(γ – 1) = 1 – 1 / 8 0.4 = 0.565 = 56.5 % (iv) Volume of air per unit mass at the beginning of compression is given by
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v1 = RT1 / p1 = 287 x 290 / (100 x 1000) = 0.8323 m3/kg. Net work output in J/kg wn 451.77 x 1000 MEP = --------------------------------------- = ---------------- = ---------------------------Stroke volume in m3 / kg (v1 – v2) [0.8323 – 0.8323 / 8] = 6.203 x 105 N/m2 = 6.203 bar. (v) Maximum cycle temperature ratio = t = Tmaximu / Tminimum = T3 / T1 = 1780.2 / 290 = 6.138. For maximum work output, Rc = Rc* = t 1/ 2 (γ – 1) = 6.138 1 / (2 x 0.4) = 9.66 (vi) Net work output corresponding to Rc * is maximum and is given by (wn)maximum = CvT1 [√ t − 1] 2 = 0.718 x 290 x [√ 6.138 – 1] 2 = 454.54 kJ/kg. Example 3.5:- An air standard Otto cycle is to be designed according to the following specifications. Pressure at the start of the compression process = 101 kPa ; Temperature at the start of compression process = 300 K; Compression ratio = 8; Maximum pressure in the cycle = 8.0 MPa; Find (i) the net work output per unit mass of air, (ii) cycle efficiency, and (iii)MEP. Solution: Refer to p-V diagram of the cycle shown in example 3.4. Given:- p1 = 101 kPa ; T1 = 300 K; V1 / V2 = Rc = 8 ; p3 = 8 MPa ; Assume Cv = 0.718 kJ/kg – K ; γ = 1.4. To find:- (i) wn ; (ii) η Otto ; (iii) MEP (i)Process 1-2 is isentropic. Hence T2 = T1 (V1 / V2) (γ – 1) = 300 x (8) 0.4 = 689.2 K. Also p2 = p1 (V1/ V2) γ = 101 x 8 1.4 = 1856.3 kPa.
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For process 2-3 we have p2V2 / T2 = p3V3 / T3 and V2 = V3. Hence
T3 = T2 (p3 / p2) = 689.2 x (8 x 103 / 1856.3) = 2970 K.
Process 3 -4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1) = T3 (V2 / V1) (γ – 1) = 2970 x (1 / 8) 0.4 = 1292.9 K. Heat supplied per unit mass of air = q 2-3 = Cv(T3 – T2) = 0.718 x (2970 – 689.2) = 1637.6 kJ/kg. Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (300 – 1292.9) = − 712.9 kJ/kg. (Negative sign for q 4-1 indicates that heat is transferred from air to the surroundings) Net work output per unit mass of air = wn = ∑q = 1637.6 – 712.9 = 924.7 kJ/kg. 1 1 (ii) Thermal efficiency = η Otto = 1 − ------------- = 1 − -------------Rc (γ – 1) 8 0.4 = 0.565 = 56.5 %. Thermal efficiency can also be calculated from the formula wn 924.7 η Otto = ------------- = -------------- = 0.565 = 56.5 % q2-3 1637.6 (ii)Specific volume at state 1 = v1 = RT1 / p1 = 287 x 300 / 101 x 10 3 = 0.8525 m3 / kg. Hence
v2 = v1 / 8 = 0.8525 / 8 = 0.10656 m3/kg. wn 924.7 x 1000 MEP = -------------- = ---------------------- = 12.396 x 10 5 N/m2 = 12.396 bar. (v1 – v2) (0.8525 – 0.10656)
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Example 3.6:- From the p – V diagram of an engine working on Otto cycle, it is found that the pressure in the cylinder after 1/8th of the compression stroke is completed is 1.4 bar. After 5/8th of the compression stroke is completed, the pressure is found to be3.5 bar. The maximum cycle temperature is limited to 1000 C.If the compression process is according to the law pV1.35= constant, find (i) the compression ratio, (ii) work output per unit mass of air, and (iii) thermal efficiency. Assume the minimum temperature in the cycle to be 27 C. Solution: Refer to p – V diagram shown below. p
3 Given:V1 – Va = (1/8)Vs pVn = Const
2 b
V1 – Vb = (5/8)Vs. pa = 1.4 bar; pb = 3.5 bar;
4 a
Vb
Va
T3 = 1000 + 273 =1273 K.
1 V1
V
T1 = 27 + 273 = 300 K Compression index = n = 1.35
Fig. E3.6: Figure for example 3.6. To find:- (i) Rc ; (ii) wn ; (iii) η (i) V1 – Va = (1/8) (V1 – V2) or Va = V1 – (1/8)(V1 – V2) = (7/8)V1 + (1/8)V2 Hence
Va / V2 = (7/8)Rc + 1/8 ………………………..(a)
Similarly
Vb / V2 = (3/8)Rc + 5/8 ………………………..(b)
From Eqs. (a) and (b) we have Va (7/8)Rc + 1/8 7Rc + 1 --- = ----------------- = ----------- ……………….(c) Vb (3/8)Rc + 5/8 3Rc + 5
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For the compression process we have paVan = pbVbn Va / Vb = (pb/pa) 1/n = (3.5 / 1.4) 1/1..35
Or
= 1.971 Substituting this value in Eq. (c) we have 7Rc + 1 --------- = 1.971 3Rc + 5 1.971 x 5 – 1 Rc = ------------------- = 8.146. (7 – 1.971 x 3)
Or
(ii)
For process 1-2 we have T2 = T1 Rc(n – 1) = 300 x 8.146 0.35 = 625 K. q2-3 = Cv(T3 – T2) = 0.718 x (1273 – 625) = 465.26 kJ/kg. For process 3-4 we have T4 = T3 (V3 / V4) (γ – 1) = 1273 x (1 / 8.146) 0.4 = 550.1 K. q4-1 = Cv(T1 – T4) = 0.718 x (300 – 550.1)
= − 179.6 kJ/kg. (Negative sign for q 4-1 indicates that during the process 4-1 heat is transferred from air to the surroundings) Since process 1-2 is not isentropic q1-2 is not equal to zero and therefore we have to find q1-2. Applying first law for process 1-2, we have, per unit mass of air q1-2 – w1-2 = (u2 – u1) …………………………(d) Now And
w1-2 = (p1v1 – p2v2) / (n – 1) = R(T1 – T2) /(n – 1) u2 – u1 = Cv(T2 – T1).Substituting these expressions in Eq.(d) we have
19
R(T1 – T2) q1-2 = ------------- + Cv(T2 – T1) = (T2 – T1) [Cv – R /(n – 1)] (n – 1) Hence
q1-2 = (625 – 300) x [ 0.718 – 0.287 / 0.35] = − 33.15 kJ/kg wn = ∑w = ∑q = q1-2 + q2-3 + q3-4 + q4-1 = − 33.15 + 465.26 + 0 − 179.6 = 252.5 kJ/kg.
Wn 252.5 Thermal efficiency = η = ----------------- = -------------- = 0.5427 = 54.27 %. q2-3 465.26 Example 3.7:- Derive an expression for thermal efficiency in terms of compression ratio and maximum cycle temperature ratio for a cycle which is similar to Otto cycle except that the compression process is isothermal.Compare the efficiency and work output of this cycle with that of an Otto cycle having the same compression ratio of 6 and same maximum cycle temperature ratio of 5.. Solution: The p – V diagram for the cycle is shown in Fig. E3.7. p 3
Isentropic process 2
Isothemal process 4
1 V Fig. E3.7: p – V diagram for example 3.7.
20
Solution: Net work output per unit mass = wn = w1-2 + w2-3 + w3-4 + w4-1. But w2-3 = w4-1 = 0 as both the processes 2-3 and 4-1 are constant volume processes. Since process 1-2 is isothermal ( for a perfect gas this is equivalent to process according to the law pv= constant) w1-2 = p1v1 ln (v2/v1) = − RT1 ln Rc where Rc = v1/v2. (p3 v3 – p4v4) R(T3 – T4) Since process 3-4 is isentropic, w3-4 = ----------------- = ---------------------(γ – 1) (γ – 1) RT1 (T3 / T1 – T4 / T1) = -----------------------------(γ – 1) Now T3 / T1 = t = maximum cycle temperature ratio, And T4 / T1 = (T4/T3) x (T3/T1) = (V3/V4)(γ – 1) t = t / Rc(γ – 1) Hence
RT1 [ t – t / Rc(γ – 1) ] w3-4 = --------------------------(γ – 1)
RT1 [ t – t / Rc(γ – 1) ] Therefore wn = − RT1 ln Rc + -----------------------------(γ – 1) ={ RT1 / (γ – 1)}{ [ t – t / Rc(γ – 1) ] − (γ – 1) ln Rc}……………..(a) Heat supplied = q2-3 = Cv(T3 – T2) = Cv(T3 – T1) = CvT1(T3/T1 – 1) = CvT1(t – 1) Thermal efficiency = η = wn / q2-3 { RT1 / (γ – 1)}{ [ t – t / Rc(γ – 1) ] − (γ – 1) ln Rc} = -----------------------------------------------------------CvT1(t – 1)
21
Since Cv = R / (γ – 1), the above expression simplifies to {t [ 1 – 1 / Rc(γ – 1) ] − (γ – 1) ln Rc} η = ----------------------------------------------------------( t – 1) (ii) Given:- Rc = 6 ; t = 5 ; To find (a) wn / (wn)Otto ; and (b) η / ηOtto (a)Work output for an Otto cycle in terms of T1, Rc and t is given by Eq.(3.14) as (wn)Otto = CvT1 [t – Rc(γ – 1) – t / Rc(γ – 1) +1] = 0.718 T1 x [ 5 – 6 0.4 – 5 / 6 0.4 + 1] = 1.0844 T1 For the given cycle, from Eq. (a) wn ={ RT1 / (γ – 1)}{ [ t – t / Rc(γ – 1) ] − (γ – 1) ln Rc} 0.287 x T1 = --------------- x { [ 5 – 5 / 60.4] – 0.4 x ln 6 } 0.4 = 1.321 T1 Hence
(wn)Otto 1.0844 T1 --------- = --------------- = 0.821. wn 1.321 T1
(b) Thermal efficiency of Otto cycle = η otto = 1 – 1 / 6 0.4 = 0.5116 = 51.16 % For the given cycle thermal efficiency is {t [ 1 – 1 / Rc(γ – 1) ] − (γ – 1) ln Rc} η = ----------------------------------------------------------( t – 1) { 5 x [ 1 – 1/ 60.4] – 0.4 x ln 6} = ------------------------------------------ = 0.4603 = 46.03 %. (5 – 1)
22
3.6. Diesel Cycle:- The diesel cycle is the ideal cycle for compression ignition engines (CI engines). CI engine was first proposed by Rudolph Diesel in 1890. The diesel engine works on the principle of compression ignition. In such an engine, only air is compressed and at the end of the compression process, the fuel is sprayed into the engine cylinder containing high pressure air, so that the fuel ignites spontaneously and combustion occurs. Since only air is compressed during the compression stroke, the possibility of auto ignition is completely eliminated in diesel engines. Hence diesel engines can be designed to operate at much higher compression ratios (between 12 and 24). Also another benefit of not having to deal with auto ignition is that fuels used in this engine can be less refined (thus less expensive). The p – V and T – s diagrams for an air-standard diesel cycle are shown in Fig. 3.4. The diesel cycle is similar to Otto cycle except that the heating process takes place at constant pressure in a diesel cycle. The various processes involved in an ideal diesel cycle are as follows. p 2
Constant Pressure Process
T
3 Isentropic Process
3
4
4
2 1
1
Constant volume process V
s
Process 1-2: Isentropic compression of air from state 1 to state 2. During this process work is done on air by the surroundings. Process 2-3: Constant pressure heating of air till the maximum permissible temperature is reached. Process 3-4: Isentropic expansion of air from state 3 to state 4. during this process work is done by air on the surroundings. Process 4-1: Constant volume cooling of air so that air comes back to its original state to complete the cycle.
23
3.6.1.Expressions for net work output and thermal efficiency Heat supplied per unit mass of air = q2-3 = Cp (T3 – T2). Heat rejected per unit mass of air = q4-1 = Cv (T1 – T4) Net work ouput per unit mass of air = wn = ∑w = ∑q = q1-2 + q2-3 + q3-4 + q4-1. But q1-2 = q3-4 = 0 as these two processes are isentropic. Hence
w n = q2-3 + q4-1 = Cp (T3 – T2).+ Cv (T1 – T4) ………………(3.18)
Therefore thermal efficiency of the diesel cycle is given by wn Cp (T3 – T2).+ Cv (T1 – T4) η Diesel = ---------- = ---------------------------------q2-3 Cp (T3 – T2). (Cv/Cp) (T4 – T1) = 1 − ----------------------- …………………………(3.19) (T3 – T2). Temperatures T2, T3 and T4 can be expressed in terms of T1, the compression ratio Rc and the cut off ratio ρ (ρ = V3 / V2) as follows. Process 1-2 is isentropic. Hence T2 = T1(V1/V2)(γ – 1) = Rc(γ – 1) T1. For process 2-3, p2V2 / T2 = p3V3 / T3 and p2 = p3. Hence
T3 = (V3 / V2) T2 = ρ Rc(γ – 1) T1.
Proocess 3-4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1). But V4 / V3 = Expansion ratio = (V4 / V2)(V2 / V3) = (V1 / V2)(V2 / V3) = Rc / ρ.
24
T4 = ( ρ / Rc) (γ – 1) T3
Therefore
= ( ρ / Rc) (γ – 1) ρ Rc(γ – 1) T1 = ρ γ T1. Substituting the expressions for T2,T3, and T4 in Eq. (3.19) ( ρ γ− 1 ) η Diesel = 1 − -----------------------------γ [ρ Rc(γ – 1) − Rc(γ – 1) ] ( ρ γ− 1 ) η Diesel = 1 − ------------------------------ ………………(3.20) γ Rc(γ – 1) [ρ −1]
or
Now substituting the expressions for T2,T3, and T4 in terms of T1 in Eq. (3.18) we get wn = CvT1[γ (ρ Rc(γ – 1) – Rc(γ – 1)) − ρ γ + 1 ] wn = CvT1[ γ Rc(γ – 1) (ρ – 1) − ρ γ + 1] …………..(3.21)
Or ηDiesel
ρ
Rc
1 Fig. 3.5 : Effect of Rc and ρ on efficiency of diesel cycle
25
It can be seen from Eq. (3.20) that the thermal efficiency of the diesel cycle depends on the compression ratio Rc and the cut-off ratio ρ. The effect of these parameters on thermal efficiency is shown in Fig. 3.5. It can be seen from this figure that for a given value of the cut-off ratio the thermal efficiency increases with Rc and for a given value of Rc, the efficiency deceases with increase in the cut-off ratio.The increase in cut-off ratio results in decrease in the expansion ratio which in turn decreases the work done during expansion process. At the same time increase in cut-off ratio also results in increase in heat supplied. The overall effect is that the efficiency decreases. 3.6.2.Illustrative examples on diesel cycle Example 3.8:- An air standard diesel cycle has a compression ratio of 14. The air condition at the beginning of compression is 1 bar and 27 C.The maximum temperature in the cycle is 2500 C. Determine (i) temperature and pressure at salient points of the cycle, (ii) net work output per unit mass of air, (iii) thermal efficiency, (iv) specific air consumption in kg/kWh, and (v) MEP. Solution: Given:- Rc = V1/V2 = 14; p1= 1 bar;
p 2
3
T1 = 27 + 273 = 300 K ;
q2-3
T3 = 2500 + 273 = 2773 K ; Assume:γ = 1.4;Cp = 1.005 kJ/kg-K Cv = 0.718 kJ/kg-K
4 q4-1 1
To find:- (i) T2,p2, p3, T4, p4 ; (ii) wn ; (iii) η Diesel ; (iv) SAC in
V kg/kWh ; (v) MEP
(i)
Process 1-2 is isentropic. Hence T2 = T1 (V1/V2) (γ – 1) = 300 x 14 0.4 = 862.13 K.
Also
p2 = p1(V1/V2) γ = 1 x 14 1.4 = 40.23 bar
26
For process 2-3 we have p2V2 / T2 = p3V3 / T3 and p2 = p3. Hence cut – off ratio = ρ = V3 / V2 = T3 / T2 = 2773 / 862.13 = 3.216. Expansion ratio = V4 / V3 = Rc / ρ = 14 / 3.216 = 4.353. Process 3-4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1) = 2773 x (1 / 4.353) 0.4 = 1539.7 K. p3V3 γ = p4V4 γ or p4 = p3 (V3 / V4) γ
Also Hence
p4 = 40.23 x (1/ 4.353) 1.4 = 5.131 bar.
(ii)
Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2) = 1.005 x (2773 – 862.13) = 1920.4 kJ/kg
Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (300 – 1539.7) = − 890.11 kJ/kg (Negative sign for q4-1 indicates that heat is transferred from air to the surroundings during this process). Net work output per unit mass of air = wn = 1920.4 – 890.11 = 1030 kJ/kg wn 1030 Thermal efficiency = η Diesel = --------- = ---------------q2-3 1920.11
(iii)
= 0.5364 = 53.64 %. .
(iv) Let m = mass of air consumed in kg / s.
27
.
Air consumption in kg/h m x 3600 Then Specific air consumption in kg/kWh = --------------------------------- = --.----------Net work output in kW m x wn 3600 3600 = ----------- = -------------- = 3.495 kg/kWh. wn 1030 (v) Specific volume at the beginning of compression = v1 = RT1 / p1 Hence
287 x 300 v1 = --------------- = 0.861 m 3/kg 1 x 10 5
Therefore
v2 = v1 / Rc = 0.861 / 14 = 0.0615 m3 / kg.
Wn 1030 x 1000 Mean effective pressure = MEP = ---------------- = ----------------------(v1 – v2) ( 0.861 – 0.0615) = 12.883 x 10 5 N / m2 = 12.883 bar Example 3.9:- A diesel cycle has a compression ratio of 16. The temperature before compression is 300 K and after expansion it is 900 K. Determine (i) Net work input per unit mass of air, (ii) the air standard efficiency and (iii) MEP if the minimum pressure in the cycle is 1 bar. Given:- Rc = V1 / V2 = 16 ; T1 = 300 K ; T4 = 900 K ; To find :- (i) wn ; (ii) η Diesel ; (iii) MEP. Solution: Refer to p-v diagram of example 3.8. (i)Process 1-2 is isentropic. Hence T2 = T1(V1 / V2) (γ – 1) = 300 x 16 0.4 Or
T2 = 909.4 K
Process 3-4 is isentropic. Hence T3 = T4 (V4/V3) (γ – 1) = T4 [V1/V3] (γ – 1) Or
T3 = T4 [(V1/V2) (V2/V3)] (γ – 1) = T4 Rc(γ – 1) (T2 / T3) (γ – 1)
Or
T3γ = T4 [Rc T2](γ – 1)
Or
T3 = {T4 [Rc T2](γ – 1) }1 / γ = { 900 x [ 16 x 909.4] 0.4} 1/1.4 = 1993.27 K
28
Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2) = 1.005 x (1993.27 – 909.4) = 1089.3 kJ/kg. Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (900 – 300) = 430.8 kJ/kg Net work output per unit mass of air = wn = 1089.3 – 430.8 = 658.5 kJ/kg. wn 658.5 (ii) Thermal efficiency = ηDiesel = --------- = --------------- = 0.6045 q2-3 1089.3 = 60.45 % (iii)Specific volume of air at the beginning of compression = v1 = RT1 / p1 Hence
287 x 300 v1 = ------------------- = 0.861 m3 / kg. 1 x 105
Therefore
v2 = v1 / 16 = 0.861 / 16 = 0.0538 m3/kg. wn 658.5 x 1000 MEP = --------------- = --------------------(v1 – v2) (0.861 – 0.0538) = 8.158 x 10 5 N/m2 = 8.158 bar.
Example 3.10:- In an air standard diesel cycle, the pressure at the end of expansion is 240 kPa and temperature is 550 C. At the end of compression process, the pressure is 4.2 MPa and temperature is 700 C. Determine (i) the compression ratio, (ii) the cut-off ratio, (iii) heat supplied per unit mass of air, and (iv) cycle efficiency. Solution: The p – V diagram for the cycle is shown in Fig. E3.10. Given:- p4 = 240 kPa ; T4 = 550 + 273 = 823 K ; p2 = 4.2 x10 3 kPa ; T2 = 700 + 273 = 973 K. To find:- (i) Rc ; (ii) ρ ; (iii) q 2-3 ;(iv) η Diesel
29
(i)
Process 3-4 is isentropic. Hence T3 = T4 (p3 /p4) (γ – 1) / γ = T4 (p2 /p4) (γ – 1) / γ T3 = 823 x (4.2 x103 / 240) 0.286
Or
= 1866 K Cut-off ratio = ρ = V3 / V2 = T3 / T2 = 1866 / 973 = 1.92. (ii) Expansion ratio = Re = V4 / V3 = (p3 / p4) 1 / γ = (4.2 x 103 / 240) 1/ 1.4 = 7.725 Compression ratio = Rc = V1/V2 = Re ρ = 7.725 x 1.92 = 14.832 (iii) Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2) = 1.005 x (1866 – 973) = 897.5 kJ/kg. (iv) For process 1-2 we have, T1 = T2 (V2 / V1) (γ – 1) = 973 x (1/14.832) 0.4 = 331 K Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (331 – 823) = − 353.3 kJ/kg (Negative sign for q4-1 indicates that heat is transferred from air to the surroundings during this process). Net work output = wn = 897.5 – 353.3 = 544.2 kJ/kg. 544.2 Hence thermal efficiency = η Diesel = wn / q2-3 = ------------------897.5 = 0.6063 = 60.63 %. Example 3.11:- An oil engine works on diesel cycle with a compression ratio of 20. Heat addition takes place up to 10 % of the stroke. Initial pressure and temperature of air are 1 bar and 27 C.The bore and stroke of the engine are 16 cm and 20 cm respectively. The
30
compression process is according to the law pV 1.32= constant and the expansion process is according to the law pV 1.30= constant. Find (i) the pressure and temperature at salient points of the cycle, (ii) the net work output per unit mass of air, (iii) MEP, (iv) thermal efficiency of the engine, (v) relative efficiency with respect to air standard efficiency. Solution: p 2
q2-3
Given:- Rc = V1/V2 = 20 3 pV ne = Constant
V3 – V2 = 0.1(V1 – V2)
pV nc = Constant
p1 = 1 bar; T1 = 27 + 273 = 300 K;
4
nc = 1.32 ; ne = 1.30 q4-1
bore = d = 0.16 m ;
1
stroke = L = 0.20 m V To find:- (i) p2, p3,p4,T2,T3,T4 ; (ii) wn ; (iii) MEP ; (iv) η engine ; (v) η engine / η Diesel (i) Stroke volume = V1 – V2 = (π / 4) d 2 L = (π / 4) x 0.16 2 x 0.20 = 4.02 x 10 − 3 m3. Hence And
20 V2 – V2 = 4.02 x 10 − 3 Or V2 = 2.116 x 10 − 4 m3. V3 = V2 + 0.1 (V1 – V2).
Hence cut-off ratio = ρ = V3 / V2 = 1 + 0.1 (V1/V2 – 1) = 1 + 0.1 x (20 – 1) = 2.9. Expansion ratio = Re = V4 / V3 = Rc / ρ = 20 / 1.9 = 6.9. Now
T2 = T1 Rc (nc – 1) = 300 x 20 0.32 = 782.44 K T3 = T2 ρ = 782.44 x 2.9 = 2269.1 K T4 = T3 (1 / Re)(ne – 1) = 2269.1 / 6.9 0.3 = 1271.6 K. p2 = p1 Rcnc = 1 x 20 1.32 = 52.163 bar = p3.
31
p4 = p3 (1/Re) ne = 52.163 / 6.9 1.30 = 4.235 bar. (ii) Since both the compression process 1-2 and the expansion process 3-4 are not isentropic, the heat transfer during these processes will not be equal to zero.It is therefore necessary to calculate the heat transfer during these processes to determine the total external heat supplied during the cycle. Now
(p1v1 – p2v2) R(T1 – T2) 0.287 x (300 – 782.44) w1-2 = ------------------ = --------------- = ----------------------------(nc – 1) (nc – 1) (1.32 – 1) = − 431.2 kJ/kg. w2-3 = p2 (v3 – v2) = p3v3 – p2v2 = R(T3 – T2) = 0.287 x (2269.1 – 782.44) = 425.2 kJ/kg. (p3v3 – p4v4) R (T3 – T4) 0.287 x (2269.1 – 1271.6) w3-4 = ----------------- = ---------------- = --------------------------------(ne – 1) (ne – 1) (1.30 – 1) = 951.4 kJ/kg. w4-1 = 0 as process 4-1 is at constant volume.
Hence net work out put = wn = − 431.2 + 425.2 + 951.4 = 945.3 kJ/kg. (iii) specific volume at state 1 = v1 = RT1 / p1 = 287 x 300 / 1 x105 = 0.861 m3/kg. Therefore
v2 = v1 / Rc = 0.861 / 20 = 0.04305 m3/kg. wn 945.3 x 1000 MEP = --------------- = ---------------------(v1 – v2) [0.861 – 0.04305] = 11.597 x 10 5 N/m2 = 11.597 bar.
(iv)
q2-3 = Cp(T3 – T2) = 1.005 x (2269.1 – 782.440 = 1494 kJ/kg. q4-1 = Cv(T1 – T4) = 0.718 x ( 300 – 1271.6) = − 698.3 kJ/kg.
Applying I law for process 1-2 we have
32
q1-2 = w1-2 + (u2 – u1) = w1-2 + Cv(T2 – T1) = − 431.2 + 0.718 x (782.44 – 300) = − 84.32 kJ/kg Applying I law for the cycle we have q1-2 + q2-3 + q3-4 + q4-1 = wn Or
q3-4 = wn − (q1-2 + q2-3 + q4-1) = 945.3 − (− 84.32 + 1494 − 698.3) = 233.92 kJ/kg
Total heat supplied during the cycle = qs = q2-3 + q3-4 = 1494 + 233.92 = 1727.9 kJ/kg 945.3 Thermal efficiency of the engine = η Engine = wn / qs = -----------1727.9 = 0.547 = 54.7 %. (v) Air standard efficiency of the diesel cycle is given by (ργ – 1) (2.9 1.4 – 1) ηDiesel = 1 – -------------------------- = 1 − ---------------------------γ Rc(γ – 1) (ρ – 1) 1.4 x 20 0.4x (2.9 – 1) = 0.6098 = 60.98 % Relative efficiency = η Engine / ηDiesel = 0.547 / 0.6098 = 0.897 = 89.7 %. Example 3.12:- In an air standard diesel cycle, air is compressed isentropically from 26 C and 105 kPa to 3.7 MPa. The entropy change during heat rejection process is − 0.6939 kJ/kg-K. Determine (i) heat supplied per unit mass of air, (ii)thermal efficiency, (iii) maximum temperature in the cycle, and (iv) temperature at the start of the heat rejection process. Given:- T1 = 26 + 273 = 299 K ; p1 = 105 kPa ; p2 = 3.7 x 10 3 kPa ; s1 – s4 = − 0.6939 kJ/(kg-K). To find :- (i) q2-3 ; (ii) η Diesel ; (iii) T3 ; (iv) T4 The p-v and T-s diagram for the air standard diesel cycle are shown in Fig. E3.12. It follows from the T-s diagram that s2 – s3 = s1 – s4
33
Process 2-3 is at constant pressure. Therefore s3 – s2 = Cp ln (T3 / T2). Or T3 / T2 = exp {(s3 – s2) / Cp} Or
T3 / T2 = exp {0.6939 / 1.005} = 1.995
Process 1-2 is isentropic. Hence T2 = T1 (p2 / p1) (γ – 1) / γ = 299 x {3.7 x 10 3 / 105 }0.286 = 828 K Hence
T3 = 1.995 x 828 = 1651.86 K
(i) Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2) = 1.005 x (1651.86 – 828) = 828 kJ/kg. (ii)For process 4-1 which is at constant volume we have s1 – s4 = Cv ln (T1 / T4) Hence
T4 = T1 exp {(s4 – s1) / Cv} = 299 x exp {0.6939 / 0.718} = 784.9 K.
Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (299 – 784.9) = − 349.4 kJ/kg. Net work output per unit mass of air = wn = 828 – 349.4 = 478.6 kJ/kg Thermal efficiency = ηDiesel = wn / q2-3 = 478.6 / 828 = 0.578 = 57.8 %. (iii) Maximum temperature in the cycle = T3 = 1651.86 K. (iv) Temperature at the start of heat rejection process = T4 = 784.9 K.
34
3.7. Dual - Combustion Cycle or Semi-Diesel Cycle In practice, the p-V diagrams taken from oil engines indicate that the combustion do not take place at constant pressure as in a diesel cycle, but is found to take place partly at constant volume and partly at constant pressure. Hence for most oil engines the ideal cycle is taken as one in which heating takes place partly at constant volume and partly at constant pressure. Such a cycle is called as “dual combustion or semi-diesel cycle”. The p-V and T-s diagrams for a dual combustion cycle are shown in Fig.3.6.
p
3
2
4
Constant pressure process 3
4 s = const.
5
5
2 1
1
V
Constant volume process
Fig. 3.6 : p-V and T-s diagrams for dual combustion cycle. Process 1-2:- Isentropic compression of air from state 1 to state 2. During this process work is done on air by the surroundings. Process 2-3:- Constant volume heating of air from state 2 to state 3. Process 3-4:- Constant pressure heating of air till maximum permissible temperature is reached. Process 4-5:- Isentropic expansion of air from state 4 to state 5. During this process work is done by air on the surroundings. Process 5-1:- Constant volume cooling of air till the air comes back to its original state to complete the cycle. 3.7.1.Expressions for net work output and thermal efficiency Heat supplied per unit mass of air = qs = q2-3 + q3-4 = Cv(T3 – T2) + Cp (T4 – T3) Heat rejected per unit mass of air = q5-1 = Cv (T1 – T5)
35
Net work out put per unit mass of air = wn = ∑w = ∑q = qs + q5-1 Or
wn = Cv(T3 – T2) + Cp (T4 – T3) + Cv (T1 – T5)
Cv(T3 – T2) + Cp (T4 – T3) + Cv (T1 – T5) Hence thermal efficiency = η Dual = wn / qs = --------------------------------------------------Cv(T3 – T2) + Cp (T4 – T3) Cv (T5 – T1) = 1 − ---------------------------------- ……………(3.22) Cv(T3 – T2) + Cp (T4 – T3) The temperatures T2, T3, T4, and T5 can be expressed in terms of T1 as follows. Process 1-2 is isentropic. Hence T2 = (V1 / V2) (γ – 1) T1 = Rc(γ – 1) T1. For process 2-3, p2V2 / T2 = p3V3 / T3, and V2 = V3. T3 = (p3 / p2) T2 = Rp Rc(γ – 1) T1.
Hence
For process 3-4, p3V3 / T3 = p4V4 / T4, and p3 = p4. T4 = (V4 / V3) T3 = ρ Rp Rc(γ – 1) T1.
Hence
For process 4-5 we have
T5 = (V4 / V5) (γ – 1) T4 = {(V4 / V3)(V3 / V1)} (γ – 1) T4 = {ρ / Rc}(γ – 1) ρ Rp Rc(γ – 1) T1 = ρ γ Rp T1
Substituting these expressions in Eq. (3.22) we have T1 [ρ γ Rp − 1] η Dual = 1 − ----------------------------------------------------------------------T1[{Rp Rc(γ – 1) − Rc(γ – 1) } + γ{ ρ Rp Rc(γ – 1) − Rp Rc(γ – 1)}] η Dual
[ρ γ Rp − 1] = 1 − --------------------------------------------- ……………………….(3.23) Rc (γ – 1) [ (Rp – 1) + γ Rp (ρ – 1) ]
It can be seen from Eq. (3.23) that the air standard efficiency of the dual combustion cycle depends on the compression ratio Rc, the pressure ratio Rp and the cutoff ratio ρ. The effect of these parameters on efficiency is shown in Fig.3.7. It can bessen from this figure that for given values of Rp and ρ , the efficiency increase with Rc, rapidly for small values of Rc, but gradually for higher values of Rc. For a given value of Rc and Rp, the
36
efficiency decreases with increase in the cut – off ratio and for given values of Rcand ρ, the efficiency increases with increase in Rp. η Dual
Rp
ρ
Rc Fig. 3.7: Effect of Rc, Rp and ρ on efficiency of dual cycle
It can also be seen from Eq.(3.23) that this expression reduces to that of the diesel cycle if Rp = 1 and to that of Otto cycle if ρ = 1. It is also clear from this expression that the efficiency of the dual combustion cycle is intermediate between those of an Otto cycle and a Diesel cycle which have the same compression ratio and same cut-off ratio. 3.7.2. Illustrative examples on dual combustion cycle Example 3.13:- The compression and expansion ratios of an oil engine working on a dual cycle are 9 and 5 respectively. The initial pressure and temperature are 1 bar and 30 C. The heat added at constant pressure is twice that added at constant volume. Determine the thermal efficiency and the mean effective pressure. Given : Rc = V1 / V2 = 9 ; Re = V5 / V4 = 5 ; p1 = 1 bar ; T1 = 30 + 273 = 303 K ; q 3-4 = 2 q 2-3 ; To find :- (i) η Dual ; (ii) MEP Solution: Refer to p – V diagram shown in Fig. E3.13. (i) Process 1-2 is isentropic. Hence T2 = T1 Rc (γ – 1) = 303 x 9 0.4 = 729.7 K.
37
Compression ratio (V1 / V2) 9 Cut off ratio = V4 / V3 = ---------------------------- = ----------------- = ------Expansion ratio (V5 / V4) 5 = 1.8. For process 3-4, p3V3 / T3 = p4V4/T4 and p3 = p4. Hence Also
T4 = (V4 / V3) T3 = 1.8 T3 …………………………….(a) q3-4 = 2 q 2-3
i.e.,
Cp (T4 – T3) = 2Cv(T3 – T2)
or
(γ / 2) [1.8 T3 – T3] = T3 – T2
Or
T2 729.7 T3 = ------------------ = ---------------------[1 – 0.8(γ / 2)] [1 – 0.8 x 1.4 / 2] = 1658.7 K.
Hence
T4 = 1.8 x 1658.7 = 2985.4 K
Process 5-1 is isentropic. Hence T5 = T4 (V4 / V5) (γ – 1) = 2985.4 x (1 / 5) 0.4 = 1568.2 K. Heat supplied per unit mass of air = qs = q2-3 + q3-4 = 3q 2-3 = 3 Cv (T3 – T2) = 3 x 0.718 x (1658.7 – 729.7) = 2003.85 kJ/kg. Heat rejected per unit mass of air = q5-1 = Cv(T1 – T5) = 0.718 x (303 – 1568.2) = − 909.7 kJ/kg. (Negative sign for q5-1 indicates that during the process heat is transferred from air to the surroundings). Net work output per unit mass of air = wn = 2003.85 – 909.7 = 1094.15 kJ/kg
38
wn 1094.15 Thermal efficiency = η Dual = ------------ = ------------------ = 0.546 = 54.6 %. qs 2003.85 287 x 303 Specific volume of air before compression = v1 = RT1 / p1 = -----------------------1 x 10 5 = 0.8666 m3 / kg. Specific volume after compression = v2 = v1 / Rc = 0.8666 / 9 = 0.0963 m3 / kg. wn 1094.15 x 1000 Mean effective pressure = MEP = ----------------- = ----------------------(v1 – v2) (0.8666 – 0.0963) = 14.2 x 10 5 N/m2 = 14.2 bar. Example 3.14:- The maximum and the compression pressures in a dual cycle are 64 bar and 32 bar respectively. The compression curve is polytropic with index n = 1.35. The pressure in the cycle after 1/3rd of the compression stroke is completed is 1.65 bar. If 60 percent of the energy addition occurs at constant volume while 40 percent occurs at constant pressure, find (i) the compression ratio, (ii) the suction pressure, (iii)work output if the expansion index is 1.34, and (iv) thermal efficiency. p
3
Given:- p3 = p4 = 64 bar
4
p2 = 32 bar; nc = 1.35 ; ne = 1.34 ;
pVne = const pVnc = const
2
pa = 1.65 bar ; q2-3 = 0.6 qs; q3-4 = 0.4 qs ; V1 – Va = (1/3) (V1 – V2)
5
To find:- (i) Rc ; (ii) p1 ; (iii) wn ;
1
(iv) η Dual
a
Va
V
V1
39
Solution: (i) V1 – Va = (1/3) (V1 – V2) or Va = V1 – (1/3) (V1 – V2) Hence Now
Va / V2 = (2/3)(V1/V2) + 1/3 ……..(a) paVanc = p2V2nc or Va / V2 = (p2/pa) 1 / nc Va / V2 = (32 / 1.65) 1 / 1.35 = 8.99
Or
Substituting this in Eq.(a) and solving for (V1 / V2) we have (8.99 x 1/3) Compression ratio = V1 / V2 = ---------------- = 13 (2/3) (ii)For process 1-2 we have p1 = (V2 / V1) nc p2 = (1 /13) 1.35 x 32 = 1.003 bar. (iii) It is not possible to calculate the temperatures at staes 2, 3, 4 and 5 unless temperature at state 1 is known. Since T1 is not given it is assumed as 300 K. Hence for process 1-2, T2 = T1 (V1 / V2) (nc – 1) = 300 x 13 0.35 = 736.21 K. For process 2-3 we p2V2 / T2 = p3V3 / T3 and V2 = V3. Hence
T3 = T2(p3 / p2) = 736.21 x (64 / 32) = 1472.42 K.
q2-3 0.6 qs Now ---------------- = ----------- = 1.5 q3-4 0.4 qs Therefore q2-3 = 1.5 q3-4 Or Cv (T3 – T2) = 1.5 Cp (T4 – T3) Or
T4 = T3 + (1/ 1.5)(1 / γ) [T3 – T2] = 1472.42 + (1/1.5) x (1/ 1.4) x [1472.42 – 736.21] = 1823 K.
Cut off ratio = ρ = V4 / V3 = T4 / T3 = 1823 / 1472.42 = 1.24
40
Compression ratio 13 Expansion ratio = V5 / V4 = ----------------------- = --------- = 10.5 Cut-off ratio 1.24 For the expansion process 4-5 we have, T5 = T4 (V4/V5) (ne – 1) = 1823 x (1/10.5) 0.34 = 819.6 K. Since the index for compression process and expansion process are not equal to γ, these processes are not isentropic. Therefore there will be heat transfers during these processes which have to be determined to know the total heat supplied during the cycle. (p1v1 – p2v2) R (T1 – T2) 0.287 x (300 – 736.21) Now w1-2 = ----------------- = ----------------------- = ----------------------------(nc – 1) (nc – 1) 0.35 = − 356.5 kJ/kg. w2-3 = 0 as the process is at constant volume. w3-4 = p3(v4 – v5) = p4v4 − p5v5 = R(T4 – T5) = 0.287 x (1823 – 1472.42) = 100.3 kJ/kg. (p4v4 – p5v5) R (T4 – T5) 0.287 x (1823 – 819.6) w4-5 = ----------------- = ----------------------- = ----------------------------(ne – 1) (ne – 1) 0.34 = 844 kJ/kg. w5-1 = 0 as the process is at constant volume. Net work output per unit mass of air = wn = w1-2 + w2-3 + w3-4 + w4-5 + w5-1 = − 356.5 + 0 + 100.3 + 844 + 0 = 587.8 kJ/kg. (ii) Now for process 4-5 by I law q4-5 = w4-5 + Cv(T5 – T4) = 844 + 0.718 x (819.6 – 1823) = 122.55 kJ/kg.
41
Total heat supplied per unit mass of air = qs = q2-3 + q3-4 + q4-5 = Cv(T3 – T2) + Cp(T4 – T3) + q4-5 = 0.718 x(1472.42 – 736.21) + 1.005 x(1823 – 1472.42) + 122.55 = 1004.25 kJ/kg 587.80 Thermal efficiency = η Dual = wn / qs = --------------- = 0.5853 = 58.53 % 1004.25 Example 3.15:- . A diesel engine works between the temperatures of 1250 C and 25 C. The energy addition during combustion is 550 kJ /kg. A dual combustion cycle operates between the same temperature limits, and has the same total energy addition as for diesel cycle except that this energy is equally divided between the constant volume and constant pressure processes. Compare the efficiencies of the two cycles p
p 2
3
3
4
2
4
5
1
1 V
V Diesel Cycle
Dual combustion cycle
(i) Analysis of diesel cycle:- Given: Tmax = T3 = 1250 + 273 = 1523 K ; Tmin = T1 = 25 + 273 = 298 K ; qs = q 2-3 = 550 kJ/kg.
42
To find: η Diesel and compare it with η Dual. Solution: q2-3 = Cp (T3 – T2). Or T2 = T3 – q2-3 / Cp = 1523 – 550 / 1.005 = 975.74 K . Cut – off ratio = ρ = V3 / V2 = T3 / T2 = 1523 / 975.74 = 1.56. Compression ratio = Rc = V1 / V2 = (T2 / T1) 1 / (γ – 1) = (975.74 / 298) 1 / 0.4 = 19.4
1 [ρ γ – 1] Thermal efficiency = η Diesel = 1 − ------------------- x ------------γ Rc(γ – 1) [ρ – 1] 1 [ 1.56 1.4 – 1] = 1 − ----------------- x -----------------1.4 x 19.4 0.4 [ 1.56 – 1] = 0.6635 = 66.35 % (ii) Analysis of Dual combustion cycle: Given:- q2-3 = q3-4 = (1/2) x550 = 275 kJ/kg ; Tmax = T4 = 1523 K ; Tmin = T1 = 298 K ; To find: η Dual and compare it with η Diesel q 3-4 = Cp(T4 – T3) or T3 = T4 – q3-4 / Cp = 1523 – 275 / 1.005 = 1249.4 K Similarly
T2 = T3 – q2-3 / Cv = 1249.4 – 275 / 0.718 = 866.5 K.
Compression ratio = V1 / V2 = (T2 / T1) (γ – 1) = (866.5 / 298) 0.4 = 14.4 Cut-off Ratio = V4 / V3 = T4 / T3 = 1523 / 1249 = 1.22
43
Compression ratio 14.4 Expansion ratio = V5 / V4 = ------------------------ = -----------Cut –off ratio 1.22 = 11.8 For expansion process 4-5 we have T5 = T4 (V4 / V5) (γ – 1) = 1523 x (1/11.8) 0.4 = 567.4 K. Heat rejected per unit mass of air = q5-1 = Cv (T1 – T5) = 0.718 x (298 – 567.4) = − 193.7 kJ/kg ( 550 – 193.7) Thermal efficiency = η Dual = ------------------- = 0.6478 = 64.78 %. 550 Comparing the two efficiencies we have η Diesel > η Dual
44
Example 3.16:- In a dual cycle, two thirds of the total energy added occurs at constant volume.. If the compression ratio is 15, and the maximum pressure in the cycle is 53 bar, compute(i)the temperatures at the salient points of the cycle, and (ii) thermal efficiency. Assume standard conditions of air at the start of the compression process. Assume the minimum temperature and pressure in the cycle to be 27 C and 1 bar. 3
p
Given:- q2-3 = (2/3) qs ;
4
q3-4 = (1/3) qs ; V1 / V2 = 15 ; pmax = p3 = p4 = 53 bar ;
2
Tmin = T1 = 27 + 273 = 300 K ; pmin = p1 = 1 bar. To find:- (i) T2, T3, T4, T5 5
(ii) η Dual
1 V Solution: (i) Process 1-2 is isentropic. Hence T2 = T1 (V1 / V2) (γ – 1) Or
T2 = 300 x 15 0.4 = 880.3 K.
Also
p2 = p1(V1/V2) γ = 1 x 15 1.4 = 44.3 bar.
For process 2-3 we have
T3 = T2 (p3 / p2) = 880.3 x (53 / 44.3) = 1053.2 K.
q2-3 = Cv(T3 – T2) = 0.718 x (1053.2 – 880.3) = 123.85 kJ/kg. Hence qs = (3/2) x 123.85 = 185.775 kJ/kg. Therefore q3-4 = (1 / 3) x 185.775 = 61.925 kJ/kg Now q3-4 = Cp (T4 – T3). Or T4 = T3 + q3-4 / Cp = 1053.2 + 61.925 / 1.005 = 1114.82 K Cut – off ratio = V4 / V3 = T4 / T3 = 1114.82 / 1053.2 = 1.06
45
Expansion ratio = V5 / V4 = 15 / 1.06 = 14.15. T5 = T4 (V4 / V5) (γ – 1) = 1114.82 x (1 / 14.15) 0.4
Hence
= 386.3 K. Heat rejected = q5-1 = Cv(T1 – T5) = 0.718 x (300 – 386.3) = 61.96 kJ/kg Net work out put = wn = 185.775 – 61.96 = 123.815 kJ/kg. Thermal efficiency = η Dual = 123.815 / 185.775 = 0.6665 = 66.65 %. 3.8. Comparison between Otto, Diesel and Dual combustion cycles:- The important variables which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat supplied, heat rejected and the net work output. In order to compare the performance of the Otto, Diesel and Dual combustion cycles some of these variables have to be fixed. 3.8.1. Comparison with same compression ratio and heat supply: The comparison of these cycles for the same compression ratio and same heat supply are shown in Fig. 3.8 on both p – V and T – s diagrams.In these diagrams, cycle 1-2-3-4-1 represents Otto
T
3
p 3’’ 3’ 2’’ 3 2’’ 2
3’’
4
3’ 4’ 4’’ 4
4”
4’
2 1
1 V 5 6 6” 6’ Fig.3.8: Comparison with same compression ratio and heat supply
46
s
Cycle, cycle 1-2-3’-4’-1 represents diesel cycle and cycle 1-2”-3”-4”-1 represents the dual combustion cycle for the same compression ratio and heat supply. From the T-s diagram, it can be seen that area 5236 = area 522”3”6” = area 523’6’ as this area represents the heat supply which is same for all the cycles.All the cycles start from the same initial point 1 and the air is compressed from state 1 to state 2 as the compression ratio is same. It is seen from the T-s diagram, that for the same heat supply, the heat rejection in Otto cycle (area 5146) is minimum and heat rejection in Diesel cycle (area 514’6’) is maximum. Consequently Otto cycle has the highest work output and efficiency. Diesel cycle has the least efficiency and dual cycle has the efficiency between the two. 3.8.2:- Same compression ratio and heat rejection:- Fig. 3.9 shows the comparison between the Otto cycle and Diesel cycle on p-V and T-s diagrams.Cycle 1-2-3-4-1 represents the Otto cycle and cycle 1-2-3’-4-1 represents the Diesel cycle. T
p
3
3
2
3’ 3’ 4 2
4 1
1 5
V
6
Fig. 3.9: Same compression ratio and same heat rejection
s
Since both the cycles start from the same state point 1 and the heat rejection is same for both the cycles state 4 is same for both the cycles. Also since the compression ratio is same for both the cycles the state after compression process (state 2) is same for both the cycles.It can be seen from the T-s diagram that the area representing the Otto cycle (area 1234) is more than that representing the Diesel cycle (area 123’4). Hence the work output for the Otto cycle is more than that for the Diesel cycle.Thermal efficiency of any power cycle can be expressed as Heat supply – Heat rejection Heat rejection Thermal efficiency = ----------------------------------- = 1 − ----------------------Heat supply Heat supply
47
Since the heat rejection is same for both the cycle and heat supply for Otto cycle (area 2365) is more than that for the Diesel cycle (area 23’65), thermal efficiency for the Otto cycle is higher than that for the diesel cycle. 3.8.3:Same peak pressure, peak temperature and heat rejection: Fig. 3.10 show the comparison, on p-V and T-s diagrams, between Otto and Diesel cycles with same peak pressure, peak temperature and heat rejection.Cycle 1-2-3-4-1 represents the Otto cycle, p
T
2’
3
4
3 2’ 2 2
4 1
1 V
5
6
s
Fig. 3.10: Same peak pressure, same peak temperature and same heat rejection Cycle 1-2’-3-4-1 represents the Diesel cycle. It can be seen from the T-s diagram that area representing the Diesel cycle (area 12’34) is more than that representing the Otto cycle (area 1234) and hence Diesel cycle has higher work output than Otto cycle.It can also be seen that the heat supply for Diesel cycle (area 52’36) is more than that for Otto cycle (area 5236). Since the heat rejection is same for both the cycles, it follows that the thermal efficiency for the Diesel cycle is more than that for the Otto cycle for the same peak pressure, peak temperature and same heat rejection. 3.9. Stirling Cycle The Carnot power cycle has a low mean effective pressure because of its low work output. Hence one of the modified forms of the cycle to produce higher mean effective pressure with the efficiency equal to Carnot efficiency is the Stirling cycle. The Stirling cycle consists of two isothermal and two constant volume processes. The heat rejection and addition take place at constant temperatures. The p-V and T-s diagrams for the Stirling cycle are shown in Fig. 3.11. It is clear from the T-s diagram that the amount of heat addition and rejection during constant volume processes is same.Hence the thermal efficiency of the cycle is given as:
48
∑w ∑q ηStirling = ------------------ = ------------------qs q3-4 q1-2 + q2-3 + q3-4 + q-41 = ------------------------------------q3-4 Now q2-3 = −q4-1. q1-2 + q3-4 RT3 ln(V4/V3) + RT1 ln(V2/V1) ηStirling = ----------------- = -------------------------------------q3-4 RT3 ln(V4/V3) Since V4 = V1 and V3 = V2, the above expression for efficiency reduces to T3 – T1 ηStirling = -----------T3 This expression is same as that for a Carnot cycle working between the same temperature limits. The Stirling cycle was used for hot air engines and became obsolete as Otto and Diesel cycle came into use. The design of stirling engines involves a major difficulty in the design of a heat exchanger to achieve the heat transfer processes at constant volume and to operate at high temperature continuously. However, with the development of new materials and intensive research on this engine, Stirling engine has staged a come back. Since the heat exchanger cannot have 100 % efficiency in practice, the thermal efficiency of a practical Stirling engine will be less than that for a Carnot engine working between the same temperature limits.
49
3.10. Gas Turbine Cycles 3.10.1. Assumptions for analysis of Ideal Gas Turbine Cycles :- Following assumptions are made to analyse ideal gas turbine cycles: (i) The working substance is air and air behaves as a perfect gas (ii) Expansion and compression processes are isentropic. (iii) There are no pressure losses in the piping connecting the various components as well as in the heat exchangers. (iv) Changes in kinetic and potential energies of the fluid are negligible (v) Flow through various components is one dimensional, steady and uniform 3.10.2. Brayton Cycle (Simple Gas Turbine Cycle):- Brayton cycle is the basic cycle for the simple gas turbine power plant.The p-v and T-s diagram for this cycle is shown in Fig.3.12.It can be seen from these diagrams that the cycle consists of two isentropic processes and two constant pressure processes. The schematic for the cycle is shown in Fig. 3.13. p
3
T
2
3
4
2
1
4
1 v
s
Fig.3.12: p-v and T – s diagrams for a Brayton cycle Process 1 – 2: Isentropic compression of air in the compressor. During this process work is done on air by the surroundings. Process 2 – 3: Constant pressure heating of air in the heater till the maximum permissible temperature is reached.
50
Process 3 – 4: Isentropic expansion of air in the turbine. During this process work is done by air on the surroundings. Part of this work is used to drive the compressor. Process 4 – 1: Constant pressure cooling of air in the intercooler till the air comes back to its original state.(Process 4-1 is executed only in a closed cycle gas turbine plant, where as in an open cycle plant air is exhausted from the turbine and fresh air is drawn in from the atmosphere by the compressor) .
Qs .
m 2
3
.
.
Wc
Wt
T
C 1
4
IC .
Qr
C: Compressor T: Turbine CC: Combustion Chamber IC: Inter cooler
Fig. 3.13: Schematic for Brayton cycle Expression for thermal efficiency in terms of the pressure ratio Rp : Applying steady flow energy equation to the compressor and neglecting the changes in kinetic and potential energies we have Energy entering the compressor = Energy leaving the compressor .
Or
.
.
Wc + m h1 = m h2 .
Or
.
.
Wc = m (h2 – h1) = m Cp[T2 – T1] ...................................3.24 .
.
Compressor work per unit mass = wc = Wc / m = Cp[T2 – T1] …………………...3.25 Similarly applying steady flow energy equation to the turbine we have .
.
.
Wt = m (h3 – h4) = m Cp[T3 – T4] ...................................3.26 .
.
And turbine work per unit mass = wt = Wt / m = Cp [T3 – T4] ……………………3.27
51
The heat supplied in the heater similarly is given by .
.
.
Qs = m (h3 – h2) = m Cp[T3 – T2] ....................................3.28 .
.
And heat supply per unit mass = qs = Qs / m = Cp[T3 – T4] ……………………….3.29 Net work out put per unit mass = wn = wt – wc = Cp[T3 – T4] – Cp[T2 – T1] Thermal efficiency of the cycle = η = wn / qs Cp[T3 – T4] – Cp[T2 – T1] = ----------------------------------Cp[T3 – T2] [T4 – T1] T1[(T4 / T1) − 1] Or η = 1 − ------------------- = 1 − --------------------- ..........................................3.30 [T3 – T2] T2[(T3 / T2) − 1] Process 1-2 is isentropic and the working substance is assumed to behave as a perfect gas. Hence
T2 / T1 = (p2/p1) (γ – 1) / γ
Similarly for process 3-4 we have T3 / T4 = (p3 / p4) (γ – 1) / γ Since it is assumed that there are no pressure losses in the piping connecting the various components and in the heat exchangers p2 / p1 = p3 / p4 = Rp ……………………….3.31 Therefore it follows that T2 / T1 = T3 / T4 = Rp(γ – 1) / γ ……………………………3.32 Substituting this expression in equation 3.30 we get after simplification 1 η = 1 − ------------------- ………………………3.33 Rp(γ – 1) / γ Equation 3.33 indicates that the thermal efficiency of the Brayton cycle depends on the pressure ratio Rp. The effect of Rp on the efficiency is shown in Fig 3.14(a).It can be seen from this figure that initially the efficiency rapidly increases with the pressure ratio but as Rp increases the rate of increase of efficiency decreases.
52
Expression for net work output in terms of pressure ratio Rp and the maximum cycle temperature ratio t:
80 η
60
40
20
0
1
2
3
4 Rp
5
6
7
8
Fig.3.14(a) : Effect of pressure ratio on efficiency of a Brayton cycle Net work output per unit mass = wn = Cp[(T3 – T4) – (T2 – T1)] = CpT1 [(T3 / T1) – (T4 / T1) – (T2 / T1) + 1] ………….3.34 T3 / T1 is the ratio of maximum temperature in the cycle to minimum cycle temperature and it is normally denoted by ‘t’. t T4 / T1 = (T4 / T3) (T3 / T1) = ----------------Rp(γ – 1) / γ. Substituting these expressions in Eq.3.34 we get wn = CpT1[ t – {t / Rp(γ – 1) / γ} – Rp (γ – 1) / γ + 1] Or
wn* = wn / (CpT1) = [ t – {t / Rp(γ – 1) / γ} – Rp (γ – 1) / γ + 1]……………3.35
53
In Eq. 3.35 wn* represents dimensionless net work output and depends not only on the pressure ratio but also on the maximum cycle temperature ratio t. This is illustrated in Fig. 3.14(b). 0.28 wn* 0.24 0.20 0.16 0.12 0.08 0.04 0
Rp 2 4 6 8 10 12 14 16 Fig. 3.14(b): Effect of Rp and t on net work output for Brayton cycle
For a given value of t we can find the pressure ratio Rp which gives maximum net work output by differentiating the expression for wn* with respect to Rp and equating it to zero. Thus dwn* ---- = CpT1[ 0 – {t (1 – γ) / γ}Rp{(1 - γ ) / γ } – 1 - {(γ – 1) / γ}Rp{(γ – 1)/ γ } – 1 ] = 0 dRp Or Or
t Rp(1 - 2γ) / γ = Rp- (1 / γ) Rp = t γ / 2 (γ – 1) ………………………………3.36
It also follows from Eq. (3.36) that for maximum work output √t = Rp(γ – 1) / γ = T2 / T1 = T3 / T4. Hence (T2 / T1) x (T3 / T4) = (√t) x (√t) = t. But T3 / T1 = t and consequently it follows that T2 = T4. Thus the specific work output is a maximum when the pressure ratio is such that the compressor and turbine outlet temperatures are equal. If this value of Rp is substituted in the expression for wn we get the expression for maximum work output from a Brayton cycle as follows:
54
(wn)max = CpT1[t – (t / √t) – (√ t) + 1] = CpT1[ t – 2 √t + 1] (wn) max = CpT1[√ t – 1]2 ……………………..3.37
Or
The efficiency corresponding to the maximum work output is given by 1 η* = 1 − ------------------- ………………….3.38 √t 3.10.3. Brayton Cycle with Regenerator (Exhaust Heat Exchanger) :- The thermal efficiency of a Brayton cycle can be increased by incorporating an heat exchanger to heat the air coming from the compressor internally before it enters the combustion chamber. By so doing the external heat supplied is decreased without any change in the net work ouput from the cycle. A schematic of a gas turbine cycle employing an exhaust heat exchanger is shown in Fig. 3.15 (a) and the corresponding T-s diagram is shown in Fig.3.15(b). Process 1-2: Isentropic compression of air in the compressor. During this process work is done on air by the surroundings. Process 2-5: Constant pressure heating of air in the regenerator. During this process the air absorbs heat from the exhaust gases coming out of the turbine. Process 5-3: Constant pressure heating of air in the combustion chamber using an external source till the maximum permissible temperature is reached. Process 3-4: Isentropic expansion of air in the turbine. During this process work is done by air on the surroundings. Part of this work is used to drive the compressor. Process 4-6: Constant pressure cooling of exhaust from the turbine in the regenerator. During this process heat is transferred from the exhaust gases to the incoming air from the compressor. Regenerator
.
Qs 6 4 .
2
Wc
5
3
1
Fig. 3.15 (a): Schematic for Brayton cycle with regenerator
55
.
Wt
3
T
1-2-5-3-4-6-1 = cycle with exhaust heat exchanger. 4
5 2
1-2-3-4-1 = Brayton cycle 6
1 s Fig. 3.15(b): T-s diagram for Brayton cycle with exhaust exchanger
Expression for Thermal Efficiency:- While deriving the expression for thermal efficiency it is assumed that the effectiveness of the exhaust heat exchanger is 100 percent i.e. the incoming air from the compressor is heated to the temperature of the exhaust gases entering the exchanger(T5 = T4).The effectiveness of a heat exchanger is defined as follows: Actual heat transferred to air in the regenerator Effectiveness = ε = --------------------------------------------------------------Maximum possible heat transfer in the regenerator Cp [ T3 – T5] [ T3 – T5] ε = ------------------ = ------------- .............................................3.39 Cp [T3 – T4] [T3 – T4] Net work ouput per unit mass = wn = wt – wc = Cp[T3 – T4] – Cp[T2 – T1] External heat supply per unit mass = qs = Cp[T3 – T5] = Cp[T3 – T4] wn Cp[T3 – T4] – Cp[T2 – T1] [T2 – T1] Thermal efficiency = η = ------- = ---------------------------------- = 1 − --------------qs Cp[T3 – T4] [T3 – T4] T1[(T2 / T1) − 1] = 1 − ----------------------T4[(T3 / T4) − 1] Since T2 / T1 = T3 / T4 = Rp(γ – 1) / γ, it follows that
56
η = 1 − (T1/T4) = 1 − (T1/T3) (T3/T4) Rp(γ – 1) / γ η = 1 − ------------------ ............................................3.40 t
Or
It can be seen from Eq. 3.40 that the thermal efficiency of this cycle depends both on the pressure ratio Rp and the maximum cycle temperature ratio t. The effects of these two parameters on efficiency are shown in Fig. 3.16. The efficiency curve for simple cycle is also shown in this figure for comparison. It can be seen from the figure and it is evident from Eq. 3.39, that for a given value of “t”, the thermal efficiency decreases with increase in Rp and for a given value of Rp the efficiency increases with increase in t.The efficiency curves for different values of t have been drawn upto a value of Rp where the curve intersects the curve for simple cycle. Beyond this value of Rp , for any Rp, the temperature of the exhaust gases from the turbine will be less than that for air coming out of the compressor and therefore adding an exhaust heat exchanger will deteriorate the performance of the cycle.
80
η
60
40 20
0
0
1
2
3
4
5
6
7
8
9
Fig. 3.16: Effects of pressure ratio and maximum cycle temperature ratio on efficiency for a cycle with exhaust heat exchanger 3.10.4. Gas Turbine Cycle With Multi-Stage Expansion:- The net work out put from a Brayton cycle can be increased either by employing multi-stage compression with intercooling between the stages or by using multi-stage expansion with reheating in between the stages. By employing multi-stage expansion with reheating in between the
57
stages, the total turbine work as compared to single stage expansion is increased resulting an increase in the net work output.A schematic of an ideal gas turbine cycle employing two stages of expansionwith reheating in between the stages is shown in Fig. 3.16(a). The corresponding T-s diagram for the cycle is shown in Fig.3.16 (b).
qs
5
4 3
T2 wt1
2 T1
wc C
wt2 1
6
Fig. 3.16(a): Schematic for a gas turbine cycle with multi stage expansion Air from atmosphere is compressed isentropically from state 1 to state 2 in the compressor C .It is then heated at constant pressure in the heater H1 till the maximum permissible temperature T3 is reached. Then the air is expanded isentropically in the first stage turbine T1 from state 3 to state 4. It is then heated at constant pressure in the second heater H2 till the maximum permissible temperature T5 (T5 = T3) is reached. The air is finally expanded isentropically in the second stage turbine from state 5 to state 6.In the open cycle the air from the turbine is exhausted to the atmosphere and fresh air is drawn in by the compressor for the next cycle of operation, where as in the closed cycle the exhaust from the turbine is cooled back to its original temperature in another heat exchanger and fed to the compressor for the cycle to be repeated.
58
5
T
6 4 3
2 1 s
Fig. 3.16(b): T – s diagram for a gas turbine cycle with two stage compression It can be seen from the T-s diagram that by employing multi-stage compression with inter cooling the area representing the cycle has increased as compared to the area for a Brayton cycle. Since area of a closed curve on a T-s diagram represents the net work transfer during the cycle, it follows that the net work output from a cycle with multi stage compression is more than that for a Bray ton cycle working between the same pressure limits and the same maximum cycle temperature. As far as heat supply is concerned, it is evident from the T-s diagram that in the case of the cycle with multi-stage compression the air has to be heated from a lower temperature (T4 < T2’) than for Brayton cycle. Therefore more heat has to be supplied in the cycle with multi-stage compression than that required for Brayton cycle. The overall effect is that the thermal efficiency of the cycle with multi-stage compression will be lower than that for a simple cycle. Expressions for Net Work Output and Condition for Maximum Work output:- For the cycle with multi-stage compression net work output per unit mass of air is given by wn = wt – (wc1 + wc2) = Cp[T5 – T6] – Cp[(T2 – T1) + (T4 – T3)] = Cp[T5 – T6] – Cp[(T2 – T1) + (T4 – T3)] = CpT1[(T5 / T1) – (T6 / T1) – (T2 / T1) – (T4 / T3) + 2] ................3.41 Let p2/p1 = R1 = Pressure ratio for first stage compression, p4/p3 = R2 = Pressure ratio for second stage compression, and p5 / p6 = Rp = Pressure ratio for expansion.
59
Now process 1-2 is isentropic. Hence T2 / T1 = (p2/p1) (γ – 1) / γ = R1(γ – 1) / γ Similarly
T4 / T3 = (p4/p3) (γ – 1) / γ = R2(γ – 1) / γ
For expansion process we have T5 / T6 = (p5 / p6) (γ – 1) / γ = Rp(γ – 1) / γ p5 / p6 = p4 / p1 = (p4/p3) (p3 / p1) = (p4 / p3) (p2 / p1) Hence
Rp = R1 R2 ………………………………….3.42
T5 / T1 = maximum cycle temperature ratio = t. t T6 / T1 = (T6 / T5)(T5 / T1) = -----------------Rp(γ – 1) / γ T4 / T3 = R2(γ – 1) / γ = [Rp / R1] (γ – 1) / γ Substituting the expressions for temperature ratios in Eq. 3.41 we have wn = CpT1[ t –{t / Rp(γ – 1) / γ } – R1 (γ – 1) / γ − {Rp / R1} (γ – 1) / γ + 2] ……………3.43 It can be seen from Eq. 3.43 that for given values of T1 (Atmospheric temperature) and the overall pressure ratio wn depends only on R1. Therefore for maximum work output dwn / dR1 = 0. Or dwn / dR1 = CpT1[0 – 0 − {(γ – 1) / γ}R1− (1/ γ) − Rp (γ – 1) / γ {(1 – γ) / γ}R1 (1 - 2γ) / γ ] = 0 Or
R1− (1/ γ) -------------- = Rp (γ – 1) / γ R1 (1 - 2γ) / γ
Or
R1 = √ Rp ……………………………………3.44
and
R2 = Rp / R1 = √ Rp …………………………..3.45
Therefore, for maximum work out put from a cycle with two stages of compression with inter cooling in between the stages, the pressure ratio has to be same for both the stages and it should be equal to the square root of the overall pressure ratio for the cycle. Since R1 = R2 = √Rp it follows that the intermediate pressure p2 = √(p1p6) i.e. the intermediate pressure is the geometric mean of the maximum and minimum pressures in the cycle for maximum work output.If this condition is substituted in Eq. 3.43 we get the expression for maximum work output as :
60
[wn]max = CpT1[ t –{t / Rp(γ – 1) / γ } – 2Rp (γ – 1) / 2γ + 2] …………………………….3.46 Since R1 = R2 and T1 = T3, it follows, that the exit temperature from the second stage is same the exit temperature from the first stage compression, which means that the work input for the first stage is same as that for the second stage. Thermal Efficiency:- Heat supply per unit mass of air is given by qs = Cp[T5 – T4] = CpT1[(T5 / T1) – (T4 / T1)] = CpT1[(T5 / T1) – (T4 / T3)] = CpT1[ t – R2(γ – 1) / γ] Substituting the condition for maximum work output (R2 = √Rp) we get qs = CpT1[t – Rp (γ – 1) /2 γ ] Hence thermal efficiency is given by wn CpT1[ t –{t / Rp(γ – 1) / γ } – 2Rp (γ – 1) / 2γ + 2] η = -------- = -------------------------------------------------------qs CpT1[t – Rp (γ – 1) /2 γ ] [ t –{t / Rp(γ – 1) / γ } – 2Rp (γ – 1) / 2γ + 2] η = -------------------------------------------------- ...............................3.47 [t – Rp (γ – 1) /2 γ ]
or
80
η
Fig. 3.17 : Effects of Rp and ‘t’ on thermal efficiency of a cycle with multi-stage compression
60
40 20
0
0
1
2
3
4
Rp
61
5
6
7
8
3.10.5.Gas Turbine Cycle Multi-Stage Expansion With Reheat:- The net work out put can also be increased by splitting the expansion process and reheating the gas back to its original temperature after expansion in the high pressure turbine. It can be shown that the work output will be a maximum if the pressure ratio is same for both the stages of expansion and equal to the square root of the overall pressure ratio. Since the gas is reheated after first stage expansion the heat supply in this cycle is more than that for a simple cycle and the overall effect is that the thermal efficiency of this cycle is less than that for a simple cycle for the same pressure ratio and maximum cycle temperature ratio. The detailed analysis of this cycle is illustrated in example 3.19. 3.10.6. Illustrative examples on Ideal Gas Turbine Cycles: Example 3.17:- An air standard Brayton cycle has air entering the compressor at 100 kPa and 27 0C. The pressure ratio is 10 and themaximum allowable temperature in the cycle is 1350 K. Determine (i)temperatures at salient points of the cycle, (ii) compressor and turbine work per unit mass of air, (iii)net work output and work ratio, (iv)thermal efficiency of the cycle, (v) specific air consumption in kg/kWh, and (vi) improvement in the thermal efficiency of the cycle if a regenerator with 100 % effectiveness is incorporated in the cycle.
Given: T1 = 27 + 273 = 300 K;
3
T
T3 = 1350 K; p2/p1 = p3/p4 = 10 Assume: γ = 1.4 ; Cp = 1.005 kJ/(kg-K);
5
4
To find : (i) T2 and T4; (ii) wc and wt ; (iii) wn and wn / wt ; (iv) ηcycle ; 2
6
(v) S A C in kg/kWh 1
(vi) ηcycle for cycle with regenerator
s
Solution:- (i) Process 1-2 is isentropic. Hence T2 / T1 = (p2/p1) (γ – 1) / γ Or
T2 = T1 (p2/p1) (γ – 1) / γ = 300 x 10 (1.4 – 1) / 1.4 = 579.6 K
62
Similarly
T4 = T3 (p4/p3) (γ – 1) / γ = 1350 x (1 / 10) (1.4 – 1) / 1.4 = 698.8 K
(ii)compressor work per unit mass = wc = Cp[T2 – T1] = 1.005 x [579.6 – 300] = 281 kJ/kg. Turbine work per unit mass = wt = Cp[T3 – T4] = 1.005 x [1350 – 698.8] = 654.5 kJ/kg (iv)
Net work output per unit mass = wn = wt – wc = 654.5 – 281 = 373.5 kJ/kg
Work ratio = wn / wt = 373.5 / 654.5 = 0.57 (v)
Heat supplied per unit mass = qs = Cp[T3 – T2] = 1.005 x [1350 – 579.6] = 774.25 kJ/kg
Thermal efficiency = ηcycle = wn / qs = 373.5 / 774.25 = 0.4824 = 48.24 % (v) Specific air consumption = SAC = 3600 / wn = 3600 / 373.5 = 9.64 kg / kWh (vi) when a regenerator is incorporated in the cycle then air has to be heated from 5 to 3 (see T-s diagram) where T5 = T4. Then heat supplied per unit mass of air is given by qs = Cp[T3 – T5] = 1.005 x [1350 – 698.8] = 654.5 kJ/kg Thermal efficiency = wn / qs = 373.5 / 654.5 = 0.57 = 57 % Example 3.18:- If the simple gas turbine cycle of example 3.17 is modified such that there are two stages of compression with intercooling in between the stages, determine the net work output per unit mass of air and the thermal efficiency of the modified cycle. Assume the pressure ratio for each stage is such that the work output from the cycle is maximum.Assume the overall pressure ratio, the minimum cycle temperature and the maximum cycle temperature to be same as that in example 3.17. Given: T1 = 27 + 273 = 300 K; T5 = 1350 K; p5 / p6 = 10 Since the cycle is designed for maximum work output, p2/p1 = p4/p3 = √(p5/p6)
63
and T3 = T1. Assume γ = 1.4 and Cp = 1.005 kJ/(kg –K) T 5
6
4
2
3
1 s
To find: (i) wn ; (ii) ηcycle Solution: p2/p1 = p4/p3 = √10 = 3.162 Process 1-2 is isentropic. Hence T2 = T1 (p2 / p1) (γ
- 1) /γ
= 300 x (3.162) 0.286
= 417 K. Similarly T4 = T3 (p4/p3) (γ
- 1) /γ
= 300 x (3.162) 0.286 = 417 K
Total compressor work = wc = wc1 + wc2 = 2wc1 = 2 Cp[T2 – T1] = 2 x 1.005 x [417 – 300] = 235.17 kJ/kg Turbine work = wt = same as in example 3.17 = 654.5 kJ/kg (i)
net work output = wn = 654.5 – 235.17 = 419.33 kJ/kg
(ii)
Heat supplied per unit mass = qs = Cp[T5 – T4] = 1.005 x [1350 – 417] = 937.7 kJ/kg ηcycle = wn / qs = 419.33 / 937.7 = 0.447 = 44.7 %
64
Example 3.19:- An ideal gas turbine cycle has an overall pressure ratio Rp. The expansion takes place in two stages with reheating in between the stages. If R1 and R2 are the pressure ratios for the first and second stages of expansion show that for maximum work output from the cycle R1 = R2 = √Rp. Also obtain expressions for the maximum net work output and for the corresponding thermal efficiency in terms of Rp and the maximum cycle temperature ratio ‘t’.Also draw the schematic and T-s diagrams for the cycle Solution: Rp = p2 / p1 = p3 / p6 = (p3 / p4) (p5 / p6) = R1 R2 Or
R2 = Rp / R1 ………………………………..(1)
Compressor work per unit mass = wc = Cp[T2 – T1] = CpT1[(T2/T1) – 1] = CpT1[Rp(γ – 1) / γ – 1] Turbine work per unit mass = wt = wt1 + wt2 = Cp[T3 – T4] + Cp[T5 – T6] = CpT1[(T3/T1) – (T4/T1) + (T5 / T1) – (T6/T1)] Now T3 / T1 = T5 / T1 = maximum cycle temperature ratio = t (It is assumed that after first stage expansion the gas is reheated back to its original temperature i.e. T5 = T3).
qs1
qs2
4 2
wc
CC 2
CC 1 5
3
C
T1
1
wt1
wt2
T2
6
Fig. : Schematic for a gas turbine cycle with two stage expansion & Reheat
65
T
3
5
4
6
2
1 s Fig: T-s diagram for a gas turbine cycle with multi stage expansion & reheat
t T4 / T1= (T4/T3)(T3 / T1) = ---------------------R1 (γ – 1) / γ t t R1 (γ – 1) / γ T6 / T1 = (T6 / T5)(T5 / T1) = ------------------- = -----------------R2 (γ – 1) / γ Rp (γ – 1) / γ Hnece the net work out put from the cycle is given by t { t R1 (γ – 1) / γ } wn = Cp T1 [ t – -------------- + t – --------------------- − Rp (γ – 1) / γ + 1 ] ......................(2) Rp (γ – 1) / γ R1 (γ – 1) / γ For given values of ‘t’ and Rp the net work output will be maximum if dwn / dR1 = 0 dwn / dR1 = CpT1[ t (1 – γ)/γ R1(1 – 2 γ) / γ − t {(γ – 1 )/γ}R1 − 1 / γ / Rp (γ – 1) / γ ] = 0 or
R1(1 – 2 γ) / γ = R1 − 1 / γ / Rp (γ – 1) / γ
66
or
R1 = √ Rp
Also
R2 = Rp / R1 = Rp / √ Rp or R2 = √Rp
Substituting these expressions for R1 and R2 in Eq. (2) and simplifying we get [wn]max = CpT1[2t –2t Rp(1 – γ) / 2γ − Rp (γ – 1) / γ + 1] Total heat supplied per unit mass = qs = qs1 + qs2 = Cp[(T3 – T2) + (T5 – T4)] Or Or
qs = CpT1[(T3 / T1) – (T2/T1) + (T5/T1) –(T4/T1)] qs = CpT1[2 t - Rp (γ – 1) / γ – t Rp(1 – γ ) / 2γ ]
CpT1[2t –2t Rp(1 – γ) / 2γ − Rp (γ – 1) / γ + 1] Hence thermal efficiency = ηcycle = ------------------------------------------------CpT1[2 t - Rp (γ – 1) / γ – t Rp(1 – γ ) / 2γ ] [2t –2t Rp(1 – γ) / 2γ − Rp (γ – 1) / γ + 1] Hence thermal efficiency = ηcycle = ------------------------------------------------[2 t - Rp (γ – 1) / γ – t Rp(1 – γ ) / 2γ ] Example 3.20:- Determine the net work output and thermal efficiency of an ideal gas turbine cycle having two stages of compression with intercooling in between the stages and two stages of expansion with reheating in between the stages. The overall pressure ratio for the cycle is 4 and the maximum cycle temperature is 900 0C Assume that the atmospheric temperature is 15 0C and the cycle is designed for maximum work output. Draw the schematic and T-s diagrams for the cycle.What would be the improvement in the thermal efficiency if an ideal regenerator is incorporated in the cycle? Given: Overall pressure ratio = p4 / p1 = p5 / p8 = 4; T5 = T7 = 900 + 273 = 1173 K; T1 = T3 = 15 + 273 = 288 K; Cycle is designed for maximum work output i.e. p2 / p1 = p4 / p3 = √ 4 = 2 ; p5 / p6 = p7 / p8 = √4 = 2 ; Assume γ = 1.4 and Cp = 1.005 kJ/(kg-K) To find:- (i) wn ; (ii) ηcycle ; (iii) improvement in efficiency when an exhaust heat exchanger is incorporated in the cycle.
67
qs1
2
3
C1 wc1
4
5
8
wt1 T1
C2
T2
wc2
wt2 6
1
7 qs2
Fig. E3.20(a) : Schematic diagram for Example E3.20 Solution: Process 1-2 is isentropic. Hence T2 = T1(p2 / p1) (γ – 1) / γ = 288 x 2 0.286 =351.1 K Since T3 = T1 and p2/p1 = p4/p3, T4 = T2 and wc1 = wc2. Therefore total compressor work per unit mass = wc = 2 wc1 = 2Cp[T2 – T1]
68
T
5 9
4
2
3
1
7 6
8
s Fig. E3.20 (b): T – s diagram for example E 3.20
Or
wc = 2 x 1.005 x [351.1 – 288] = 126.8 kJ / kg.
For expansion process 5-6 we have T6 = T5 (p6/p5) (γ – 1) / γ = 1173 x 2 0.286 = 962 K Since T7 = T5 and p7/p8 = p5/p6 we have T8 = T6 and wt1 = wt2. Therefore total turbine work per unit mass = wt = 2 wt1 = 2 Cp[T5 – T6] Or
wt = 2 x 1.005 x [1173 – 962] = 424 kJ/kg
Hence net work output = wn = wt – wc = 424 – 126.8 = 297.2 kj?kg Total heat supplied per unit mass = qs = qs1 + qs2 = Cp[T5 – T4] + Cp[T7 – T6] = 1.005 x [1173 – 351.1 + 1173 – 962] = 1038.1 kJ/kg Thermal efficiency = ηcycle = wn / qs = 297.2 / 1038.1 = 0.2863 = 28.63 %
69
When an ideal exhaust heat exchanger is incorporated in the cycle then air is heated in the heat exchanger up to a temperature T9 = T8. Hence total heat supply = qs = Cp[T5 – T9] + Cp[T7 – T6] = 2 Cp{T5 – T9] = 2 x 1.005 x [1173 – 962 ] = 424.11 kJ/kg Thermal efficiency = 297.2 / 424.11 = 0.70 = 70 %. 3.10.7. Practical Gas Turbine Cycles: Deviations of Practical cycles from Ideal Cycles: 1. It was assumed in the analysis of ideal cycles that the compression and expansion processes are isentropic (reversible adiabatic). But in actual cycles because of friction the actual processes will not be isentropic. Therefore in the analysis of practical cycles the frictional losses during these processes are accounted by defining a parameter called “isentropic efficiency”. Fig. 3.18 illustrates the actual and isentropic processes on the T-s diagram. For the compression process the isentropic efficiency is defined as follows: Isentropic work of compression Isentropic efficiency of compressor = η c = ---------------------------------------Actual work of compression
T
2 2’
1 s Fig. 3.18: Representation of isentropic and actual compression processes
70
The isentropic efficiency in terms of enthalpies is given by [h2’ – h1] Cp [T2’ – T1] ηc = ------------- = ------------------[h2 – h1] Cp [T2 – T1] Or
[T2’ – T1] ηc = ------------- ...........................................................................3. 47 [T2 – T1]
Fig 3.19 illustrates the actual and isentropic expansion processes on the T-s diagram. The isentropic efficiency for an expansion process is given by Actual work of expansion Isentropic efficiency of expansion = -------------------------------------------Isentropis work of expansion
i.e.
[h3 – h4] Cp [T3 – T4] ηt = ------------- = -------------------[h3 – h4’] Cp [T3 – T4’] T 3
4 4’
s Fig.3.19: Actual and isentropic expansion processes on the T-s diagram
Or
[T3 – T4] ηt = ---------------- ..........................................................3.48 [T3 – T4’]
71
Because of frictional losses in the compressor and in the turbine, the actual work of compression is more than that required for ideal compression and the actual work output from the turbine is les than that from an ideal turbine and therefore the net work output from the actual cycle is less than that from an ideal cycle. For the same heat supply the thermal efficiency of the actual cycle will be less than that for the ideal cycle. 2. In the practical cycle, there will be pressure losses in the piping connecting the various components as well as in the heat exchanger. Therefore the pressure at inlet to the turbine will be less than that at the exit of the compressor and the pressure at the exit of the turbine will be more than that at the inlet to the compressor with the result the pressure ratio for expansion process is reduced which in turn decreases the turbine work and hence the net work output from the cycle. 3. In the practical cycle only air is compressed in the compressor where as the products of combustion undergoes expansion in the turbine. Our assumption of constant specific heat throughout the cycle is not valid while analyzing the actual cycle. Therefore for practical cycles it is assumed that γ = 1.4 and Cp = 1.005 kJ/(kg-K) for the compression process whereas γ = 1.33 and Cp = 1.148 kJ/(kg-K). 4. Air-Fuel Ratio: In an actual cycle,because fuel is mixed with air in the combustion chamber, the mass flow rate during combustion and expansion will be more than the mass flow rate during compression. An expression for air fuel ratio by mass can be obtained by applying the steady flow energy equation to the combustion chamber as follows
Fuel (mf)
Air (ma)
CC 2
Products of 3 combustion (ma + mf)
mah2 + mf ηcomb hf = (ma + mf) h3 ,
72
where hf is the calorific value of the fuel and η comb is the combustion efficiency.ηcomb is introduced in the above equation as all the energy in the fuel will not be released in an actual combustion process. Assuming that both air and the products of combustion behave as perfect gases the above equation can be written as ma CpT2 + mf ηcomb hf = (ma + mf) Cp T3 Dividing throughout by mf and solving for the air-fuel ratio ma/mf we have ma [ ηcomb hf – CpT3] ----- = --------------------- ........................................3.49 mf Cp[T3 – T2] since L >> CpT3, the above equation can be approximated as follows: ma ηcomb hf ----- = --------------------- ........................................3.50 mf Cp[T3 – T2]
3.10.8.Illustrative Examples on Practical Gas Turbine Cycles: Example 3.21:-The pressure ratio of an open cycle gas turbine cycle is 6. The compressor inlet conditions are 1 bar and 15 0C. The maximum temperature in the cycle is 800 0C. The isentropic efficiency of compressor is 85 % and that of the turbine is 90 %. The combustion efficiency is 95 %. There is a pressure drop of 2 % of the inlet pressure in the combustion chamber. The calorific value of the fuel used is 42,000 kJ/kg.Assuming that the values of γ and Cp remain same throughout the cycle and equal to 1.4and 1.005 kJ/(kg-K) respectively determine (i)net work output per unit mass of air, (ii) air-fuel ratio, (iii) thermal efficiency of the plant, (iv) specific fuel combustion in kg/kWh, and (v) power output from the plant for a mass flow rate of air of 1.0 kg/s.
73
Given : p2/p1 = 6 ; T1 = 15 + 273 =288 K ;
T
3
ηc = 0.85 ; ηt = 0.90 ; ηcomb = 0.95 ; T3 = 800 + 273 = 1073 K ;
4
p2 – p3 = 0.02 p2 ; p1 = 1 bar ; 2
hf = 42,000 kJ/kg ; γ = 1.4
2’
Cp = 1.005 kJ/(kg-K) To find: (i) wn; (ii) ma/mf; (iii) ηcycle
1
.
(iv) SFC; (v) Wn if ma = 1 kg / s.
s
Solution: Process 1-2’ is isentropic. Hence T2’ = T1(p2/p1) (γ – 1) / γ Or
Or
T2’ = 288 x 6 0.286 = 480.8 K. (T2’ – T1) ηc = --------------. (T2 – T1) T2 = T1 + (T2’ – T1) / ηc = 288 + (480.8 – 288) / 0.85 = 514.8 K
Compressor work per unit mass of air = wc = Cp[T2 – T1] = 1.005 x [514.8 – 288] = 227.96 kJ / kg Pressure at inlet to the turbine = p3 = p2 – 0.02(p2 – p3) Or
p3 = (1- 0.02)p2 / (1 +0.02) = .98 x 6 / 1.02 = 5.765 bar
Process 3-4’ is isentropic. Hence T4’ = T3 (p3 / p4’) (γ – 1) / γ = 1073 x (1/ 5.765) 0.286
74
= 650.15 K Now ηt = (T3 – T4) / (T3 – T4’) or T4 = T3 – ηt(T3 – T4’) = 1073 – 0.90 x (1073 – 650.15) = 692.75 K (ii) To find air – fuel ratio: Using Eq. 3.49 we have ma [ ηcomb hf – CpT3] ----- = --------------------- ........................................3.49 mf Cp[T3 – T2] [ 0.95 x 42,000 – 1.005 x 1073] = --------------------------------------1.05 x [1073 – 514.8] = 69.2 (i) Turbine work per unit mass of air = wt = (ma + mf)Cp (T3 – T4) / ma = (1 + mf / ma) Cp[T3 – T4] = (1 + 1 / 69.2) x 1.005 x [1073 – 692.75] = 387.7 kJ/kg Net work out put per unit mass of air = wn = wt – wc = 387.7 – 227.96 = 159.74 kJ/kg (iii)Heat supplied per unit mass of air =qs = (1 + mf / ma) Cp[T3 – T2] = (1 + 1 / 69.2) x 1.005 x [1073 – 514.8] Or
qs = 569.1 kJ/kg
Thermal efficiency = ηcycle = wn / qs = 159.74 / 569.1= 0.281 = 28.1 % Mass of fuel in kg /h (iv) Specific fuel consumption in kg/kWh = ---------------------------------Net work output in kW
75
3600 mf 3600 = ------------------ = ------------------ma wn 69.2 x 159.74 = 0.326 kg / kWh .
(v) Power out put = Wn = ma wn = 1.0 x 159.74 = 159.74 kW Example 3.22:-The isentropic discharge temperature of air flowing out of a compressor is 195 0C, while the actual temperature is 240 0C. The conditions of air at compressor inlet are 1 bar and 170C. If the air fuel ratio in the combustion chamber is 75:1 and net power output is 650 kW, compute (i)the isentropic efficiencies of the compressor and turbine and (ii) the overall cycle efficiency. Assume that the plant consumes 5.2 kg/min of fuel and the calorific value of the fuel used is 42,000 kJ/kg. Assume Cp = 1.005 kJ/(kg-K) and γ = 1.4 for air and Cp = 1.148 kJ/(kg-K) and γ = 1.148 for products of combustion. 3
Given: T2’ = 195 + 273 = 468 K ; T2 = 240 + 273 = 513 K ; p1 = 1 bar ;
4
T1 = 17 + 273 = 290 K ; ma / mf = 75;
2
.
Wn = 650 kW ; hf = 42,000 kJ / kg ;
2’
mf = 5.2 / 60 = 0.087 kg / s ; For compression γa = 1.4 ;
1
Cpa = 1.005 kJ / (kg – K) For heating & expansion γg = 1.33 and Cpg = 1.148 kJ / (kg – K) ;
To find : (i) ηc ; ηt ; (ii) ηcycle Solution :
(i)
[T2’ – T1] ( 468 – 290 ) ηc = ------------- = ----------------- = 0.798 = 79.8 % [T2 – T1] (513 – 290)
76
4’
[T3 – T4] ηt = ------------[T3 – T4’] To find T3 :- Energy balance equation for the combustion chamber gives ma Cpa T2 + mf ηcomb hf = (ma + mf) Cpg T3 (ma / mf) Cpa T2 + ηcomb hf T3 = ----------------------------------Cpg [ma / mf + 1]
Or
75 x 1.005 x 290 + 1.0 x 42,000 = ---------------------------------------1.148 x [75 + 1] = 924.6 K To find T4’:- Pressure ratio for expansion = p3 / p4 = p2 / p1 = (T2’ / T1) γ / (γ – 1) Hence p3 / p4 = [468 / 290] 1.4 / 0.4 = 5.34 Process 3-4’ is isentropic. Therefore T4’ = T3 [p4 / p3] (γ – 1 ) / γ = T4’ = 924.6 x [1 / 5.34] (1.33 – 1 ) / 1.33
Or
= 608.4 K .
To find (T3 − T4):- Compressor work per unit time = Wc = ma Cpa [T2 – T1] .
Or
Wc = 0.087 x 75 x 1.005 x [513 – 290] = 1462 kW .
.
.
Hence Turbine work per unit time = Wt = Wn + Wc = 650 + 1462 = 2112 kW .
But
Wt = (ma + mf) Cpg [T3 – T4]
.
Or
.
Wt Wt (T3 – T4) = ------------------ = --------------------------(ma + mf) Cpg mf (ma / mf + 1) Cpg
77
2112 = --------------------------------0.087 x ( 75 + 1 ) x 1.148 = 278.2 K
Therefore
[T3 – T4] 278.2 ηt = -------------------- = --------------------[T3 – T4’] [924.6 – 608.4] = 0.879 = 87.9 %
(iii)
Thermal efficiency is given by .
Wn ηcycle = -----.------Qs .
Qs = mf hf = 0.087 x 42,000 = 3654 kW. Therefore
650 ηcycle = ------------- = 0.1778 = 17.78 % 3654
Example 3.23:- Determine the thermal efficiency of a gas turbine cycle having two stages of compression and two stages of expansion with an overall pressure ratio of 4 and a maximum cycle temperature of 900 0C. The compressor inlet temperature is 15 0C. The compression stages have efficiencies of 80% each and the turbine stages have efficiencies of 85% each. Assume that the pressure ratio for the two stages of compression and expansion are chosen for maximum work output from the cycle. What would be the improvement in thermal efficiency if a regenerator of 85% effectiveness is incorporated in the cycle? Draw the schematic and T-s diagram for the cycle. Given: Overall pressure ratio = p4 / p1 = p5 / p8 = 4 ; T5 = 900 + 273 = 1173 K ; T1 = 15 + 273 = 288 K ; ηc1 = ηc2 = 0.80 ; ηt1 = ηt2 = 0.85 ; Cycle designed for maximum thermal efficiency i.e. p2 / p1 = p4 / p3 = √(p4 / p1) = √ 4 = 2 ; P5 / p6 = p7 / p8 = √ (p5 / p8) = √ 4 = 2 ; T3 = T1 ; T7 = T8 ; ε = 0.85 To Find : (i) ηcycle ; (ii) Improvement in cycle efficiency when a regenerator is Incorporated in the cycle.
78
Solution: For schematic diagram see Fig.3.20(a). The T –s diagram for the given cycle is shown in Fig. E 3.23 7
5
T
6
9 6’ 4’
4 2’
3
1
8 8’
2
s Fig. E 3.23: T – s diagram for example 3.23 For isentropic process 1 – 2’ we have T2’ = T1(p2 / p1) (γ – 1) / γ = 288 x 2 0.286 = 351.1 K. [ T2’ – T1] ηc1 = ------------- Or T2 = T1 + [ T2’ – T1] / ηc1 = 288 + [351.1 – 288] / 0.80 [T2 – T1] Or
T2 = 366.9 K.
Since p2 / p1 = p4 / p3 and T3 = T1, it follows that T4’ = T2’.Further since ηc1 = ηc2 andT3 = T1 it follows that
T4 = T2 = 366.9 K
Also work is equally divided between the two compressor stages. Hence Total compressor work per unit mass = wc = 2 wc1 = 2 Cp [T2 – T1] = 2 x 1.005 x [366.9 – 288] = 158.6 kJ/kg. For expansion process 5 – 6’ we have T6’ = T5 (p6 / p5) (γ – 1) / γ = 1173 x (1 /2 0.286) = 962 K
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(T5 – T6) Now ηt1 = ------------------. Or (T5 – T6’ )
T6 = T5 - ηt1(T5 – T6’ )
Or
T6 = 1173 – 0.85 x [1173 – 962] = 993.65 K
Since the pressure ratio for the two stages of expansion are same, the inlet temperatures for both stages are same and the efficiencies of the two stages are same, it follows that the exit temperature of gas for both the stages are same and the work output from both the stages are equal; i.e.
T8 = T6 = 993.65 K.
and
wt = 2 Cp [T5 – T6] = 2 x 1.005 x [1173 – 993.65] = 360.5 kJ/kg
Hence net work output = wn = 360.5 – 158.6 = 201.9 kJ/kg Total heat supply per unit mass = qs = Cp[T5 – T4] + Cp[T7 – T6] = 1.005 x [1173 – 366.9 + 1173 – 993.65] = 990.4 kJ/kg Thermal efficiency = ηcycle = 201.9 / 990.4 = 0.204 = 20.4 % When a regenerator is incorporated in the cycle:- When a regenerator is incorporated in the cycle then the total heat supplied per unit mass of air is given by qs = q9-5 + q6-7 = Cp[T5 – T9] + Cp[T7 – T6} [T9 – T4] Now effectiveness of the regenerator = ε = -------------[T8 – T4] Therefore T9 = T4 + ε [T8 – T4] = 366.9 + 0.85 x [993.65 – 366.9] = 899.6 K Hence qs = 1.005 x [1173 – 899.6] + 1.005 x [1173 – 993.65] = 452.75 kJ/kg Thermal efficiency = ηcycle = 201.9 / 452.75 = 0.446 = 44.6 %
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Example 3.24:- Determine the specific work output, specific fuel consumption and cycle efficiency for a gas turbine power plant using a regenerator and having the following specifications: Compressor pressure ratio = 4.0 Turbine inlet temperature = 1100 K; Isentropic efficiency of the compressor = 0.85; Isentropic efficiency of the turbine = 0.87; Mechanical transmission efficiency = 0.99; Combustion efficiency = 0.98; Heat exchanger effectiveness = 0.80; Combustion chamber pressure loss = 2 % of compressor delivery pressure; Heat exchanger air side pressure loss = 3 % of compressor delivery pressure; Heat exchanger gas side pressure loss = 0.04 bar Calorific value of the fuel used = 42,000 kJ/kg Compressor inlet conditions = 1 bar and 300 K Given: p2 / p1 = 4 ; T3 = 1100 K ; ηc = 0.85 ; ηt = 0.87 ; ηTrans = 0.99 ; ηcomb = 0.98 ; ε = 0.80 ; p5 – p3 = 0.02 p2 ; p2 – p5 = 0.03 p2 ; p4 – p6 = 0.04 bar ; p1 = p6 = 1 bar ; T1 = 300 K ; hf = 42,000 kJ /kg . To find : (i) work output per unit mass of fuel ; (ii) SFC ; (iii) ηc Solution
81
T
3 4 4’
5 2 6
2’
1 s
For process 1-2’ we have T2’ = T1(p2 / p1) (γ – 1) / γ = 300 x 4 0.286 = 446 K . ηc = (T2’ – T1) / (T2 – T1) or T2 = T1 + (T2’ – T1) / ηc or
T2 = 300 + (446 – 300) / 0.85 = 471.8 K.
Compressor work per unit mass of air = wc = Cp[T2 – T1] = 1.005 x [471.8 – 300] = 172.7 kJ/kg. Because of transmission losses, actual compressor work per unit mass of air is (wc)actual = wc / ηTrans = 172.7 / 0.99 = 174.4 kJ/kg Turbine inlet pressure = p3 = p5 – 0.02p2 = (p2 – 0.03p2) – 0.02p2 = 0.95 p2 = 0.95 x 4 = 3.8 bar Turbine exit pressure = p4 = p6 + 0.04 = 1 + 0.04 = 1.04 bar. Hence pressure ratio for turbine = p3 / p4 = 3.8 / 1.04 = 3.654 For process 3 – 4’ we have T4’ = T3 (p4 / p3) (γ – 1) / γ = 1100 x (1 / 3.654) 0.286= 759.3 K. Therefore
T4 = T3 – ηt[T3 – T4’] = 1100 – 0.87 x [1100 – 759.3]
82
= 803.6 K Turbine work per unit mass of air = wt = Cp[T3 – T4] = 1.005 x [1100 – 803.6] = 297.9 kJ/kg Net work output per unit mass of air = wn = 297.9 – 174.4 = 123.5 kJ/kg Effectiveness of the regenerator = ε = (T5 – T2) / (T4 – T2) Hence
T5 = T2 + ε (T4 – T2) = 471.8 + 0.80 x (803.6 – 471.8) = 732.2 K
Energy balance equation for the combustion chamber can be written as : ma Cp T5 + mf ηcomb hf = (ma + mf) Cp T3 Or
[ ηcomb hf – Cp T3] [0.98 x 42,000 – 1.005 x 1100] ma / mf = ----------------------- = -------------------------------------------Cp [T3 – T5] 1.005 x [1100 – 732.2] = 108.4
mf ηcomb hf 0.98 x 42,000 Heat supplied per unit mass of air = qs = ------------------ = -------------------- = 379.7 kJ/kg ma 108.4 Thermal efficiency = ηcycle = 123.5 / 379.7 = 0.325 = 32.5 % Specific work out put = work output per unit mass of fuel = ma wn / mf = 123.5 x 108.4 = 13387.4 kJ / kg of fuel
mf x 3600 3600 Specific fuel consumption in kg / kWh = ---------------------- = -----------------ma wn 108.4 x 123.5 = 0.269 kg / kWh. Example 3.25:- In a closed cycle gas turbine plant, the compressor inlet and exit pressures are 5 bar and 32.5 bar respectively. After passing through a regenerator with an effectiveness of 0.83, the air is heated in a nuclear reactor to a temperature of 945 K. the pressure drop in the regenerator and the reactor reduces the air pressure at turbine inlet to 31.5 bar. After expansion to5 .25 bar in the turbine with an efficiency of 0.88, the air passes through the regenerator and a cooler before being ready to enter the
83
compressor whose efficiency is 0.80.The temperature of air at compressor inlet is 20 0C. Calculate (i) cycle thermal efficiency, (ii) the turbine and compressor power, (iii) the heat transfer in the reactor and the mass flow rate of air if the net power output from the plant is 650 kW. Draw the schematic diagram for the plant.
C : Compressor; T : Turbine; NR:Nuclear Reactor: R : Regenerator; IC : Inter cooler
1 coolant out
IC
C
T
.
.
Wc
Qs 2
coolant in
5
.
Wt 4
NR 3
R 6
Given : p1 = 5 bar ; p2 = 32.5 bar ; p3 = 31.5 bar ; p4 = 5.25 bar ; ε = 0.83 ; ηc = 0.80 ; .
ηt = 0.88 ; T3 = 945 K ; T1 = 20 + 273 = 293 K ; Wn = 650 kW ; .
To find : (i) ηcycle ; (ii) wc and wt; (iii) Qs and ma
84
Solution: 3
T
4 5
4’
2 2’
6
1 s
For process 1 – 2’, T2’ = T1 (p2 / p1) (γ – 1 ) / γ = 293 x (32.5 / 5) 0.286 = 500.45 K. Hence
T2 = T1 + (T2’ – T1) / ηc = 293 + (500.45 – 293) / 0.80 = 552.3 K.
For process 3 – 4’, T4’ = T3 (p4 / p3) (γ – 1 ) / γ = 945 x (5.25 / 31.5) 0.286 = 566.1 K. Hence
T4 = T3 – ηt [T3 – T4’] = 945 – 0.88 x [945 – 566.1] = 611.5 K
T5 = T2 + ε [T4 – T2] = 552.3 + 0.83 x [611.5 – 552.3] = 601.44 K. Compressor work per unit mass of air = wc = Cp[T2 – T1] = 1.005 x [552.3 – 293] = 260.6 kJ/kg Turbine work per unit mass of air = wt = Cp[T3 – T4] = 1.005 x [945 – 611.5] = 335.17 kJ/kg Net work out put = wn = 335.17 – 260.6 = 74.57 kJ/kg Heat supplied per unit mass of air = qs = Cp[T3 – T5] = 1.005 x [945 – 601.44]
85
= 345.3 kJ/kg (i) Thermal efficiency = ηcycle = 74.57 / 345.3 = 0.2159 = 21.59 % .
Mass flow rate of air = ma = Wn / wn = 650 / 74.57 = 8.72 kg /s .
(ii) Hence Turbine power = Wt = ma wt = 8.72 x 335.17 = 2922.7 kW .
Compressor Power = Wc = ma wc = 8.72 x 260.6 = 2271.6 kW .
(iv)Heat supplied in the reactor per unit time = Qs = ma qs = 8.72 x 345.3 = 3011 kW
86
Chapter 4 Vapour Power Cycles 4.1. Introduction:Gas power cycles with the possible exception of the Gas turbine cycle are not suitable for the use of fuels like coal. Since the fuel has to be burnt in the engine cylinder, the internal combustion cycle cannot use fuels that leave large amounts of refuse in the combustion chamber. Even in the open Brayton cycle, it is not possible to use coal, because, the gases after combustion pass into the turbine, and these gases have to be totally free from abrasive materials like ash. But the Vapour power cycles use a working substance which does not contact the fuel. So, impurities in the fuel will not affect the working substance or the machine through which the hot fluid has to pass in doing work. Secondly, in gas power cycles it is extremely difficult to achieve an isothermal process where as this can be easily achieved in a vapour power cycle using constant pressure phase change process.Vapour power cycle has the further advantage that it can use high speed, light weight turbines to produce work output instead of the bulky reciprocating piston engines that are used in internal combustion engines. Another advantage of vapour power cycle over the Brayton cycle is that compression work is very small as to neglect it in comparison with the net work output. But vapour power cycles suffer from poor thermal efficiencies as compared to gas power cycles. High efficiencies in vapour power cycles can be achieved only by using very high pressure or super-critical pressure system with multi stage feed water heating and reheating. Power plants employing vapour power cycles have water as the working substance, and pressures around 250 bar have to be used to reach the super-critical state, with temperatures around 600 C. Even under such conditions the overall thermal efficiency can be as high as 40 %, but the life of the equipment will be seriously affected, unless high quality and expensive materials are used. In Brayton cycle temperatures of the order of 1100 K are reached as the equipment is not subjected to steady high pressure continuously. In reciprocating engine cycles, even temperatures greater than 2500 K may be reached, since there is no steady high temperature and pressure maintained during the operation of the engine. 4.2 The Rankine Cycle (Ideal simple vapour power cycle) Assumptions made in the analysis of Ideal Vapour power cycles:- (i) The expansion process in the turbine and the compression process in the pump are isentropic. (ii) There are no pressure losses in the piping connecting various components as well as in the heat exchangers like boiler, condenser, re-heaters and feed water heaters. (iii) Changes in kinetic and potential energies of the working fluid as it flows through the various components are negligible. (iv) Fluid flow is steady and one-dimensional A schematic diagram for the Rankine cycle along with the corresponding T – s and h – s are shown in Fig.4.1.
87
P: Feed water pump; B: Boiler; T: Turbine; C: Condensor
3 T
wt
B
qs
4 C
qc
2 wp
P
1
Fig.4.1(a) Schematic diagram for Rankine cycle T
h
s
s
Fig.4.1(b) T – s and h – s diagrams for Rankine cycle Process 1 -2 : Isentropic pumping of feed water. During this process the feed water pressure is raised from condenser pressure to boiler pressure by doing work on it. Process 2 – 3: Constant pressure heating of water in the boiler till it becomes a saturated dry steam. Process 3 – 4: Isentropic expansion of steam in the turbine from boiler pressure to the condenser pressure. During this process work is done by the steam on the surroundings.
88
Process 4 – 1: Constant pressure condensation of steam in the condenser till it becomes a saturated liquid. During this process heat is rejected by the working substance to the surroundings 4.2.1 Expressions for net work output and thermal efficiency: Applying steady-state, steady flow energy equation to the feed water pump we have .
.
Wp = m (h2 – h1) …………………….(4.1) .
Where m is the mass flow rate of feed water through the pump. or per unit mass of water the pump work is given by .
.
wp = Wp / m = (h2 – h1) ………………(4.2) .
Similarly for the turbine we have
.
wt = Wt / m = (h3 – h4) ……………..(4.3) .
.
qs = Qs / m = (h3 – h2) ………………(4.4) .
and
.
qc = Q / m = (h4 – h1)……………….(4.5)
Net work output per unit mass of steam = wn = wt - wp = (h3 – h4) – (h2 – h1) Therefore thermal efficiency is given by wn (h3 – h4) – (h2 – h1) (h4 – h1) η Rankine = --------------- = --------------------------- = 1 − ----------------……...(4.6) qs (h3 – h2) (h3 – h2) It can be seen from Eq. (4.5) and (4.6) the net work output and thermal efficiency of the Rankine cycle depends on the enthalpies h1,h2, h3, and h4, which in turn depends on the boiler pressure, the condenser pressure and the temperature at which steam is entering the turbine. 4.2.2.Expression for Thermal efficiency of the Rankine cycle in terms of temperatures wn w1-2 + w2-3 + w3-4 + w4-1 q1-2 + q2-3 + q3-4 + q4-1 η Rankine = --------------- = ------------------------------- = -----------------------------qs q2-3 q2-3 q2-3 + q4-1 Or η Rankine = ---------------q2-3
89
s1 Now q4-1 = ∫Tds = T1(s1 – s4) s4 s3 and q2-3 = ∫Tds = Tm(s3 – s2). s2 Tm is defined such that when it is multiplied by the change in entropy during the heating process gives heat supplied during the process. Tm is called as the “thermodynamic mean temperature” at which heat is supplied to the working substance. Substituting these expressions for q4-1 and q2-3 in the expression for thermal efficiency for the Rankine cycle ,we have Tm(s3 – s2) + T1(s1 – s4). Tm(s3 – s2) − T1(s4 – s1). η Rankine = ------------------------------- = --------------------------------Tm(s3 – s2). Tm(s3 – s2). Since s1 = s2 and s4 = s3, the above expression reduces to (Tm – T1) T1 η Rankine = ---------------- = 1 − ------- ………………………………………(4.7) Tm Tm It can be seen from Eq.(4.7) that higher the value of Tm higher will be the thermal efficiency for a given value of the condensation temperature T1.Also if the entire heat is supplied at one constant temperature then the Rankine efficiency will be equal to the Carnot efficiency between the same temperature limits. Therefore in present day steam power plants efforts are being made to see that major portion of the heat supply takes place at one constant temperature. 4.3. Effects of pressure and temperature on the performance of the Rankine cycle 4.3.1. Effects of condsensor pressure: The effects of decreasing the condenser pressure on the performance of the Rankine cycle are illustrated both on T – s diagram and h – s diagram shown in Fig. 4.2. In the figure cycle 1-2-3-4-1 represents Rankine cycle with higher condenser pressure and cycle 1’-2’-3-4’-1’ represents the cycle with lower condenser pressure. It can be seen from both the diagrams that area of the cycle 1’-2’-34’-1’ is more than that for the cycle 1-2-3-4-1. Since area of a closed curve on a T-s diagram represents the net work transfer during the cycle, it follows that cycle with lower condenser pressure will have a higher work output as compared to the cycle with higher condenser pressure. However with decrease in condenser pressure the water has to be heated from 2’ to 3 instead of from 2 to 3. This means that lowering the condenser pressure will result in an increase in heat supply. But the increased work output will more than compensate the additional heat supply with the result the thermal efficiency of the
90
cycle also increases with decrease in condenser pressure. But lower condenser pressure has certain disadvantages. It can be seen from
T
h
s
s Fig.4.2.Effects of Condenser pressure on performance of Rankine cycle
The T – s diagram that as the condenser pressure decreases the quality of steam during the last stages of expansion also decreases (x4’ < x4) which in turn decreases the efficiency of the turbine. For good turbine efficiency and long turbine life the quality of steam during expansion should not be less than 0.9.Also if the condenser pressure is below the atmospheric pressure then there is a tendency for the atmospheric air to leak into the condenser which in turn affects the performance of the condenser in the form of decrease in ts efficiency. In such cases it becomes necessary to have air extractors fitted to the condenser at extra cost to remove air from the condenser. 4.3.2.Effects of Boiler Presure: The effects of increasing the boiler pressure on the performance of the Rankine cycle are illustrated both on T – s and h – s diagrams in Fig. 4.3.It can be seen from these diagrams that a rise in boiler pressure will result in an increase in the area representing the cycle by an amount A1 and a decrease in the area by an amount A2.Up to a certain boiler pressure say about 170 bar, A1 > A2, indicating that there is a net increase in the work out put from the cycle. Also up to this pressure the thermodynamic mean temperature at which heat is supplied increases with increase in boiler pressure there by increasing the thermal efficiency of the cycle. Beyond this pressure the thermodynamic temperature starts decreasing with increase in boiler pressure and hence the thermal efficiency decreases with increase in boiler pressure.This can be seen clearly from Fig. 4.4, which is a graph of cycle efficiency versus boiler pressure.
91
A rise in boiler pressure will also results in decrease in the quality of steam during the last stages of expansion thereby affecting the performance and life of the turbine. 4.3.3. Effects of Superheating of steam: The effects of superheating the steam before it enters the turbine on the performance of the Rankine cycle are shown both on T – s and h – s diagrams in Fig.4.5. It can be seen from these diagrams that by super heating the T
h
s
s
Fig. 4.3: Effects of boiler pressure on Rankine cycle
60
η
40
20
0 0
50
100
150 Boiler pressure (bar)
Fig. 4.4: Rankine cycle efficiency as a function of boiler pressure
92
200
steam the area representing the cycle has increased indicating that the net work output from the cycle has increased. The diagram also shows that the quality of steam during the last stages of expansion has increased thereby improving the performance of the turbine. Superheating of steam needs additional heat supply.But the increased heat supply is more than compensated by the increased work output, with the result the thermal efficiency increases with the superheating of steam before it enters the turbine. But with the present day materials used in the manufacture of steam turbines, the maximum temperature that the turbine blades can withstand is about 650 C.Fig. 4.6 shows the effect of superheating of steam on Rankine cycle efficiency.Thus superheating has two beneficial effects on the T
h
s
s
Fig. 4.5: Effects of superheating of steam on Rankine cycle
0.5 0.4 0.3 0.2 0.1 0
0
50
100
150
200
250
Fig. 4.6. Efficiency of Rankine cycle with superheated steam
93
Overall cycle efficiency. First it helps reduce moisture in the exhaust steam and friction at the turbine blades, thus increasing the internal efficiency of the turbine. Secondly, superheating increases the thermodynamic mean temperature at which heat is supplied which in turn increases the thermal efficiency of the cycle. Also with superheating, the Rankine cycle thermal efficiency continuously increases with increase in boiler pressure. 4.4. Modifications of simple Rankine cycle :- In a simple Rankine cycle, as the boiler pressure goes higher and higher, a stage is reached when superheating of steam once to 600 C is not sufficient to maintain a sufficiently low moisture level in the steam during last stages of expansion. Further very low condenser pressure will also result in poor quality of steam during last stages of expansion. Therefore, in order to take the advantages of high boiler pressure and low condenser pressure in terms of higher work output and higher thermal efficiency, the simple Rankine cycle is modified by adding additional components. One modification is the “Reheat cycle” which increases the net work output and the second modification is the “Regenerative cycle” which increases the thermal efficiency as compared to the simple Rankine cycle. 4.4.1. Reheat Cycle:- The net work output of a simple Rankine cycle can be increased by employing multistage expansion with reheating between the stages of expansion. A schematic diagram of a Reheat cycle employing two stages of expansion with reheating in between the stages is shown in Fig. 4.7(a). The corresponding T – s and h – s diagrams are shown in Fig. 4.7 (b) and Fig. 4.7 (c) respectively.
wt1 3
wt2
6
4 qs
qc
5 2 wp
Fig.4.7(a) Schematic
1
For Reheat cycle
Superheated steam at high pressure and temperature (state point 3) enters the first stage turbine T1 and expands to an intermediate pressure (state point 4) and is then reheated at constant pressure back to its original temperature (state point 5 ) by passing the steam
94
through the re-heater coils provided in the boiler. Then the steam is expanded for the second time from the intermediate pressure to the condenser pressure in the second stage
T
h
s
s
Fig.4.7(b): T – s diagram for Reheat Cycle
Fig. 4.7(c) h – s diagram for Reheat cycle
turbine. It is then condensed in the condenser at constant pressure and pumped back to the boiler where the water is converted in to superheated steam to complete the cycle. The pressure at which the steam has to be withdrawn for reheating should be chosen such that the work output from the turbine is a maximum for the given boiler exit conditions and the condenser pressure. If this reheating pressure p4 is too high, the benefit obtained by reheating will be limited; on the other hand, if p4 is too low, the pressure losses occurring in the pipe line carrying steam to be reheated will be so great as to off-set the advantages gained by reheating. It is best to choose p4 such that the thermodynamic mean temperature for the entire heating process, including the boiler, the superheater and the reheater becomes the highest attainable. If T5 = T3, p4 may be approximated as 25 % of p3. Expression for net work output and thermal efficiency in terms of enthalpies Pump work per unit mass = wp = (h2 – h1). Turbine work per unit mass = wt = wt1 + wt2 = (h3 – h4) + (h5 – h6) Net work out put = wn = wt – wp = (h3 – h4) + (h5 – h6) – (h2 – h1) Heat supplied per unit mass = qs = q2-3 + q4-5 = (h3 – h2) + (h5 – h4)
95
wn [(h3 – h4) + (h5 – h6) – (h2 – h1)] Thermal efficiency = ηReheat = -------- = --------------------------------------- …….(4.8) qs [(h3 – h2) + (h5 – h4)] 4.5.Illustrative examples on Ideal Rankine and Reheat Cycles Example 4.1:- In a simple Rankine cycle, dry saturated steam at 20 bar expands to a pressure of 1 atmosphere. Determine (i) the pump work, (ii) turbine work, (iii) network output, (iv) thermal efficiency, (v) quality of steam entering the condenser, and (vi) specific steam consumption in kg/kWh. What would be the (i) network output, (ii) cycle efficiency, (iii) specific steam consumption in kg/kWh and (iv) quality of steam entering the condenser if the condenser pressure is reduced to 0.06 bar and compare the performance of the two cycles. Case 1:- When the condenser pressure is 1 atmosphere Given:- Boiler exit pressure = p2 = p3 = 20 bar; Condenser pressure= p4 = p1 = 1.01325 bar; To find:- (i) wp; (ii) wt ;(iii) wn ; (iv) ηRankine ; (v) x4 ; (vi) s.s.c
h
s
96
Solution:- From Mollier chart, h3 = 2798 kJ/kg ; h 4 = 2305 kJ/kg ; x 4 = 0.834 ; From steam tables: h1 = hf at 1.01325 bar = 419.1 kJ/kg ; v1 = vf = 0.0010437 m3 / kg. (i) For an isentropic process we have dh – vdp = 0. p2 Hence wp = h2 – h1 = ∫vdp . For a liquid v = constant. p1 Therefore wp = h2 – h1 = v1 (p2 – p1) = 0.0010437 x {20 – 1.01325} x 10 5 = 1982 J/kg Or
wp = (h2 – h1) = 1.982 kJ/kg.
(ii) Turbine work = wt = (h3 – h4) = (2798 – 2305) = 493 kJ/kg. (iii) Net work output = wn = wt – wp = 493 – 1.982 = 491.02 kJ/kg (iv) h2 = h1 + wp = 419.1 + 1.982 = 421.08 kJ/kg. Hence heat supplied per unit mass = qs = (h3 – h2) = (2798 – 421.08) = 2376.92 kJ/kg (v) ηRanine = wn / qs = 491.02 / 2376.92 = 0.2066 = 20.66 %. (vi) Specific steam consumption in kg/kWh = 3600 / wn = 3600 / 491.02 = 7.332 kg/kWh. Case 2: When the condenser pressure is 0.06 bar For this case from Mollier chart h4 = 1955 kJ/kg ; x4 = 0.745 ; (i) h1 = hf at 0.06 bar = 151.5 kJ/kg ; v1 = vf at 0.06 bar = 0.0010064 m3 / kg. Pump work = wp = 0.0010064 x {20 – 0.06} x 105 / 1000 = 2.007 kJ/kg Turbine work = wt = 2798 – 1955 = 843 kJ/kg. Net work output = wn = 843 – 2.007 = 841 kJ/kg (ii) h2 = h1 + wp = 151.5 + 2.007 = 153.51 kJ/kg Heat supplied = qs = h3 – h2 = 2798 – 153.51= 2644.5 kJ/kg. ηRankine = wn / qs = 841 / 2644.5 = = 31.80%. (iii) Specific steam consumption = 3600 / wn = 3600 / 841 = 4.281 kg/kWh
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Comparison between the two cycles Cycle
wn (kJ/kg)
qs(kJ/kg)
η (%)
Cycle with high condensor pressure (1 atm)
491.02
2376.92
20.66
7.332
0.834
2644.50
31.80
4.281
0.745
Cycle with 841.0 Low condenser Pressure (0.06 bar)
SSC (kg/kWh)
quality of steam at condenser inlet
From the above table the following conclusions can be drawn: (i) Decreasing the condenser pressure has resulted in an increase in work output as well as the heat supplied but increase in the thermal efficiency of the cycle. (ii) Decreasing the condenser pressure also resulted in decrease in specific steam consumption an advantage. (iii) Decreasing the condenser pressure resulted in poor quality of steam during the last stages of expansion which is a disadvantage. Example 4.2:- Compare the performance of simple Rankine cycle with boiler exit steam conditions of 20 bar and dry saturated with that of another simple Rankine cycle with boiler exit steam conditions of 30 bar and dry saturated in terms of (i) net work output, (ii) heat supply, (iii) thermal efficiency, (iv) steam rate and (v) quality of steam entering the condenser. Assume the condenser pressure to be 0.06 bar for both the cycles. Solution: Case 1: Cycle with low boiler pressure of 20 bar
From Mollier chart : h3 = 2798 kJ/kg; h4 = 1955 kJ/kg; x4 = 0.745 (i) Net work out put = wn = 841 kJ/kg [ see case 2 of example 4.1] (ii) Heat supply = qs = 2644.5 kJ/kg [ see case 2 of example 4.1] (iii) Thermal efficiency = 841 / 2644.5 = .3180 = 31.8 % (iv) steam rate = 3600 / 841 = 4.281 kg/kWh (v) quality of steam at condenser inlet = x4 = 0.745. Case 2: Cycle with high boiler pressure of 30 bar
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From Mollier chart : h3 = 2800 kJ/kg ; h4 = 1900 kJ/kg ; x = 0.725 ; h1 = hf at 0.06 bar = 151.5 kJ/kg ; v1 = vf at 0.06 bar = 0.0010064 m3/kg. (i) Pump work per unit mass = wp = v1(p2 – p1) = 0.0010064 x [30 – 0.06] x 105 / 1000 = 3.013 kJ/kg Hence h2 = h1 + wp = 151.5 + 3.013 = 154.513 kJ/kg Turbine work per unit mass = wt = (h3 – h4) = (2800 – 1900) = 900 kJ/kg. Net work output = wn = wt – wp = 900 – 3.013 = 897 kJ/kg (ii) Heat supply per unit mass = qs = (h3 – h2) = (2800 – 154.513) = 2645.5 kJ/kg Hence thermal efficiency = η = wn / qs = 897/ 2645.5 = 0.3390 = 33.9 % (iii) Steam rate = 3600 / 897 = 4.0133 kg/kWh. Comparison between the two cycles Cycle
wn (kJ/kg)
qs (kJ/kg)
η (%)
steam quality of steam rate at condenser (kg/kWh) inlet
Cycle with low boiler pressure (20 bar) 841 2644.5 31.8 4.281 0.745 -----------------------------------------------------------------------------------------------------------Cycle with high boiler pressure (30 bar) 897 2645.5 33.9 4.013 0.725 -----------------------------------------------------------------------------------------------------------Example 4.3:- Compare the performance of an ideal reheat cycle with that of a simple Rankine cycle in terms of (i)net work output, (ii) thermal efficiency, (iii) steam rate, and (iv) quality of steam entering the condenser assuming the following data. Boiler exit conditions are 15 bar and 300 C. Condenser pressure is 0.1 bar. Reheater pressure is 4 bar. The steam is reheated at constant pressure back to its original temperature in the reheater. Solution: Analysis of Reheat cycle:- The h – s diagram for the Reheat cycle is shown in Fig. E4.3. Given:- p2 = p3 = 15 bar ; T3 = 300 C ; p4 = p5 = 4 bar ; p6 = 0.1 bar.
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From Mollier chart: h3 = 3035 kJ/kg ; h4 = 2750 kJ/kg ; h5 = 3060 kJ/kg ;
h
s
Fig. E4.3 : h –s diagram for example 4.3 h6 = 2405 kJ/kg ; x6 = 0.925. From steam tables, h1 = hf at 0.1 bar = 191.8 kJ/kg ; v1 = vf at 0.1 bar = 0.0010102 m3/kg. (i)Pump work per unit mass = wp = v1[p2 – p1] = 0.0010102 x [ 15 – 0.1] x 105/1000 = 1.505 kJ/kg. Turbine work per unit mass = wt = (h3 – h4) + (h5 – h6) = [3035 – 2750] + [3060 – 2405] = 940 kJ/kg. Net work output per unit mass = wn = 940 – 1.505 = 938.5 kJ/kg (ii) h2 = h1 + wp = 191.8 + 1.505 = 193.305 kJ/kg Heat supplied per unit mass = qs = (h3 – h2) + (h5 – h6) = [3035 – 193.305] + [3060 – 2750] = 3151.7 kJ/kg
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Thermal efficiency = η = wn / qs = 938.5 / 3151.7 = 0.2978 = 29.78 %. (iii) Steam rate = 3600 / wn = 3600 / 938.5 = 3.836 kg/kWh. (iv) Quality of steam entering the condenser = x6 = 0.925 Analysis of simple Rankine cycle:- In Fig. E4.3 cycle 1 – 2 – 3 – 4’ – 1 is the simple Rankine cycle. From Mollier chart : h4’ = 2195 kJ/kg ; x4’ = 0.835 (i) Pump work is same as in the Reheat cycle. Turbine work per unit mass = wt = (h3 – h4’) = 3035 – 2195 = 840 kJ/kg. Net work out put = wn = 840 – 1.505 = 838.5 kJ/kg (ii) Heat supplied per unit mass = qs = (h3 – h2) = 3035 – 193.305 = 2841.7 kJ/kg Thermal efficiency = η = 838.5 / 2841.7 = 0.2950 = 29.5% (iii) Steam rate = 3600 / 838.5 = 4.293 kg/kWh. (iv) quality of steam at condenser inlet = x4’ = 0.835 Comparison: Cycle
wn(kJ/kg)
qs(kJ/kg)
η (%)
SSC quality of steam (kg/kWh) at condenser Inlet
Reheat
938.5
3151.7
29.78
3.836
0.925
Rankine
838.5
2841.7
29.50
4.293
0.835
4.6. Deviations of the actual Rankine cycle from the ideal cycle In analysing the ideal Rankine cycle as well as the ideal reheat cycle it was assumed that (i) there is no pressure drop during steam flow through pipes, superheater and reheater colis, in the boiler as well as in the condenser., (ii) the expansion in the turbines and compression in the pump are isentropic processes, and (iii) all heat transfer processes are internally reversible. This type of cycle can never be realised in practice. The effect of losses is to distort the cycle and to reduce the net work output and at the same time increasing the external heat input.
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Because of pressure losses in the piping connecting the boiler exit and turbine inlet the turbine inlet pressure will be less than the boiler exit pressure and this reduces the expansion ratio for the turbine which in turn reduces the work output from the turbine.Similarly because of losses in the piping, bends etc between the pump exit and boiler inlet the pressure at which boiling takes place reduces which in turn reduces the expansion ratio and hence the work out put from the turbine. Because of frictional losses during expansion the actual work output from the turbine is reduced. To account for the frictional losses during expansion a parameter called “isentropic efficiency of turbine” is defined as follows: Actual work of expansion Isentropic efficiency of turbine = ηt = --------------------------------------Isentropic work of expansion The effect of frictional losses during expansion is illustrated on the T-s diagram shown in Fig.4.8. Due to friction pumping process also will not be isentropic and we define a parameter called “Isentropic efficiency of the pump” to account for the frictional losses in the pump.This is illustrated inFig. 4.8
h
s
With reference to the above diagram , isentropic efficiency for the turbine can be written as: (wt)actual (h3 – h4) ηt = ------------------- = ---------------- ……………………………….(4.9) (wt)Isentropic (h3 – h4’)
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And the isentropic efficiency of the pump is given by (wp)Isentropic (h2’ – h1) ηp = ------------------ = ------------ ………………………………….(4.10) (wp)Actual (h2 – h1) 4.7. Illustrative examples on Practical Rankine and Reheat Cycles Example 4.4:- In a simple Rankine cycle, steam conditions at the boiler exit are 10 bar and 300 C. In the pipe line between the boiler exit and turbine inlet, there is an energy loss of 50 kj/kg and a drop in pressure of 0.5 bar. The steam expands in the turbine to a pressure of 0.09 bar. The isentropic efficiency of the turbine is 0.86 and that of the pump is 0.70. Determine (i) the condition of steam entering the turbine, (ii) actual pump work per unit mass of water, (iii) turbine work per unit mass of steam (iv) net work output and thermal efficiency of the cycle, and (v) quality of steam entering the condenser. Given: Boiler exit pressure = p3’ = 10 bar; boiler exit temperature = T3’ = 300 C; q3’-3 = h3’ – h3 = 50 kJ/kg ; p3’ – p3 = 0.5 bar; p4 = 0.09 bar; ηt = 0.86; ηp = 0.70. To find:- (i) p3 and T3 ; (ii) (wp)Actual ; (iii) (wt)Actual ; (wn)Actual and η Thermal ; (v) x4 Solution: h
s
From Mollier chart ; h3’ = 3050 kJ/kg .
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Therefore h3 = 3050 – 50 = 3000 kJ/kg and p3 = 10 – 0.5 = 9.5 bar. From Mollier chart, T3 = 275 C (by interpolation) ; h4’ = 2220 kJ/kg (i)From steam tables h1 = hf at 0.09 bar = 183.3 kJ/kg, and v1 = vf at 0.09 bar = 0.0010094 m3/kg. (wp)Isentropic = v1(p2 – p1) = 0.0010094 x [10 – 0.09] x 10 5 / 1000 = 1.00 kJ/kg. (wp)Actual = (wp)Isentropic / η p = 1.00 / 0.7 = 1.4286 kJ/kg. Hence h2 = h1 + (wp)Actual = 183.3 + 1.4286 = 184.73 kJ/kg. (ii) Now ηt = (h3 – h4) / (h3 – h4’) or h4 = h3 – ηt (h3 − h4’) Or
h4 = 3000 – 0.86 x [3000 – 2220] = 2329.2 kJ/kg.
Therefore (wt)Actual = h3 – h4 = 3000 – 2329.2 = 670.8 kJ/kg. Heat supply = qs = h3’ – h2 = 3050 – 184.73 = 2865.27 kJ/kg Thermal efficiency = ηThermal = (wn)actual / qs = [670.8 – 1.4286] / 2865.27 = 0.2336 = 23.36 %. (iii) Since h4 and p4 are known, state point ‘4’ can be located on the Mollier chart and x4 can be read. Hence from Mollier chart x4 = 0.894. Example 4.5: In a reheat steam cycle, the boiler exit conditions are 25 bar and 300 C. The exit pressure of steam at the end of first stage is 5 bar. The steam is then reheated to 300 C before expanding in the second turbine to 0.05 bar. Assuming the high and low pressure turbines to have efficiencies of 87% and85 % respectively, find (i) the thermal energy input in the reheater, (ii) the cycle efficiency, (iii) specific steam consumption and (iv) power output for a mass flow rate of 2 kg/s. Given:- Boiler exit pressure =p3 = 25 bar; boiler exit temperature = T3 = 300 C; Reheater Pressure = p4 = p5 = 5 bar; Temperature of steam after reheating = T5 = 300 C; Condenser pressure = p6 = 0.05 bar; Turbine efficiency for first stage = ηt1 = 0.87; .
Turbine efficiency for second stage = ηt2 = 0.85; mass flow rate = m = 2 kg/s.
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.
To find:- (i) q4-5 ; (ii) ηthermal ; (iii) SSC ; (iv) Wn Solution: The h – s diagram for the given problem is shown below.
h
s
From Mollier chart we have: h3 = 3010 kJ/kg ; h4’ = 2680 kJ/kg ; h5 = 3060 kJ/kg ; h6’ = 2285 kJ/kg. (h3 – h4) (i) Efficiency of I stage turbine = ηt1 = -----------(h3 – h4’) Hence
h4 = h3 – ηt1 (h3 – h4’) = 3010 – 0.87 x [3010 – 2680] = 2722.9 kJ/kg.
Also turbine work for I stage per unit mass of steam = wt1 = (h3 – h4) Or
wt1 = [ 3010 – 2722.9] = 287.1 kJ/kg
Similarly, Efficiency of II stage turbine = ηt2 = Hence
(h5 – h6) -----------(h3 – h6’)
h6 = h5 – ηt2 (h3 – h6’) = 3060 – 0.85 x [3010 – 2285]
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= 2443.75 kJ/kg Turbine work of II stage = wt2 = (h5 – h6’) = [3060 – 2443.75] = 616.25 kJ / kg. Thermal energy input in the reheater per unit mass = q4-5 = (h5 – h4) Or
q4-5 = [3060 – 2722.9] = 337.1 kJ/kg.
(ii) Total heat supplied per unit mass of steam = qs = q2-3 + q4-5 Or
qs = (h3 – h2) + q4-5
Neglecting pump work,
h2 = h1 = hf at 0.05 bar = 137.8 kJ/kg.
Therefore
qs = [3010 – 137.8] + 337.1 = 3209.3 kJ/kg.
Total turbine work per unit mass = wt = wt1 + wt2 = 287.1 + 616.25 = 903.35 kJ/kg. Since pump work is neglected , wn = wt = 903.3 kJ/kg. Hence thermal efficiency = ηthermal = wn / qs = 903.3 / 3209.3 = 0.2815 = 28.15 %. (iii) Specific steam consumption = 3600 / wn = 3600 / 903.3 = 3.985 kg/ kWh. .
.
(iv) Net power output = Wn = m wn = 2 x 903.3 = 1806.6 kW. Example 4.6:- Steam at 50 bar and 350 C expands to 12 bar in a high pressure stage, and is dry saturated at the stage exit. This is now reheated to 280 C without any pressure drop. The reheated steam expands in an intermediate stage and again emerges as a dry saturated steam at a lower pressure. The steam is once again reheated to 280 C at constant pressure before it is finally expanded in the low pressure stage to 0.05 bar. Assuming the work output is the same for the high and intermediate stages, and the
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efficiencies of the high pressure stage and low pressure stage are equal, find (i) efficiency of the high pressure stage, (ii) pressure of steam at the exit of the intermediate stage, (iii) total power output from the three stages for a mass flow rate of 1 kg/s, (iv) condition of the steam entering the condenser, and (v) thermal efficiency of the cycle. T
s Fig. E4.6: h – s diagram for example 4.6. Solution: (i) On the Mollier diagram, the state points 3,4,4’ and 5 can be located and the corresponding enthalpies can be read. Therefore h3 = 3070 kJ/kg ; h4 = 2780 kJ/kg ; h4’ = 2755 kJ/kg ; h5 = 3000 kJ/kg ; h1 = hf at 0.05 bar = 137.8 kJ/kg (from steam tables) Pump work is neglected. Hence h2 = h1 Now wt1 = h3 – h4 = 3070 – 2780 = 290 kJ/kg. But wt1 = wt2 = h5 – h6. Hence h6 = h5 – wt1 = 3000 – 290 = 2710 kJ/kg. Now state point 6 can be located on the Mollier chart since h6 and the steam is dry saturated at 6. Hence from Mollier chart p6 = 2.25 bar = p7. Since T7 = 280 C state point 7 can be located on the Mollier chart. From Mollier chart h7 = 3020 kJ/kg ; h8’ = 2380 kJ/kg .
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Efficiency of high pressure stage = η t1 = (h3 – h4) / (h3 – h4’) = (3070 – 2780) / (3070 – 2755) = 0.92 = 92 %. Now η t1 = η t3 = (h7 – h8) / (h7 – h8’). Hence
h8 = h7 – η t3 (h7 – h8’) = 3020 – 0.92 x (3020 – 2380) = 2431.2 kJ/kg.
Turbine work from the low pressure stage = wt3 = h7 – h8 = 3020 – 2431.2 = 588.8 kJ/kg .
.
.
(iii) Total power output neglecting pump work = Wn = Wt = m (wt1 + wt2 + wt3) .
Hence
Wn = 1.0 x (290 + 290 + 588.8) = 1168.8 kW.
(iv) Since h8 and p8 are known, the state point 8 can be located on the Mollier digram and the quality x8 corresponding to point 8 can be read. Therefore from Mollier chart x8 = 0.9425 (v) Total heat supplied per unit mass of steam = qs = q2-3 + q 4-5 + q 6-7 Or
qs = (h3 – h2) + (h5 – h4) + (h7 – h6) = (3070 – 137.8) + (3000 – 2780) + (3020 – 2710) = 3462.2 kJ/kg. .
.
Net workoutput per unit mass of steam = wn = Wn / m = 1168.8 / 1.0 = 1168.8 kJ/kg Thermal efficiency of the cycle = η thermal = wn / qs = 1168.8 / 3462.2 = 0.3376 = 33.76 %
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4.8. Regenerative Vapour Power Cycle The thermal efficiency of a steam power plant cycle is quite low even with superheating and reheating as compared to a Carnot power cycle between the same temperature limits. For example, a steam power plant cycle with a maximum temperature of 550 C ,a boiler pressure of 80 bar and a condenser pressure of 0.08 bar will have a thermal efficiency of about 41 % as against an efficiency of 63 % for a Carnot power cycle working between the same temperature limits. By employing regenerative feed water heating, the thermodynamic mean temperature at which heat is supplied can be increased thereby increasing the thermal efficiency of the cycle. 4.8.1. Theoretical Regenerative Vapour Power cycle :- To illustrate the principle of regenerative feed water heating, let us consider a simple Rankine cycle as shown in Fig. 4.9.During the heating process 1-2, the subcooled liquid water at boiler pressure is
Fig. 4.9 : Theoretical Regenerative Feed Heating Cycle
brought to the saturation temperature. Hence the mean temperature of the fluid during this process is lower than the maximum temperature of the cycle, with the result the thermal efficiency of the Rankine cycle will be lower than that for the Carnot cycle. If the amount of heat required for this process is supplied internally instead of externally, then the cycle efficiency would approach that of Carnot cycle. This process of supplying heat internally is called regeneration.Fig.4.9 also shows the T-s diagram for a theoretical regenerative cycle. If it is possible to expand the steam in a reversible manner so that the area under the path 3 – 3’ would be exactly equal to the area under the path 1 – 2’, then we would have all the heat supplied externally at T3 (T3 = T2’), and all heat rejected another constant temperature T1. Consequently, the thermal efficiency of the ideal
109
regenerative cycle will be equal to that of a Carnot cycle working between the same temperature limits. To carry out the ideal regenerative process, the liquid leaving the feed water pump at state 1 is made to circulate through the coils around a regenerative turbine through which the steam is expanding from state 3.The feed water is heated regeneratively by absorbing heat from the expanding steam to reach saturated liquid state at 2’.The feed water then enters the boiler where it is vapourised to become dry steam.Steam, after non-adiabatic expansion to state 3’ , finally expands isentropically to state 4’ before condensing in the condenser. The ideal cycle described above cannot be achieved in practice because of the complexities involved in the construction of a regenerative turbine. Also, even if it is possible to build a regenerative turbine, the quality of steam during the last stages of expansion will be very poor so that there is considerable decrease in the efficiency and life of the turbine. Therefore, in practice, the regenerative feed water heating is achieved by bleeding a small fraction of steam from the turbine and use this steam to heat the feed water in a feed water heater. 4.8.2. Practical Regenerative Feed Heating Cycle : A schematic diagram of a practical regenerative cycle employing a single feed water heater is shown in Fig. 4.10 (a). The corresponding T – s and h – s diagrams are shown in Figures 4.10 (b) and 4.10 (c) respectively.
Turbine
.
m
.
3
Boiler
Qs
. WT 4
.
2 . Wp2
m1 .
P2
m m
5 .
.
(m – m1) 1 Condenser 7
Feed Water Heater
P1 6 Fig. 4.10 (a) : Schematic of a Practical Regenerative cycle employing a single feed water heater
110
Fig. 4.10 (b) : T – s diagram Fig. 4.10 (c): h – s diagram for regenerative cycle employing a single feed water heater
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CHAPTER 5 RECIPROCATING COMPRESSORS 5.1.Introduction :- This chapter deals with thermodynamic analysis (energy analysis) of reciprocating gas compressors. Gas compressors are devices in which mechanical work is done on the gas, raising its pressure. Energy analysis is one of the many major aspects to be considered in order to design a gas compressor. Compressed gas has many industrial applications. It is used to transport solid material, to provide control air for pneumatic systems, to drive tools in construction industries and so on. Compressors are also part of refrigeration and air conditioning systems. Compressors are broadly classified into two types: (i) positive displacement type and (ii) turbomachine type. In a positive displacement type of compressor, the interaction between the moving part and the fluid involves a change of volume, translation of the fluid or both. Fluid compression or expansion thus occurs without an appreciable displacement of the mass centre of gravity of the contained fluid. Hence changes in kinetic energy and momentum may be neglected in such machines. In the positive displacement machine the fluid cannot escape the boundaries except for leakage. The action is therefore nearly static. In the case of a turbomachine, the fluid undergoes change in momentum and kinetic energy due to dynamic action between the flowing fluid and a rotor. Positive displacement compressors are further classified as (i) reciprocating type and (ii) rotary type. This chapter deals with analysis of reciprocating type of compressors.In a reciprocating air compressor, air is sucked in to the compressor cylinder, it is then compressed to the required pressure and then exhausted into a receiver where it is stored for further use. 5.2. Classification of Reciprocating Compressors:Reciprocating compressors are classified according to (i) the number of surfaces (one or two) of the piston which actually participates in the compression process, (ii) the number of stages employed to get the desired pressure ratio and (iii) the number of cylinders employed. According to the first classification, compressors are classified into single acting compressor and double acting compressor. In the case of single acting compressor, only one side of the piston face (see Fig. 5.1) participates in the compression process.One revolution of the crank is required to complete one cycle of operation. Hence if N is the speed of the compressor say in RPM then the number of cycles per minute executed by the compressor, Nc is equal to N. On the other hand, in the case of a double acting compressor, since both sides of the piston are participating in the compression process (i.e. if compression is taking place on one side of the piston, suction is taking place on the other side) the number of cycles per unit time will be equal to twice the speed of the compressor, i.e. Nc = 2N if the cross sectional area of the piston rod is negligible.
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Otherwise Nc = 2 { 1 – (a / A)}N, wher a = area of cross section of the piston rod and A is the area of cross section of the cylinder.
(a) single acting compressor
(b) double acting compressor
Fig. 5.1 : Single acting and double acting compressors Compressors are also classified as single - stage compressors and multi-stage compressors. In the case of a single stage compressor the entire compression process from the surroundings pressure to the required delivery pressure is carried out in a single stage, where as in a multi stage compressor the whole compression process is divided into a number of stages, so that after the compression in each stage, the gas is cooled in an inter cooler at constant pressure back to its original temperature and then compressed
113
Wc
(Wc)I stage p1,T1
p1,T1
First stage Compression
p2, T2
pi,Ti
Coolant out Single stage Compressor
Intercooler Coolant in
p2,T2
pi,T1 Second stage Compression
Two-stage compressor
(Wc)II stage
Fig. 5.2 : Difference between single stage and two stage compressor in the next stage. Fig. 5.2 shows the difference between a single stage compressor and a two-stage compressor. The total work required in a two stage compressor is equal to the sum of work required in I stage and II stage, which will always be less than the work required in a single stage compressor for the same delivery pressure. There are other advantages of multi-stage compressors over a single stage compressor which will be discussed later. Whenever a large quantity of gas are required at high pressure it is not advisable to employ a single cylinder as in that case the size of the cylinder will be very bulky. Under such circumstances, the required quantity of gas to be compressed is divided into smaller quantities compressed in separate cylinders and then discharged into a common receiver. The suction and delivery pressures for all the cylinders are one and the same.Fig. 5.3 shows the difference between a compressor with one cylinder and a compressor with two cylinders.
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Wc .
m,p1,T1
Single cylinder compressor
m,p2,T2 Wc1 .
m/2,p1,T1 Cylinder 1
m/2,p2,T2 .
m,p2,T2 Wc2 = Wc1 m,p1,T1
Cylinder 2 .
m/2,p2,T2
m,p1,T1
Fig. 5.3 : Difference between a single cylinder compressor and a two-cylinder compressor. 5.3. Expression for work done on a gas in a single stage compressor without clearance The p – V diagram for a single stage compressor without clearance volume is shown in Fig.5.4. Line 0 – 1 represents suction of air when the piston moves from its inner most position to outer most position. During this process both pressure and temperature of the gas remains constant.Line 1 – 2 represents the compression process where in the gas is compressed from the suction pressure to the required delivery pressure according to the law pV n = Constant. Line 2 – 3 represents the delivery of the gas from the cylinder to a receiver where the gas is stored at this pressure.During the delivery stroke both pressure and temperature of the gas remains constant.The wokdone on the gas per cycle is given by
115
p 3
2 pV n= C
0
1
V Fig. 5.4: p – V diagram for a single stage compressor without clearance 1
2
3
Wc = W0-1 + W1-2 + W2-3 = ∫pdV + ∫pdV + ∫pdV 0
1
2
[p1V1 – p2V2] = p1[V1 – V0] + -------------------- + p2[V3 – V2} (n – 1) For a compressor without clearance V0 = V3 = 0. Therefore [p1V1 – p2V2] n Wc = p1V1 – p2V2 + -------------------- = --------- p1V1 [ 1 – (p2V2) / (p1V1)] (n – 1) (n – 1) For process 1-2 we can write p1V1n = p2V2 n or V2 / V1 = (p2/p1) – 1/n. n Therefore Wc = ---------- p1V1[ 1 – (p2 / p1) (n – 1)/n ] ………………………..(5.1) (n – 1) If mc = mass of air compressed per cycle, assuming that the gas behaves as a perfect gas We have p1V1 = mcRT1. Substituting this expression in Eq. (5.1) we get
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n Therefore Wc = ---------- m cRT1[ 1 – (p2 / p1) (n – 1)/n ] ………………………..(5.2) (n – 1) Work done per unit mass of gas in order to raise its pressure from p1 to p2 is given by n w = Wc / mc = ---------- m cRT1[ 1 – (p2 / p1) (n – 1)/n] ………………..(5.3) (n – 1) It can be seen from Eq. (5.3) that work input required to compress unit mass of a gas from pressure p1 to p2 depends on (i) the index of compression ‘n’, (ii) the desired pressure ratio p2/p1 and (iii) the suction temperature T1.Therefore for given suction conditions (T1 is fixed), w depends on ‘n’ and p2/p1. The effects of these two parameters on w is shown in Fig. 5.5 where absolute value of w (non dimensionalized) is plotted
4 |w| /(mcRT1) 2
1 0 1
3
5
7
9
11
13
p2 / p1 Fig. 5.5. Variation of work input with compression index against p2/p1 for different values of the index ‘n’.It is clear from this plot that for a given pressure ratio, the work input increases with increase in ‘n’ and is a minimum for n=1(a compression index of n < 1 cannot be attained in practice as a gas cannot be cooled below its original temperature, unless a refrigeration process is used). Therefore for minimum work of compression , the index n should be unity which means that the compression process should be according to the law pV = constant which for a perfect gas means that T = constant or the compression process should be isothermal.
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Therefore (w) minimum = RT1 ln (p2 / p1) ………………………………………..(5.4)
5.4. Expression for work done on a gas in single stage compressor with clearance In practice a compressor without clearance does not exist. Some clearance space between the piston face and the cylinder head, when the piston is in its innermost position is necessary to prevent the piston from colliding with the cylinder head and therby providing a “cushioning effect” when the piston moves in. When the clearance has been provided, the gas in the clearance space alternately expands and contracts as the piston moves back and forth and has no effect otherwise on the compressor cycle. The gas in the clearance space after each cycle has to be first expanded to the suction pressure before fresh gas can enter the cylinder during suction stroke of the next cycle. Therefore the ideal p – V diagram for a compressor with clearance will be as shown in Fig. 5.6. Process 3 – 0 is the expansion of gas in the clearance space from delivery pressure to suction pressure. Fresh gas will enter the compressor at 0 and stops at 1.Process 1 -2 is the compression of the gas to the desired pressure and process 2 – 3 is the delivery of high pressure gas from the cylinder to a receiver in which the high pressure gas is stored and used as and when required. p 3
2 pV ne = const pV nc = const 1 0 V Vs = V1 – V3 Clearance volume = V3
Fig. 5.6: p – V diagram for a compressor with clearance The work done on the gas per cycle is given by 1
2
3
0
Wc = W0-1 + W1-2 + W2-3 + W3-0 = ∫pdV + ∫pdV + ∫pdV + ∫pdV 0
1
2
3
[p1V1 – p2V2] [p3V3 – p0V0] = p1[V1 – V0] + ------------------ + p2[V3 – V2] + --------------------(nc – 1) (ne – 1)
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Since p0 = p1 and p3 = p2, the above equation can be simplified as [p1V1 – p2V2] [p3V3 – p0V0] Or Wc = [p1V1 – p0V0] + -------------------- + [ p3V3 – p2V2] + ------------------(nc – 1) (ne – 1) 1 1 = [p1V1 – p2V2] { 1 + ---------- } – [p0V0 – p3V3] { 1 + ------------ } (nc – 1) (ne – 1) nc ne = --------- p1V1[ 1 – (p2V2/p1V1) ] − ------------- p0V0 [ 1 – (p3V3 / p0V0] (nc – 1) (ne – 1) For process 1-2 we have p1V1nc = p2V2nc or V2 / V1 = (p2 / p1) −1/nc. Similarly for process 3 -0 we have
V3/V0 = (p3/p0) − 1/ne.
Substituting these expressions in the expression for Wc and simplifying we get nc ne (nc – 1)/nc Wc = ----------- p1V1 [ 1 – (p2/p1) − ------------- p0V0 [1 – (p2/p1) (ne – 1)/ne] (nc – 1) (ne – 1) ……………………(5.5) Normally the index for compression is taken as equal to the index for expansion, as the error involved in such an assumption is negligible (this is illustrated in example 5.1) Hence assuming that nc = ne = n, Eq.(5.5) reduces to n n (n – 1)/n Wc = ----------- p1V1 [ 1 – (p2/p1) ] − ------------- p0V0 [1 – (p2/p1) (n – 1)/n] (n – 1) (n – 1) Noting that p0 = p1, the above expression for Wc can be written as n Wc = ---------- p1(V1 – V0) [ 1 – (p2/p1) (n – 1)/n] ………………………………(5.6) (n – 1) (V1 – V0) is the volume of air sucked and compressed per cycle as measured at suction conditions. Therefore if mc is the mass of gas compressed per cycle then
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p1(V1 – V0) = mcRT1. Therefore Eq. (5.6) can also be written as n Wc = ---------- mc RT1 [ 1 – (p2/p1) (n – 1)/n] ………………………………….(5.7) (n – 1) n Work done per unit mass of gas = w = Wc / mc = ---------- RT1 [1 – (p2/p1) (n – 1)/n] (n – 1) This expression is same as the one obtained for a compressor without clearance, indicating that the work input per unit mass of gas to achieve the desired pressure ratio is independent of the clearance volume. 5.5. Volumetric Efficiency for a reciprocating compressor For a compressor with clearance, the volume of air sucked as measured at suction conditions is less than the swept volume of the piston. In order to denote how efficiently gas is drawn into the compressor the term “volumetric efficiency” is used. The ideal volumetric efficiency is defined as the ratio of the actual mass of gas drawn to the mass of gas filling the stroke volume at suction conditions.
i.e.
mc p1(V1 – V0) / RT1 (η v)ideal = --------- = --------------------------mswept p1(V1 – V3) / RT1
or
(V1 – V0) (η v)ideal = ------------- ……………………………..(5.8) (V1 – V3)
Normally the compressor specifications are given as the pressure ratio that has to be developed and the volume of gas that has to be compressed per unit time as measured at free air conditions (surrounding conditions). In that case the volumetric efficiency is defined as the ratio of volume of gas compressed as measured at surroundings conditions to the swept volume of the piston. This is the actual volumetric efficiency of the compressor. Thus Vf (η v) actul = ---------------- ………………………….(5.9) (V1 – V3) If the surroundings pressure is pf and temperature is Tf and Vf is the volume of gas compressed per cycle as measured at these conditions then we have p1(V1 – V0) pfVf mc = ----------------- = ------------ Or Vf = (p1/pf)(Tf/T1) (V1 – V0) RT1 RTf
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Substituting this expression for Vf in Eq. (5.9) we get p1Tf (V1 – V0) (η v)actual = ------------------- = (p1Tf / pfT1) (η v) ideal ………….(5.10) pf T1 (V1 – V3) Expression for Ideal Volumetric Efficiency in Terms of Clerance Ratio, Pressure Ratio and index of expansion
(V1 – V0) (η v)ideal = ------------------(V1 – V3) (V1 – V3) + (V3 – V0) = --------------------------(V1 – V3) V3 = 1 + ------------ [ 1 – V0 / V3] (V1 – V3) V3 / (V1 – V3) = Clearance ratio = C ; V0 / V3 = (p3 / p0) 1/ne = (p2 / p1) 1/ne. Substituting these expressions we have (η v)ideal = 1 + C [ 1 – (p2 / p1) 1/ne] ………………….(5.11) The above equation is represented graphically in Fig. 5.7. It can be seen from this figure that the volumetric efficiency diminishes very rapidly as the clearance ratio and pressure ratio increases. 5.6. Actual p – V diagram for a single stage compressor While deriving the expression for the work done on a gas in order to compress it from pressure p1 to pressure p2, it was assumed that during the suction stroke the pressure and temperature of the gas inside the cylinder remains constant and similarly during the delivery stroke also the pressure and temperature of the gas inside the cylinder remains constant.The use of spring loaded valves causes a defect called “valve flutter” because of which there will be fluctuations in suction and delivery pressures.Hence the intake and delivery lines on the p – v diagram will be more or less sinuous rather than straight lines.. Further it is necessary that there exists a finite pressure difference between the surroundings and the cylinder interior so that the inlet valve opens and gas is sucked into the cylinder. Similarly, compressed gas will flow out of the cylinder only if its pressure is
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(η v)Ideal 1.0 1 / C = 20 = 15 = 10
0
p2/p1 1
10
20
30
40
50
60
70
Fig.5.7: Effect of pressure ratio and clearance on volumetric eficiency more than the receiver pressure. These effects will make the actual p – V diagram bigger than the ideal diagram.The process of compression can never be isothermal nor it can be isentropic. Usually it will be polytropic with index ‘n’ anywhere between 1 and 1.4.The work input to compressor will be minimum if the compression process is isothermal. In practice, the isothermal compression is approached either by cooling the cylinder head by spraying cold water or by circulating cold water in jacket surrounding the cylinder. The actual p – V diagram which includes the pressure drops across the valves, the valve flutter and non-adiabatic compression is shown in Fig.5.8.The average pressure during intake p1, is less than the atmospheric pressure by δp1 (δp1 is the inlet valve pressure drop), while the average pressure at exit,p2 is greater than the receiver pressure by δp2 which is the exit valve pressure drop. The real indicator diagram has its corners rounded at the end of compression and expansion processes as compared with the ideal diagram. Further the actual area of the diagram is greater than the theoretical area so that the actual work input is greater than that for an ideal compressor given by Eqs (5.6) and (5.7).In order to account for these deviations in the actual p – V diagram, a factor called “diagram factor,K” is introduced, which is defined as the ratio of the area of the actual p – V diagram to the area of the ideal p – V diagram so that the actual work don per cycle is given by (Wc)actual = K(Wc)ideal ………………………………..(5.12)
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p p2 pr
δp2
p1 δp1
patm
V
Fig.5.8: Actual p – V diagram for a reciprocating compressor
5.7. Performance Factors for Reciprocating Compressors In an actual compressor, because of irreversibilities, the required compressor work will not be equal to the ideal work of compression. To account for these irreversibilitis a parameter called “compressor efficiency” is defined. Compressor efficiency is an indication of how closely the actual compression process The compressor efficiency is defined as follows: Ideal work of compression ηc = ---------------------------------- ……………………(5.13) Actual work of compression
If the ideal work of compression is based on isentropic work of compression then the efficiency is called as “Isentropic efficiency of compressor”. If the ideal process is isothermal then the efficiency is called as “ Isothermal efficiency of compressor”. For a reciprocating compressor the ideal work of compression is the isothermal work of compression ,and hence Isothermal efficiency of compressor is used as one of the performance parameters.In the case of turbo compressors, the ideal compression is the isentropic process and hence isentropic efficiency of compressor is used as a performance parameter for turbo compressors. The actual work of compression in Eq.(5.13) is the work transferred to air in the cylinder. This will not be the work supplied by the external agency which is driving the compressor. There are
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mechanical losses due to friction which will diminish the amount of work supplied to the gas in the cylinder. To account for mechanical losses, a parameter called “mechanical efficiency” is defined as follows: Actual work of compression ηmech = -------------------------------------- ………………..(5.14) Shaft work supplied The overall compressor efficiency is defined as follows: Ideal work of compression ηoverall = ----------------------------------- = ηmech ηc ……………………………..(5.15) Shaft work supplied 5.8.Illustrative Examples on Single stage reciprocating compressors Example 5.1:- An ideal compressor has a displacement volume (stroke volume) of 14 litres and a clearance volume of 0.7 litre. It receives air at 100 kPa and discharges at 500 kPa. The compression is polytropic with index equal to 1.3 and expansion is isentropic. Assuming that air behaves as a perfect gas, determine (i) work done on air per cycle and (ii) the error involved in calculation of work done if the index for compression and for expansion are both equal to 1.3.. Solution:Given :V1 – V3 = 14 x 10 − 3 m3
p 3
2
V3 = 0.7 x 10 − 3 m3; p1 = 100 kPa ; p2 = 500 kPa;
0
1
nc = 1.3 ; ne = 1.4 To find Wc:-
V Solution:- When the index for expansion is not equal to the index for compression the work done per cycle is given by Wc = {nc/(nc – 1)}p1V1[1 – (p2/p1) (nc – 1)/nc] – {ne/(ne – 1)}p0V0[1 – (p2/p1) (ne – 1)/ne] Now V1 = (14.7) x 10 − 3 m3. V0 = (p3/p0) 1/neV3 = (500/100) 1/1.4 x 0.7 x 10 − 3
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Or V0 = 2.21 x 10 − 3 m3. Substituting the numerical values in the expression for Wc we get Wc = {1.3 /0.3) x 100 x 103 x 14.7 x 10 − 3 [1 – (500/100) 0.3/1.3] − {1.4/0.4} x 100 x 10 3 x 2.21 x 10 − 3 x {1 – (500/100)0.4/1.4] = − 2.414 x 10 3 J = − 2.414 kJ (Negative sign indicates that work is done on air by the surroundings) (ii) When nc = ne = n, work done per cycle is given by Wc = {n/(n – 1)}p1(V1 – V0) [1 – (p2/p1) (n – 1)/n] Now V0 = (500/100) 1/1.3x 0.7 x 10 −3 = 2.414 x 10 − 3 = (1.3/0.3) x 100 x 10 3 x (14.7 – 2.414) x 10 − 3 x [1 – (500/100) 0.3/1.3] = − 2.385 x 10 3 J = − 2.385 kJ Percent error in assuming the same index for both compression and expansion processe is given by % error = (2.414 – 2.385) x 100 / 2.414 = 1.2 %. Example 5.2:-A double acting compressor, with a piston displacement of 0.05 m 3 per stroke, operates at 500 rpm. The clearance is 5 percent and it receives air at 100 kPa and discharges at 600 kPa. The compression is polytropic according to the law pV 1.35 = constant. Determine the power required to drive the compressor and the mass of air delivered in kg/s if the suction temperature is 27 C Given:- Double acting,hence Nc = 2N; N = 500 rpm; V1 – V3 = 0.05 m3; V3 = 0.05(V1 – V3); p1 = 100 kPa; p2 = 600 kPa; nc = ne = n = 1.35; T1 = 27+273 = 300 K .
To find :- (i) Power input ,P ; (ii) m Solution:- Refer to the p – V diagram shown in example 5.1 V3 = 0.05 x 0.05 = 2.5 x 10 − 3 m3. Hence V1 = 0.05 + 2.5 x 10 − 3 = 0.0525 m3. For the expansion process 3 – 0 we have p0V0n = p3V3n or V0 = (p3/p0) 1/nV3 Therefore V0 = (600 / 100) 1/1.35 x 0.0025 = 9.4265 x 10 − 3 m3.
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Mass of air compressed per cycle = mc = p1(V1 – V0) / (RT1) = 100 x 103 x [0.0525 – 9.4265 x 10 − 3] / (286 x 300) Or
mc = 0.0502 kg/cycle .
Mass of air compressed per second = m = mcNc = 0.0502 x (2 x 500)/60 = 0.837 kg/s (ii) Power = P = WcNc = {n/(n – 1)}mcNcRT1[1 – (p2/p1) (n – 1) / n] = (1.35/0/35) x 0.837 x 0.286 x 300 x [1 – (600/100) 0.35 / 1.35] = − 163.8 kW Example 5.3:- A single acting air compressor has a cylinder of bore 15 cm and the piston stroke is 25 cm. The crank speed is 600 rpm.Air is taken from atmosphere (1 bar and 27 C) and is delivered at 11 bars.Assuming polytropic compression of the type pV 1.,25 = C, find the power required to drive the compressor if its mechanical efficiency is 80 %. The compressor has a clearance which is 1/20th of the stroke volume. How long will it take to deliver 1 m 3 of air at the compressor inlet conditions. Also find the volumetric efficiency of the compressor. Given: Bore = d = 0.15 m; stroke = l = 0.25 m; N = 600 rpm; p1 = 1 bar; T1 = 27 + 273 = 300 K; Single acting and hence Nc = N; p2 = 11 bar; η mech = 0.80; nc = ne = n = 1.25; V3 = (1/20)[V1 – V3] To find:- (i) time to deliver 1 m3 of air as measured at suction conditions, (ii) ηv; (iii) Power required, P Solution:- Refer to the p – V diagram shown in example 5.1. (i) Stroke volume = (V1 – V3) = (πd2/4)l = π x (0.15)2 x 0.25 / 4 = 4.42 x 10 − 3 m3. Therefore
V3 = (1/20) x 4.42 x 10 − 3 = 2.21 x 10 − 4 m3 V1 = 4.42 x 10 − 3 + 2.21 x 10 − 4 = 4.641 x 10 − 3 m3
Also p0V0n = p3V3n. Or V0 = (p3 / p0) 1/n V3= (p2/p1) 1/n V3
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V0 = (600/100) 1/ 1.35 x 2.21 x 10 − 4 = 1.51 x 10 − 3 m3.
Or
.
Volume of air compressed per minute as measured at suction conditions ,Vs is given by .
Vs = (V1 – V0)Nc= [4.641 – 1.51] x 10 − 3 x 600 = 1.88 m3/min. Hence time required to compress 1 m3 of air as measured at suction conditions is given By
t = 1 /{(V1 – V0)Nc} = 1/1.88 = 0.532 min = 31.91 s.
(ii) Volumetric efficiency = (η v)ideal
(V1 – V0) (4.641 – 1.51) x 10 − 3 = ------------- = --------------------------- = 0.7084 (V1 – V3) 4.42 x 10 − 3 = 70.84 %.
(iii) Power required to drive the compressor is given by P = (1/ηmech){n/(n – 1)}p1(V1 – V0)[1 – (p2/p1) (n – 1)/n]Nc 1 1.25 1 x 10 5 = ------- x -------- x ------------ x (3.131 x 10 − 3) x [1 – (600/100)0.25/1.25] x (600/60) kJ/s 0.80 0.25 1000 = − 12 kW. Example 5.4:- A reciprocating compressor has a 5 % clearance with a bore and stroke of 25 x 30 cm. The compressor operates at 500 rpm. Air enters the cylinder at 27 C and 95 kPa and discharges at 2000 kPa. If the indices for both comprssion and expansion are equal to 1.3 Determine (i) volumetric efficiency, (ii) the volume of air handled at inlet conditions in m3/s, (iii) the power required to drive the compressor if the mechanical efficiency is 90 %, (iv) the mass of air delivered in kg/s, (v) the mass of air in the clearance space. Given:- V3 = 0.05 (V1 – V3); d = 0.25 m; l = 0.3 m; N = 500 rpm; p1 = 95 kPa; T1 = 27 + 273 = 300 K; p2 = 2000 kPa; η mech = 0.90; nc = ne = n = 1.3 Since nothing has been mentioned whether the compressor is single acting or double acting, it is assumed that it is single acting. Hence Nc = N. .
To find:- (i) (ηv)ideal ; (ii) (V1 – V0)Nc; (iii) Power, P; (iv) m (v) m3 Solution:- Refer to the p – V diagram shown in example 5.1.
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V1 – V3 = (πd2/4)l = (π x 0.252/4) x 0.30 = 0.01473 m3. Therefore V3 = 0.05 x 0.01473 = 7.365 x 10 − 4 m3. Hence V1 = 0.01473 + 7.365 x 10 − 4 = 0.0155 m3. For process 3 – 0 we have p0V0n = p3V3n. Or V0 = (p3/p0) 1/nV3. V0 = (2000/95) 1/1.3 x 7.365 x 10 − 4 = 4.6585 x 10 − 3 m3.
Hence
(V1 – V0) [0.0155 – 4.6585 x 10 − 3] Therefore (ηv)ideal = -------------- = --------------------------------- = 0.736 = 73.6 %. (V1 – V3) 0.01473 (ii) Volume of air delivered per second as measured at suction conditions is given by .
Vs = (V1 – V0)Nc = {0.0155 – 4.6585 x 10 − 3} x 500 /60 = 0.0903 m3 / s. 1 n (iii) P = WcNc / η mech = ------ x ------------- p1(V1 – V0)Nc [1 – (p2/p1) (n-1)/n] η mech (n – 1) 1 1.3 95 x 10 3 = ---------- x --------- x ---------- x 0.0903 x [ 1 – (2000/95) 0.3/1.3] 0.90 0.3 1000 = − 42 kW .
.
(iv) Mass of air compressed per second = m = Vs p1 /(RT1) 0.0903 x 95 x 10 3 = ----------------------- = 0.01 kg/s 286 x 300 (v) Mas of air in the clearance space = m3 = p3V3/(RT3) Now T3 = T2 = (p2/p1) (n – 1)/nT1 = (2000/95) 0.3 / 1.3 x 300 = 606 K 2000 x 10 3 x7.365 x 10 − 4 Therefore m3 = ----------------------------------- = 0.0085 kg 286 x 606 Example 5.5:- A single cylinder single acting air compressor takes air from atmosphere (1.0315 bar and 25 C) and delivers at 9 bar.The compressor running at 900 rpm, delivers
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1 kg of air per minute. The compression index is 1.25. The stroke to bore ratio is 1.25 and the mechanical efficiency is 83 %. Calculate : (i) the cylinder dimensions, (ii) the power required to drive the compressor, and (iii) the heat transfer during compre assumptions needed to solve this problem. Given:- p1 = 1.03125 bar; T1 = 25 + 273 = 298 K; p2 = 9 bar; N = 900 rpm; Single acting .
and hence Nc = N; m = 1.0 kg/min; ηmech = 0.83; l/d = 1.25. .
To find:- (i) d,l; (ii) P; (iii) Q1-2 Solution:- Since clearance ratio is not given, it is not possible to determine the stroke volume for the given mass of air to be compressed and hence it is necessary to assume that clearance is negligible. In that case the p – V diagram will be as shown below. p 3
2
0
1 V
.
(i) mc = m / Nc = 1/900 = 1.11 x 10 − 3kg / cycle. 1.11 x 10 − 3 x 286 x 298 p1V1 = mcRT1 or V1 = (mcRT1)/p1 = ------------------------------- = 9.337 x 10 − 4 m3 1.01325 x 10 5 Now V1 = (πd2/4)l = (πd2x 1.25d)/4. Hence d = [(4V1) / (1.25 π) ] 1/3 = [(4 x 9.337 x 10 − 4) / (1.25 x π)] 1/3
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= 0.0983 m. Hence l = 1.25 x 0.0983 = 0.123 m. .
(ii) P = (1/ηmech){n /(n – 1)}m RT1[1 – (p2/p1)(n – 1)/n] 1 1.25 1 = -------- x ----------- x ------ x 0.286 x 298 x [1 – (9/1.01325) 0.25/1.25] 0.83 0.25 60 = − 2.344 kW. (iii) Applying first law for the compression process1-2 we have .
.
.
Q1-2 − W1-2 = m (u2 – u1) .
.
.
Or Q = W1-2 + m Cv [T2 – T1] .
.
= m [p1v1 – p2v2] / (n – 1) + m Cv [T2 – T1] .
.
= m [RT1 – RT2] / (n – 1) + m Cv [T2 – T1] .
= m (T2 – T1) [ Cv – {R /(n – 1)}] Now T2 = T1(p2/p1)(n-1)/n = 298 x (9/1.01325)0.25/1.25 = 461.23 K .
Hence Q1-2 = (1/60) x [461.23 – 298] x [0.719 – {0.286/0.25}] = − 1.156 kW Example 5.6:- A single-cylinder double acting air compressor is to deliver air at 10 bar, starting from atmospheric air at 20 C. The cylinder bore is 15 cm and the stroke is 20 cm. The compressor speed is 600 rpm.Assuming the temperature of the delivered air to be 200 C, find, (i) the mass of air delivered in kg/min, (ii) the index of compression, and (iii) the power needed to drive the compressor. Assume that the effective cylinder area on the piston side is 90 % of the total area. Given: Double acting compressor with Nc = (1 +0.9)N = 1.9N; N= 600 rpm; p1 = 1 atm T1 = 20 + 273 = 293 K; p2 = 10 bar; d = 0.15m; l = 0.2 m; T2 = 200 + 273 = 473 K .
To find: (i) m ; (ii) n; (iii) P Solution: Clearance is neglected. Hence p – v diagram will be as shown in example 5.5. π x 0.15 2 x 0.2 Stroke volume = V1 = (πd /4)l = ---------------------- = 3.534 x 10 − 3 m3 4 2
.
m = mcNc = (p1V1/RT1)Nc = [1.01325 x 105 x 3.534 x 10− 3 x 1.9 x 600] / (286 x 293)
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.
Or m = 4.8714 kg/min. (ii) T2 / T1 = (p2/p1) (n – 1)/n or (n – 1) / n = ln(T2/T1) / ln(p2/p1) ln(473/293) Hence (n – 1) / n = ----------------------- = 0.21 ln(10/1.01325) Or n = 1.266. .
.
(iii) P = (1/ηmech){n/(n – 1)}mRT1 [1 – (p2/p1) (n – 1)/n] = (1/ηmech){n/(n – 1)}mR(T1-T2) 1 1.266 4.8714 = ------ x --------- x ------------- x 0.286 x [293 – 473] = − 19.9 kW 1.0 0.266 60 Example 5.7:- A single acting single cylinder air compressor runs at 750 rpm. The heat transfer during compression is 1/3rd of the total compression work. The working pressure limits are 1 bar and 10 bar.If the clearance volume is 1/30th of the stroke volume and if the stroke to bore ratio is 1.2, compute (i) the index of compression, (ii) the volumetric efficiency, (iii) the cylinder dimensions to deliver 1 kg/min of air taken from atmosphere at 30 C and (iv) power required to drive the compressor if the mechanical efficiency is 80 percent. Given: Single acting and hence Nc = N; N = 750 rpm; Q1-2 = (1/3)W1-2; p1 = 1 bar; .
p2 = 10 bar;V3 = (1/30)(V1 – V3); l/d = 1.2; m = 1.0 kg/min; T1 = 30 + 273 = 303 K; ηmech = 0.80. To find: (i) n; (ii) (ηv)ideal; (iii)d,l; (iv)P Solution: The p – V diagram will be as shown in example 5.1. (i) For the compression process the first law equation is Q1-2 − W1-2 = U2 – U1 Hence (1/3)W1-2 – W1-2 = U2 – U1 Or
− (2/3)W1-2 = U2 – U1
Per unit mass the above equation can be written as
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− (2/3) [RT1 – RT2}/(n – 1) = Cv{T2 – T1] Solving for ‘n’ we have
2 x 0.286 n = 1 + (2/3) (R/Cv) = 1 + ------------- = 1.265 3 x 0.719
.
(ii) mc = m / Nc = 1/750 kg/cycle. (1/750) x 286 x 303 Hence V1 – V0 = (mcRT1) / p1 = -------------------------- = 1.155 x 10 − 3 m3 /cycle. 1.0x 10 5 But V0 = (p3/p0) 1/n V3 = (10/1) 1/1.265 V3 = 6.173 V3. Hence Also
V1 – 6.173V3 = 1.155 x 10 − 3……………………………(a) V3 = (1/30) (V1 – V3) or V1 = 31V3.
Substituting this in Eq.(a) we have (31 – 6.173)V3 = 1.155 x 10 − 3 Or
V3 = 1.155x 10 − 3 / (31 – 6.173) = 4.652 x 10 − 5 m3
Hence V1 = 31 x 4.652 x 10 − 5 = 1.442 x 10 − 3 m3 And V0 = 6.173 x 4.652 x 10 − 5 = 2.872 x 10 − 4 m3. (V1 – V0) 1.155 x 10 − 3 Therefore (ηv)ideal = ------------- = --------------------------------------(V1 – V3) [1.442 x 10 − 3 – 4.652 x 10 − 5 ] = 0.8277 = 82.77 % (iii) Stroke Volume = Vs = V1 – V3 = (1.442 – 0.04652) x 10 − 3 = 1.395 x 10 − 3 m3. But Vs = (πd2/4)l = (πd2/4) x 1.2 d Or d = (4Vs / 1.2 π) 1/3. Hence d = { (4 x 1.395 x 10 − 3) /(1.2 x π) }1/3 = 0.114 m. Hence l = 1.2 x 0.114 =0.137 m Example 5.8:- A single cylinder single acting air compressor uses “isothermal” compression to compress 0.7 kg/min of air from 1 bar and 25 C to 7 bar, while running at 600 rpm. The clearance volume is 1/25th of the stroke volume which is 1.2 litres. If the
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actual area of the indicator diagram is 10 % greater than the theoretical and the mechanical efficiency is 81%, calculate (i) the volumetric efficiency, (ii) the index of expansion and (iii) actual power required to drive the compressor. Neglect all pressure drops. .
Given: compression index = nc = 1 (because process is isothermal); m = 0.7 kg/min; p1 = 1 bar; p2 = 7 bar; T1 = 25 + 273 = 298 K; N = Nc = 600 rpm;V3/ Vs = 1/25; Vs = (V1 – V3) = 1.2 x 10 − 3 m3; Diagram factor = K = 1.1;ηmech = 0.81; To find: (i) (ηv)ideal; (ii) ne; (iii) Pactual Solution: The p – V diagram for this problem will be as shown in example 5.1. .
(i) mc = m / Nc = 0.7 / 600 = 1.167 x 10 − 3 kg/cycle. Volume of air compressed per cycle as measured at suction conditions is given by mcRT1 1.167 x 10 − 3 x 286 x 298 (V1 – V0) = --------- = -------------------------------- = 0.995 x 10 − 3 m3 p1 1 x 10 5 (V1 – V0) 0.995 x 10 − 3 Therefore (ηv)ideal = -------------- = ----------------------- = 0.8292 = 82.92 %. Vs 1.2 x 10 − 3 (ii) Now V3 = (1/25) x 1.2 x 10 − 3 = 4.8 x 10 − 5 m3. Hence V1 = 1.2 x 10 − 3 + 4.8 x 10 − 5 = 1.248 x 10 − 3 m3 And therefore V0 = 1.248 x 10 − 3 – 0.995 x 10 − 3 = 0.253 x 10 − 3. For the expansion process 3 – 0 we have p0V0ne = p3V3ne ln(p3/p0) ln(7/1) Or ne = -------------------- = --------------------- = 1.171 ln(V0/V3) ln(0.253/0.048) (iii) Referring to the p – V diagram shown in example 5.1 work done per cycle is given 1
2
3
0
Wc = W0-1 + W1-2 + W2-3 + W3-0 = ∫pdV + ∫pdV + ∫pdV + ∫pdV 0
1
2
3
= p1(V1 – V0) + p1V1 ln(p1/p2) + p3(V3 – V2) + (p3V3 – p0V0) / (ne – 1) Now V2 – V3 = (mcRT2) / p2. But T2 = T1 as process 1-2 is isothermal.
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1.167 x 10 − 3 x 286 x 298 Hence V2 – V3 = -------------------------------- = 1.421 x 10 − 4 m3. 7 x 10 5 Therefore Wc = [1x 0.995 x10 − 3 + 1 x 1.248 x 10 − 3 x ln(1/7) − 7 x 1.421 x 10 − 4 + {7 x 4.8 x 10 − 5 - 1 x 0.253 x 10 − 3.}/(1.171 – 1)] x 10 5 = [0.995 − 2.429 − 0.9947 + 0.4854] x 102 = − 194.36 J/cycle Ptheoretical = WcNc = − 194.36 x 600 / (60 x 1000) kW = − 1.9436 kW Pactual = K Ptheoretical / η mech = − 1.9436 x 1.1 / 0.81 = − 2.639 kW Example 5.9:- The following data refer to a single acting air compressor: Suction pressure = 1 bar; Receiver pressure = 10 bar; Loss of pressure due to valve resistance at inlet = 0.05 bar; pressure drop at the exit valve = 0.33 bar; Cylinder bore = stroke = 12.0 cm; Clearance volume = 1/25th of stroke volume; Index for expansion and compression = 1.25; Mechanical efficiency = 80 %. If the receiver capacity is 600 litres and if it takes 8 minutes to fill it to 10 bar starting from 1 bar, find the maximum power required to drive the compressor. Assume that the receiver temperature remains at 25 C throughout the filling process. Also determine the mass of air compressed the volumetric efficiency and the speed of the compressor. Given: p1 = 1 bar; pR = 10 bar; pf – p1 = 0.05 bar; p2 – pR = 0.33 bar; d = l = 0.12 m; V3 = (1/25)Vs; nc = ne = n = 1.25; η mech = 0.80; VR = 0.6 m3; pR = 10 bar; Time taken to fill the receiver = t = 8 min;TRi = TRf = 25 + 273 = 298 K; T1 = TRi .
To find: (i) P;(ii) m ;(iii) (ηv)actual Solution:The p – V diagram for this problem will be as shown in the figure below. 1 x 105 x 0.6 (ii) Mass of air initially in the receiver = mi = pRiVR /(RTRi) = ------------------- = 0.704 kg 286 x 298 10 x 105 x 0.6 Mass of air finally in the receiver = mf = pRfVR /(RTRf) = ---------------------- = 7.04 kg 286 x 298 (7.04 – 0.704) . Mass of air delivered to the receiver per minute = m = (mf – mi) / t = -----------------8 = 0.792 kg/min.
134
p
p2
3
2
pR
pf p1
0
1 V
.
(i) P = (1/ηmech){n/(n – 1)}m RT1[1 – (p2/p1)(n-1)/n] 1 1.25 0.792 = --------- x ---------- x -------- x 0.286 x 298 x [1 – (10.33/1)0.25/1.25] 0.80 0.25 60 = − 4.185 kW. (iii) (ηv)ideal = 1 + C { 1 – (p2/p1)1/n} = 1 + (1/25) x [1 – (10.33 / 1)1/1.25} = 0.781 1.0 x 298 (ηv)actual = [(p1Tf) / (pfT1)] (ηv)ideal = ----------------- x 0.781 = 0.744 = 74.4% 1.05 x 298 Example 5.10:- A single stage double acting air compressor running at 1000 rev/min delivers air at 25 bar. The suction conditions are 1 bar and 40 C. The free air conditions are 1.013 bar and 15 C and the free air delivered is 0.25 m 3 / min. The clearance volume is 3 % and the stroke to bore ratio is 1.2 : 1. Calculate the cylinder dimensions and the volumetric efficiency. Assume that n = 1.3 for both compression and expansion processes. Also calculate the indicated power and isothermal efficiency.
135
Given: Double acting and hence Nc = 2N; N = 1000 rpm; p2 = 25 bar; p1 = 1 bar; .
T1 = 40 + 273 = 313 K; pf = 1.013 bar; Tf = 15+273 = 288 K; Vf = 0.25 m3/min; Clearance Ratio = V3 / (V1 – V3) = 0.03; l / d = 1.2; nc = ne = n = 1.3 To find: (i) d,l; (ii) (ηv)actual; (iii) P; (iv) ηisothermal The p – V diagram for this problem will be as shown in example 5.1 .
pf Vf 1.013 x 10 5 x 0.25 Mass of air delivered / min = m = ------------- = ---------------------------- = 0.3075 kg/min RTf 286 x 288 .
.
Hence
mc = m / Nc = 0.3075 / (2 x 1000) = 1.5375 x 10 − 4 kg/cycle
Volume of air compressed /cycle = (V1 – V0) = mcRT1 / p1 1.5375 x 10 − 4 x 286 x 288 = ----------------------------------1.0 x 10 5 = 1.263 x 10 − 4 m3. Now V0 = (p3 / p0) 1/n V3 = (25/1) 1/1..3 V3 = 11.89 V3. Hence
V1 – 11.89V3 = 1.263 x 10 − 4 ………………(a)
But V3 = 0.03 (V1 – V3) or V1 = 34.333V3. Substituting this expression in Eq.(a) we get (34.333 – 11.89)V3 = 1.263 x 10 − 4 Or
V3 = 5.63 x 10 − 6 m3.
Hence V0 = 11.89 x 5.63 x 10 − 6, V1 = 34.333 x 5.63 x 10 − 6 = 1.933 x 10 − 4 And Vs = V1 – V3 = 1.933 x 10 − 4 − 5.63 x 10 − 6 = 1.877 x 10 − 4. Now (πd2/4)l = Vs or (πd2/4) 1.2d = Vs. Therefore d = {(4Vs) / (1.2π) }1/3 = { (4 x 1.877 x 10 − 4) / (1.2 x π)}1/3
136
Or
d = 0.0584 m and l = 1.2 x 0.0584 = 0.07 m
(V1 – V0) (1.933 x 10 − 4 − 6.694 x 10 − 5) (ii) (ηv)ideal = ------------- = --------------------------------------- = 0.6732 Vs 1.877 x 10 − 4 p1Tf 1 x 288 (ηv)actual = ------- (ηv)ideal = -------------------- x 0.6732 = 0.6115 PfT1 1.013 x 313 .
(iii) Indicated power = P = (1/ηmech){n/(n – 1)}m RT1[1 – (p2/p1) (n – 1)/n] 1 1.3 0.3075 = ------ x ---------- x ---------- x 0.286 x 313 x [1 – (25/1)0.3/1.3] 1.0 0.3 60 = − 2.2 kW. (iv) Isothermal work done on air cycle is given by (Refer to the p – V diagram shown below)
p
3
2
pV = constant or T = constant
0
1 V
1
2
3
(Wc)Isothermal = W0-1 + W1-2 + W2-3 = ∫pdV + ∫pdV + ∫pdV 0
1
137
2
= p1V1 + p1V1 ln(p1/p2) – p2V2 = p1V1 ln(p1/p2) = mcRT1 ln(p1/p2) .
Hence Pisothermal = (Wc)Isothermal Nc = m RT1 ln(p1/p2) = (0.3075 / 60) x 0.286 x 313 x ln(1/25) = − 1.477 kW ηisothermal = Pisothermal / Pactual = 1.477 / 2.2 = 0.6714 = 67.14 %. 5.9. Disadvantages of single stage compressors (need for multi-stage compressors) Following are the disadvantages of a single stage compressor: (i) In a single stage compressor, as the delivery pressure increases the volume of air compressed per cycle decreases ( see Fig. 5.9) thereby decreasing the volumetric efficiency of the compressor. Low volumetric efficiency means that the speed of the compressor has to be increased in order to compress a given mass of gas per unit time. There is a practical limitation beyond which the speed of the compressor cannot be increased. (ii) Increase in delivery pressure will also result in increase in the temperature of the gas and this may cause lubrication problem (There is a possibility of the lubricating oil getting vaporized and getting ignited). (iii) If the pressure ratio for a single stage compressor is very large then the work required to drive the compressor will also be very large All these disadvantages are overcome by employing multistage compression with intercooling in between the stages. 5.10. Work done in a Two-stage compressor with intercooling in between the stages The schematic diagram and the corresponding p – V diagram for a two stage compressor with intercooling in between the stages are shown in Fig 5.10(a) and 5.10(b) respectively. It is assumed that the index for compression and expansion are equal for both the stages of compression.Work done per cycle is given by Wc = (Wc)I + (Wc)II = {n1 /(n1 – 1)}mcRT1[1 – (p2/p1)(n – 1)/n] + {n2 /(n2 – 1)}mcRT5[1 – (p6/p5)(n – 1)/n] …………………………(5.16)
138
Where n1 = index for I stage and n2 = index for II stage compression. If n1 = n2 = n, then
p
V3 = Clearance Volume for I stage V1 – V3 = Stroke volume for I stage
7
6
3
V7 = Clearance volume for II stage V5 – V7 = stroke volume for II stage V1 – V0 = Volume of air compressed per cycle as measured at suction conditions of I stage.
2
4
5
V5 – V4 = volume of air compressed per cycle as measured at suction conditions of II stage.
0
1 Fig.5.10(a): p-V diagram for a two stage compressor
V
QIC (Wc)I
p1,T1
First stage Compressor
Inter cooler p2,T2
p5,T5
(Wc)II
Fig. 5.10(b): Schematic for a two stage compressor
Second stage compressor
p6,T6
Wc = {n /(n – 1)}mcRT1[1 – (p2/p1)(n – 1)/n] + {n /(n – 1)}mcRT5[1 – (p6/p5)(n – 1)/n] …………………………….(5.17)
139
It can be seen from the above equation that for given inlet conditions to the compressor and for given overall pressure ratio p6 / p1, Wc depends on the intermediate pressure p2. The intermediate pressure p2 is chosen such that the work required to dive the compressor is a minimum. 5.11. Optimum intermediate pressure for a two-stage compressor (condition for minimum work input for a two stage compressor) In order to obtain the condition for minimum work input the following assumptions are made. (i) There is no pressure loss in the inter cooler; i.e., p5 = p2; (ii) the index for compression and expansion are equal and are same for both the stages of compression. With these assumptions the Eq.(5.17) reduces to Wc = {n /(n – 1)}mcRT1[1 – (p2/p1)(n – 1)/n] + {n /(n – 1)}mcRT5[1 – (p6/p2)(n – 1)/n] For given values of p1,T1, and p6, Wc will be optimum if dWc / dp2 = 0. Now dWc/dp2 = − {n /(n – 1)}mcRT1 (p1)(1 – n)/n p2(n – 1)/n – 1 + {n /(n – 1)}mcRT5 (p6)(n - 1)/n p2(1 - n)/n – 1= 0 Or Or
or
T1 (p1)(1 – n)/n p2− 1/n = T5 (p6)(n - 1)/n p2(1 – 2n)/n p2− 1/n ----------- = (T5 /T1) (p6/p1) (n - 1)/n p2(1 – 2n)/n ___________________ p2 = √[(T5 /T1) n/(n – 1) p6 p1 ] ……………………..(5.18)
Eq.(5.18) gives the optimum intermediate pressure for a two stage compressor. Special case: Perfect intercooling:- When the intercooling is perfect, the gas will be cooled back to its original temperature in the inter cooler, i.e., T5 = T1. Then Eq.(5.18) reduces to ______ p2 = √ (p6p1) ……………………..(5.19) _____ Therefore pressure ratio for first stage for minimum work = p2 / p1 = √ (p6p1) / p1
140
Or
p2 / p1 = √ (p6 / p1)
Similarly for second stage pressure ratio = p6/p5 = p6/p2 = p6 /(√(p6p1) = √ (p6 / p1) Thus for minimum work of compression, if the inter cooling is perfect, the pressure for both the stages of compression are same and equal to the square root of the overall pressure ratio for the compressor. This can be extended to a compressor with ‘N’ stages of compression with inter cooling between the stages. For such a compressor with perfect inter cooling, the work required will be minimum if the pressure ratio is same for all the stages and equal to the Nth root of the overall pressure ratio Now for a two stage compressor with perfect intercooling, the work done in both the stages will be same, because the inlet temperature and the pressure ratio for both the stages are same.Hence (Wc)minimum = {2n/(n – 1)}mcRT1[1 – (p6/p1) (n – 1)/2n] …………………………(5.20) Eq.(5.20) is valid only if the intercooling is perfect and the compressor is designed for minimum work input. 5.12. Illustrative examples on Multi-stage compressors Example 5.11:- A two-cylinder, two stage air compressor delivers 2 kg/min of air at 25 bar, taking in air at 1 bar and 30 C. The compression index is 1.25. Neglecting clearance calculate (i) the intermediate pressure for minimum power, (ii) heat transfer during intercooling and (iii) power required to drive the compressor. Given: Number of cylinders = 2; No.of stages = 2; mass of air delivered = 2 kg/min; p6 = p7 = 25 bar; p1 = 1 bar; T1 = 30 + 273 = 303 K; n = 1.25 ; Clearance neglected. .
To find: (i) p2 for minimum power; (ii) QIC; (iii)P. Solution: It is assumed that there is no pressure loss in the inter cooler; i.e., p5 = p2 . And the inter cooling is perfect ,i.e. T5 = T1. The p – V diagram for the given problem will be as shown in the next page. (i)With perfect inter cooling for minimum power p2 = √ (p6p1) = √(25 x 1) = 5 bar (ii) Temperature at the exit of I stage = T2 = T1(p2/p1)(n – 1)/n = 303 x (5/1)0.25/1.25 = 418 K.
141
p 7
0 – 1 – 2 – 3 : p – V diagram for first stage
6
4 – 5 – 6 – 7 : p – V diagram for second stage (4)3
5 2
0
1 V
Fig. E5.11: Figure for example 5.11 .
.
Mass flow rate through each cylinder = m = mtotal / number of cylinders = 2/2 = 1 kg/min Applying first law for the inter cooler and neglecting changes in kinetic and potential energies we have . . QIC = m C p(T2 – T5 ) = (1/60) x 1.005 x (418 – 303) kW = 1.926 kW. (iii) For two stage compressor with perfect inter cooling the total power required is given by .
P = {2n/(n – 1)} mtotal RT1[ 1 – (p6 / p1)(n
- 1) / 2n
]
2 x 1.25 2.0 = --------------- x------- x 0.286 x 303 x [ 1 – (25/1)0.3/ (2 x1..3] 0.25 60 = − 13.0 kW.
142
Example 5.12: A two stage, double acting air compressor operates at 150 rpm. The conditions of air at the beginning of compression are 97.9 kPa and 27 C. The low pressure cylinder with a bore and stroke of 35 X 38 cm discharges the air at 379 kPa to the intercooler. The air in the intercooler suffers a pressure drop of 17.2 kPa and enters the high pressure cylinder at 29 C. The discharge pressure is 2000 kPa. Compression and expansion processes in both the stages are according to the law pV 1.3 = C. The surroundings are at 100 kPa and 20 C. The percent clearance is 5 % for each cylinder. Determine (i) “free air” capacity in m3/s, (ii) the heat loss in the intercooler, (iii) the total power required, (iv) optimum interstage pressure, (v) diameter of the hp cylinder if the stroke is same for both the stages (vi) the heat loss in the low-pressure and highpressure compression processes. Given: No. of atages = 2; Double acting and hence Nc = 2N; N = 150 rpm; p1 = 97.9 kPa; T1 = 27 + 273 = 300 K; d1 = 0.35 m; l1 = 0.38 m; p2 = 379 kPa; p2 – p5 = 17.2 kPa; T5 = 29 + 273 = 302 K; p6 = p7 = 2000 kPa; n = 1.3 throughout; pf = 100 kPa; Tf = 20 + 273 = 293K; V3 = 0.05 (V1 – V3) ; V7 = 0.05 (V5 – V7); .
.
To find: (i) Vf ; (ii) QIC ; (iii) P ; (iv) Optimum value of p2; (v) d2 if l2 = l1; (vi) Q1-2 and Q5-6. Solution: The p – V and the schematic diagrams are shown below. p V3 = Clearance Volume for I stage V1 – V3 = Stroke volume for I stage 7
6
3
V7 = Clearance volume for II stage V5 – V7 = stroke volume for II stage V1 – V0 = Volume of air compressed per cycle as measured at suction conditions of I stage.
2
4
5
0
V5 – V4 = volume of air compressed per cycle as measured at suction conditions of II stage. 1 V
143
Fig.E5.12(a): p-V diagram for the example 5.12
QIC (Wc)I
p1,T1
First stage Compressor
Inter cooler p2,T2
p5,T5
(Wc)II
Fig. E5.12(b): Schematic for example 5.12
Second stage compressor
p6,T6
π x 0.35 2 x 0.38 (i) Stroke volume of I stage = V1 – V3 = (πd12/4)l1 = -------------------------4 = 0.0366 m Hence V3 = 0.05 x 0.0366 = 1.83 x 10 − 3 m3 ;V1 = 0.0366 + 1.83 x 10 − 3 = 0.03843 m3 Therefore V0 = (p3/p0) 1/n V3 = (379/97.9) 1/1.3 x 1.83 x 10 − 3 = 5.184 x 10 − 3 m3. (V1 – V0) = 0.03843 − 5.184 x 10 − 3 = 0.03325 m3. p1(V1 – V0) 97.9 x 10 3 x 0.03325 Mass of air compressed per cycle = mc = ------------------- = --------------------------RT1 286 x 300 = 0.0379 kg/cycle .
Hence
m = mcNc = 0.0379 x 2 x 150 = 11.37 kg/min. .
m RTf 11.37 x 286 x 293 Volume of free air delivered / minute = Vf = ------------ = ----------------------pf 100 x 10 3 .
= 9.53 m3/min
144
(ii) Applying I law for the inter cooler we have .
.
QIC = m Cp (T2 – T5). But T2 = (p2/p1)(n – 1)/nT1 = (379 / 97.9) 0.3 / 1.3 x 300 .
Or T2 = 410 K. Therefore Q IC = (11.37 / 60) x 1.005 x [410 – 302] = 20.6 kW ____________________ (iii) Optimum value of p2 = √ [ (T5 / T1) n / (n – 1) p6p1 ] ___________________________ = √ [(302/300) 1.3 / 0.3 x 2000 x 97.9 ] = 449 kPa. (iv) p5 = p2 – 17.5 = 379 – 17.5 = 361.5 kPa. Power = P = PI stage + PII stage n n . . (n – 1) / n = -------- m RT1 [ 1 – (p2/p1) ] + ------- m RT5 [ 1 – (p6/p5) (n – 1) / n] (n – 1) (n – 1) 1.3 11.37 = ---------- x -------- x 0.286 x [ 300 x {1 – (379/97.9) 0.3 / 1.3} 0.3 60 + 302 x {1 – (2000/361.5)0.3 / 1.3} ] = − 60.17 kW mc RT5 0.0379 x 286 x 302 (v) Now (V5 – V4) = ------------ = ------------------------- = 9.05 x 10 − 3 m3 p5 361.5 x 10 3 Now V4 = (p7/p4) 1 / n V7 = (2000 /361.5) 1/1.3 V7 = 3.73 V7. Hence
V5 – 3.73 V7 = 9.05 x 10 − 3 ………………………………(a)
But V7 = 0.05 (V5 – V7 ) or V5 = 21 V7 Substituting this in Eq. (a) we get
(21 – 3.73) V7 = 9.05 x 10 − 3 V7 = 5.24 x 10 − 4 m3
Or And
V5 = 21 x 5.24 x 10 − 4 = 0.0110 m3.
Hence stroke volume for II stage = Vs2 = (V5 – V7) = (0.0110 − 5.24 x 10 − 4 ) = 0.0105 m3
145
_____________ _______________________ d2 = √ [ (4Vs2) / (πl2)] = √ [ (4 x 0.0105) / (π x0.38)]
Hence
= 0.1875 m. .
.
.
(vi) Applying I law for process 1-2 we have Q1-2 = W1-2 + m(u2 – u1) .
Or
.
.
Q1-2 = mR(T1 – T2) / (n – 1) + m Cv(T2 – T1) 11.37 = -------- x [300 – 410] x { 0.286/ 0.3 – 0.719 } 60 = − 4.885 kW .
Similarly
.
.
Q5-6 = mR(T5 – T6) / (n – 1) + m Cv(T6 – T5)
Now T6 = (p6 / p5)(n – 1) / n T5 = (2000 / 361.5) 0.3/1.3 x 302 = 448.35 K .
Hence
Q5-6 = (11.37/60) x [302 – 448.35] x {0.286 / 0.3 – 0.719} = − 6.5 kW.
Example 5.13:- A two-stage air compressor is required to take in 1500 litres of free air per minute at 1 bar and 25 C. The delivery pressure is 20 bar. The heat transfer during compression, which may be assumed to be polytropic, is double that in the intercooler for the first stage, and 1/3rd of the total compression work for the second stage. Assuming the intercooler effectiveness to be 0.83 and an intermediate pressure as the geometric mean of the suction and delivery pressures of the compressor, find the power required to drive the compressor assuming a mechanical efficiency of 80 %. .
Given: No. of stages = 2; Vf = 1.5 m3/min; pf = p1= 1 bar;Tf = T1= 25 + 273 = 298 K; p6 = 20 bar; Q1-2 = 2QIC; Q5-6 = (1/3) W5-6; ε = 0.83; p2 = √(p6p1); ηmech = 0.80. To find: (i) Pactual ; Solution: First step is to find the compression index for the two stages of compression. It is given that for the first stage Q1-2 = 2QIC …………………………(a) Now QIC = 2mcCp[T5 – T2] = 2mcCp ε [T1 – T2] Applying first law for the process 1-2 we have Q1-2 = W1-2 + (U2 – U1)
146
Or
Q1-2 = mcR(T1 – T2) / (n1 – 1) + mcCv(T2 – T1)
Substituting the expressions for QIC and Q1-2 in Eq. (a) and simplifying we get 2Cp ε = R / (n1 – 1) - Cv R 0.286 n1 = 1 + --------------------- = --------------------------------(Cv + 2Cp ε) [0.719 + 2 x 1.005 x 0.83]
Or
= 1.12. For the second stage it is given that Q5-6 = (1/3)W5-6 First law equation for process 5-6 is Q5-6 – W5-6 = U6 – U5 Hence
(1/3)W5-6 – W5-6 = U6 – U5
Or − (2/3) W5-6 = (U6 – U5) .
Or − (2/3) mcR(T5 – T6) / (n2 – 1)= mcCv(T6 – T5)
Or n2
2R 2 x 0.286 =1+ ------ = 1 + --------------- = 1.265 3Cv 3 x 0.719
Now p2 = √ [p6p1] = √ [20 x 1] = 4.472 bar Pactual = (1/ηmech)P = (1/ηmech) [PI stage + PII stage] .
.
= (1/ηmech) [{n1/(n1 – 1)}mRT1{1 – (p2/p1)(n1 – 1)/n1} + mRT5{1 – (p6/p5)(n2 – 1)/n2}] .
= (1/ηmech) [{n1/(n1 – 1)}mR {T1 – T2} + {n2/(n2 – 1)}mR {T5 – T6}] Now T2 = (p2 / p1) (n1 – 1)/n1T1 = (4.472/1)(1.12 – 1) / 1.12 x 298 = 350 K ε = (T2 – T5) / (T2 – T1) or T5 = T2 – ε(T2 – T1) = 350 – 0.83 x (350 – 298) = 306.84 K .
.
m. = (pfVf) /(RTf) = (1 x 10 5 x 1.5) / (286 x 298) = 1.76 kg/min = 0.0293 kg/s Similarly T6 = (p6 / p5) (n2 – 1)/n2 T5 = (20/4.472) 0.265 / 1.265x 306.84 = 420 K Hence Pactual = (1/0.8) x 0.0293 x 0.286 x [ (1.12/0.12)x(298 – 350) + (1.265/0/265)x (306.84 – 420)] Or Pactual = − 10.742 kW
147
Example 5.14:- A multistage single acting compressor compresses air from 1 bar and 25 C to 30 bar. The maximum temperature in each stage is limited to 100 C. The cylinder heads are cooled so that the heat transfer during compression is 0.7 times that in the intercooler, where the air returns to its initial temperature. Find the index of compression, the number of stages (assume perfect intercooling), the compression pressures, the temperature at the end of compression and the power required to drive the compresso to compress1000 litres per minute of air,for minimum work input. Given: p1 = 1 bar; T1 = 25 + 273 = 298 K; No.of stages = S; Delivery pressure from the last stage = ps+1 = 30 bar;Exit temperature from each stage ≤ (100 + 273) = 373 K; Q1-2 = 0.7 QIC; To find: (i)n; (ii) S; (iii) delivery pressure from each stage; (iv) delivery temperature from each stage; (v) P Solution: (i) Applying I law for the compression process in stage 1 we have Q1-2 = W1-2 + (U2 – U1) = mcR(T1 – T2) / (n – 1) + mcCv(T2 – T1) Or
Q1-2 = mc(T1 – T2)[R /(n – 1) − Cv] ……………………….(a)
Also
Q1-2 = 0.7 QIC = 0.7mcCp(T1 – T2).
Substituting this expression in Eq.(a) we have 0.7mcCp(T1 – T2) = mc(T1 – T2)[R /(n – 1) − Cv] Or
0.7 Cp = [R /(n – 1) − Cv]
Or
R 0.286 n = 1 + ------------------- = 1 + --------------------------[Cv + 0.7Cp] {0.719 + 0.7 x 1.005] = 1.2
(ii) Pressure ratio for the first stage = (p2 / p1) ≤ (T2/T1) n/(n – 1) Or
(p2 / p1) ≤ (373 / 298) 1.2 / 0.2 = 3.85
With perfect intercooling for minimum work pressure ratio should be same for all stages. Therefore ln(pS+1/p1) ln(30 / 1) ps+1 / p1 = (p2/p1) S or S = ---------------- = --------------- = 2.52 ln(p2/p1) ln(3.85)
148
Since the number of stages cannot be a fraction S should be rounded off to the next integer. Hence S = 3. Then pressure ratio developed per stage is given by Pr = (30 /1) 1/3 = 3.11 (iv) Therefore delivery temperature from each stage = T2 = T1(Pr) (n – 1)/n Or
T2 = 298 x (3.11) 0.2 / 1.2 = 360 K
(iii) Delivery pressure from I stage = 1 x 3.11 = 3.11 bar Delivery pressure from II stage = 3.11 x 3.11 = 9.6721 bar Delivery pressure from III stage = 9.6721 x 3.11 = 30 bar (as given in the problem) (v) Minimum power required with perfect intercooling is given by .
(n – 1) / n
Pminimum = S {n / (n – 1)}p1Vf [ 1 – (Pr)
]
3 x 1.2 1 x 105 1000 x 10 − 3 = ---------- x ----------- x ------------------ x [ 1 – (3) 0.2 / 1.2 ] 0.2 1000 60 = − 6.03 kW
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CHAPTER 6 REFRIGERATION CYCLES 6.1. Introduction:- Refrigeration is a term used to denote the process of maintaining a space or a body at a temperature lower than that of its surroundings. To produce and maintain the low temperature, it is necessary to transfer heat from the space to be refrigerated or the cold body. A refrigerator is a device that is employed to accomplish refrigeration by the expenditure of external energy in the form of work or heat or both. For the refrigerator to operate continuously, it must reject heat to an external sink, usually the atmosphere. The working substance used in the refrigerator, which absorbs the heat from the refrigerated space and rejects to the sink, is called a refrigerant. 6.2. Capacity and Coefficient of performance of a refrigerator:- The heat removed from the refrigerated space or the cold body is called “refrigeration effect” or “ capacity of the refrigerator”. The refrigeration effect is normally expressed in tons of refrigeration. The term “ton” is derived from the old British system of units and it means the amount of heat that has to be removed from 1 ton of water at 32 0F to convert it into ice at 32 0F in 24 hours. In SI system of units this will be equal to 211 kJ/min or 3.517 kW. Thus 1 ton of refrigeration = 3.517 kW………………….6.1 The performance of a refrigerator is also expressed in terms of power required to produce 1 ton of refrigeration and expressed in kW / ton. The “coefficient of performance” is another parameter used to measure the effectiveness of a refrigerator and is defined as the ratio of the heat removed from the refrigerated space to the external energy input; that is Refrigeration effect COP = -------------------------------- ……………………………6.2 External energy input 6.3. Refrigeration Cycles:- The cyclic process executed by the refrigerant to produce the required refrigeration effect is called the refrigeration cycle / heat pump cycle. Refrigeration cycles are classified into two types; namely (i) gas refrigeration cycles and (ii) vapour refrigeration cycles. As the name suggests, in the case of a gas refrigeration cycle the working substance will be in gaseous phase throughout out the cyclic process, where as in vapour refrigeration cycles the working substance will undergo a change of phase from liquid phase to vapour phase in one pert of the cycle and from vapour phase to liquid phase in another part of the cycle. Vapour refrigeration cycles are further classified into two types, namely (a) Vapour compression cycle and (b) vapour absorption cycle.
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6.4. Gas Refrigeration Cycles 6.4.1. Carnot Gas Refrigeration Cycle (Carnot Refrigerator):- Fig.6.1(a) and 6.1 (b) represent the schematic diagram and T – s diagram for a Carnot refrigeration cycle.
Compressor
qIC
Inter cooler Expander
wc
we
Cold Chamber
qR
Fig. 6.1: Physical layout and T-s diagrams for a Carnot refrigerator Process 1-2 :- Isentropic compression of the working substance from state 1 to state 2. During this process work is done on the substance by the surroundings. Process 2-3:- Reversible isothermal cooling of the working substance in the intercooler. During this process(process 2-3), heat is rejected by the working substance to the sink at temperature TH. Process 3-4:- Isentropic expansion of the working substance from state 3 to state 4. During this process work is done by the working substance on the surroundings. Process 4-1 :- Reversible isothermal heating of the working substance from the place to be refrigerated (cold chamber) maintained at temperature TL. Expression for Refrigeration effect and COP Refrigeration effect per unit mass = Heat removed from the refrigerated space = qR = q 4-1
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1
q R = ∫ Tds = TL (s1 – s4) 4
Similarly heat rejected in the intercooler = q2-3 = TH(s3 – s2) Net work input per unit mass = w N = w1-2 + w2-3 + w3-4 + w4-1 By first law for the Carnot cycle we have w1-2 + w2-3 + w3-4 + w4-1 = q1-2 + q2-3 + q3-4 + q4-1 and q1-2 = q3-4 = 0 as both the processes are isentropic. Therefore
wN = q2-3 + q4-1 = TH(s3 – s2) + TL(s1 – s4)
Therefore
TL(s1 – s4) (COP)Carnot = qR / wN = -------------------------------TH(s3 − s2) + TL(s1 – s4)
But
Therefore
s1 – s4 = s2 – s3 TL (COP)Carnot = ------------- ………………………………6.3 TH - TL
Eq.6.3 indicates that COP increases as TH – TL decreases. For a given cold body temperature, the lower the temperature at which heat is rejected, greater will be the COP. Disadvantages of a Carnot Refrigerator:- The Carnot refrigeration cycle is the most efficient refrigeration cycle operating between the two specific temperature levels.But it is not a suitable model for refrigeration cycles because of the following reasons. It is difficult to maintain isothermal conditions during heat absorption and heat rejection processes. Further both the expansion process and the compression process must take place very slowly if the processes have to be quasi-static with the result the time required to complete each cycle of operation is very large and hence impracticable. Nevertheless, the Carnot refrigerator is valuable as a standard of comparison, since it represents perfection. 6.4.2. Ideal Air Refrigeration Cycle ( Reversed Brayton Cycle or Bell – Coleman Cycle) Assumptions made in the analysis of the ideal cycle (i) The working fluid is air
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(ii) Air behaves as a perfect gas. (iii) All processes that the working substance undergoes are reversible. (iv) There are no pressure losses in the piping connecting the various components and also in the heat exchanger. (v) Flow is steady and one dimensional. (vi) Changes in kinetic and potential energies of the working substance are negligible. Description of an ideal air refrigeration cycle: Figure 6.2 (a) and 6.2 (b) shows the schematic and T – s diagram for an air refrigeration cycle. T QIC 2 2
3
Wc
p = const
We
4
1
3
4
1
s
QR Fig. 6.2 : Schematic and T-s diagrams for air refrigeration cycle Process 1-2:- Isentropic compression of air from state 1 to state 2. During this process work is done on air by the surroundings. Process 2-3:- Constant pressure cooling of air in the intercooler. Process 3-4:- Isentropic expansion of air from state 3 to state 4. During this process work is done by air on the surroundings. Process 4-1:- Constant pressure heat removal by air in the cold chamber so that it comes back to original state to complete the cycle. Expressions for Refrigeration effect and COP .
Let QR = Refrigeration effect per unit time = Heat removed from the cold chamber per unit time
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Applying steady-state steady- flow energy equation to the cold chamber and noting that changes in kinetic and potential energies are assumed to be neglected we have .
.
.
.
m h4 = QR + m h1, where m is the mass flow rate of refrigerant through the cold chamber. . . . or QR = m (h1 – h4) = mcp(T1 – T4) …………….6.4 .
.
.
Similarly Compressor Work input = Wc = m (h2 – h1) = mcp(T2 – T1) …………...6.5 .
And
.
Expansion work = We = m (h3 – h4) = mcp(T3 – T4)…………..……………..6.6 .
.
.
Therefore net work input to the cycle = Wn = Wc − We .
Or
.
.
Wn = mcp(T2 – T1) – mcp(T3 – T4) …………..6.7 .
.
Coefficient of performance = COP = QR / Wn .
mcp(T1 – T4) = ------------------------------------[mcp(T2 – T1) – mcp(T3 – T4)] 1 or COP = ------------------------------------ ………………..6.8 (T2 – T3) ----------- − 1 (T1 – T4) Process 1-2 is isentropic. Therefore T2 / T1 = (p2 / p1)(γ – 1) / γ. Similarly,
T3 / T4 = (p3 / p4)(γ – 1) / γ = (p2 / p1)(γ – 1) / γ
Therefore T2 / T1 = T3 / T4 = (T2 – T3) / (T1 – T4) = (p2 / p1)(γ – 1) / γ. Substituting this in Equation 6.8 we get 1 COP = ---------------------------- …………………….6.9 [ (p2 / p1)(γ – 1) / γ − 1 ] It can be seen from equation 6.9 that COP for an ideal gas refrigeration cycle depends only on the compressor pressure ratio. The variation of COP with respect to the pressure ratio is shown in Fig.6.3. as well as is given in table 6.1.
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COP
p2 /p1 Fig. 6.3 : Variation of COP with pressure ratio for an ideal gas refrigeration cycle
Table 6.1 : COP for an ideal air(γ = 1.4) refrigeration cycle for different compressor pressure ratios ___________________________________________________________________ Pressure Ratio 1 2 3 4 5 6 7 ___________________________________________________________________ COP ∞ 4.56 2.71 2.05 1.72 1.5 1.34 ___________________________________________________________________ It should be noted that the pressure ratio p2 / p1 have limitations on account of the operating temperatures, viz., (i) T1 as the highest refrigeration temperature and (ii) T3 as the lowest ambient temperature. The effect of discharge pressure p2 on the performance of an ideal gas refrigeration cycle is illustrated on the T-s diagram shown in Fig. 6.4. Point 1 on the diagram is fixed by the temperature T1 and pressure p1 (p1is generally atmospheric). Point 3 is fixed because of the limitations of the ambient temperature T3 to which the gas can be cooled in the intercooler. The discharge pressure can however be varied within wider limits, starting from a minimum discharge pressure p2,min onwards as shown in the figure. With the compressor discharge pressure equal to p2,min the refrigeration effect is zero. The air is alternately compressed and expanded between points 2min and 1. The net work input is also zero and hence the COP is indeterminate. However as p2 is increased, though the
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T
2’ 2 2”
p2’ p2 p2” 3’
3
2 min
3”
1
4’
4
4”
pmin
Ambient Temp, T3
p1 s
Fig. 6.5. Effect of Discharge pressure p2 on the performance of ideal cycle refrigeration effect (area under curve 4-1) increases, the discharge pressure also increases. For example when the discharge pressure is p2, the refrigeration effect is equal to the area under the curve 4-1 and the net work input is area 1-2-3-4-1. Whenthe discharge pressure is increased to p’2, the refrigeration effect is equal to the area under the curve 4’-1 and the net work input will be equal to the area 1-2’-3’-4’-1. It is evident that the increase in work input is much more than the increases in refrigeration effect and hence the COP decreases as the pressure ratio increases. 6.4.3. Comparison between Carnot cycle and Ideal Refrigeration Cycle The comparison between the Carnot cycle and the ideal gas refrigeration cycle is illustrated on the T-s diagram shown in Fig. 6.6. It can be seen from this diagram, that, for the same compressor inlet conditions (state 1) and same ambient conditions (state 3), the ideal air refrigeration cycle requires an additional work input of ∆wn and there is a decrease in the refrigeration effect by an amount ∆qR, thereby decreasing the COP of the cycle
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T 2 p = const. Area = + ∆WN 3
2’
4’
1 Area = − ∆qR
4
p = const.
s Fig.6.6 : Comparison between Carnot Refrigeration cycle and air Refrigeration cycle Example 6.1:- A reversed Carnot cycle is used for heating and cooling. The work supplied is 10 kW. If the COP is 3.5 for cooling determine (a) the ratio of maximum temperature to minimum temperature in the cycle , (b) refrigeration effect in tons and (c) COP if the cycle is used as a heat pump. Solution: The schematic for this example is shown in Fig. E6.1 (a) If the cycle is used for cooling purpose, then .
.
(COP)Ref = QR / WN = TL / (TH – TL) = 1 / (TH / TL – 1) Therefore
3.5 = 1 / (TH/TL – 1)
Or
TH / TL = 1 + 1 / 3.5 = 1.286 .
.
(b) Refrigeration effect = QR = (COP)Ref WN = 3.5 x 10 / 3.517 ton = 9.952 ton (c) (COP)HP = (COP)Ref + 1 = 3.5 + 1 = 4.5
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High Temperature reservoir at TH K
Q1 W
Q2 Low Temperature reservoir at TL K Fig. E6.1: Schematic for example 6.1 Example 6.2:- An ideal air refrigeration cycle has the following specifications: Pressure of air at compressor inlet = 101 kPa; Pressure of air at turbine inlet = 404 kPa; Temperature of air at compressor inlet = −6 C; Temperature of air at turbine inlet = 27 C; Determine (i) The COP of the cycle, (ii) Power required to produce 1 ton of refrigeration, and (iii) air circulation rate per ton of refrigeration. Solution: The T-s diagram for the example is shown in Fig. E6.2 Data:- p1 = p4 = 101 kPa ; p2 = p3 = 404 kPa ; T1 = − 6 + 273 =266 K; cp = 1.005 kJ / kg (assumed) T3 = 27 + 273 = 300 K ; Refrigeration effect = QR = 1 ton = 3.9 kW. (i) For the ideal cycle, COP = 1 / [(p2 / p1)(γ – 1) / γ − 1] 1 = --------------------------------- = 2.055 [ (404 / 101)(1.4 − 1) / 1.4 − 1 ] .
.
(ii) Power input = WN = QR / COP = 3.517 / 2.055 = 1.711 kW / ton
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T 2
3
1
4 s Fig. E6.2: T-s diagram for example 6.2
(iii) Applying steady state steady flow energy equation to the cold chamber and neglecting the kinetic and potential energies of the fluid we have .
.
.
QR = m(h1 – h4) = mcp(T1 – T4) .
Therefore
.
m = QR / cp(T1 – T4) ……………………….(1)
Now, process 3-4 is isentropic. Therefore T4 = T3(p4 / p3)(γ – 1) / γ = 300 x [101 / 404 ](1.4 – 1) / 1.4 = 201.8 K Substituting in equation (1) we get 1 x 3.9 . m = --------------------------1.005 x ( 266 – 201.8) =0.0545 kg / s.
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Example 6.3:- In an air refrigerating machine, the compressor takes in air at 1 bar and 10 C. After compression to 5.5 bar, the air is cooled to 30 C before expanding it back to 1 bar. Assuming ideal conditions, determine (i) refrigeration effect per unit mass of air,(ii)heat rejected by air per unit mass in the intercooler, and (ii) COP of the cycle, In an actual plant using the above cycle, the air flow rate is 1700 kg / h and the relative COP of the actual plant is 0.65. Determine the power required for the actual plant for the same refrige Solution: T 2
3
1
4 s
Fig. E6.3 : T-s diagram for example 6.3 Data :- T1 = 10 + 273 = 283 K; T3 = 30 + 273 = 303 K; p1 = p4 = 1 bar; p2 = p3 = 5.5 bar Cp = 1.005 kJ / kg (assumed) (6) Process 1-2 is isentropic. Therefore T2 = T1 (p2 / p1)(γ − 1) / γ or
T2 = 283 x [5.5 / 1](1.4 – 1) / 1.4 = 460.82 K
Similarly
T4 = T3(p4 / p3)(γ – 1) / γ = 303 x (1 / 5.5)(1.4 – 1 ) / 1.4 = 18 6.0 K
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.
.
Refrigeration effect per unit mass = qR = QR / m = cp(T1 – T4) = 1.005 x (283 – 186 ) = 97.5 kJ / kg. (ii) Heat rejected in the intercooler per unit mass = qc = cp (T2 – T3) = 1.005 x (460.82 – 303) = 158.61 kJ / kg. (iii) Net work input per unit mass = wN = qc – qR = 158.61 – 97.50 = 61.11 kJ / kg. COP = qR / wN = 97.50 / 61.11 = 1.595 For the actual plant,
(COP)actual = Relative COP x (COP)ideal = 0.65 x 1.595 = 1.0335 .
.
QR = m qR = (1700 / 3600) x 97.50 = 46.04 kW .
Therefore
.
(WN)actual = QR /(COP)actual = 46.04 / 1.0335 = 44.55 kW
6.4.5. Practical Air Refrigeration Cycles Deviations of Practical cycles from an Ideal Cycle:- (i) In any practical air refrigeration cycle, there will always be pressure drops as the air flows through the various components. There will also be pressure drops in the piping connecting the various components. Because of these pressure drops, the pressure ratio for expansion process , p3 / p4 will be less than that for the compression process, p2 / p1.This results in the reduction of the expansion work, which in turn increases the net work input thereby decreasing the COP of the cycle. (ii)The compression and expansion processes in a practical cycle are not isentropic but adiabatic with frictional losses. This results in an increase in compression work and a
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decrease in expansion work, thereby increasing the net work input to the cycle. Hence the there will be a decrease in the COP of the cycle. The frictional losses in compression and
T
2
2’
3’
3
1 4
4”
4’ s
Fig.6.7: T-s diagram for a practical refrigeration cycle and expansion processes are taken into account in the analysis by using a parameter called “Isentropic efficiency”. The isentropic efficiency for compression process is defined as the ratio of the isentropic work of compression to the actual work of compression for the same inlet conditions and exit pressure and the isentropic work of expansion is defined as the ratio of the actual work of expansion to the isentropic work of expansion for the same inlet conditions and the exit pressure. The deviations of the actual cycle from the ideal cycle is shown on the T-s diagram in Fig. 6.7, where 1 – 2’ – 3’ −4” − 1 represents the ideal cycle and 1–2–3– 4–1 is the actual cycle. Expression for COP of a Practical Air Refrigeration Cycle Let ηc = Isentropic efficiency of the compressor, ηt= Isentropic efficiency of the turbine, ∆p1 = pressure loss in the cold chamber = p4 – p1, ∆p2 = pressure loss in the inter cooler = p2 – p3.
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(p2 − ∆p2) p2[1 − ∆p2 / p2] Now expansion ratio = p3 / p4 = ---------------- = ---------------------(p1 + ∆p1) p1 [1 + ∆p1 / p1]
or
p2 [1 − ∆p2 / p2] [1 − ∆p1 / p1] p3 / p4 = ----- ----------------------------------p1 [ 1 − (∆p1 / p1)2 ]
Since (∆p1 / p1)2 << 1, the above expression for the expansion ratio can be written as p3 / p4 = (p2 / p1) ( 1 − ∆p2 / p2) ( 1 − ∆p1 / p1) …………… 6.10 ∆p / p is normally referred to as “ the relative pressure drop” for the heat exchanger circuit. Defining ‘β’ as the pressure drop factor equal to ( 1 – ∆p / p ), the expansion ratio given in equation 6.10 can be written as P3 / p4 = β23 β41 (p2 / p1) …………………………………..6.11(a) Where
β23 = ( 1 − ∆p2 / p2), ………………………………….6.11(b)
And
β41 = ( 1 − ∆p1/ p1)……………………………………6.11(c)
Refrigeration effect per unit mass = qR = cp(T1 – T4) Net work input per unit mass = wN = wc – wt = cp(T2 – T1) – cp(T3 – T4) Therefore
COP = qR / wN cp(T1 – T4) = --------------------------------------cp(T2 – T1) – cp(T3 – T4)
or
(T1 – T4) COP = ------------------------------ ………………….6.12 (T2 – T1) – (T3 – T4)
The above expression for COP can be expressed in terms of (p2 / p1), ηc,ηt, β23, β41, and
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T1(1 – T4 / T1) COP = ------------------------------------------T1(T2 / T1 – 1) – T3(1 – T4 / T3)
(1 – T4 / T1) = ----------------------------------------------- …….. (T2 / T1 – 1) – (T3 / T1) [1 – T4 / T3] { 1 − (T4 / T3)(T3 / T1)} = ---------------------------------------------------- …6.13 (T2 / T1 – 1 ) – (T3 / T1) { 1 – (T4 / T3)} Now
Therefore
(T2’ – T1) (T2’ / T1 – 1 ) ηc = ------------------ = ----------------------(T2 – T1) (T2 / T1 – 1) (T2’ / T1 – 1) T2 / T1 = 1 + -------------------ηc
Process 1-2’ is isentropic. Therefore T2’ / T1 = (p2 / p1)(γ – 1) / γ Therefore
T2 / T1 = 1 + (1/ ηc) [(p2 / p1)(γ – 1) /γ – 1 ]………….6.14(a) (T3 – T4) (1 – T4 / T3) ηt = ------------- = -----------------(T3 – T4’) (1 – T4’ / T3)
Therefore
T4 / T3 = 1 – ηt (1 – T4’ / T3)
Process 3-4’ is isentropic. Therefore T4’ / T3 = (p4’ / p3)(γ – 1) /γ = [β23 β41 (p2 / p1)](γ – 1) / γ …………..6.14(b) Substituting the expressions for T2 / T1 and T4 / T3 from equations 6.14 (a) and 6.14 (b) in equation 6.13 we get
164
1 1 − (T3 / T1) [ 1 − ηt { 1 − ------------------------------- } {β23 β41(p2 / p1)}(γ – 1) / γ COP = ------------------------------------------------------------------------------------------1 (γ – 1) / γ (1/ ηc) [(p2 / p1) − ] − ηt(T3 / T1) [ 1 − ------------------------------ ] {β23 β41(p2 / p1)}(γ – 1) / γ
Example 6.4:- An air refrigeration system is to be designed according to the following specifications: Pressure of air at compressor inlet = 101 kPa; Pressure of air at compressor exit = 404 kPa; Temperature of air at compressor inlet = − 6 C; Temperature of air at turbine inlet = 27 C; Isentropic efficiency of compressor = 85 %; Isentropic efficiency of turbine = 85 %; Relative pressure drop in each heat exchanger = 3 % Capacity of the plant = 1 ton Determine (a) COP of the cycle, (ii) Power required in kW, and (iii) air circulation rate r Solution: The T-s diagram for the given cycle is shown in Fig. E6.4. T 2’
∆p2
2
3 1 4’
4
∆p1
s
Fig.E6.4 : T-s diagram for example 6.4
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Data:- p1 = 101 kPa; p2 = 404 kPa; T1 = − 6 + 273 = 266 K; T3 = 27 + 273 = 300 K; .
ηc = 0.85; ηt = 0.85; ∆p2 / p2 = ∆p1 / p1 = 0.03; QR = 1 ton = 3.517 kW. The first step is to know the temperature at the salient points of the cycle, namely, T1, T2,T3 and T4. T1 and T3 are given and T2and T4 are determined as follows. Process 1-2’ is isentropic. Therefore T2’ = T1(p2 /p1)(γ – 1) / γ = 266 x (404 / 101)0.4 / 1.4 = 395.4 K T2’ – T1 ηc = ---------------T2 – T1 Therefore
T2 = T1 + (T2’ – T1) / ηc (395.4 – 266) = 266 + ------------------0.85 = 418.2 K ∆p2 / p2 = (p2 – p3) / p2 = 0.03.
Therefore Similarly, Therefore
p3 = (1 – 0.03) p2 = 0.97 x 404 = 391.9 kPa. ∆p1 / p1 = (p4 – p1) /p1. p4 = (1 + 0.03) p1 = 1.03 x 101 = 104.03 kPa
Hence expansion ratio for the turbine = p3 / p4 = 391.9 / 104.03 = 3.77 Process 3-4’ is isentropic. Therefore T4’ = T3 (p4 / p3)(γ -1) / γ = 300 x (1 / 3.77 )0.4 / 1.4 = 205.2 K. ηt = (T3 – T4) / (T3 – T4’)
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T4 = T3 – ηt (T3 – T4’) = 300 – 0.85 x (300 – 205.2)
Therefore
= 219.4 K. Refrigeration effect per unit mass = qR = cp(T1 – T4) = 1.005 x (266 – 219.4) = 46.83 kJ / kg. Net work input per unit mass of air = wN = wc – wt = cp(T2 – T1) – cp(T3 – T4) Or
wN = 1.005 x (418.2 – 266 ) – 1.005 x (300 – 219.2)
Or
wN = 152.96 – 81.2 = 71.76 kJ / kg. COP = qR / wN = 46.83 / 71.76 = 0.6526
1 x 3.517 (ii) Power Required = WN = QR / COP = ----------0.6526 = 5.389 kW .
.
.
.
(iii)Mass flow rate of air = m = QR / qR = 1 x 3.517 / 46.83 = 0.0751 kg / s
Example 6.5:- An air refrigerator unit uses a reciprocating compressor and a reciprocating expander. 5 kg / min of air at 30 C (ambient temperature is 25 C) and 4.8 bar expand behind a piston to 1 bar. The expansion is according to the law pv1.35= constant. After expansion, the air enters a cold chamber where its temperature rises to 0 C and the it is compressed back to 4.8 bar according to the law pv1.28 = constant. Determine (a) the power required to drive the unit if the mechanical efficiencies of the expander and the compressor are both equal to 85 %, (b) capacity of the refrigerator in tons, (c) energy rejected by air to the ambient during the cooling process at 4.8 bar and (d) the actual COP of the plant. Solution: A schematic for the example is shown in Fig. E6.5(a) and the T-s diagram in Fig. E6.5(b). .
Data:- m = 5 kg / min = (1 / 12) kg /s ; T3 = 30 + 273 = 303 K ; p3 = p2 = 4.8 bar ;
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p1 = p4 = 1 bar ; ne = index for expansion process = 1.35 ; T1 = 0 + 273 = 273 K ; nc = index for compression process = 1.28 ; (ηm)comp = (ηm)exp = 0.85 ; For compression process 1-2 we have T2 = T1(p2 / p1)(nc – 1) / nc = 273 x (4.8/1)(1.28 – 1) / 1.28 = 384.75 K Similarly for process 3-4 we get
T4 = T3 (p3 / p4) (ne – 1) / ne = 303 x (1/4.8) 0.35 / 1.35 = 201.75 K.
Actual compression work per unit mass = (wc)actual = (wc)ideal / (ηm)comp nc = {1 / (ηm)comp} ----------- RT1 [(p2 / p1)(nc – 1) / nc − 1] (nc – 1) nc = {1 / (ηm)comp} ----------- R[T2 – T1] = (1/ 0.85) x (1.28 / 0.28) x 0.287 x {384.75 – 273} (nc – 1) Or Similarly
(wc)actual = 172.49kJ / kg. (we)actual = (ηm)e (we)ideal ne = (ηm)e ------------ RT3 [ 1 – (p4 / p3)(ne − 1) / ne ] ( ne – 1 ) ne = (ηm)e ------------ R[T3 – T4] (ne – 1) = 0.85 x (1.35 / 0.35) x 0.287 x [ 303 – 201.75] = 95.27 kJ/kg
Therefore
(wN)actual = 172.49 – 95.27 = 77.22 kJ / kg.
Therefore
(wN)actual = 172.78 – 95.29
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= 77.49 kJ / kg. .
.
(a) Actual power required = (WN) = m (wN)actual = (1 / 12 ) x 77.49 kW = 6.46 kW (b) Capacity of the refrigerator = QR = mcp (T1 – T4) = (1 / 12) x 1.005 x (273 – 201.75) = 5.97 kW = 5.97 / 3.517 = 1.6975 ton .
.
(c) Energy rejected by air in the intercooler = QIC = mcp (T2 – T3)
Intercooler 2
3
compressor
expander
Cold chamber
1
4
Fig. E6.5 : Figure for example 6.5(a) .
.
For compression process we have T2 = T1 (p2 / p1)(nc – 1) / nc = 273 x (4.8 / 1.0 )0.28 / 1.28 = 384.75 K .
Therefore
QIC = (1 / 12) x 1.005 x (384.75 – 303) = 6.85 kW
169
.
(d)
.
(COP)actual = QR / (WN)actual = 5.97 / 6.46 = 0.924
Example 6.6:- In an ideal air refrigeration cycle, air after compression in the compressor is first cooled in an intercooler and then passed through a regenerative heat exchanger. It is then expanded in a turbine and after expansion the air flows through the regenerative heat exchanger where it exchanges heat with the air coming from the intercooler. Then the cold air is passed through the cold chamber before it enters the compressor.(a) Draw the schematic layout of the plant.(b) obtain an expression for the COP of the cycle in terms of the pressure ratio of the compressor and the temperature ratio of the compressor inlet temperature to the turbine inlet temperature. Solution: The schematic for the given example is shown in Fig. E6.6(a) and the corresponding T-s diagram is shown in Fig. E6.6(b). Assumption:- The regenerator is perfect i.e., its effectiveness is 100 % (T3 = T1, and T6 = T4). Refrigeration effect per unit mass = qR = cp(T1 – T6) = cp(T1 – T4) = cpT1( 1 – T4 / T1) ………………………(1) Net work input per unit mass = wN = wc − wT = cp(T2 – T1) − cp(T4 – T5) = cpT1[(T2 / T1) – 1 – (T4 / T1) + (T5 /. T1)] ………………………(2) Now process 1-2 is isentropic. Therefore T2 / T1 = (p2 / p1)(γ – 1) /γ T5 / T1 = (T5 / T4) (T4 / T1) = (p1 / p2)(γ – 1) /γ (T4 / T1) Substituting these expressions in Eq.(2) we have wN = cpT1[(p2 / p1)(γ – 1) / γ – 1 – (T4 / T1) + (T4 / T1)(p1 / p2)(γ – 1) / γ ] ………………..(3) cpT1{1 – (T4 / T1) } Therefore COP = qR / wN = ---------------------------------------------------------------------CpT1{(p2 / p1)(γ − 1) / γ − 1 − (T4 / T1) + (T4 / T1)(p1/p2)(γ − 1)/γ }
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3
qH
4
2 6
5
wc
wT
1
qR Fig. E6.6(a): Schematic diagram for example 6.6
2
T
3
4
1
6
5 s Fig. E6.6(b): T-s diagram for example 6.6
Example 6.7 :- An air refrigeration unit takes in air from a cold chamber at 5 C and compresses it from 1 bar to 6.5 bar. The index of compression is 1.25. The compressed air is cooled to a temperature which is 10 C above the ambient temperature of 30 C
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before being expanded isentropically in an expander. Neglecting the clearance volume of the compressor and expander find the COP and the amount of air circulation per minute if 2000 kg of ice at 0 C is to be formed per day from water at 25 C. What will be the tonnage of the unit? Solution: Data:- T1 = 5 + 273 =278 K ; p1 = p3 = 1 bar ; p2 = p4 = 6.5 bar ; T3 = 40 + 273 = 313 C ; compression index = nc = 1.25 ; expansion index = ne = γ = 1.4 Mass of ice to be formed = m = 2000 kg ; Time duration for making the ice = t = 24 h ; Temperature of water available = Tw = 25 C.
3 2
we
wc 1 4 qR
Fig. E6.7: Schematic for example E6.7 Refrigeration effect = Amount of heat to be removed from water at 25 C to convert it into ice at 0 C .
Or
QR ={mcp(Tw – 0) + m x latent heat of ice} / t 2000 x{ 4.192 x (25 – 0) + 335.35} = -----------------------------------------24 x 3600
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6.24 = 10.2 kW = 10.2 / 3.517 = 2.9 ton For compression process 1-2, we have T2 = (p2 / p1)(n – 1)/ n T1 = (6.5 / 1)(0.25 / 1.25 x 278 = 404.23 K. For expansion process 3-4, we have
T4 = (p4 / p3)(γ – 1) / γ T3 = (1 / 6.5)0.4 / 1.4 x 313 = 183.3 K
n Compression work per unit mass = wc = ------------ RT1 {(p2 / p1)(n – 1) / n – 1 } (n–1) = {n / ( n – 1)}R [ T2 – T1] = (1.25 / 0.25) x 0.287 x [404.23 – 278] = 181.14 kJ / kg. γ Expansion work per unit mass = we = ---------------- RT4 {(p3 / p4)(γ – 1 ) / γ – 1 } (γ–1) = [γ / (γ – 1) ] R [T3 – T4] = Cp[T3 – T4] = 1.005 x [313 – 183.3] =130.36 kJ / kg Net work input = wN = wc – we = 181.14 – 130 .36 = 50.78 kJ / kg. Refrigeration effect per unit mass = qR = cp(T1 – T4) = 1.005 x (278 – 183.3) = 95.17 kJ / kg.
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COP = qR / wN = 95.17 / 50.78 = 1.87. .
.
Mass flow rate of air required = m = QR / qR = 10.2 / 95.17 = 0.1072 kg / s. 6.4.6. Practical Applications of Gas Refrigeration Cycles :- The gas refrigeration cycle is exclusively used in air conditioning systems of military and commercial aircrafts. It is more appropriate to call it as “air cycle refrigeration”, since only air has found application as a working substance in this cycle. The COP of this cycle is lower than that of the vapour compression cycle between the same temperature limits. Nevertheless, the air cycle continues to be favoured for aircraft refrigeration because of its many advantages. The present day jet aircrafts have very high cooling loads because of their large occupancy, electronic equipment and high velocity and consequent heat generation due to skin friction. The air cycle can either work as an open system or as a closed system. A closed system also called as a “dense air machine” has many thermodynamic advantages. It can work at a suction pressure higher than atmospheric pressure. This reduces the volume handled by the compressor and expander. Also the operating pressure ratio can be reduced resulting in higher COP. In an open air-cycle system, the air after expansion is directly led to the conditioned space. It is therefore necessary to expand air to one atmospheric pressure. This requires larger volumes to be handled. The open air system has an advantage over the closed system, in respect that, it does not require a heat exchanger for the refrigeration process. This saves the weight and cost of the equipment. But it has one disadvantage, viz., when the air drawn from the refrigerated space is humid, it might produce fog and ice at the end of the expansion process and clog the line. A drier in the circuit is required in such a case. 6.5. Mechanical Vapour Compression Refrigeration Cycle 6.5.1. Limitations of Carnot Refrigeration Cycle With Vapour as a Refrigerant:The Carnot refrigeration cycle ( The T-s diagram for the cycle is shown in Fig. 6.7) can be used as a practical cycle with certain modifications. Though the isothermal process of
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T
3
4
2
1
s
Fig. 6.7 : T – s diagram for Carnot Refrigeration cycle with vapour as a refrigerant heat rejection (process 2-3) and heat absorption (process 4-1) can be achieved in practice, it is extremely difficult to achieve isentropic compression (process 1-2) and isentropic expansion (process 3-4) in practice when the vapour is wet. Therefore in a mechanical vapour compression refrigeration cycle, the isentropic expansion is replaced by a throttling expansion or by an expansion through a capillary and the compression of wet vapour is avoided for reasons explained later in this chapter. 6.5.2. Analysis of an Ideal Mechanical Vapour Compression Refrigeration Cycle Assumptions:- (i) The compression process in the compressor is isentropic (ii) The refrigerant enters the compressor as a dry saturated vapour. (iii)There are no pressure losses in the piping connecting the various components as well as in the heat exchangers, viz., the condenser and the evaporator. (iv)The flow of the refrigerant is steady and one dimensional. (v)Changes in kinetic and potential energies of the refrigerant, as it flows through the various components, are negligible. A schematic diagram of a mechanical vapour compression refrigeration cycle is shown in Fig. 6.8(a). The corresponding T – s and p – h diagrams are shown in Fig. 6.8(b).
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Process 1-2: Isentropic compression of the refrigerant from state 1 to state 2. During this process work is done on the refrigerant by the surroundings.At the end of the process the refrigerant will be in superheated vapour state.
Coolant In
Coolant out Condenser qc Throttle valve
wc
Compressor qR
Evaporator
Fig. 6.8 (a) : Schematic for a vapour compression refrigeration cycle
s
Process 2-3: Constant pressure condensation of the refrigerant in the condenser till it becomes a saturated liquid.
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Process 3-4: Throttling expansion of the refrigerant from condenser pressure to the evaporator pressure. Process 4-1: Constant pressure vapourisation of the refrigerant in the evaporator till it becomes a dry saturated vapour. During this process heat is absorbed by the refrigerant from the place to be refrigerated. Expression for Refrigeration effect and COP in terms of Enthalpies Applying steady flow steady state energy equation to the evaporator and neglecting the changes in kinetic and potential energies we have .
.
Refrigeration effect = QR = m (h1 – h4) …………………………………(6.16) Since process 3-4 is a throttling process, h4 = h3. .
Hence
.
QR = m (h1 – h3) …………………………………(6.17)
Similarly, by applying steady flow, steady state energy equation to compressor we get .
.
Compressor work input = Wc = m (h2 – h1) ………………………………(6.18) .
Hence
QR (h1 – h4) COP = -----.------ = ------------ …………………(6.19) Wc (h2 – h1)
6.5.3. Effect of sub-cooling the refrigerant on the performance of a Vapour Compression Cycle The effect of sub cooling the refrigerant before it enters the expansion valve on the performance of a vapour compression refrigeration cycle is shown on the T-s diagram in Fig. 6.9. In this diagram 1-2-3-4-1 represents the ideal cycle without sub-cooling, while 1-2-3’-4’-1 is the cycle with sub-cooling. It can be seen form the T-s diagram that by subcooling the refrigerant say up to 3’, the refrigeration effect per unit mass has increased from (h1 – h4) to (h1 – h4’). Therefore, theoretically, it is desirable to sub-cool the refrigerant before it enters the throttle valve. But for almost all refrigerants, the constant pressure line in the sub-cooled liquid region almost coincides with the saturated liquid line. Therefore it is sufficient to compress the refrigerant up to 2’ (p2’ is the saturation pressure corresponding to T3’) and then cool it in the condenser till it becomes a saturated liquid at state 3” (h3’ ≈ h3”). Therefore it is not necessary to compress the refrigerant up to 2 and then sub-cool it to 3’.
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T
s Fig. 6.9: Effect of sub-cooling the refrigerant on Vapour compression cycle But in practice, the condensers are always designed assuming certain amount of subcooling of the refrigerant to ensure that the refrigerant coming out of the condenser is at least in saturated liquid state. 6.5.4. Dry compression versus Wet compression In the analysis of the ideal vapour compression refrigeration cycle it was assumed that the refrigerant enters the compressor as a dry saturated vapour.As the compression process proceeds the vapour becomes a superheated vapour. In the case of wet compression it will be assumed that the refrigerant leaves the compressor as a dry saturated vapour.The comparison of these two compression processes are shown on the T-s diagram in Fig. 6.10. When a reciprocating compressor is used for compression, wet compression is not found suitable due to the following reasons. (i) Liquid refrigerant may be trapped in the head of the compressor cylinder because, the speed of the compressor is quite high and the liquid present in the wet vapour cannot readily evaporate. If enough liquid is collected in the compressor, then the compression may not be possible and the system may break down. (ii) Liquid refrigerant droplets may wash away the lubricating oil from the walls of the compressor cylinder thus increasing wear. It is therefore desirable to have compression with refrigerant in dry saturated vapour state or even slightly in superheated vapour state at entry to the compressor.This type of compression is known as dry compression. In dry compression the state of refrigerant at the end of compression will therefore be at 2, pressure at 2 being the saturation pressure corresponding to the condensing temperature T2’. For wet compression the state point after the end of compression will be at 2”, which would be
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the state point if Carnot cycle were to be employed. Because of dry compression, the discharge temperature T2 will be higher than the condensing temperature.Because of dry compression, the work of compression increases by an amount shown by the shaded area and this area is normally referred to as “superheat horn”. In some cases, wet compression is indeed desirable and also practicable with the use of screw compressor instead of a reciprocating compressor. 6.5.5. Practical Vapour Compression Refrigeration Cycle Deviations of Practical Cycle from ideal cycle:- (i)In the ideal cycle it has been assumed that the refrigerant leaving the compressor is in dry saturated vapour state. Since the cooling load is not constant, it is extremely difficult to ensure that the fluid is dry at compressor inlet. Moreover in high speed reciprocating compressors, the compression occurs so fast that the liquid in the vapour cannot readily evaporate and may collect inside the compressor. If enough liquid collects, vapour compression will not be possible and the system may break down. The practice, therefore, is to design the evaporator to superheat the vapour emerging from it as shown by state point 1’ on T-s diagram in Fig. 6.11.With this arrangement, the net work input increases to achieve the same compression ratio, the volumetric efficiency decreases and the overall COP of the cycle also decreases. (ii) In the ideal cycle analysis it was assumed that the compression process is isentropic. But in a practical cycle, the process will not be isentropic but adiabatic with frictional losses. Therefore, in the practical cycle, the frictional losses during compression are T
s Fig. 6.11: T-s diagram for a practical vapour compression refrigeration cycle
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accounted by using a parameter called “isentropic efficiency of compressor”. This efficiency is defined as the ratio of isentropic work of compression required to the actual work of compression between the same pressure limits. Because of frictional losses the net work input increases, which in turn decreases the overall COP. (iii) For a reciprocating compressor, there will be pressure drop across the inlet and delivery valves. This increases the operating pressure ratio for the compressor, thereby increasing the net work input to the compressor and hence the COP decreases. 6.5.6. Desirable properties of refrigerants to be used in a vapour compression cycle Following are the desirable properties for refrigerants to be used in a vapour compression cycle. (i) When the evaporator works under vacuum conditions, it is difficult to make the system completely leak-proof. Hence the saturation pressure corresponding to the evaporator temperature should be close to the atmospheric pressure. Infact it is preferable to have the evaporator pressure slightly above atmospheric pressure in order to prevent air leaking into the system. Even the presence of a small amount of air in the evaporator reduces the heat transfer coefficient, which in turn decreases the refrigeration effect. (ii) The pressure difference between the evaporator and condenser should be moderate to reduce the leakage losses. Also if the pressure difference is large then it will be necessary to use multi-stage compression with inter cooling in order to have a high compression efficiency. (iii) The triple point and the critical point of the refrigerant should be far away from the operating range of the refrigerating system. (iv) The density of the vapour should be as high as possible to avoid large flow velocities which result in high pressure drops in pipe lines. For a vapour with low density it is preferable to use centrifugal compressors. (v) Higher the enthalpy of vapourization of the refrigerant lower will be the mass flow rate of refrigerant required to achieve the desired refrigeration effect. (vi) The refrigerant should have a high thermal conductivity and low viscosity so that the evaporator and condenser can be reasonably small in size.A low viscosity ensures a low pressure drop. (vii) The specific heats of the liquid and vapour refrigerant should be as low as possible. A fluid, which has a low specific heat, will have nearly vertical sides for the saturation dome in the T – s diagram. In that case the real cycle approaches towards Carnot cycle. (viii) The refrigerant should be easily available and safe to handle. (ix) The refrigerant vapour should be inert, stable and should not react with materials of the compressor, condenser, evaporator coils and the valves. (x) The vapour should be insoluble in water.
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CHAPTER 8 Measurements and Testing of I.C.Engines 8.1. Introduction: - The basic task in the design and development of I.C.Engines is to reduce the cost of production and improve the efficiency and power output. In order to achieve the above task, the engineer has to compare the engine developed by him with other engines in terms of its output and efficiency. Hence he has to test the engine and make measurements of relevant parameters that reflect the performance of the engine. In general the nature and number of tests to be carried out depend on a large number of factors. In this chapter only certain basic as well as important measurements and tests are described. 8.2. Important Performance Parameters of I.C.Engines:- The important performance parameters of I.C. engines are as follows: (i) Friction Power, (ii) Indicated Power, (iii) Brake Power, (iv) Specific Fuel Consumption, (v) Air – Fuel ratio (vi) Thermal Efficiency (vii) Mechanical Efficiency, (viii) Volumetric Efficiency, (ix) Exhaust gas emissions, (x) Noise 8.3. Measurement of Performance Parameters in a Laboratory 8.3.1. Measurement of Friction Power:- Friction power includes the frictional losses and the pumping losses. During suction and exhaust strokes the piston must move against a gaseous pressure and power required to do this is called the “pumping losses”. The
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friction loss is made up of the energy loss due to friction between the piston and cylinder walls, piston rings and cylinder walls, and between the crank shaft and camshaft and their bearings, as well as by the loss incurred by driving the essential accessories, such as water pump, ignition unit etc. Following methods are used in the laboratory to measure friction power: (i) Willan’s line method; (ii) From the measurement of indicated power and brake power; (iii) Motoring test; (iv) Retardation test; (v) Morse Test. 8.3.1.1. Willan’s Line Method:- This method is also known as fuel rate extrapolation method. In this method a graph of fuel consumption (vertical axis) versus brake power (horizontal axis) is drawn and it is extrapolated on the negative axis of brake power (see Fig. 8.1).The intercept of the negative axis is taken as the friction power of the engine at 1.0 Fuel 0.8 consumption (g/s) 0.8 0.6 0.4 0.2 5
10
15 20 25 Brake power (kW)
Friction Power Fig. 8.1. Willan’s line method that speed. As shown in the figure, in most of the power range the relation between the fuel consumption and brake power is linear when speed of the engine is held constant and this permits extrapolation. Further when the engine does not develop power, i.e. brake
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power = 0, it consumes a certain amount of fuel. This energy in the fuel would have been spent in overcoming the friction. Hence the extrapolated negative intercept of the horizontal axis will be the work representing the combined losses due to friction, pumping and as a whole is termed as the frictional loss of the engine. This method of measuring friction power will hold good only for a particular speed and is applicable mainly for compression ignition engines. The main draw back of this method is the long distance to be extrapolated from data between 5 and 40 % load towards the zero line of the fuel input.The directional margin of error is rather wide because the graph is not exactly linear. 8.3.1.2.From the Measurement of Indicated Power and Brake Power:- This is an ideal method by which friction power is obtained by computing the difference between the indicated power and brake power. The indicated power is obtained from an indicator diagram and brake power is obtained by a brake dynamometer. This method requires elaborate equipment to obtain accurate indicator diagrams at high speeds. 8.3.1.3.Morse Test:- This method can be used only for multi – cylinder IC engines. The Morse test consists of obtaining indicated power of the engine without any elaborate equipment. The test consists of making, in turn, each cylinder of the engine inoperative and noting the reduction in brake power developed. In a petrol engine (gasoline engine), each cylinder is rendered inoperative by “shorting” the spark plug of the cylinder to be made inoperative. In a Diesel engine, a particular cylinder is made inoperative by cutting off the supply of fuel. It is assumed that pumping and friction are the same when the cylinder is inoperative as well as during firing. In this test, the engine is first run at the required speed and the brake power is measured. Next, one cylinder is cut off by short circuiting the spark plug if it is a petrol engine or by cutting of the fuel supply if it is a diesel engine. Since one of the cylinders is cut of from producing power, the speed of the engine will change. The engine speed is brought to its original value by reducing the load on the engine. This will ensure that the frictional power is the same. If there are k cylinders, then Total indicated power k when all the cylinders are working = ip1 + ip2 + ip3 + …………...+ ipk = ∑ipj j=1
k
We can write ∑ipj = Bt + Ft ………………………………………..(8.1) j=1
where ipj is the indicated power produced by j th cylinder, k is the number of cylinders,
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Bt is the total brake power when all the cylinders are producing power and Ft is the total frictional power for the entire engine. If the first cylinder is cut – off, then it will not produce any power, but it will have frictional losses. Then k
we can write ∑ipj = B1 - Ft………………………………………..(8.2) j=2
where B1 = total brake power when cylinder 1 is cut - off and Ft = Total frictional power. Subtracting Eq. (8.2) from Eq. (8.1) we have the indicated power of the cut off cylinder. Thus ip1 = Bt – B1 ………………………………………..(8.3). Similarly we can find the indicated power of all the cylinders, viz., ip2, ip3, …..ipk. Then the total indicated power is calculated as k
(ip)total = ∑ipj ……………………………………….(8.4) j=1
The frictional power of the engine is therefore given by Ft = (ip)total – Bt ……………………………………(8.5) The procedure is illustrated by some examples worked out at the end of the chapter. 8.3.2. Measurement of Indicated Power: Indicated power of an engine tells about the health of the engine and also gives an indication regarding the conversion of chemical energy in the fuel into thermal energy. It is an important variable because it is the potential output of the cycle.Hence the measurement of indicated power must be very accurate. For obtaining indicated power the cycle pressure must be determined as a function of cylinder volume. It is of no use to determine pressure accurately unless volume or crank angle is accurately measured. Following methods are usually adopted to estimate the indicated power of an IC engine: (i) using the indicator diagram (ii) by measuring brake power and friction power and adding them to give indicated power
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(iii) By conducting Morse test (applicable only for multi-cylinder engines). (i) Method using Indicator diagram:- Indicator is a device which measures the variation of the pressure in the cylinder over a part or full cycle and the plot of such information obtained is called an “indicator diagram”. Indicator diagram gives a very good indication of the process of combustion. Also the losses occurring in the suction and exhaust stroke can be studied using this diagram. Indicated power = indicated work per cycle x number of cycles / second pim Vs ne K ip = ---------------1000 x 60 pim LA n K = --------------- kW…………………………………….(8.6) 60,000 Where ip = indicated power (kW) pim = indicated mean effective pressure (N / m2) L = Length of the stroke of the piston (m) A = Area of the piston (m2) ne = number of power strokes per minute or number of explosions per minute K = number of cylinders. Now ne = N / 2 for a four stroke engine and = N for a two – stroke engine, where N is the speed of the engine in RPM. The indicated mean effective pressure is calculated by using the equation Area of the indicated diagram in m2 pim = ----------------------------------------------- x indicator spring stiffness in (N /m2) /m Length of the indicator diagram in m (ii) Method by measuring Brake Power and Friction Power:- In this method the brake power is measured by using a brake dynamometer and the friction power is measured by using Willan’s line method or by Motoring test or by retardation test and then the indicated power is calculated as ip = bp + fp where bp is the brake power of the engine and fp is the friction power.
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(iii) Morse Test:- This method is already described in section 8.3.1.3. 8.3.3. Measurement of Brake Power: Indicated power is based on indicated net work and is thus a measure of the forces developed within the engine cylinder.The rotational force available at the engine crank shaft and the power corresponding to this is of more practical interest. This power is interchangeably referred to as brake power or shaft power. Measurement of brake power is one of the most important measurements in the test schedule of an engine. It involves the determination of the torque and the angular speed of the engine output shaft. The torque measuring device is called the dynamometer. Brake power is usually measured by attaching a power absorbing device to the output shaft of the engine. Such a device will set up forces (which can be measured) counteracting the forces delivered by the engine. The measured force is usually referred to as the braking force, W. If ‘r’ is the distance from the centre of the output shaft and the point of acting of Fb ( r is also called the lever arm), then braking torque = T = W r …………………………………………….(8.7) If N is the speed of the engine in RPM, then brake power in KW is given by Brake power = bp = 2πNT / (60 x 1000) Or
bp = 2πNT / 60,000
kW ………………………………..(8.8)
In the above equation T should be in N – m. Since in Eq. (8.7) only the brake load and speed can vary for a given engine, the equation can be written as WN bp = --------------- …………………………………………..(8.9) C where C is called the dynamometer constant. Eq. (8.7) or Eq. (8.8) can be used to calculate the brake power depending on the information available for a given engine. Brake mean effective pressure: - Brake mean effective pressure (bmep) can be considered as that portion of indicated mean effective pressure which goes into the development of useful work output from the engine. Hence bmep is related to brake power by the following equation. (bmep) LA ne K bp = ----------------------- ……………………………………….(8.10) 60000
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8.3.4. Measurement of Fuel Consumption: There are two basic types of fuel measurement methods namely (i) volumetric method and (ii) gravimetric method. In volumetric method the time taken for the engine to consume a known volume of fuel is measured and knowing the specific gravity of the fuel the mass flow rate of fuel consumed is calculated as follows: Let t = time in seconds for the engine to consume ‘y’ cc of fuel, and s = specific gravity of the fuel. y x 10 − 6 Mass of fuel consumed per second = m = -------------- x s x 1000 t .
ys m = ----------- kg/s………………….(8.11) 1000 t .
Or
Indicated Specific Fuel Consumption: This is defined as the mass of fuel consumption per hour in order to produce an indicated power of one kilo watt. .
3600 m Thus, indicated specific fuel consumption = isfc = --------------- kg/kWh …..(8.12) ip Brake Specific fuel consumption:- This defined as the mass of fuel consumed per hour, in order to develop a brake power of one kilowatt. .
3600 m Thus, brake specific fuel consumption = bsfc = --------------- kg/kWh ……..(8.13) bp
8.3.5. Thermal Efficiency : There are two definitions of thermal efficiency as applied to IC engines. One is based on indicated power and the other on brake power.The one based on indicated power is called as ‘indicated thermal efficiency”, and the one based on brake power is known as “brake thermal efficiency”. Indicated thermal efficiency is defined as the ratio of indicated power to the energy available due to combustion of the fuel. Indicated Power in kW Thus ηith = ------------------------------------------------------------------------------------(Mass flow rate of fuel in kg/s) x (Calorific value of fuel in kJ/kg )
Or
ip ηith = --------------- ………………………………………………..(8.14) m x CV
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Similarly brake thermal efficiency is defined as the ratio of brake power to energy available due to combustion of the fuel.
Or
bp ηbth = --------------- ………………………………………………..(8.15) m x CV
8.3.6.Mechanical Efficiency: Mechanical efficiency takes into account the mechanical losses in an engine. The mechanical losses include (i) frictional losses, (ii) power absorbed by engine auxillaries like fuel pump, lubricating oil pump, water circulating pump, magneto and distributor, electric generator for battery charging, radiator fan etc., and (iii) work requited to charge the cylinder with fresh charge and work for discharging the exhaust gases during the exhaust stroke. It is defined as the ratio of brake power to indicated power. Thus bp ηmech = -------- ……………………………(8.16) ip 8.3.7. Volumetric efficiency: Volumetric efficiency is the ratio of the actual mass of air drawn into the cylinder during a given period of time to the theoretical mass which should have been drawn in during the same interval of time based on the total piston displacement, and the pressure and temperature of the surrounding atmosphere. .
Thus
mactual ηv = ----------------- …………………………(8.17) mth
where
m = ρa n Vs ………………………………….(8.18)
.
where n is the number of intake strokes per minute and Vs is the stroke volume of the piston. 8.4. Illustrative examples: Example 8.1:- The following observations have been made from the test of a four cylinder, two – stroke petrol engine. Diameter of the cylinder = 10 cm; stroke = 15 cm; speed = 1600 rpm; Area of indicator diagram = 5.5 cm2; Length of the indicator diagram = 55 mm; spring constant = 3.5 bar/cm; Determine the indicated power of the engine. Given:- d = 0.1 m; L = 0.15 m ; No. of cylinders = K = 4; N = 1600 rpm; n = N (two – stroke); a = 5.5 cm2; length of the diagram = ld = 5.5. cm; spring constant = ks = 3.5 bar/cm ; To find: indicated power, ip.
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a ks Solution: Indicated mean effective pressure = pim = --------------ld or
pim
5.5 x 3.5 = -------------- = 3.5 bar = 3.5 x 10 5 N / m2 5.5
pim LAnK 3.5 x 105 x 0.15 x (π/4) x 0.12 x 1600 x 4 Indicated power = ip = ---------------- = --------------------------------------------------60,000 60,000 = 43.98 kW Example 8.2:- A gasoline engine (petrol engine) working on Otto cycle consumes 8 litres of petrol per hour and develops 25 kW. The specific gravity of petrol is 0.75 and its calorific value is 44,000 kJ/kg. Determine the indicated thermal efficiency of the engine Given:- Volume of fuel consumed/hour = y/t = 8 x 10 3 / 3600 cc/s ; ip = 25 kW; CV = 44,000 kJ/kg; Specific gravity of petrol = s = 0.75 To find: ηith ;
ys 8 x 10 3 x 0.75 Solution: Mass of fuel consumed = m = ----- = --------------------- = 1.67 x 10 − 3 kg/s. 1000 t 1000 x 3600 .
ip 25 Indicated thermal efficiency = ηith = -------------- = ---------------------------m CV 1.67 x 10 − 3x 44000 = 0.3402 = 34.02 %. Example 8.3:- The bore and stroke of a water cooled, vertical, single-cylinder, four stroke diesel engine are 80 mm and 110 mm respectively.The torque is 23.5 Nm.Calculate the brake mean effective pressure. What would be the mean effective pressure and torque if the engine rating is 4 kW at 1500 rpm? Given:- Diameter = d = 80 x10 − 3 = 0.008 m ; stroke = L = 0.110 m; T = 23.5 N-m; To find (i) bmep ; (ii) bmep if bp = 4 kw and N= 1500 rpm.
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Solution: (i) Relation between brake power (bp) and brake mean effective pressure (bmep) is given by 2πNT (bmep)LAn bp = ------------ = --------------60,000 60,000 Hence bmep = (2πNT) / (LAn) = (2πNT) / {(Lπd2 /4) N/2} 16T 16 x 23.5 = ------------- = ---------------- = 5.34 x 10 5 N / m2 = 5.34 bar d2L 0.08 2 x 0.11 (ii) when bp = 4 kw and N = 1500 rpm, we have 60,000 bp 60,000 x 4 bmep = -------------- = ------------------------------------------LAn 0.110 x (π/4) x 0.08 2 x (1500 / 2) = 5.79 x 10 5 N/m2 = 5.79 bar.
Also bp = 2πNT / 60,000
60,000 bp 60,000 x 4 or T = ---------------- = -------------------- = 25.46 N – m. 2πN 2 x π x 1500
Example 8.4:-Find the air fuel ratio of a four stroke, single cylinder, air cooled engine with fuel consumption time for 10 cc is 20.4 s and air consumption time for 0.1 m3is 16.3 s. The load is 7 N at the speed of 3000 rpm. Find also the brake specific fuel consumption in kg/kWh and brake thermal efficiency.Assume the density of air as 1.175 kg/m3 and specific gravity of the fuel to be 0.7. The lower heating value of the fuel is 43 MJ/kg and the dynamometer constant is 5000. Given:- y = 10 cc ; t = 20.4 s ; Va = 0.1 m3; ta = 16.3 s ; W = 7 N ; N = 3000 rpm; ρa = 1.175 kg/m3; s = 0.7 ; CV = 43 x 10 3 kJ/kg; Dynamometer constant = C = 5000. To find:- (i) ma / mf ; (ii) bsfc ; (iii) ηbth. 0.1 x 1.175 Solution: (i) Mass of air consumed = ma = ---------------- = 7.21 x 10 − 3 kg/s. 16.3 ys 10 x 0.7 Mass of fuel consumed = mf = --------- = ------------------ = 0.343 x 10 − 3 kg/s 1000 t 1000 x 20.4
190
8.11 −3
ma 7.21 x 10 Air fuel ratio = --------- = ------------------- = 21 mf 0.343 x 10 − 3 7 x 3000 (ii) Brake power = bp = WN / C = ------------- = 4.2 kW 5000 mf x 3600 0.343 x 10 −3 x 3600 bsfc = ------------------- = ---------------------------- = 0.294 kg/kWh bp 4.2
(iii) bith
bp 4.2 = --------------- = ----------------------------------- = 0.2848 = 28.48 %. mf CV 0.343 x 10 −3 x 43 x 10 3
Example 8.5:- A six cylinder, gasoline engine operates on the four stroke cycle. The bore of each cylinder is 80 mm and the stroke is 100 mm. The clearance volume in each cylinder is 70 cc. At a speed of 4000 rpm and the fuel consumption is 20 kg/h. The torque developed is 150 N-m. Calculate (i) the brake power, (ii) the brake mean effective pressure, (iii) brake thermal efficiency if the calorific value of the fuel is 43000 kJ/kg and (iv) the relative efficiency if the ideal cycle for the engine is Otto cycle. Given:- K = 6 ; n = N /2 ; d = 8 cm ; L = 10 cm ; Vc = 70 cc ; N = 4000 rpm ; mf = 20 kg/h ; T = 150 N-m ; CV = 43000 kJ/kg ; To find:- (i) bp ; (ii) bmep ; (iii) ηbth ; (iv) η Relative. Solution: (i)
2π NT 2 x π x 4000 x 150 bp = ------------ = -------------------------60,000 60,000 = 62.8 kW
60,000 bp 60,000 x 62.8 (ii) bmep = -------------------- = -------------------------------------------LAnK 0.1 x (π / 4) x 0.08 2 x (4000/2) x 6 = 6.25 x 10 5 N/m2 = 6.25 bar bp 62.8 (iii) ηbth = ----------------------- = ------------------------------ = 0.263 = 26.3 %. mf CV (20 / 3600) x 43,000
191
(iv) Stroke volume = Vs = (π / 4) d2 L = (π / 4) x 8 2 x 10 = 502.65 cc Vs + Vc 502.65 + 70 Compression Ratio of the engine = Rc = ------------ = --------------------- = 8.18 Vc 70 Air standard efficiency of Otto cycle = η Otto = 1 – (1/ Rcγ – 1) 1 = 1 − ---------------------- = 0.568 = 56.8 % 8.18 0.4 Hence Relative efficiency = ηRelative = η bth / η Otto = 0.263 / 0.568 = 0.463 = 46.3 %. Example 8.6:- An eight cylinder, four stroke engine of 9 cm bore, 8 cm stroke and with a compression ratio of 7 is tested at 4500 rpm on a dynamometer which has 54 cm arm. During a 10 minute test, the dynamometer scale beam reading was 42 kg and the engine consumed 4.4 kg of gasoline having a calorific value of 44,000 kJ/kg. Air at 27 C and 1 bar was supplied to the carburetor at a rate of 6 kg/min. Find (i) the brake power, (ii) the brake mean effective pressure, (iii) the brake specific fuel consumption, (iv) the brake specific air consumption, (v) volumetric efficiency , (vi) the brake thermal efficiency and (vii) the air fuel ratio. Given:- K = 8 ; Four stroke hence n = N/2 ; d = 0.09 m; L = 0.08 m; Rc = 7; N = 4500 rpm; Brake arm = R = 0.54 m ; t = 10 min ; Brake load = W = (42 x 9.81) N .
mf = 4.4 kg ; CV = 44,000 kJ/kg ; Ta = 27 + 273 = 300 K ; pa = 1 bar; ma = 6 kg/min; .
.
To find:- (i) bp ; (ii) bmep ; (iii) bsfc ; (iv) bsac ; (v) η v ; (vi) ηbth ; (vii) ma / mf Solution: 2π NT 2π NWR 2 x π x 4500 x (42 x 9.81) x 0.54 (i) bp = ----------- = ------------ = -------------------------------------------60,000 60,000 60,000 = 104.8 kW 60,000 bp 60,000 x 104.8 (ii) bmep = ------------- = -----------------------------------------------LAnK 0.08 x (π / 4) x 0.09 2 x (4500 / 2) x 8 = 6.87 x 10 5 N/m2 = 6.87 bar.
192
.
(iii) mass of fuel consumed per unit time = mf = mf / t = 4.4 x 60 / 10 kg/h = 26.4 kg/h .
mf 26.4 Brake specific fuel consumption = bsfc = ----------- = ------------- = 0.252 kg / kWh bp 104.8 .
ma 6 x 60 (iv) brake specific air consumption = bsac = ------------- = -----------------bp 104.8 = 3.435 kg / kWh (v) ηbth
bp 104.8 = --.-------------- = ----------------------------- = 0.325 = 32.5 %. mf CV (26.4 / 3600) x 44,000 .
(vi) Stroke volume per unit time = Vs = (πd2/4) L n K π = ------ x (0.092) x 0.08 x (4500 / 2) x 8 4 = 9.16 m3 / min. .
ma Ra Ta 6 x 286 x 300 Volume flow rate of air per minute = Va = ------------- = -------------------pa 1 x 10 5 .
= 5.17 m3 / min .
.
Volumetric efficiency = ηv = Va / Vs = 5.17 / 9.16 = 0.5644 = 56.44 %. .
.
(vii) Air fuel ratio = ma / mf = 6 /(4.4 / 10) = 13.64 Example 8.7:- A gasoline engine working on four- stroke develops a brake power of 20.9 kW. A Morse test was conducted on this engine and the brake power (kW) obtained when each cylinder was made inoperative by short circuiting the spark plug are 14.9, 14.3, 14.8 and 14.5 respectively. The test was conducted at constant speed. Find the indicated power, mechanical efficiency and brake mean effective pressure when all the cylinders are firing. The bore of the engine is 75mm and the stroke is 90 mm. The engine is running at 3000 rpm.
193
Given:- brake power when all cylinders are working = Bt = 20.9 kW ; Brake power when cylinder 1 is inoperative = B1 = 14.9 kW ; Brake power when cylinder 2 is inoperative = B2 = 14.3 kW ; Brake power when cylinder 3 is inoperative = B3 = 14.8 kW ; Brake power when cylinder 4 is inoperative = B4 = 14.5 kW ; N = 3000 rpm ; d = 0.075 m ; L = 0.09 m ; To find:- (i) (ip)total ; (ii) ηmech ; (iii) bmep ; Solution: (i) (ip)total = ip1 + ip2 + ip3 + ip4 = (Bt – B1) + (Bt – B2) + (Bt – B3) + (Bt – B4) = 4Bt – (B1 + B2 + B3 + B4) = 4 x 20.9 – (14.9 + 14.3 + 14.8 + 14.5) = 25.1 Kw Bt 20.9 (ii) ηmech = ------------- = --------- = 0.833 = 83.3 % (ip)total 25.1 60,000 Bt 60,000 x 20.9 (iii) bmep = --------------- = -------------------------------------------------LAnK 0.09 x (π / 4) x 0.075 2 x (3000 / 2) x 4 = 5.25 x 10 5 N / m2 = 5.25 bar. Example 8.8:- The following observations were recorded during a trail of a four – stroke, single cylinder oil engine. Duration of trial = 30 min ; oil consumed = 4 litres ; calorific value of oil = 43 MJ/kg ; specific gravity of fuel = 0.8 ; average area of the indicator diagram = 8.5 cm2; length of the indicator diagram = 8.5 cm; Indicator spring constant = 5.5 bar/cm; brake load = 150 kg; spring balance reading = 20 kg; effective brake wheel diameter = 1.5 m ; speed = 200 rpm ; cylinder diameter = 30 cm ; stroke = 45 cm ; jacket cooling water = 10 kg/min ; temperature rise of cooling water = 36 C. Calculate (i) indicated power, (ii) brake power, (iii) mechanical efficiency, (iv) brake specific fuel consumption, (v) indicated thermal efficiency, and (vi) heat carried away by cooling water.
194
Given:- t = 30 min ; y = 4000 cc; CV = 43 x10 3 kJ/kg; s = 0.8 ; area of the diagram = a = 8.5 cm2; length of the diagram = ld = 8.5 cm ; indicator spring constant = ks = 5.5 bar / cm; W = 150 x 9.81 N ; Brake radius = R = 1.5 / 2 = 0.75 m; N = 200 rpm ; d = 0.3 m ; .
L = 0.45 m ; mw = 10 kg/min ; ∆Tw = 36 C; Spring Balance Reading = S = 20 x 9.81 N .
To find:- (i) ip ; (ii) bp ; (iii) ηmech ; (iv) bsfc ; (v) ηith ; (vi) Qw Solution: a 8.5 (i) pim = ------- ks = ----------- x 5.5 = 5.5 bar = 5.5 x 105 N/m2 ld 8.5 pim L A n K 5.5 x 10 5 x 0.45 x (π / 4) x 0.32 x (200 / 2) x 1 ip = ------------------- = -----------------------------------------------------------60,000 60,000 = 29.16 kW 2π N(W – S) R 2 x π x 200 x (150 – 20)x 9.81 x 0.75 (ii) bp = ----------------------- = -----------------------------------------------60,000 60,000 = 20.03 kW (iii) ηmech = bp / ip = 20.03 / 29.16 = 0.687 = 68.7 %. ys 4000 x 0.8 . (iv) Mass of fuel consumed per hour = mf = ------------- x 60 = ------------------ x 60 1000 t 1000 x 30 = 6.4 kg / h. 6.4 bsfc = mf / bp = ------------- = 0.3195 kg/kWh 20.03 .
ip 29.16 (v) ηith = ---.--------- = -------------------------------- = 0.3814 = 38.14 %. mf CV (6.4 / 3600) x 43 x 10 3 .
.
(vi) Qw = m Cp ∆Tw = (10 / 60) x 4.2 x 36 = 25.2 kW
195
Example 8.9:- A four stroke gas engine has a cylinder diameter of 25 cm and stroke 45 cm. The effective diameter of the brake is 1.6 m.The observations made in a test of the engine were as follows. Duration of test = 40 min; Total number of revolutions = 8080 ; Total number of explosions = 3230; Net load on the brake = 80 kg ; mean effective pressure = 5.8 bar; Volume of gas used = 7.5 m3; Pressure of gas indicated in meter = 136 mm of water (gauge); Atmospheric temperature = 17 C; Calorific value of gas = 19 MJ/ m3 at NTP; Temperature rise of cooling water = 45 C; Cooling water supplied = 180 kg. Draw up a heat balance sheet and find the indicated thermal efficiency and brake thermal efficiency. Assume atmospheric pressure to be 760 mm of mercury. Given:- d = 0.25 m ; L = 0.45 m; R = 1.6 / 2 =0.8 m; t = 40 min ; Ntotal = 8080 ; Hence N = 8080 / 40 = 202 rpm ntotal = 3230 ; Hence n = 3230 / 40 = 80.75 explosions / min; W = 80 x 9.81 N; pim = 5.8 bar ; .
Vtotal = 7.5 m3; hence V = 7.5 / 40 = 0.1875 m3/min; pgauge = 136 mm of water (gauge); Tatm = 17 + 273 = 290 K; (CV)NTP = 19 x 10 3 kJ/ m3 ; ∆Tw = 45 C; .
mw = 180 / 40 = 4.5 kg/min; patm = 760 mm of mercury To find:- (i) ηith ; (ii) ηbth ; (iii) heat balance sheet Solution: (i)
pim L A n K 5.8 x 10 5 x (π / 4) x 0.252 x 0.45 x 80.75 ip = ------------------ = ---------------------------------------------------60,000 60,000 = 17.25 kW. 2π N W R 2 x π x 202 x (80 x 9.81) x 0.8 bp = --------------------- = -----------------------------------------60,000 60,000 = 13.28 kW
Pressure of gas supplied = p = patm + pgauge = 760 + 136 / 13.6 = 770 mm of mercury .
.
Volume of gas supplied as measured at NTP = VNTP = V (TNTP / T)(p / pNTP) 0.1875 x 273 x 770 = -------------------------- = 0.17875 m3 / min 290 x 760
196
.
.
Heat supplied by fuel = Qf = VNTP (CV)NTP = 0.17875 x 19 x 103 = 3396.25 kJ/min Heat equivalent of bp in kJ/min = 13.28 x 60 = 796.4 kJ/min .
Heat lost to cooling water in kJ/min = mw Cp ∆Tw = 4.5 x 4.2 x 45 = 846.5 kJ/min Friction power = ip – bp = 17.25 – 13.28 = 3.97 kW Hence heat loss due to fiction, pumping etc. = 3.97 x 60 = 238.2 kJ/min Heat lost in exhaust, radiation etc (by difference) = 3396.25 – (896.4 + 796.4 + 238.2) = 1465.15 kJ/min Heat Balance Sheet: Item No.
Heat Energy Input (kJ/min) (percent)
1
Heat supplied by fuel
2
Heat equivalent of bp
896.4
26.4
3
Heat lost to cooling Water
796.4
23.4
4
Heat equivalent of fp
238.2
7.0
5
Heat unaccounted (by difference)
1465.15
43.2
3396.25
100.0
Total
3396.25
Heat Energy spent (kJ/min) (percent)
100.00
3396.25
100.0
Example 8.10:- A test on a two-stroke engine gave the following results at full load. Speed = 350 rpm; Net brake load = 65 kg ; mean effective pressure = 3 bar ; Fuel consumption = 4 kg/h ; Jacket cooling water flow rate = 500 kg/h ; jacket water temperature at inlet = 20 C ; jacket water temperature at outlet = 40 C ; Test room temperature = 20 C ; Temperature of exhaust gases = 400 C; Air used per kg of fuel = 32 kg ; cylinder diameter = 22 cm ;stroke = 28 cm; effective brake diameter = 1 m ; Calorific value of fuel = 43 MJ/kg ; Mean specific heat of exhaust gases = 1 kJ/kg –K. Find indicated power, brake power and draw up a heat balance for the test in kW and in percentage.
197
Given:- Two stroke engine. Hence n = N ; N = 350 rpm ; W = (65 x 9.81) N ; .
.
pim = 3 bar ; mf = 4 kg/h ; mw = 500 kg/h ; Twi = 20 C ; Two = 40 C ; Tatm = 20 C ; .
.
Teg = 400 C ; ma / mf = 32 ; d = 0.22 m ; L = 0.28 m ; Brake radius = R = ½ m ; CV = 43,000 kJ/kg ; (Cp) eg = 1.0 kJ/(kg-K) ; To find:- (i) ip ; (ii) bp ; and (iii) heat balance; Solution: pim LAn 3 x 10 5 x 0.28 x (π/4) x 0.22 2 x 350 (i) ip = -------------- = ---------------------------------------------------60,000 60,000 = 18.63 kW. 2π N WR 2 x π x 350 x (65 x 9.81) x 0.5 (ii) bp = ------------- = --------------------------------------60,000 60,000 = 11.68 kW. .
(iii) Heat supplied in kW = mf CV = (4 / 3600) x 43,000 = 47.8 kW .
Heat lost to cooling water = mw (Cp)w [Two – Twi] = (500 / 3600) x 4.2 x [40 – 20] = 11.7 kW. .
.
Heat lost in exhaust gases = (ma + mf) (Cp)eg [Teg – Tatm] (32 + 1 ) x 4 = --------------- x 1.0 x [400 – 20] 3600 = 13.9 kW
198
Heat balance sheet: Heat Input
kW
%
Heat supplied by fuel
47.8
100
Total
47.80
100
Heat Expenditure
kW
%
Heat in bp
11.68
24.4
Heat lost to cooling Water
11.70
24.5
Heat lost to exhaust Gases
13.90
29.1
Unaccounted heat (by difference)
10.52
22.0
Total
47.80
100.0
199
200
