Advance Thermodynamics Project Report on Boiler Energy Analysis Submitted to: Dr. Hafiz Muhammad Ali
IMRAN SAJID SHAHID 12-MS-PT-TSE-12
Problem Description:
Perform energy analysis on boiler of educational power plant (TH 135M compact Stem Power Plant) installed in HITEC university, shown in figure below:-
Solution:
Typical thermodynamic process diagram of the training boiler is shown below. The area of interest is boxed in red. Superheated portion is not included at the time of purchase. So, superheated r egion is not included in our analysis.
For thermodynamics theoretical consideration, as shown below, is generally used by engineers as a standard of reference for comparing the performance of actual steam engines and steam turbines. The system contains of following apparatus: 1) A steam generating unit 2) A prime mover 3) A condenser and cooling tower 4) A boiler feed water pump
T-S Diagram
p-v Diagram
Above figure shows the following ideal thermodynamics processes:
Process line C-D: the phase of pumping feed water into the boiler
Process line D-E: the process of heating of feed water in the boiler representing the work input into the system
Process line E-A: represents evaporation
Process line A-B: assumed to be isentropic expansion, a constant entropy process, in the steam turbine
Process line B-C: condensation of vapor in the condenser at constant pressure.
Following is the data measured under the steady condition, for simplicity and time limitation only four cycle data is collected.
Tested by: …I.S.Shahid… Date:3/6/12 At Steady state Condition
Fuel pump
FUEL Running time
Vol.
status
(m's'')
(mL)
Feed water pump status
Start
15:26
874
Stop
15:46
Start
FEED WATER Running time Vol.
Water tank temp
Steam pressure
BOILER Steam outlet temp
o
o
Stack temp o
(m's'')
(Ltr)
T1( C)
P2(kg/cm2)
T 4( C)
T9( C)
Start
0
37.1
37
5
159
159
1189
Stop
0:11
39.2
37
5
159
159
17:31
1189
Start
1:49
39.2
37
5
158
159
Stop
17:57
1469
Stop
1:58
41.3
37
5
159
159
Start
19:33
1469
Start
3:33
41.3
37
5
159
159
Stop
19:59
1728
Stop
3:43
45.5
37
5
159
159
Start
21:29
1728
Start
5:19
45.5
37
5
159
159
Stop
22:01
1969
Stop
5:29
47.6
37
5
159
159
Start
23:18
1969
Start
7:05
47.6
37
5
159
159
Differences:
7:52
1095
7:05
10.5
1509.33333
0.134799383
42.7
37
5
158.8888889
159
Average:
In calculation only average of four cycle data is used. Specific gravity of fuel is 0.842, given in instruction manual. Avg. feed water flow rate from 4 cycles, mc=Volume/t= 10.5Ltr / (4X60+5) = 0.024706 kg/ sec Avg. fuel flow rate from 4 cycles,
mf =Volume/t= 1.095Ltr X 0.842 / (7X60+52) = 0.001953 kg/ sec
Energy Analysis:-
Now, performed energy analysis will be performed on each process shown on Rankine cycle. Process C-D:
For general steady flow process the energy balance will be; EI - Eo =E Q - W = U + KE + PE Since Q=0 and W=0. So, the equation reduces to: 0=U=U2-U1 0 = h2 - p2v2 - h1 - p1v1 0 = (h2 - h1) - (p2 - p1)v
In our system change subscript 2 with1 and 1 with 0, the resulting enthalpy of feed water after pumping is: h1 = h0 - (p1 - p0)v 0
From steam table at T=37 C, saturated feed water, after performing linear interpolation: h0=154.996 kJ/kg p0=101.325 kPa 3
v=0.0010068 m /kg 2
And, p1 is the average pressure of min and max boiler set pressure, which are 7 and 8 kg/cm gauge. Substituting the value in above equation yield: h =155.73 kJ/ kg 1
Process D-A:
From Data Sheet: Boiler pressure,
p2 = 5kg/cm2 guage = 5.9bars abs. = 0.59 MPa abs. 0
Steam temperature, T4 = 158.89 C 0
From steam table at p2 = 0.59MPa and T4 = 158.89 C, superheated vapor, after performing linear interpolation: h2=2755.44 kJ/kg h2=h3 (because there is no superheater) For general steady flow process the energy balance will be;
Neglecting the KE and PE terms the Equation reduce to; Q’in - Q’out = m’ (h2-h1) And, Q’in = m’ (h2-h1) + Q’out
Now, change the subscript to the current process; m’f qf = m’c (h2-h1) + mc hlb qf = the heating value/ energy value/calorific value of a substance is the amount of heat released during the combustion of a substance hlb= Enthalpy loss due to radiation, convection and flue gas at the b oiler Above equation can be solved for h lb. After substituting the values: h = 817.623 kJ/ kg lb
Efficiency of boiler is calculated by b = mc (h2-h1) X 100 / m f qf = 76.08%
Uncert aint y Analysis is also performed on feed water flow rate only for this matlab code is provided but I
am not able to understand the output:-
**************************************************************************** Cl c ; c l ear al l syms vol _f eedw t _ f _ f eedw t _ i _ f eedw m_c = vol _f eedw / ( t _f _f eedw - t _i _f eedw ) %Def i ne val ues f or t he uncer t ai nt i es i n these val ues u_vol _f eedw = 0. 01; u_t _f _f eedw = . 1; u_t _i _f eedw = . 1; % uni t i s +- [ m^3] u_m_c = sqr t ( ( di f f ( m_c, vol _f eedw) *u_vol _f eedw) ^2+. . . ( di f f ( m_c, t _f _f eedw) *u_t _f _f eedw) ^2+. . . ( di f f ( m_c, t _i _f eedw) *u_t _i _f eedw) ^2) pr et t y( u_m_c) vol _ f eedw = 10. 5; t _f _ f eedw = 425; t _i _f eedw = 0; m_c = subs( m_ c) u_m_ c = subs( u_m_c) % ant i ci pat ed possi p_er r or_ vol _f eedw p_err or _t _f _f eedw p_err or _t _i _f eedw
bl e er r or due t o = ( u_vol _f eedw = ( u_t _f _f eedw = ( u_t _i _f eedw
uncer t ani t y i n measur i ng i nst r ument / vol _f eedw) * 100 / t _f _f eedw) * 100 / t _i _f eedw) * 100
% per cent age er r or as l ar ge as p_er r or _m_c = ( u_vol _f eedw/ vol _f eedw) * 100 **************************************************************************** MATALB OUTPUT:
m_c = vol_feedw/(t_f_feedw-t_i_feedw)
u_m_c = 1/100*(1/(t_f_feedw-t_i_feedw)^2+200*vol_feedw^2/(t_f_feedw-t_i_feedw)^4)^(1/2)
/ 2 \1/2 | 1 vol_feedw | 1/100 |------------------------ + 200 ------------------------| | 2 4| \(t_f_feedw - t_i_feedw) (t_f_feedw - t_i_feedw) / m_c = 0.0247
u_m_c = 2.4924e-005
p_error_vol_feedw = 0.0952
p_error_t_f_feedw = 0.0235 Warning: Divide by zero. p_error_t_i_feedw = Inf
p_error_m_c = 0.0952