Name:----------------------------
Spring 2008 ME21 ME210/ 0/Adv Adv anced Thermodyn amics Final Exam, Tuesd ay, May 20, 2008 2008 (19:45(19:45-22 22:00, :00, 2 hour s and 15 minu tes, open bo ok & notes)
Problem # 1 (10 poi nts )
An object of mass 0.8 lbm traveling traveling at velocity velocity 200 ft/sec ft/sec enters enters a viscous viscous fluid and is essentially brought to rest before it strikes bottom. What is the increase in internal energy of the system, taking the object and the liquid as the system? Solution:
Neglecting the changes in potential energy, conservation of energy requires that the sum of the kinetic energy and internal energy remains constant, i.e., 1 m o v i2 2
1 2
+ Ui = m o v 2f + Uf 1 2
1 2
1 2
ΔU = Ui − Uf = m o v i2 − m o v f2 = m o ( v i2 − v f2 ) =
1 (0.8 )(200 2 − 0 2 ) 2gc
= 497 lbf-ft
1
Problem # 2 (20 Poin ts)
A mixture of ideal gases consists of 2 kmoles of CH 4, 1 kmole of N 2, and 1 kmole of CO2, all at corresponding temperatures 293 K and pressure 2x10 4 Pa. Heat is added until the temperature increases to 673 K while the pressure remains constant. Calculate: (a) the heat transfer (b) the work done, and (c) the change in entropy.
Solution:
(a)
Assuming constant cp Q = mmix cp (Tf -Ti)
mmix =
3
∑ N M = 2x16 + 1x28 + 1x44 = 104 kg i
i
i=1
3
cp cv
3
= ∑ mi cpi = ∑ i=1
i =1
3
3
= ∑ mi c vi = ∑ i =1
i =1
NiMi m mix Ni Mi m mix
cpi
= 0.3077 x2.254 + 0.2692 x1042 + 0.423 x0 .842 = 133 . .
c vi
= 0.3077 x1735 + 0. 2692x0.745 + 0.423x0.653 = 1011 . .
Q = 104 x 1.33 (673 - 293) = 5.257x10 4 kJ (b)
W = Q - Δu = Q - m mix cv (Tf - Ti) = 1.263x10 4 kJ (c)
Δs = mmix [ cp ln (Tf /Ti) - (cp - cv )ln (pf /pi)] = 115.05 kJ/K
2
kJ kg − K
kJ kg − K
Problem # 3 (15 Poin ts)
An important reaction in the production of sulfuric acid is the oxidation of SO 2(g) to SO3(g): 2SO2(g)+O2(g) --------- 2SO3(g) At 298 K, ΔG0= -141.6 kJ; ΔH0= -198.4 kJ; and (a) (b)
ΔS0= -187.9 J/K
Use the data to decide if this reaction is spontaneous at 25 0C, and predict how ΔG0 will change with increasing temperature. Assuming ΔH0 and ΔS0 are constant with increasing T, is the reaction spontaneous at 900 0C?
Solution:
a) ΔG0<0, so the reaction is spontaneous at 298 K: a mixture of SO 2(g), O2(g), and SO3(g) in their standard state (1 atm) will spontaneously yield more SO 3(g). With ΔS0<0, the term –T ΔS0>0 and becomes more positive at higher T. Therefore, ΔG0 will be less negative, and the reaction less spontaneous, with increasing T. b) Calculating ΔG0 at 9000C, ΔG0=ΔH0-TΔS0= -198.4 kJ – [(273+900)K](-187.9 J/K)(1 kJ/1000J) = 22 kJ ΔG0>0, so the reaction is nonspontaneous at the higher T.
3
Problem # 4 (20 Poin ts)
Assuming the ideal-gas turbine and regenerator shown below, draw the p-v and T-s diagrams and find Q in and the ratio of compressor to turbine work (i.e., back work ratio).
2
3
4 1660oR
1
5
6
Solution:
The pressure ratio is:
r p =
P2 P1
η = 1−
= T1 T4
75 = 5102 . 14.7 k −1
r p
k
540 5.102 = 1− 1660
1.4 −1 1.4
= 0.482
o
Wout
800 x0.70626 Btu . = 11722 0.482 sec η Assuming isentropic compression and expansion: o
Q in
=
T2 T5
=
⎛ P ⎞ = T1 ⎜ 2 ⎟ ⎝ P1 ⎠
= T4 rp
w com w turb
k
k
75 ⎞ = 540⎜⎛ ⎝ 14.7 ⎠⎟
1.4 −1 1.4
= 860.214 o R
1−1.4
= 1660 x5.102
1.4
= 1042 o R
kj Btu = 76.85 kg lbm kj Btu = c p ( T4 − T5 ) = 1(1660 − 1042) = 618 = 148.304 kg lbm
= c p ( T2 − T1 ) = 1(860.214 − 540 ) = 320.2
w comp w turb
1−k
k −1
=
76.85 148.304
= 0.518
4
Problem # 5 (15 Poin ts)
A Stirling cycle operates on air with a compression ratio of 10 between low temperature 660 oR and high temperature 1460 oR. Given that the low pressure is 30 psia, draw the p-v and T-s diagrams and calculate the work output and the heat input. [Hint: compressor & turbine have a common shaft]
Solution:
w out
= w 3−4 + w 1−2 = Rair T3 ln
v4
+ Rair T1 ln
v2
= Rair T3 ln(r) + Rair T1 ln
v3 v1 wout = 53.34 (1460)ln(10)+53.34(660)ln(0.1) = 9.826x10 4 ft-lbf/lb The ideal efficiency is:
η = 1−
qin
=
TL TH
w out
η
= 1−
=
T1 T3
= 1−
9.826 x10 4 0.548 x778
660 1460
= 0.548
= 230.47 Btu / lbm
5
1 r
Problem # 6 (20 Poin ts)
Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 160 kPa, (a) (b) (c) (d)
Exit temperature Exit specific volume The sign of the Joule-Thomson coefficient for this process and if it is consistant with your answer in (a) What should we do if we want the opposite of answer in (c) without changing the upstream and exit pressures?
Solution:
(a) The inlet temperature and enthalpy of R-134a, from the refrigerant tables (Tables A-11 through 13, pg. 902-905) is: For P1=0.7 MPa === T1=Tsat=26.69 oC, h1=hf =88.82 kJ/kg Assuming a throttling process, i.e., •
•
Q ≅ W = ΔKE ≅ ΔKE ≅ 0
At the final state, P2=160 kPa, the enthalpy is the same, i.e. h=88.82 kJ/kg. Again from the tables @ p=160 kPa: Tsat= -15.60 oC, hf =31.21 kJ/kg, h g=241.11 kJ/kg Note that h f
ΔT=T2-T1= -15.6 - 26.69= -42.3 oC (b) The quality at this state is determined from x 2
=
h2
v 2 = v f
−h
f
h fg
=
88.82 − 31.21 209.9
= 0.2745
+ x v fg = 0.0007437 + 0.2745 x (0.12348 − 0.0007437 ) = 0.0344 m 3 / kg
6
(c) μ
=
− 15.6 − 26.69 = 0.08 K/kPa 160 − 700
J.T. coefficient is positive, reflecting cooling. (d)
Search for a different fluid.
7
