Transcript Reflection of Buddhism in Contemporary Cinema

ENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Increase-in-Entropy Principle
Closed systems
ΔStotal = ΔSsystem + ΔSsurroundings ≥ 0
Qk
 (S2  S1)  
0
Tk
Qk
 m(s2  s1)  
0
Tk
Example 1
Water in a tank, initially a saturated liquid at
100 ºC, undergoes a process to the corresponding
saturated vapor state. There is no heat transfer
with the surroundings during the process. If the
change of states is brought about by the action of
a paddle wheel, determine the work required per
mass and the total entropy change for the process.
Example 1 (continued)
ΔU = Q – W
W
 (ug  uf )
m
Table A-4, T = 100 ºC,
uf = 418.94 kJ/kg, ug = 2506.5 kJ/kg
sf = 1.3069 kJ/kg K, sg = 7.3549 kJ/kg K
= - (2506.5 – 418.94)
= - 2087.56 kJ/kg
Qk
Stotal  (S2  S1 )  
Tk
Δs = sg – sf = 7.3549 – 1.3069 = 6.048 kJ/kg K
Increase-in-Entropy Principle
Control volumes
(a) Steady-Flow Processes
Stotal  SCV  Ssurroundings
Qsurr
 (  mese   misi ) 
0
Tsurr
Qsurr
 m(se  si ) 
0
Tsurr
Example 2
Feedwater Heater:
Inlet 1 T1 = 200 ºC, p1 = 700 kPa,
Inlet 2 T2 = 40 ºC, p2 = 700 kPa,
Exit
sat. liquid, p3 = 700 kPa.
Find S  ?
Inlet 2
Inlet 1
Exit
Example 2 (continued)
Stotal
Qsurr
 (  mese   misi ) 
0
Tsurr
m2
 4.06
m1
Inlet 1: Superheated vapor
Table A-6,
s1 = 6.8912 kJ/kg K
Inlet 2: compressed liquid
Table A-4,
s2 = 0.6387 kJ/kg K
Example 2 (continued)
Exit:
saturated liquid
Table A-5,
s3 = 2.0200 kJ/kg K
Stotal  (1 + 4.06)(2.0200)
– [(4.06)(0.6387) + (1)(6.8912)]
= 0.7369 kJ/kg K
Increase-in-Entropy Principle
Control volumes
(b) Uniform-Flow Processes
ΔStotal = ΔSCV + ΔSsurroundings ≥ 0
 ( mese   misi )  (m2s2  m1s1)CV
Qsurr

0
Tsurr
Example 3
A 0.4-m3 rigid tank is filled with saturated liquid water
200 ºC. A valve at the bottom of the tank is opened,
and liquid is withdrawn from the tank. Heat is
transferred to the water from a source at 250 ºC so that
the temperature in the tank remains constant.
Determine
(a) the amount of heat transfer by the time one-half of
the water (in mass) has been withdrawn.
(b) the total entropy change for this process.
Example 3 (continued)
Discharging Process
Uniform-Flow Process
Sat. liquid
T1 = 200 ºC
Q
Example 3 (continued)
No inlet, one outlet
me = (m1 – m2)CV
Initial state: Sat. liquid at T1 = 200 ºC
Table A-4, v1 = 0.001157 m3/kg, h1 = 850.65 kJ/kg
s1 = 2.3309 kJ/kg K
Exit: Sat. liquid at T1 = 200 ºC
Table A-4, h1 = 850.65 kJ/kg, s1 = 2.3309 kJ/kg K
V
0.4
m1 

 345.72 kg
v1 0.001157
Example 3 (continued)
Final state: Sat. mixture at T2 = 200 ºC
m2 = ½ m1, v2 = 2v1
1
m2  m1  0.5(345.72)  172.86 kg
2
v2 = 2v1 = 2(0.001157) = 0.002314 m3/kg
v 2  v f 0.002314  0.001157
x2 

 0.00917
vg  vf
0.12736  0.001157
u2 = uf + x2ufg = 850.65 + (0.00917)(1744.7)
= 866.65 kJ/kg
Example 3 (continued)
s2 = sf + x2sfg = 2.3309 + (0.00917)(4.1014)
QCV
= 2.3685 kJ/kg K




Ve2
Vi2
 WCV   me  he 
 gze    mi  hi 
 gzi 
2
2




+ (m2u2 – m1u1)CV
QCV = mehe + (m2u2 – m1u1)CV
= (172.86)(852.45) + (172.86)(866.65)
- (345.72)(850.65) = 3077 kJ
Example 3 (continued)
Stotal  ( mese   misi )  (m2s2  m1s1)CV
Qsurr

0
Tsurr
ΔStotal = (172.86)(2.3309) + (172.86)(2.3685)
3077
- (345.72)(2.3309) 
523
= 0.616 kJ/K
Example 4
A quantity of air undergoes a thermodynamic cycle
consisting of three processes in series.
Process 1-2: constant-volume heating from p1 = 0.1
MPa, T1 = 15 ºC, and V1 = 0.02 m3, to
p2 = 0.42 MPa.
Process 2-3: constant-pressue cooling.
Process 3-1: isothermal heating to the initial state.
Employ the ideal gas model with cp = 1 kJ/kg·K, calculate
the change of entropy for each process.
Example 4 (continued)
p
3
2
1
v
Example 4 (continued)
State 1: p1 = 0.1 MPa, T1 = 15 ºC, and V1 = 0.02 m3
State 2: p2 = 0.42 MPa, V2 = V1 = 0.02 m3
State 3: p3 = p2 = 0.42 MPa, T3 = T1 = 15 ºC
 T2 
 v2 
s2  s1  c v ln    Rln  
 T1 
 v1 
cv = cp – R = 1 – 0.287 = 0.713 kJ/kg·K
p1V1 p2 V2

T1
T2
p2
0.42
T2  T1 
288  1209.6 K
p1
0.1
Example 4 (continued)
 T2 
s2  s1  c v ln  
 T1 
 1209.6 
= 1.023 kJ/kg·K
 0.713 ln 

 288 
 T3 
 p3 
s3  s2  cp ln    Rln  
 T2 
 p2 
 288 
 1 ln 
= - 1.435 kJ/kg·K

 1209.6 
Example 4 (continued)
 T1 
 p1 
s1  s3  cp ln    Rln  
 T3 
 p3 
 0.1  = 0.412 kJ/kg·K
 0.287 ln 

 0.42 
Δstotal = Δs12 + Δs23 + Δs31
= 1.023 – 1.435 + 0.412 = 0