Termodinamica

10-1

CHAPTER 10
The correspondence between the new problem set and the previous 4th edition
chapter 8 problem set.
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

Old
new
new
1
2
3
5
6
7
9
15
16
10
11
49
47
13
14
38
52
8

New
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

Old
new
new
18
new
19
2021
new
new
23
24
new
new
25
26
27
32
28
33
34

New
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60

Old
35
41
42
46
54
56
39
55
53
new
43
51
17
29
30
31
36
40
44
57

The problems that are labeled advanced starts at number 53.
The English unit problems are:
New
61
62
63
64
65
66
67
68
69
70

Old
new
58
60
6361
80
62
new
65
66

New
71
72
73
74
75
76
77
78
79
80

Old
67
new
68
new
69
70
72
new
79
84

New
81
82
83
84
85
86
87

Old
75
83
77
64
71
73
78

10-2
10.1

Calculate the reversible work and irreversibility for the process described in
Problem 5.18, assuming that the heat transfer is with the surroundings at 20°C.
C.V.: A + B. This is a controlmass.
Continuity equation:
Energy:

m2 - (mA1 + mB1) = 0 ;

m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2

System: if VB ≥ 0 piston floats ⇒

PB = PB1 = const.

if VB = 0 then P2 < PB1 and v = VA/mtot see P-V diagram
P
State A1: Table B.1.1, x = 1
a
vA1 = 1.694 m3/kg, uA1 = 2506.1 kJ/kg
PB1

2

mA1 = VA/vA1 = 0.5903 kg

V2

State B1: Table B.1.2 sup. vapor
vB1 = 1.0315 m3/kg, uB1 =2965.5 kJ/kg
mB1 = VB1/vB1 = 0.9695 kg =>
At (T2 , PB1)

m2 = mTOT = 1.56 kg

v2 = 0.7163 > v a = VA/mtot = 0.641 so VB2 > 0

so now state 2: P2 = PB1 = 300 kPa, T2 = 200 °C
=> u 2 = 2650.7 kJ/kg and V2 = m2 v2 = 1.56×0.7163 = 1.117 m3
(we could also have checked Ta at: 300 kPa, 0.641 m3/kg => T = 155 °C)
ac
1W2 =
1Q 2

⌠PBdVB = PB1(V2 - V1) = PB1(V2 - V1) = -264.82 kJ
⌡
tot
B= m2u2 - mA1uA1 - mB1uB1 + 1W2 = -484.7 kJ

From the results above we have :
sA1 = 7.3593, sB1 = 8.0329, s2 = 7.3115 kJ/kg K
rev
1W2 =

To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH)
ac

= To(m2s2 - mA1sA1 - mB1sB1) + 1W2 - 1Q2To/TH
= 293.15 (1.5598 × 7.3115 - 0.5903 × 7.3593 - 0.9695 × 8.0329)
+ (-264.82) - (-484.7) × 293.15 / 293.15
= -213.3 - 264.82 + 484.7 = 6.6 kJ
1I 2

rev

ac= 1W2 - 1W2 = 6.6 - (-264.82) = 271.4 kJ

10-3
10.2

Calculate the reversible work and irreversibility for the process described in
Problem 5.65, assuming that the heat transfer is with the surroundings at 20°C.

P
2

1

Linear spring gives
1W2

v

= ⌠PdV = 2(P1 + P2)(V2 - V1)
⌡
1

1Q 2

= m(u2 - u1) + 1W2
Equation of state: PV = mRT

State 1: V1 = mRT1/P1 = 2 x0.18892 x 673.15 /500 = 0.5087 m3
State 2: V2 = mRT2/P2 = 2 x 0.18892 x 313.15 /300 = 0.3944 m3
1W2

1

= 2(500 + 300)(0.3944 - 0.5087) = -45.72 kJ

From Figure 5.10: Cp(Tavg) = 45/44 = 1.023 ⇒ Cv = 0.83 = C p - R
For comparison the value from Table A.5 at 300 K is Cv = 0.653 kJ/kg K
1Q 2
rev
1W2 =

= mCv(T2 - T1) + 1W2 = 2 x 0.83(40 - 400) - 45.72 = -643.3 kJ
To(S2 - S1) - (U2 - U1) +1Q2(1 - To/TH)
ac

= Tom(s2 - s1)+ 1W2 - 1Q2To/To
ac

= Tom[CP ln(T2 / T1) − R ln(P2 / P1)] + 1W2 - 1Q2
= 293.15 x 2 [ 1.023 ln(313/673) - 0.1889 ln(300/500)] - 45.72 + 643.3
= -402.6 - 45.72 + 643.3 = 195.0 kJ
1I 2

10.3

rev

ac

= 1W2 - 1W2 = 195.0 - (-45.72) = 240.7 kJ

The compressor in a refrigerator takes refrigerant R-134a in at 100 kPa, − 20°C
and compresses it to 1MPa, 40°C. With the room at 20°C find the minimum
compressor work.
Solution:
C.V. Compressor out to ambient. Minimum work in is the reversible work.
SSSF, 1 inlet and 2 exit energy Eq.:

wc = h1 - h2 + qrev

Entropy Eq.: s2 = s1 + ⌠dq/T + sgen = s1 + qrev/T0 + 0 => qrev = T0(s2- s1)
⌡
wc min = h1 - h2 + T0(s2- s1) = 387.22 - 420.25 + 293.15 × (1.7148 - 1.7665)
= -48.19 kJ/kg

10-4...