文档介绍：10-1
CHAPTER 10
The correspondence between the new problem set and the previous 4th edition
chapter 8 problem set.
New Old New Old New Old
1 new 21 new 41 35
2 new 22 new 42 41
3 1 23 18 43 42
4 2 24 new 44 46
5 3 25 19 45 54
6 5 26 20 46 56
7 6 27 21 47 39
8 7 28 new 48 55
9 9 29 new 49 53
10 15 30 23 50 new
11 16 31 24 51 43
12 10 32 new 52 51
13 11 33 new 53 17
14 49 34 25 54 29
15 47 35 26 55 30
16 13 36 27 56 31
17 14 37 32 57 36
18 38 38 28 58 40
19 52 39 33 59 44
20 8 40 34 60 57
The problems that are labeled advanced starts at number 53.
The English unit problems are:
New Old New Old New Old
61 new 71 67 81 75
62 58 72 new 82 83
63 60 73 68 83 77
64 63 74 new 84 64
65 61 75 69 85 71
66 80 76 70 86 73
67 62 77 72 87 78
68 new 78 new
69 65 79 79
70 66 80 84
10-2
Calculate the reversible work and irreversibility for the process described in
Problem , assuming that the heat transfer is with the surroundings at 20°C.
.: A + B. This is a control mass.
Continuity equation: m2 - (mA1 + mB1) = 0 ;
Energy: 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
State A1: Table , x = 1 P
a 2
v = m3/kg, u = kJ/kg
A1 A1 PB1
mA1 = VA/vA1 = kg
State B1: Table sup. vapor V2
3
vB1 = m /kg, uB1 = kJ/kg
mB1 = VB1/vB1 = kg => m2 = mTOT = kg
At (T2 , PB1) v2 = > va = VA/mtot = so VB2 > 0
°
so now state 2: P2 = PB1 = 300 kPa, T2 = 200 C
× 3
=> u2 = kJ/kg and V2 = m2 v2 = = m
(we could also have checked Ta at: 300 kPa, m3/kg => T = 155 °C)
ac
W = ⌡⌠P dV = P (V - V ) = P (V - V ) = - kJ
1 2 B B B1 2 1 B B1 2 1 tot
1Q2 = m2u2 - mA1uA1 - mB1uB1 + 1W2 = - kJ
From the results above we have :
sA1 = , sB1 = , s2 = kJ/kg K
rev
1W2