HOMEWORK SET 6
8.17
Find the missing properties and give the phase of the ammonia, NH
3
.
a.
T
=
65
°
C,
P
=
600 kPa
s
=
?
v
=
?
b.
T
=
20
°
C,
P
=
100 kPa
v
=
?
s
=
?
x
=
?
c.
T
=
50
°
C,
v
=
0.1185 m
3
/kg
s
=
?
x
=
?
P
=
?
a)
B.2.2 average between 60
°
C and 70
°
C
v = (0.25981 + 0.26999)/2 = 0.26435 m
3
/kg
s = (5.6383 + 5.7094)/2 = 5.6739 kJ/kgK
b)
B.2.1:
P < P
sat
= 857.5 kPa
=>
B.2.2
superheated vapor so x is undefined
v = 1.4153 m
3
/kg,
s = 6.2826 kJ/kgK
c)
B.2.1:
v > v
g
= 0.06337 m
3
/kg
=>
B.2.2
superheated vapor so x is undefined
very close to 1200 kPa,
s = 5.1497 kJ/kgK
8.28
Consider a Carnotcycle heat pump with R22 as the working fluid. Heat is
rejected from the R22 at 40
°
C, during which process the R22 changes from
saturated vapor to saturated liquid. The heat is transferred to the R22 at 0
°
C.
a.
Show the cycle on a
T–s
diagram.
b.
Find the quality of the R22 at the beginning and end of the isothermal heat
addition process at 0
°
C.
c.
Determine the coefficient of performance for the cycle.
Solution:
a)
1
2
3
4
40
0
T
s
b)
From Table B.4.1, state 3 is
saturated liquid
s
4
= s
3
= 0.3417 kJ/kg K
= 0.1751 + x
4
(0.7518)
=>
x
4
=
0.2216
State 2 is saturated vapor so from Table B.4.1
s
1
= s
2
= 0.8746 kJ/kg K = 0.1751 + x
1
(0.7518)
=>
x
1
=
0.9304
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β′
=
q
H
w
IN
=
T
H
T
H
– T
L
=
313.2
40
=
7.83
8.45
A piston cylinder contains 0.25 kg of R134a at –20
o
C, 100 kPa. It is compressed
in a reversible adiabatic process to 400 kPa. How much work is needed?
Solution:
C.V.
R134a,
Control mass, adiabatic
1
q
2
= 0
Energy Eq.5.11:
u
2
−
u
1
=
1
q
2
−
1
w
2
= 
1
w
2
Entropy Eq.8.3:
s
2
−
s
1
=
∫
dq/T
Process:
Adiabatic and reversible
=>
s
2
= s
1
State 1:
(T, P)
B.5.2
u
1
= 367.36 kJ/kg,
s
1
= 1.7665 kJ/kg K
State 2:
(P, s)
B.5.2
P
2
= 400 kPa,
s
2
= s
1
= 1.7665 kJ/kg K
interpolate to get
T
2
= 22.6
o
C,
u
2
= 394.36 kJ/kg
1
w
2
= u
1
 u
2
= 367.36 – 394.36 =
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 Spring '07
 Borgnakke
 Thermodynamics, kPa, Adiabatic process, reversible adiabatic process

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