4.11
-20C is -4F, and 50C is 122F
Use subscript I to refer to interstage state. There is no interstage cooler or
economizer between the compressors. Stream numbers from figure 4.9 in text.
(a) condenser (P
sat
at 122F) = 1.7 MPa =P
3
Evaporator (P
sat
at -4F) = 0.24 MPa =P
2
Interstage (P
I
/P
2
) = (P
3
/P
I
) so P
I
= sqrt(1.7*0.24) =0.64 MPa
(b) cooling in evaporator, Q
C
=H
2
-H
1
= 877-665 = 212 kJ/kg
m = (10,000,000 kJ/hr)(kg/212 kJ) = 47,200 kg/hr
flow must be the same through both compressors.
(c) Take efficiency as per stage,
W’
compressorI
= H’
I
-H
2
=945-877 = 68 kJ/kg, T’
I
= 82F
W
compressorI
= 68/0.8 = 85 kJ/kg, H
I
= 877+85 =962 kJ/kg, T
I
= 105F
W
I
=(85 kJ/kg)(47,200 kg/hr) = (4.012 kJ/hr)(hr/3600s) = 1,114 kW = 1500 hp
W’
compressorII
= 1020-962 =58 kJ/kg
W
compressorI
=58/0.8=72.5 kJ/kg, H
3
=962+72.5=1035 kJ/kg, T
3
=192F
W
II
= (72.5)(47,200)=(3.422e6 kJ/hr)(hr/3600s)=950 kW
(d) H
4
=665 kJ/kg
Q
H
=m(H
4
-H
3
)47,200(665-1035)=-1.75e7 kJ/hr
COP=1e7/((4.0+3.42)e6)=1.35
4.12
Stream numbers from figure 4.11 in text. Solved with propane chart ini the
appendix.
-20C is -4F, -50C is 122F
(a) condenser (P
sat
at 122F) = 1.7 MPa=P
5
Evaporator (P
sat
at -4F) = 0.24 MPa = P
2
Interstage (P
3
/P
2
) = (P
5
/P
3
), so P
3
=sqrt(1.7(0.24))=0.64 MPa
Shaded values from problem statement.
State
T(F)
P(MPa)
H(kJ/kg)
S(BTU/lb-R)
m(kg/hr)
1
-4
0.24
565
32,051
2
-4
0.24
877
1.36
3’
0.64
924
1.36
3
0.64
936
4
0.64
929
1.37
45,787
5’
1.7
977
1.37
5
1.7
989
6
122
1.7
665
45,787
7
0.64
912
13,736
8
0.64
565
32,051
(b) Fill in states on table for H
6
, H
2
, H
8
(H
1
), H
7
Cooling duty of evaporator, Q