1. A refrigerating plant for an air-conditioning system removes
10,000 Btu/min from the air. The plant circulates 170 lb of
refrigerant/min and the internal power delivered by its
compressor is 60 horsepower. The refrigerant evaporation
temperature is 40F, and its condensation temperature is
100·F. Calculate :
From: Freon table Attached :Appendix
P1= P2 = 26.51 psia
P3 = P4 = 107.9 psia
t 2 = 14F

A. the capacity of the plant, tons;
B. the refrigerating effect, Btuflb;
C. the coefficient of performance of the actual plant; and
D . the coefficient of performance of the equivalent Carnot cycle.
B. m' =

C. Wnet
heat absorbed(Btu/min)
10,000
=
= 50 tons
200 Btu/min-ton
200
heat absorbed(Btu/min)
10, 000
B. RE =
=
= 58.8 Btu/lb
mass of Refrigerant (lb/min)
170
D. COP =
J x RE
58.8
COP =
=
= 3.93
Wnet
14.97
T
(40  460)
D. COPcarnot = L 
 8.33
TH -TL (100  460)  (40  460)
2. A r e f r i g e r a t i n g plant circulates 23 lb Freon-12 per
minute and is assumed to operate on a cycle similar to that
of the figure . The pressure in the evaporator coil is 50 psia,
the temperature of the Freon-12 entering the compressor is
50" F, the pressure in the condenser is 120 psia and the
temperature of the liquid refrigerant entering the
expansion valve is 86F. Calculate or determine:
A. the evaporation temperature, • F;
B. the condensation temperature, F;
C. the refrigerating effect, Btu/lb;
D. the capacity of the plant, tons;
E. the power required to compress the Freon-12, hp; and
F. the coefficient of performance.
.
heat absorbed 200xCapacity 50(200)


 183.6
RE
RE
54.48
= h3 - h2 = 91.13 - 80.04 = 11.09 Btu/lb
h -h
RE
54.48
= 2 1 =
= 4.91
Wnet
h3 - h 2
11.09
m'freon-12   Wnet 
E. Wnet/ton =
x 60
2545 x capacity
F. QR   m'freon-12  ( h3 - h4 ) = 183.6 (91.13 - 25.56) = 12,040 Btu/min
m'freon-12  2   where : 
G. PD =
Capacity
183.6  1.516 
PD =
2
 1.516 ft 3 /lb
= 5.57 ft 3 /min.ton
50
4. An air compression refrigeration system is to have an air
pressure of 100 psia in the brine tank and an allowable air
temperature increase of 60F. For standard vapor
compression cycle temperatures of 77F entering the
expansion cylinder and 14 F entering the compression
cylinder, calculate:
A. the coefficient of performance;
B. the mass of air circulated per ton of refrigeration;
C. the required piston displacement of the compressor
cylinder, neglecting volumetric efficiency.
P

P=C

S=C
S=
S=C


T
P=C 
P

183.6 x 11.09 x 60
 0.96
2545 x 50
=
E
Wnet
(Hp)(2545 Btu/Hp-min)
60 x2545

=
= 14.97Btu/min 
J
mass of Refrigerant x 60
170(60)
C.
A. RE = h2 - h1 = 80.04 - 25.56 = 54.48 Btu/lb
PRESSUR
A. Capacity =
t 4 = 77F
S2  S3  0.17317 Btu/lb ; h2 = 80.04 and h3 = 91.13 Btu/lb

P=C
C
P=

C


P=C
V



h

h4 = h1 = 27.72 Btu/lb
h2 = 84.24 Btu/lb
S2 =S3 =0.17187
h3 =91.31 Btu/lb
See Attached Appendix:
A. Evaporation temperature = 38.3F
B. Condensation temperature = 93.4 F
C. RE = h2 – h1 = 84.24 – 27.72 = 56.52 Btu/lb
m'(RE)
23(56.52)
=
= 6.5 tons
200
200
m'  h3 - h2 
23  56.52 
E. Wnet =
=
= 3.83 Hp
42.42
42.42
h -h
RE
84.24 - 27.72
F. COP =
= 2 1 =
= 7.99
Wnet
h3 - h2
91.31 - 84.24
D. Capacity ' =
3. A cooling plant using Freon-12 as the refrigerant is to have
a capacity of 50 tons when operating on the refrigerant
rating cycle. For this ideal plant cycle determine:
D.
E.
F.
G.
.
T4 = 77 + 460 = 537R
T2 = 14 + 460 = 474R
T2 - T1 = 60F
T1 = 474 - 60 = 414R
Since: Processes 3-4 and 1-2 are constant pressure:
Use:Attached Figure
A.
B.
C.
S
the refrigerating effect, Btu/lb;
the rate of Freon-12 circulation, lb/min;
the net work required per:pound of Freon-12 circulated,
Btu/lb;
the coefficient of performance;
the power required per ton of refrigeration, hp/ton;
the heat rejected by the condenser, Btu/min; and
the compressor
piston displacement, ft3/min·ton of
refrigeration.
P3 P4

and
P2 P1
 P4 
 
 P1 
k 1
k
=
T
T4
= 3
T1
T2
T 
 537 
T3 = T2  4  = 474 
 = 615°R
 414 
 T1 
Cp  T2 -T1 
60
A. COP =
=
= 3.33
Cp  T3 -T4  - Cp  T2 - T1   615-537  - 60
B. RE - Cp  T2 -T1  = 0.24(60) = 14.4 Btu/lb
200 Btu/min.ton
=13.9 lb/min.ton
14.4 Btu/lb
m'RT2
C. Piston Displacement = V2 =
P2
m'air =

13.9(53.3)(474)
= 24.4 ft 3 /min.ton
144 x 100
5. A simplified line diagram and TS plot for one section of a
cooling system for a large aircraft are shown below. When it
is used for cooling on the ground, the following Fahrenheit
temperatures are experienced at the numbered points on the
diagrams:
T
1
3
8. A refrigerating plant for an air-conditioning system is to have a
capacity of 10 tons and a coefficient of performance of 2.50
when operating With a refrigerating effect of 61.4 Btu/lb of
refrigerant. Calculate :
A. the refrigerant flow rate, lb/min;
B. the work done on the refrigerant by the compressor, Btu/lb
C. the compressor internal horsepower, hp; and
D. the rate of heat rejection from the system, Btu/min.
2
5
S
t (• F)
1
2
3
4
5
342
142
252
145
35
For a situation where the air flow rate through the system
is 65 lb/min, specific heat of the air is assumed constant
and the compressor and expander processes are assumed
isentropic, calculate:
(A) the heat transferred to the atmospheric air supply,
Btu/min;
(B) the power developed by the expander, hp; and
(C) the heat transferred from auxiliary power unit compressor
bleed, expressed in tons of refrigeration.

A. Q1-4 =m'Cp  t1 - t 2    t 3 - t 4  
= 65(0.24)  342 - 142    252 - 145  
= 4789.2 Btu/min
m'  h4 - h5 
65(0.24)(145 - 35)
B. Wnet =
=
= 40.45 hp
42.42
42.42
Q
4789.2
C. REFRIG = 1-4 =
= 23.95 tons
200
200
6 . An ideal Freon-12 refrigerating system has a capacity of 50
tons.
The condenser pressure is 180 psia, and the Freon12 temperature leaving the condenser is 120° F.
The
pressure leaving the expansion valve is 44 psia, and the
temperature of the Freon-12 leaving the succeeding coil is 40°
F. Circulating water enters the condenser at a temperature
of 100° F and leaves it at 1 1 5  F. Determine:
the mass of Freon-12 circulated, lb/hr;
the compressor power for isentropic compression, Btu/hr
the heat capacity of the system, Btu/hr
the mass of water circulated through the condenser and
heating system, lb/hr
E. the useful heat furnished per Btu of compressor work (heating
performance ratio)
A.
B.
C.
D.

See Freon-12 table s attached Appendix/Figure:
h1 = h4 = 36.16 Btu/lb
h2 = 83.03 Btu/lb
S2 = S3 0.17142 Btu/lb
h3 = 94.31 Btu/lb
12,000 x capacity
A. m =
h2 -h1
'
12,000(50)
= 12,800lb/hr
83.03-36.16
B. Compressor Power = m'Wnet '
=
= m'  h3 - h2  = 12,800(94.31-83.03)
= 144,400 Btu/hr
C. Heating Effect : QR = h3 - h4
= 94.31 - 36.16 = 58.15 Btu/lb
Heat Capacity = HC = m'(RE)
= 12,800(58.15) = 744,300 Btu/hr
HC
744,300
D. mcirculating water =
=
= 49,620 lb/hr
C  t out - tin 
1(115  100)
E. Heating Performance =
A. 50 percent;
B. 25 percent; and
C. 12.5 percent.
Answers: (A) 4.71 hp; (B) 1.57 hp; (C) 0.673 hp
4
point
7. Calculate the horsepower required per ton of refrigeration
produced by the reversal of a Carnot cycle having a thermal
efficiency of
HC
744, 300

 5.15
Compressor Power 144, 400
Answers: (A) 32.6lb/min; (B) 24.6 Btu/lb; (C) 18.9 hp; (D) 2800
Btu/min
9. A refrigeration system has a capacity of 25 tons and rejects heat
at the rate of 6560 Btu/min. Calculate:
(A) the rate of heat absorption by the refrigerant, Btu/min;
(B) the power required as input to the system, Btu/min; and
(C) the coefficient of performance for the system.
Answers: (A) 5,000 Btu/min; (B) 1560 Btu/min; (C) 3.2
10. A modified Rankine refrigerating cycle operates with an
evaporator pressure of 21.4 psia and a condenser pressure of
141 psia. Refrigerant is Freon-12 circulating through the
system at 30 lb/min. Liquid refrigerant at 141 psia and 100"
F enters the expansion valve, and surerheated vapor at 21.4
psia and 5" F enters the compressor.
Calculate:
(A) the refrigerating effect, Btu/lb;
(B) the plant capacity in tons of refrigeration;
(C) the power required to compress the refrigerant, hp; and
(D)
the plant coefficient of performance.
Answers: (A) 47.86 Btu/lb; (B) 7.18 tons; (C) 10.66 hp; (D) 3.17
11. In an ideal (reversed Joule cycle) air-refrigerating system
the temperature of the air entering the compression
cylinder is 50F, the temperature entering the after-cooler
is 160° F. and the temperature entering the brine tank is
0F. Calculate:
(A) the temperature of the air leaving the after-cooler
(B) the coefficient of performance;
(C) the mass of air which must be circulated per ton of
refrigeration, lb/min.
Answers: (A) 99F (B) 4.54 (C) 16.7 lb/min
CNS 04
MDSP/MESL