1.

A refrigerating plant for an air-conditioning system removes From: Freon table Attached :Appendix
10,000 Btu/min from the air. The plant circulates 170 lb of
P1= P2 = 26.51 psia P3 = P4 = 107.9 psia
refrigerant/min and the internal power delivered by its
compressor is 60 horsepower. The refrigerant evaporation t 2 = 14F t 4 = 77F
temperature is 40F, and its condensation temperature is S2  S3  0.17317 Btu/lb ; h2 = 80.04 and h3 = 91.13 Btu/lb
100·F. Calculate :

A. the capacity of the plant, tons;
B. the refrigerating effect, Btuflb; A. RE = h2 - h1 = 80.04 - 25.56 = 54.48 Btu/lb
C. the coefficient of performance of the actual plant; and
D . the coefficient of performance of the equivalent Carnot cycle. heat absorbed 200xCapacity 50(200)
B. m' =    183.6
RE RE 54.48
 C. Wnet = h3 - h2 = 91.13 - 80.04 = 11.09 Btu/lb
RE h -h 54.48
heat absorbed(Btu/min) 10,000 D. COP = = 2 1 = = 4.91
A. Capacity = = = 50 tons Wnet h3 - h 2 11.09
200 Btu/min-ton 200
heat absorbed(Btu/min) 10, 000 E. Wnet/ton =
m'freon-12   Wnet  x 60
=
183.6 x 11.09 x 60
 0.96
B. RE = = = 58.8 Btu/lb 2545 x capacity 2545 x 50
mass of Refrigerant (lb/min) 170
Wnet (Hp)(2545 Btu/Hp-min) 60 x2545 F. QR   m'freon-12  ( h3 - h4 ) = 183.6 (91.13 - 25.56) = 12,040 Btu/min
C.  = = 14.97Btu/min 
J mass of Refrigerant x 60 170(60)
G. PD =
m'freon-12  2   where :   1.516 ft 3 /lb
2
J x RE 58.8 Capacity
COP = = = 3.93
Wnet 14.97
PD =
183.6  1.516  = 5.57 ft 3 /min.ton
T (40  460) 50
D. COPcarnot = L   8.33
TH -TL (100  460)  (40  460)

4. An air compression refrigeration system is to have an air

2. A r e f r i g e r a t i n g plant circulates 23 lb Freon-12 per pressure of 100 psia in the brine tank and an allowable air
minute and is assumed to operate on a cycle similar to that temperature increase of 60F. For standard vapor
of the figure . The pressure in the evaporator coil is 50 psia, compression cycle temperatures of 77F entering the
the temperature of the Freon-12 entering the compressor is expansion cylinder and 14 F entering the compression
50" F, the pressure in the condenser is 120 psia and the cylinder, calculate:
temperature of the liquid refrigerant entering the A. the coefficient of performance;
expansion valve is 86F. Calculate or determine: B. the mass of air circulated per ton of refrigeration;
C. the required piston displacement of the compressor
A. the evaporation temperature, • F; cylinder, neglecting volumetric efficiency.
PRESSUR

B. the condensation temperature, F; P

C. the refrigerating effect, Btu/lb; 
D. the capacity of the plant, tons; T
 P=C 
E

E. the power required to compress the Freon-12, hp; and P=C

F. the coefficient of performance.
. P S=C  P=

S=C S= C
 
C
 P=C   P=C

V
S
 


h 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 k 1
P3 P4  P4  k T4 T
h4 = h1 = 27.72 Btu/lb h2 = 84.24 Btu/lb  and   = = 3
P2 P1  P1  T1 T2
S2 =S3 =0.17187 h3 =91.31 Btu/lb
T   537 
T3 = T2  4  = 474   = 615°R
See Attached Appendix:  T1   414 
Cp  T2 -T1 
A. Evaporation temperature = 38.3F
B. Condensation temperature = 93.4 F
60
A. COP = = = 3.33
C. RE = h2 – h1 = 84.24 – 27.72 = 56.52 Btu/lb Cp  T3 -T4  - Cp  T2 - T1   615-537  - 60
B. RE - Cp  T2 -T1  = 0.24(60) = 14.4 Btu/lb
m'(RE) 23(56.52)
D. Capacity ' = = = 6.5 tons 200 Btu/min.ton
200 200 m'air = =13.9 lb/min.ton
m'  h3 - h2  23  56.52  14.4 Btu/lb
E. Wnet = = = 3.83 Hp m'RT2
42.42 42.42 C. Piston Displacement = V2 =
P2
RE h -h 84.24 - 27.72
F. COP = = 2 1 = = 7.99 13.9(53.3)(474)
Wnet h3 - h2 91.31 - 84.24  = 24.4 ft 3 /min.ton
144 x 100
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:

A. the refrigerating effect, Btu/lb;

B. the rate of Freon-12 circulation, lb/min;
C. the net work required per:pound of Freon-12 circulated,
Btu/lb;
D. the coefficient of performance;
E. the power required per ton of refrigeration, hp/ton;
F. the heat rejected by the condenser, Btu/min; and
G. the compressor piston displacement, ft3/min·ton of
refrigeration.
.
5. A simplified line diagram and TS plot for one section of a 7. Calculate the horsepower required per ton of refrigeration
cooling system for a large aircraft are shown below. When it produced by the reversal of a Carnot cycle having a thermal
is used for cooling on the ground, the following Fahrenheit efficiency of
temperatures are experienced at the numbered points on the
diagrams: A. 50 percent;
B. 25 percent; and
C. 12.5 percent.
T
3 1
Answers: (A) 4.71 hp; (B) 1.57 hp; (C) 0.673 hp
4
2 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
5 when operating With a refrigerating effect of 61.4 Btu/lb of
refrigerant. Calculate :
A. the refrigerant flow rate, lb/min;
S 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.
point 1 2 3 4 5
Answers: (A) 32.6lb/min; (B) 24.6 Btu/lb; (C) 18.9 hp; (D) 2800
t (• F) Btu/min
342 142 252 145 35
For a situation where the air flow rate through the system 9. A refrigeration system has a capacity of 25 tons and rejects heat
is 65 lb/min, specific heat of the air is assumed constant at the rate of 6560 Btu/min. Calculate:
and the compressor and expander processes are assumed (A) the rate of heat absorption by the refrigerant, Btu/min;
isentropic, calculate: (B) the power required as input to the system, Btu/min; and
(C) the coefficient of performance for the system.
(A) the heat transferred to the atmospheric air supply,
Btu/min;
Answers: (A) 5,000 Btu/min; (B) 1560 Btu/min; (C) 3.2
(B) the power developed by the expander, hp; and
(C) the heat transferred from auxiliary power unit compressor 10. A modified Rankine refrigerating cycle operates with an
bleed, expressed in tons of refrigeration. 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
A. Q1-4 =m'Cp  t1 - t 2    t 3 - t 4   psia and 5" F enters the compressor.
= 65(0.24)  342 - 142    252 - 145   Calculate:

= 4789.2 Btu/min (A) the refrigerating effect, Btu/lb;

m'  h4 - h5 
65(0.24)(145 - 35) (B) the plant capacity in tons of refrigeration;
B. Wnet = = = 40.45 hp (C) the power required to compress the refrigerant, hp; and
42.42 42.42 (D) the plant coefficient of performance.
Q 4789.2
C. REFRIG = 1-4 = = 23.95 tons
200 200 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

6 . An ideal Freon-12 refrigerating system has a capacity of 50 the temperature of the air entering the compression
tons. The condenser pressure is 180 psia, and the Freon- cylinder is 50F, the temperature entering the after-cooler
12 temperature leaving the condenser is 120° F. The is 160° F. and the temperature entering the brine tank is
pressure leaving the expansion valve is 44 psia, and the 0F. Calculate:
temperature of the Freon-12 leaving the succeeding coil is 40°
F. Circulating water enters the condenser at a temperature (A) the temperature of the air leaving the after-cooler
of 100° F and leaves it at 1 1 5  F. Determine: (B) the coefficient of performance;
(C) the mass of air which must be circulated per ton of
A. the mass of Freon-12 circulated, lb/hr; refrigeration, lb/min.
B. the compressor power for isentropic compression, Btu/hr
C. the heat capacity of the system, Btu/hr Answers: (A) 99F (B) 4.54 (C) 16.7 lb/min
D. 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)


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)
HC 744, 300
E. Heating Performance =   5.15
Compressor Power 144, 400
 CNS 04
MDSP/MESL