TITLE.PM5

754 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th14-3.pm5

or ln

T 3

313

F

HG

I

KJ =

0 7082 0 682

0 747

..

.

−

= 0.03507

or

T

(^3) e0 03507
313
=. = 1.0357
∴ T 3 = 313 × 1.0357 = 324.2 K
Now, h 3 = h 3 ′ + cp(324.2 – 303)
= 203.05 + 0.747(324.2 – 313) = 211.4 kJ/kg
Power required :
Power required by the machine = mh&() 32 −h
= 0.192(211.4 – 178.61) = 6.29 kW. (Ans.)
Piston displacement, V :
Volumetric efficiency, ηvol. = 1 + C – C
p
p
d
s
F n
HG
I
KJ
1/
= 1 + 0.03 – 0.03 9.607
1.509
F
HG
I
KJ
1
113.
= 0.876 or 87.6%
The volume of refrigerant at the intake conditions is
m& × vg = 0.192 × 0.1088 = 0.02089 m^3 /s
Hence the swept volume =
0 02089 0 02089
0 876
..
ηvol..
= = 0.02385 m^3 /s
∴ V = 0 02385 60
300
. × = 0.00477 m (^3). (Ans.)
Example 14.20. A food storage locker requires a refrigeration capacity of 50 kW. It works
between a condenser temperature of 35°C and an evaporator temperature of – 10°C. The refriger-
ant is ammonia. It is sub-cooled by 5°C before entering the expansion valve by the dry saturated
vapour leaving the evaporator. Assuming a single cylinder, single-acting compressor operating at
1000 r.p.m. with stroke equal to 1.2 times the bore.
Determine : (i) The power required, and
(ii) The cylinder dimensions.
Properties of ammonia are :
Saturation Pressure Enthalpy, kJ/kg Entropy, kJ/kg K Specific volume, Specific heat
temperature, °C bar m^3 /kg kJ/kg K
Liquid Vapour Liquid Vapour Liquid Vapour Liquid Vapour

• 10 2.9157 154.056 1450.22 0.82965 5.7550 — 0.417477 — 2.492
  35 13.522 366.072 1488.57 1.56605 5.2086 1.7023 0.095629 4.556 2.903
  (U.P.S.C. 1997)
  Solution. Given : (From the table above)
  h 2 = 1450.22 kJ/kg ; h 3 ′ = 1488.57 kJ/kg ; hf 4 = 366.072 kJ/kg ;
  hf 4 ′ = h 1 = hf 4 – 4.556(308 – 303)
  = 366.07 – 4.556(308 – 303) = 343.29 kJ/kg