Question

A vapor-compression refrigeration system for a household refrigerator has a refrigerating capacity of 1650 Btu/h. Refrigerant enters the evaporator at -15°F and exits at 20°F. The isentropic compressor efficiency is 75%. The refrigerant condenses at 125°F and exits the condenser subcooled at 100°F. There are no significant pressure drops in the flows through the evaporator and condenser.


For Refrigerant 134a as the working fluid, determine:

(a) the evaporator and condenser pressures, each in lbf/in.²

(b) the mass flow rate of refrigerant, in lb/min.

(c) the compressor power input, in horsepower.

(d) the coefficient of performance.

Answer 1

For R-134a, from the saturation tables, the following are the properties:

	Temp (F)	Saturation pressure (psia)	hf (BTU/lb)	hg (BTU/lb)	sf (BTU/lbF)	sg (BTU/lbF)
Evaporator	-15	14.68	7.574	100.863	0.01749	0.2273
Condenser	125	199.4	54.2	118.601	0.1074	0.2175

A)

From the table above:

• Evaporator pressure = 14.68 psia = 14.68 lb(f)/in2.
• Condenser pressure = 199.4 psia = 199.4 lb(f)/in2.

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B)

We need to first find enthalpy at state 1, h1, for this.

where

• hg = hg at evap pressure = 100.863 BTU/lb (from the table above)
• c = specific heat at constant pressure for R134a = 0.1929 BTU/lbF
• T1 = 20 F (given)
• T(sat) = saturation temp at evap pressure = -15 F (given)

And

where

• h4 = h3 [isenthalpic expansion]
• hf = hf at condenser pressure = 54.2 BTU/lb (from the table above)
• T(sat) = saturation temp at cond pressure = 125 F
• T3 = 100 F
• c = specific heat at constant pressure for R134a = 0.3149 BTU/lbF

So,

where

• m(ref) = mass flow rate of refrigeration = ?
• h1 = 107.5 BTU/lb
• h4 = 45.1 BTU/lb
• Q = refrigeration capacity = 1650 BTU/hr = 27.5 BTU/min

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C)

Now, for this, we will first find entropy at state 1, s1 = ?

where

• sg = sg at evap pressure = 0.2273 BTU/lbF
• c = specific heat at constant pressure for R134a = 0.1929 BTU/lbF
• T1 = 20 F = 266.483 K (have to convert to thermodynamic scale)
• T(sat) = saturation temp at evap pressure = -15 F = 247.039 K (have to convert to thermodynamic scale)

Now, s1 = s2' = 0.2417 BTU/lbF

But,

where

• c = specific heat at constant pressure for R134a = 0.2942 BTU/lbF
• T2' = ?
• T(sat) = saturation temp at condenser pressure = 125 F = 324.817 K
• sg = sg at condenser pressure = 0.2175 BTU/lbF

So,

where

• hg = hg at cond pressure = 118.601 BTU/lb
• c = specific heat at constant pressure for R134a = 0.2942 BTU/lbF
• T2' = 178.3 F
• T(sat) = saturation temp at condenser pressure = 125 F

So, the ideal compressor Power =

But. the compressor is 75% efficient.

Therefore,

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D)

where

• Refrigeration capacity = 27.5 BTU/min = 0.64847 HP
• P(actual) = 0.357 HP

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