A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal vapor-compression refrigeration cycle between

Question

4 years ago

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A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal vapor-compression refrigeration cycle between

0.14 and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s, determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the rate of heat rejection to the environment, and (c) the COP of the refrigerator.

Physics

2 answers:

valentinak56 [21] · Answer 1 · 2022-01-12T18:55:47+03:00

Answer:

(a) The rate of heat removal from the refrigerated space is 10.06KJ/s and the power input to the compressor is 7.57KJ/s

(b) The rate of heat rejection to the environment is 2.49KJ/s

(c) The COP of the refrigerator is 1.33

Explanation:

Mass flow rate of refrigerant = 0.05kg/s

From the steam table, enthalpies of the saturated liquid and vapor refrigerant at 0.14MPa and 0.8MPa are

0.14MPa: hf = 59.3KJ/kg, hg = 1410.4KJ/kg

0.8MPa: hf = 260.4KJ/kg, hg = 1460.2KJ/kg

(a) Rate of heat removal from the refrigerated space = 0.05(260.4 - 59.3) = 10.06KJ/s

Power input to the compressor = 10.06 - 2.49 = 7.57KJ/s

(b) Rate of rejection of heat to the environment = 0.05(1460.2 - 1410.4) = 2.49KJ/s

(c) COP of the refrigerator = 10.06 ÷ 7.57 = 1.33

Mice21 [21] · Answer 2 · 2022-01-12T17:43:25+03:00

Answer:

a) Ql = 7.18kW and Win = 1.81kW.

b) Qh = 8.99kW.

c) COPr = 3.96.

Explanation:

This is an ideal vapor-compression refrigeration cycle, and thus compressor is isotropic and the refrigerant leaves the condenser as a saturated liquid and enters the compressor as saturated vapor

From the refrigerant-134a tables, enthalpies are determined as:

P1 = 0.14MPa => h1 = 239.16kJ/kg

P2 = 0.8MPa => h2 = 275.39kJ/kg

P3 = 0.8MPa => h3 = 95.47kJ/kg

h4 ≅ h3 =95.47kJ/kg

a) Rate of heat removal from refrigerated space and the power input to compressor:

Ql = m(h1 - h4) = 0.05[239.16 - 95.47]

Ql = 7.18kW

Win = m(h2 - h1) = 0.05[275.39 - 239.16]

Win = 1.81kW

b) Rate of heat rejection from the refrigerant to the environment:

Qh = m(h2 - h3) = 0.05[275.39 - 95.47] = 8.99kW

And, It can also be determined from the formula:

Qh = Ql + Win = 7.18 + 1.81

Qh = 8.99kW

c) Coefficient of performance of the refrigerator is:

COPr = Ql / Win

COPr = 7.18 / 1.81

COPr = 3.96.

There is no any unit of the coefficient of the performance of the refrigerator.