Respuesta :
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.
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
