Example 2 A plant using R134a refrigerant has a cooling capacity of 352 kW of refrigeration. It evaporates at 0 C and condenses at 35 C, with the 5 K of superheat at the evaporator outlet and 5 K of sub-cooling at the condenser outlet. Assuming isentropic compression and ignoring pressure drops in the piping, the evaporator and condenser, determine the following; a) The refrigeration effect b) The mass flow rate of refrigerant c) The compressor power d) The rate of heat rejection at the condenser e) The COP f) The corresponding Carnot COP g) The percentage Carnot efficiency

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Analysis of the Refrigeration System
Cooling effect: Q = (h1 – h4)
Compressor work: Win = (h2 – h1)
Condenser load: Qn = (h2 – h3)
QL
(h1-h4
Theoretical COP:
COPref =
Win
Qh.
(h2-h3
COPhp =
W in
Win
Assuming total cooling capacity is : Q
Mass flow rate of refrigerant in circulation: m =
h1-h4
Compressor power: Þ = m(h2 – h,) =
(h2-h;)
(h1-h4)
Condenser capacity:: Q. = m(h, – hạ) = 0 h3-12)
(h1-h4)
Example 2
A plant using R134a refrigerant has a cooling capacity of 352 kW of
refrigeration. It evaporates at 0 C and condenses at 35 C, with the 5 K of
superheat at the evaporator outlet and 5 K of sub-cooling at the
condenser outlet. Assuming isentropic compression and ignoring
pressure drops in the piping, the evaporator and condenser, determine the
following;
a) The refrigeration effect
b) The mass flow rate of refrigerant
c) The compressor power
d) The rate of heat rejection at the condenser
e) The COP
The corresponding Carnot COP
f)
g) The percentage Carnot efficiency
Example 2
A plant using R134a refrigerant has a cooling capacity of 352 kW of
refrigeration. It evaporates at 0 C and condenses at 35 C, with the 5 K of
superheat at the evaporator outlet and 5 K of sub-cooling at the
condenser outlet. Assuming isentropic compression and ignoring
pressure drops in the piping, the evaporator and condenser, determine the
following;
a) The refrigeration effect
b) The mass flow rate of refrigerant
c) The compressor power
d) The rate of heat rejection at the condenser
e) The COP
f) The corresponding Carnot COP
g) The percentage Carnot efficiency
Transcribed Image Text:Analysis of the Refrigeration System Cooling effect: Q = (h1 – h4) Compressor work: Win = (h2 – h1) Condenser load: Qn = (h2 – h3) QL (h1-h4 Theoretical COP: COPref = Win Qh. (h2-h3 COPhp = W in Win Assuming total cooling capacity is : Q Mass flow rate of refrigerant in circulation: m = h1-h4 Compressor power: Þ = m(h2 – h,) = (h2-h;) (h1-h4) Condenser capacity:: Q. = m(h, – hạ) = 0 h3-12) (h1-h4) Example 2 A plant using R134a refrigerant has a cooling capacity of 352 kW of refrigeration. It evaporates at 0 C and condenses at 35 C, with the 5 K of superheat at the evaporator outlet and 5 K of sub-cooling at the condenser outlet. Assuming isentropic compression and ignoring pressure drops in the piping, the evaporator and condenser, determine the following; a) The refrigeration effect b) The mass flow rate of refrigerant c) The compressor power d) The rate of heat rejection at the condenser e) The COP The corresponding Carnot COP f) g) The percentage Carnot efficiency Example 2 A plant using R134a refrigerant has a cooling capacity of 352 kW of refrigeration. It evaporates at 0 C and condenses at 35 C, with the 5 K of superheat at the evaporator outlet and 5 K of sub-cooling at the condenser outlet. Assuming isentropic compression and ignoring pressure drops in the piping, the evaporator and condenser, determine the following; a) The refrigeration effect b) The mass flow rate of refrigerant c) The compressor power d) The rate of heat rejection at the condenser e) The COP f) The corresponding Carnot COP g) The percentage Carnot efficiency
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