Mechanical Engineering - Heat Transfer, Refrigeration and Air Conditioning - Discussion
Discussion Forum : Heat Transfer, Refrigeration and Air Conditioning - Section 7 (Q.No. 46)
46.
In actual air-conditioning applications for R-12 and R-22, and operating at a condenser temperature of 40° C and an evaporator temperature of 5° C, the heat rejection factor is about
Discussion:
13 comments Page 1 of 2.
Vivek mishra said:
1 decade ago
Heat rejection factor = (Heat rejected/Heat absorbed).
Here, COP = 7.942.
Implies, Q(rejected) = 1.12*Q (absorbed).
Hence, HRF = 1.12.
Here, COP = 7.942.
Implies, Q(rejected) = 1.12*Q (absorbed).
Hence, HRF = 1.12.
Hafiz said:
10 years ago
How COP = 7.92 and HRF = 1.12? Please explain.
Dhiraj mehta said:
9 years ago
COP = t2/(t1 - t2) where t2 is low temperature.
HRF = 1 + (1/cop).
HRF = 1 + (1/cop).
Vijay said:
8 years ago
COP = 278k/35k = 7.94.
HRF = 1 +1/COP.
HRF = 1 +1/COP.
Anurup said:
8 years ago
According to me, it comes 1.126.
DURGESH KUMAR DUBEY said:
8 years ago
Heat rejection factor:- 1+ 1/COP.
It will be 1.126.
It will be 1.126.
Shibunath said:
7 years ago
Actually, the answer should come 1.125.
Cop: 7.94.
HRF: 1+(1/cop).
Cop: 7.94.
HRF: 1+(1/cop).
Chra said:
7 years ago
T1=5+273=278 k.
T2=40+273=313 k.
(COP)R = T1/(T2-T1) = 7.94.
HRF = 1+( 1/COP) = 1.135.
T2=40+273=313 k.
(COP)R = T1/(T2-T1) = 7.94.
HRF = 1+( 1/COP) = 1.135.
Paul said:
7 years ago
The Answer should be 1.125
Himmat said:
7 years ago
Please Give a clear explanation.
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