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 2 of 2.
Dhiraj mehta said:
9 years ago
COP = t2/(t1 - t2) where t2 is low temperature.
HRF = 1 + (1/cop).
HRF = 1 + (1/cop).
Hafiz said:
10 years ago
How COP = 7.92 and HRF = 1.12? Please explain.
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers