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
1
1.25
2.15
5.12
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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.

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.

Milan said:   6 years ago
cop= t1/t1-t2 = 313/313-278 = 8.94.

hrf = 1+1/cop= 1 + 1/8.94 = 1.11 so its 1.25.

Girija Sankar Murmu said:   6 years ago
HRF = (heat rejected/heat absorbed) = 1+1/COP = T1/T2 = 313k/278k = 1.125.

Dhiraj mehta said:   9 years ago
COP = t2/(t1 - t2) where t2 is low temperature.

HRF = 1 + (1/cop).

Shibunath said:   7 years ago
Actually, the answer should come 1.125.
Cop: 7.94.
HRF: 1+(1/cop).

DURGESH KUMAR DUBEY said:   8 years ago
Heat rejection factor:- 1+ 1/COP.

It will be 1.126.

Hafiz said:   10 years ago
How COP = 7.92 and HRF = 1.12? Please explain.

Vijay said:   8 years ago
COP = 278k/35k = 7.94.
HRF = 1 +1/COP.

Anurup said:   8 years ago
According to me, it comes 1.126.


Post your comments here:

Your comments will be displayed after verification.