Mechanical Engineering - Heat Transfer, Refrigeration and Air Conditioning - Discussion
Discussion Forum : Heat Transfer, Refrigeration and Air Conditioning - Section 8 (Q.No. 34)
34.
In figure shown, E is a heat engine with efficiency of 0.4 and R is a refrigerator. It is given that Q2 + Q4 = 3Q1. The C.O.P. of the refrigerator is
Discussion:
5 comments Page 1 of 1.
Joe Burden said:
5 years ago
COPr = Q3/W or COPhp= Q3/W?
Please explain the right one.
Please explain the right one.
Sanjib Mondal said:
5 years ago
0.4= W/Q1.
W= Q2- Q1 for Heat Engine.
So, W = 0.4Q1.
(COP)r = Q3/W = (Q4 - W)/0.4Q1.
= Q4 - (Q1-Q2)/0.4Q1,
= Q4+Q2-Q1/0.4Q1,
= 3Q1- Q1/0.4Q1,
= 2Q1/0.4Q1,
= 5.
W= Q2- Q1 for Heat Engine.
So, W = 0.4Q1.
(COP)r = Q3/W = (Q4 - W)/0.4Q1.
= Q4 - (Q1-Q2)/0.4Q1,
= Q4+Q2-Q1/0.4Q1,
= 3Q1- Q1/0.4Q1,
= 2Q1/0.4Q1,
= 5.
(1)
Sudheer said:
6 years ago
How Q2 = 0.6 Q1?
(1)
AMAL said:
10 years ago
Q1-Q2 = Q4-Q3.
Rearranging, we get Q1+Q3 = Q2+Q4.
Q1+Q3 = 3Q1.
So, Q3 = 2Q1,
Als0 Q2 = 0.6 Q1.
Q4+Q2 = 3Q1 so, Q4 = 3Q1-0.6Q1 (because Q2 = 0.6Q1) = 2.4Q1.
Now COP = Q3/(Q4-Q3).
= 2Q1/(2.4Q1 - 2Q1).
= 2/0.4 = 5.
Rearranging, we get Q1+Q3 = Q2+Q4.
Q1+Q3 = 3Q1.
So, Q3 = 2Q1,
Als0 Q2 = 0.6 Q1.
Q4+Q2 = 3Q1 so, Q4 = 3Q1-0.6Q1 (because Q2 = 0.6Q1) = 2.4Q1.
Now COP = Q3/(Q4-Q3).
= 2Q1/(2.4Q1 - 2Q1).
= 2/0.4 = 5.
(1)
Shailesh said:
1 decade ago
For refrigerator - c.o.p=Q3/w ........(1).
For engine - n=w/Q1.
0.4= w/Q1 => w=0.4xQ1 ........(2).
Because Q4 - Q3 = Q1 - Q2.
Q4 + Q2 = Q1 + Q3.
3xQ1 = Q1 + Q3 => Q3=2xQ1.
Put value of Q3 and w in eq. (1)
Hence c.o.p = (2xQ1)/(0.4xQi).
c.o.p = 5.
For engine - n=w/Q1.
0.4= w/Q1 => w=0.4xQ1 ........(2).
Because Q4 - Q3 = Q1 - Q2.
Q4 + Q2 = Q1 + Q3.
3xQ1 = Q1 + Q3 => Q3=2xQ1.
Put value of Q3 and w in eq. (1)
Hence c.o.p = (2xQ1)/(0.4xQi).
c.o.p = 5.
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