Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 4 (Q.No. 47)
47.
One reversible heat engine operates between 1600 K and T2 K and another reversible heat engine operates between T2 K and 400 K. If both the engines have the same heat input and output, then temperature T2 is equal to
Discussion:
14 comments Page 1 of 2.
Hizal said:
1 decade ago
How it is solution?
ASHOK KUMAR said:
1 decade ago
T1-T2/Ti = T2-T3/T2.
Ti = 1600k.
T2 = ?k.
T3 = 400k.
1600-T2/1600 = T2-400/T2.
1600T2-T2*T2 = 1600T2-640000.
T2 suqare = 640000.
T2 = 800k.
Ti = 1600k.
T2 = ?k.
T3 = 400k.
1600-T2/1600 = T2-400/T2.
1600T2-T2*T2 = 1600T2-640000.
T2 suqare = 640000.
T2 = 800k.
(1)
Gaurav said:
1 decade ago
Efficiency is same for both.
T2/1600 = 400/T2.
T2 = 800.
T2/1600 = 400/T2.
T2 = 800.
(2)
RAVI CHAVAN said:
1 decade ago
For same heat input and output.
Shortcut formula T2 = geometric mean mean of T1 and T3.
Shortcut formula T2 = geometric mean mean of T1 and T3.
Shiva said:
10 years ago
The geometric mean is for heat engine in series. In this case the heat rejected by the first engine is given as input to second engine.
The question says same heat supplied as input.
The question says same heat supplied as input.
Pavan said:
8 years ago
Work done is also same in both the cases, so option B might be correct.
Murali said:
8 years ago
Thank you so much for the given explanation.
Bibin Emmanuvel said:
8 years ago
I think Both A and B are correct.
M S NAYAK said:
8 years ago
If the efficiency of two engines are same,
Intermediate temperature T2 = root(T1*T3).
If work output is same.
T2 = (T1+T2)/2.
Intermediate temperature T2 = root(T1*T3).
If work output is same.
T2 = (T1+T2)/2.
(1)
Mani said:
7 years ago
Thank you all for the explanation.
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