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.
Gaurav said:
1 decade ago
Efficiency is same for both.
T2/1600 = 400/T2.
T2 = 800.
T2/1600 = 400/T2.
T2 = 800.
(2)
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)
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)
Mori bhavdip said:
7 years ago
Both are the reversible engines and efficiency of all reversible engines are same so we can compare the efficiency equation and get the answer.
T1-T2/Ti = T2-T3/T2.
T1-T2/Ti = T2-T3/T2.
(1)
SAIROHIT SANDELA said:
7 years ago
Same heat input and output.
Means efficiencies are the same.
For the same efficiencies, the intermediate temperature is the geometric mean of the other two extreme temperatures.
If work done in both cases is the same then that would be the average of the two temperatures.
Means efficiencies are the same.
For the same efficiencies, the intermediate temperature is the geometric mean of the other two extreme temperatures.
If work done in both cases is the same then that would be the average of the two temperatures.
(1)
Girish baswa said:
7 years ago
In question it is not mentioned that both heats engines are in series, so for same heat input and workput the efficiency will remain same and the answer is geometric mean of it.
(1)
GATE 2018 guy said:
5 years ago
If heat input and output are same, then the work done must also be the same right?
Then T2 = (1600+400) /2.
T2 = 1000 K can also be the right answer.
Then T2 = (1600+400) /2.
T2 = 1000 K can also be the right answer.
(1)
Hizal said:
1 decade ago
How it is solution?
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.
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