A Carnot engine whose efficiency is 40% takes in the heat from a source maintained at a temperature of 500 k. It is desired to have an engine of efficiency of 60%. Then the intake temperature from the same exhaust (sink) temperature must be :
600 k
The efficiency of Carnot cannot be made larger than 50%
1200 k
750 k
Answers
For carnot engine-
η(efficiency)=1−
T
H
T
L
T
L
= low temperature(sink)
100
40
=1−
500
T
L
⇒
500
T
L
=
100
60
⇒T
L
=300
Now,
100
60
=1−
T
H
300
T
H
300
=1−
100
60
T
H
300
=
100
40
T
H
=
40
300×100
=750K.
A Carnot engine whose efficiency is 40% takes in the heat from a source maintained at a temperature of 500 k. It is desired to have an engine of efficiency of 60%.
To find : The intake temperature from the same exhaust (sink) temperature must be ...
solution : using formula, efficiency , η = 1 - T₁/T₂
where T₁ is temperature of sink and T₂ is temperature of source.
case 1 : η = 40% , T₂ = 500K
so, 40/100 = 1 - T₁/500K
⇒-0.6 = T₁/500K
⇒T₁ = 300K
so temperature of sink = 300K
case 2 : η = 60% , T₁ = 300K
so, 60/100 = 1 - 300K/T'₂
⇒T'₂ = 300/0.4 = 750 K
therefore the intake temperature from the same exhaust (sink) temperature must be 750K