Two Carnot engines are connectedin series with extreme temp as 2000k & 200k resp. What is the efficiency of first Carnot engine?
Answer (1)
Student Expert
2nd Aug, 2019
Hi Digambar,
Given:
T1 = 2000 K
T2 is unknown.
T3 = 200 K
As carnot engines are connected in series work done will be equal in both:
W1 = Q2 - Q3
W2 = Q1 - Q2
Since, W1 = W2
Q1 - Q2 = Q2 - Q3
Q2 = (Q1 + Q3) / 2
In carnot cycle, Q ∝ T, hence
T2 = (T1 + T3) / 2
T2 = (2000+200) / 2
T2 = 1100K.
Efficiency of 1st engine -
nA = (Q1 - Q2) / Q1
nA = (T1 - T2) / T1
nA = (2000-1100) / 2000
nA = 0.45
Hope it helps.
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