Question

Two carnot engines a and b are operated in series. Engine a receives heat from a reservoir at 600k and rejects heat to a reservoir heat to reservoir at temperaturet. Engineb receives heat rejected by engine a and in turn rejects it to a reservoir at100k. If the efficiencies of the two engines a and b are represented by a and b respectively, then what is the value of ab

Answer 1

Hey buddy,

◆ Answer-
ab = 25/84

◆ Explanation-
# Given-
T1 = 600 K
T2 = ?
T3 = 100 K

# Solution-
As carnot engines are connected in series work done will be same -
W1 = Q2 - Q3
W2 = Q1 - Q2

Therefore,
W1 = W2
Q1 - Q2 = Q2 - Q3
Q2 = (Q1 + Q3) / 2

In carnot cycle, Q ∝ T, hence
T2 = (T1 + T3) / 2
T2 = (600 + 100) / 2
T2 = 350.

Efficiency of 1st engine -
a = (Q1 - Q2) / Q1
a = (T1 - T2) / T1
a = (600 - 350) / 600
a = 5/12

Efficiency of 2nd engine -
b = (Q2 - Q3) / Q2
b = (T2 - T3) / T2
b = (350 - 100) / 350
b = 5/7

Value of ab -
ab = 5/12 × 5/7
ab = 25/84

Hope this helps you...