Question

A reversible heat engine operates between temperature 800°C and 500°C of thermal reservoir. Engine drives a generator and a reversed carnot engine using the work output from the heat engine for each unit equality. Reversed Carnot engine abstracts heat from 500°C reservoir and rejected that to a thermal reservoir at 715°C. Determine the heat rejected to the reservoir by the reversed engine as a fraction of heat supplied from 800°C reservoir to the heat engine. Also determine the heat rejected per hour for the generator output of 300KW

Answer 1

Given that

T1 = 800°C = 1073K

T2 = 500°C = 773K

T3 = 800°C = 988K

ηrev = (Q1 – Q2)/Q1

= (T1 – T2)/T1 = W/Q

= (1073 – 773)/ 1073 = W/Q1

W = 0.28Q1 ...(i)

Now for H.P

Q4/(Q3 – Q4) = T3/(T3 – T2)

Q4/(W/2) = T3/(T3 – T2)

Q4 = (W/2)[T3/(T3 – T2)]

Q4 = (0.28Q1/2) [988/(988 – 773)]

= 0.643Q1

Q4 = 0.643Q1

Now if W/2 = 300; W = 600KW

0.28Q1 = 600

Q1 = 2142.8KJ/sec

Since Q1 = W + Q2

Q2 = 2142.8 – 600 = 1542.85 KJ/sec

Q2 = 1542.85 KJ/sec