Question

A reversible steady-state device receives a flow of 1 kg/s air at 400 K, 450 kPa and the air leaves at 600 K, 100 kPa. Heat transfer of 900 kW is added from a 1000 K reservoir, 50 kW is rejected at 350 K, and some heat transfer takes place at 500 K. Find the heat transferred at 500 K and the rate of work produced.

Answer 1

41 kW

690.2 kW

Step-by-step explanation:

Given the following data

T1 = 400 K

P1 = 350 kPa

T2 = 600 K

P2 = 100 kPq

m = 1 kg/s

Q1 = 900 kW

Q2 = 50 kW

T(amb1) = 1000 K

T(amb2) = 350 K

T(amb3) = 500 K

We also will be using tables to solve the question.

R = 0.287 kJ/kgk

C(p) = 1.004 kJ/kgk

To find Q3, we'd use the formula

Q3 = m.T(amb3).[C(p).In(T2/T1) - R.In(P2/P1)] - [T(amb3)/T(amb1) * Q1] + [T(amb3)/T(amb2) * Q2]

Q3 = 1 * 500 * [1.004 * In(600/400) - 0.287 * In(100/450)] - 500/1000 * 900 + 500/350 * 50

After simplification, we find that

Q3 = 41 kW

Next, we use the equation

Q3 - Q2 + Q1 = W + m.C(p).(T2 - T1)

41 - 50 + 900 = W + 1 * 1.004 * (600 - 400)

891 = W + 1.004 * 200

891 = W + 200.8

W = 891 - 200.8

W = 690.2 kW

If we make W the subject of the formula, we have