Question

The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200 K and 400 K, respectively. For each case, evaluate the net power developed by the cycle, in kW, and the thermal efficiency. Also in each case apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible. (a) Q . H 5 600 kW, Q . C 5 400 kW (b) Q . H 5 600 kW, Q . C 5 0 kW (c) Q . H 5 600 kW, Q . C 5 200 kW

Answer 1

Answer:

a. Irreversible

b. Impossible

c. Impossible

Explanation:

to determine if a case is reversible, irreversible or impossible we compare the actual efficiency to the maximum efficiency of each case

maximum efficiency of the power cycle

nmax = 1 - TC/TH

nmax = 1 - 400/1200

Check attachment for complete solution

Answer 2

Answer:

a) N=0.04 (4%)

b)N=0.99 (99%)

c)N=0.07 (7%)

Explanation:

The maximum power efficiency of any power cycle operating between two reservoirs is

Nmax=1-Tc/Th

Nmax=1-400k/1200k

Nmax=1-0.3333

Nmax=0.667 (66.7%)

Given Q . H= 5 600 kW, Q . C= 5 400 kW

a) Wcycle= Qh-Qc

Wcycle=5600-5 400

Wcycle=200KW

N=Wcycle/Qh

N=200KW/5 600 kW

N=0.04 (4%)

b) Q . H =5 600 kW, Q . C= 5 0 kW

Wcycle= Qh-Qc

Wcycle= 5 600 kW-5 0 kW

Wcycle=5550 kW

N=Wcycle/Qh

N=5550/5 600

N=0.99 (99%)

C) Q . H= 5 600 kW, Q . C= 5 200 kW

Wcycle= Qh-Qc

Wcycle=5 600-5 200

Wcycle=400 kW

N=Wcycle/Qh

N=400kW/5 600 kW

N=0.07 (7%)