Question

A power cycle operates between a lake’s surface water at a temperature of 300 K and water at a depth whose temperature is 285 K. At steady state the cycle develops a power output of 10 kW, while rejecting energy by heat transfer to the lower temperature water at the rate 14,400 kJ/min. Determine (a) the thermal efficiency of the power cycle and (b) the maximum thermal efficiency for any such power cycle.

I have no idea on how to work this out, so if you could please provide detailed answers - that aren't just a copy of an answer key that would be greatly appreciated! Thanks.

Answer 1

Given data :
Surface temperature T1 =300K
Under water temperature T2 = 285K
Power out put W = 10KW
Heat rejection at a rate of Q2 = 14400KJ/min
Q2 = 14400 X 60KJ/sec
Q2 = 240 KJ/sec
Q2 = 240KW
(a)The thermal efficiency of the power cycle is:
We know that as per thermodynamics second law

Thermal efficiency η =  (T1 - T2)/T1

η = (300-285)/300

  η = 0.05

   η = 5%

We also know the another way to determine the themal efficiency is

Thermal efficiency η = W/Q1 ………….(1)
But here Q1 = heat input
We know W = Q1 - Q2

Q1 = W + Q2

Q1 = 10 + 240

Q1 = 250KW

Put this value in abouve equation (1)
Thermal efficiency η = W/Q

  η = 10/250

   η = 0.04%

   η = 4%

These are the thermal efficiencies of the power cycle

(b)The maximum thermal efficiency of the power cycle:

The maximum thermal efficiency of the power cycle is the efficiency which is obtained by the temperatures i.e 5%
Maximum thermal efficiency is η = 5%