Aheat engines is operating on a carnot cycle and has a thermal efficiency of 55 percent. the waste heat from this engine is rejected to a nearby lake at 15°c at a rate of 800 kj/min. determine: (a) the power output of the engine, and (b)the temperature of the source.

a) P = 1776.97 kJ/min

b) Source temperature, T = 640 K

Given:

T =

Q' = 800 kJ/min

Solution:

Refer to the given fig.:

a) We first calculate the value of Q from the eqn:

Solving for Q, we get:

0.55 Q = Q - 800000

Q = 1777.778 kJ/min

Now,

Output Work, W = Q - Q' = 1777.778 - 800 = 1776.97 kJ/min = power output, P

b) Temperature of source, T':

We know that:

T ∝ Q

Therefore,

T = 640 K

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: