Question

A heat 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.

Answer 1

Answer:

a) P = 1776.97 kJ/min

b) Source temperature, T = 640 K

Given:

$\eta = 55%$

T = $15^{\circ} = 288 K$

Q' = 800 kJ/min

Solution:

Refer to the given fig.:

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

$\eta = 1 -\frac{Q'}{Q}$

$0.55 = 1 -\frac{800\times 1000}{Q}$

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,

$\frac{T}{T'} = \frac{Q}{Q'}$

$\frac{T}{288} = \frac{1777.78\times 1000}{800}$

T = 640 K