Thermodynamic Question (power plant problem)

[gotrilum]

I am wondering if anyone can confirm this answer I got for the following problem;

A coal burning electrical power plant burns at 706 degrees celsius. Heat is exhausted into the river near the power plant; the average river temperature is 19 degrees celsius. What is the minimum possible rate of thermal pollution (heat exhausted into the river) if the station generates 125 MW of electricity?

*First, I found the efficiency to be .7 (1 - 292.15K / 979.15 K). I took .7 and plugged it into the equation Wasted power Pw = Po[(1/η) - 1] = 125E6*[(1/.7) - 1] = 53.6 MW

Another question I have is where does the equation above come from? Someone insisted that I use this equation but I have no ideal from where it is derived.

Thanks

 

[rl.bhat]

Power generation of the station = ( heat input - heat output) per second.= Qi - Qo
efficiency = n = (Qi -Qo )/Qi
Qi = (Qi - Qo)/n
Qo = Qi - Qi + Qo = (Qi - Qo)/n - ( Qi - Qo) = ( Qi - Qo)[ 1/n - 1]

 

[xxChrisxx]

The best efficiency a set cycle can achieve between two temperatures is the efficiency of the carnot cycle.

If you always think of efficiency as:
what you get / what you pay for.

By this definition the efficiency of the cycle is:
Work out / Heat In

Work is the difference in heats between the hot and cold reservoirs. and heat in is of the hot reservoir.

So from the efficiency above we put in out Q values

(Qh - Qc)/Qh

Which reduces to:

Qh/Qh - Qc/Qh
=1-(Qc/Qh)

For a reversible cycle such as this we can write the heat transfer as a functuion of temperatre of the reservoir so it becomes:

efficiency = 1-(Tc/Th)

As Lord Kelvin came up with the above relation about reversible cycles, he called the scale the Kelvin scale (how modest of him) so naturally as you have used above the temperatures are in K.

The biggest thing to take from this is how to think of efficiecny in your mind, it really helped me visualise what the right efficieny is. especially for heat pump and refirigiration cycles.