Ideal Carnot Engine Homework | 5kW Output, 35% Efficiency, 30deg Heat Rejection

[cathy88]

Homework Statement

An ideal Carnot engine produces a power output of 5kW at a thermal efficiency of 35% while rejecting heat reversibly to a reservoir at 30degrees. It achieves this output while executing 10 cycles per second.

a) What temperature is the engine receiving its heat supply?
b)What are the rates of heat supply to and heat rejection from the engine?
c)Hence determine the change in the total entropy of the working fluid during each of the two heat transfer process.

Homework Equations

Not really given but I got first part n_carnot=(Th-TL) / TH

The Attempt at a Solution

I use that about equation for part a to find Th which is where the temperature receiving the heat supply also which is process 1-2 isothermal heat addition.
But I am totally stoned about part b and c. How can I calculate the rates of heat supply etc, would someone be able gimme ahint on how to even start it?

Thanks

 

[Mapes]

Power is energy transferred per unit time. The relationship between the rates of heat supply, heat rejection, and power output are exactly the same as the relationship between heat transferred in (Q) and out and work done (W) per cycle. And this relationship is in turn related to the efficiency. Look for one more equation to tie these variables together.

 

[cathy88]

n carnot = Wcycle / Qin
this would help me solve rate for power supply into system, given i know thermal efficency as .35 / 35% but I do not know the W cycle, I am only given the Work output which is 5KW. No volume is given so I can't go through the first law and calculate the total work for the cycle. What should I be doing? Thanks

 

[Mapes]

From this equation, you can calculate the rate of heat supply immediately. Apply the first law, and you can calculate the rate of heat rejection.

 

[cathy88]

How do I calculated the rate heat supplied? I thought the 5KW output is not the total Work Cycle?

 

[Mapes]

If you take a look at my post #2 again, the ratio between the work per cycle and the heat input per cycle is the same as the ratio between the (average) work per second and the heat input per second, which is the same as the ratio between the work power (which you're given) and the heat supply rate.

 

[cathy88]

ahh, I get it, so I do 5kw/.35= q in?
another question how could i calculate heat reject. It is process 3-4 in a carnot cycle which is isothermal heat rejection. If I apply First Law it wud be typically that q3-4 or qout= RT3*ln(v4/v3) but I can not get this volume , or should I be using somthing else instead of doing this? Thanks

edit: can't use pV=mRT as I don't have the mass or do I use pv=RT where its going to be a Pressure vs Specific volume graph in that case I could where T=the temperature I found in part a, but lol I don't have the pressure.

Thank you

 

[Mapes]

Yet another way of looking at it is that, out of the heat energy that comes in, what isn't output as work must be rejected.