Calculating Maximum Cycle Temperature in Carnot Cycle | Homework Solution

[db725]

Homework Statement

An ideal engine operating on the Carnot cycle with a minimum temperature of 25°C develops 20 kW and rejects heat a rate of 12 kW. Calculate the maximum cycle temperature and the required heat supply of the cycle.

Homework Equations

Qhot/Thot =Qcold/Tcold

The Attempt at a Solution

Since this is ideal entropy must be conserved. Δs hot =Δs cold is as far as I got.

Really stuck with this not sure how to go about it, any help would be greatly appreciated

 

[rude man]

What about the first law of thermodynamics?

 

[navier1120]

Carnot Efficiency is n=1-(Qc/Qh), Qc is heat rejected, Qh is heat supplied
also Carnot efficiency is n=1-(Tc/Th)

Drawing a control volume around the device, (20 KW + 12 leaving) means 32 input or Qh=32, thus you have an efficiency, and then you can use this efficiency to solve for Th

So n=1-(12/32)=62.5% efficient

.625=1-(tc/th)
Tc/.375=Th thus 25/.375=66.67 C

hope that helps

 

[db725]

Thanks very much. Makes complete sense now