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

 

[DeeNos]

.


I would approach this problem by using the fundamental principles of thermodynamics. The Carnot cycle is a theoretical cycle that represents the most efficient way to convert heat into work, and it consists of two isothermal and two adiabatic processes.

First, we can use the equation for efficiency in a Carnot cycle: η = 1 - Tcold/Thot. We know that the efficiency of an ideal Carnot cycle is 1, so we can rearrange this equation to solve for the maximum cycle temperature, Thot. This gives us Thot = Tcold / (1 - η). Plugging in the given values, we get Thot = (273+25) K / (1 - 1) = 298 K.

Next, we can use the equation for heat transfer in an isothermal process: Q = nRTln(V2/V1). In this case, we are given the work (20 kW) and heat rejection (12 kW) rates, so we can solve for the heat supply, Qhot. This gives us Qhot = (20 kW + 12 kW) / ln(V2/V1). We know that in an isothermal process, the volume ratio is equal to the temperature ratio, so V2/V1 = Thot/Tcold. Plugging in the values, we get Qhot = (32 kW) / ln(298 K / 298 K) = 32 kW.

Therefore, the maximum cycle temperature is 298 K (or 25°C) and the required heat supply is 32 kW. This shows that the Carnot cycle is able to convert all of the supplied heat into work, and it also highlights the importance of temperature difference in the efficiency of a heat engine.