Carnot engine with radiative heat loss minimizing radiator area

[nineeyes]

Problem :
Consider an engine in outer space which operates on the Carnot cycle. The only way in which heat can be transferred from the engine is by radiation. The rate at which heat is radiated is proportional to the fourth power of the absolute temperature and the area of the radiating surface ($$Q_L$$ is proportional to $$A(T_L)^4$$). Show that for a given power output and a given $$T_H$$ the area of the radiator will be an minimum when $$T_L/T_H=3/4$$ .

I was guessing I need to try to show Q_L is a minimum using the given ratio. I can find the efficiency but after fooling around with it a few times in some equations I haven't come up with much, I generally have problems when few numbers are provided.

Any hints that can be provided would be great, Thanks!

 

[Dr.Brain]

OK This is simple.

First of all I would like to tell you that , if a body is at temperature T ,it radiates heat energy(E) given by:

$E=esAT_L^4<br />$

where T_L is the temperature of the engine.

Now outside temperature is T_H

Now amount of energy radiated by the engine reduces because Outside region also supplies some energy into the engine.Therefore now the net Energy radiated becomeS:

$E=esA ( T_L^4 - T_H^4)$
s in above equation is the stefan's constant.And the above equation is the Stefan's Law.

Now differentiate it to get the minima...You will get the answer.Easy isn't it?

 

[nineeyes]

Hi,
Sorry, but we have not yet encountered this equation in my class. I was wondering do I differentiate with respect to A? If I do, doesn't that just eliminate the A from the equation? I was thinking I would need to somehow solve for A , in terms of T_H and the Power Output.
Sorry if I misunderstood what you meant.
Thanks for the help.

 

[Dr.Brain]

Differentiate it w.r.t $T_H$ or $T_L$.

Do you know we can find the maxima or minima of an expression by simply differentiating it ?...The same concept we apply to the above problem.