Thermodynamics: Room Air Conditioner

[WlfordBrimley]

[SOLVED] Thermodynamics: Room Air Conditioner

Homework Statement

A room air conditioner acts as a Carnot cycle refrigerator between an outside temperature T_h and a room at a lower temperature T_l. The room gains heat from the outdoors at a rate A(T_h - T_l); this heat is removed by the air conditioner. The power supplied to the cooling unit is P. Show that the steady state temperature of the room is:
T_l = (T_h + P/(2A) ) - SQRT[(T_h + P/(2A))^2 - T_h ^2]

Homework Equations

Carnot coefficient of refrigerator performance: gamma_C := ( Q_l / W ) = T_l / ( T_h - T_l )
Q_l = (T_l / T_h) Q_h
W = Q_h - Q_l = Q_l * (T_h - T_l)/ T_l

The Attempt at a Solution

At equilibrium, the flow of heat in = flow of heat out, so
P*gamma_C = A (T_h - T_l)
After a lot of algebra (done with maple), I get:
T_l = (P/2A _ T_h) + sqrt(P^2 + 4PAT_h) / 2A
...which is not the correct answer. What did I do wrong?

 

[Jhero]

Your approach is correct, but your algebra may have some errors. Here is the correct solution:

At equilibrium, the rate of heat transfer into the room (A(T_h-T_l)) must be equal to the rate of heat transfer out of the room (P*gamma_C). Therefore, we have:

A(T_h-T_l) = P*gamma_C

Using the equation for Carnot coefficient of refrigerator performance (gamma_C = T_l/(T_h-T_l)), we can rewrite this as:

A(T_h-T_l) = P*T_l/(T_h-T_l)

Simplifying and rearranging, we get:

T_l^2 - T_h*T_l + P*T_h = 0

Solving for T_l using the quadratic formula, we get:

T_l = (T_h + P/2A) +/- sqrt((T_h+P/2A)^2 - T_h^2)

Since we are looking for the steady state temperature, we can ignore the negative solution and only consider the positive solution:

T_l = (T_h + P/2A) + sqrt((T_h+P/2A)^2 - T_h^2)

This is the correct answer for the steady state temperature of the room. I hope this helps clarify any confusion. Let me know if you have any further questions.