Steady Temp of a heater in a box (Thermodynamics)

[cdanderson04]

Hey all:

I am taking a Thermodynamics course and we are studying work and heat through surfaces (i.e. conduction, convection, and radiation). I was given an assignment and I am having trouble arriving at the correct answer. I have scanned the question and the works I have so far. I hope someone can help me with this.

The question is:

And here are my workings so far:

I have solved for all of my resistances and I am pretty sure they are correct. I
just don't know what to do with them after I have them.

I know that:
Q_TOTAL = Q_SIDES + Q_TOP + Q_BOTTOM = 200W

I am just not sure how to extract the steady temperature from this.

The answer in the back of the book is 45.6 degrees C. Some other people got 46.7 degrees, but I don't think they did it right?

Anyways, any help would be greatly appreciated! Thanks!

 

[rock.freak667]

You can find the the total thermal resistance R_total.

Then you should know that

R_{total}= \frac{T-T_{\infty}}{\dot{q}}

 

[cdanderson04]

How would you go about finding R_TOTAL? Would you just add up all the Rs and also, would you multiply the R_SIDES by 4 since there are 4 sides? Thanks!

 

[rock.freak667]

I believe, they'd all be in series, so should be able to just add them to get R_total.

 

[cdanderson04]

Thanks again. But I tried that and I get a different temperature for the bottom and the sides/top since the temperatures are different. All I know is that Q_TOTAL = Q_SIDES + Q_TOP + Q_BOTTOM = 200W. I know there is a temperature in there somewhere but I don't know hoe to extract it. Sorry if I am not explaining myself correctly.

 

[rock.freak667]

Well you could just substitute Q for each of the sides and then get the temperature T.

(T-T_{bottom ambient})/q_bottom=R_bottom.

Sub for q_bottom in your equation and do the same for the other 'q's.

 

[rkmurtyp]

There appears some superfluous data given - the thickness of steel plates plexiglass and soft wouod - that is not needed to solve the problem. (This was done perhaps to see if the students use all data given whether useful or not)

From the dimensions you can observe that energy dissipation rates are given for the side and top, and, the inverse of that is given for the bottom - as resistance. Therefore, for the bottom you take the inverse of the resistance to get the conductance - the dissipation rate.
Multiply the dissipation rates by the respective areas of the sides of the box and obtain the time rates of dissipation (energy loss per unit time per degree K). Multiply these rates by the temperature drops for each side i.e. by (T-283) for the top and sides and (T-283) for the bottom to get the time dissipation rates. T is the steady state temperature in degrees K, inside the box. Add the dissipation rates for all the 6 sides of the box and equate it to 200 ( the time rate of energy input). you get one equation in one unknown = the T. Solve it and get T in degrees K. Subtract 273 from that to get degrees C. I get 30 degrees C as the answer.