# Heat Transfer Problem Conduction and Convection

1. Nov 9, 2009

### charan1

1. The problem statement, all variables and given/known data
A tent being considered for cold-weather conditions has insulation on the top and sides with a conductivity, k = 0.040 W/(m*K) and a thickness, l = 0.50 cm. The tent (and surrounding medium) has the following conditions: a heater (inside the tent) producing heat at a rate of 1500 W, the total surface area of the tent is 33 m2, an average heat transfer coefficient between the inside air and inside surface of hi = 7 W/(m2*K), an average heat transfer coefficient between the outside air and outside surface of ho = 10 W/(m2*K), and an outside temperature of tao = 5*C. Determine the tent temperature, tai, with the insulated tent. Also find the inside surface temperature of the tent insulation material, tsi, and the outside surface temperature, tso.

Thermal Conductivity of tent material (k) = .040 W/m*k
Surface Area of Tent (A) = 33m2
Outside air Temp (tao) = 5*C + 273.15 = 278.15K
Convective heat transfer coefficient inside tent(hi) = 7 W/(m2*K)
Convective heat transfer coefficient outside tent(ho) = 10W/(m2*K)
q inside tent = 1500W
Tent Thickness ($$\Delta$$x) = .005m
Inside air Temp (tai) = ?
Inside Tent Surface Temp tsi = ?
Outside Tent Surface Temp tso = ?

2. Relevant equations
Convection:
q=h*A*$$\Delta$$T

Conduction:
q=-k*A*$$\Delta$$T / $$\Delta$$x

3. The attempt at a solution

For Outside Air to outside tent Surface (Which is convection) so $$\Delta$$T = tso - tao :
q=(10W/m2*K)(33m2)(tso - 278.15K)

For the Heat transfer between the inner and outer surface of the tent (Which is Conduction):
q= -(.04W/m*k)(33m2)(tso - tsi) / .005m

For Heat transfer between the inside air and inner tent surface (which is convection) so $$\Delta$$T = tsi - tai :
1500W=(7W/m2*k)(33m2)($$\Delta$$T)
$$\Delta$$T=6.4935k
6.4935k=tsi - tai

I got this far, I know I have to set some equations equal to each other but every time I attempted that I end up having about 2-3 unknowns in one equation...please point me in the right direction to solve this.
Thank you

2. Nov 9, 2009

### Mapes

If you assume steady state, can you say anything about all three heat flux values?

3. Nov 9, 2009

### charan1

I'm not exactly sure what that is...we have never gone over anything that has to do with heat flux this is a basic Engineering Sciences 101 homework. We only covered convection and conduction and have these formulas.

4. Nov 9, 2009

### Mapes

The heat flux is q, and it's measured in J/s, or W... Know what I'm getting at?

5. Nov 10, 2009

### charan1

Oh I see so, delta U - Work = Q

Delta U would be the temperature change? And W is equal to 0? because there is no mechanical energy?

What would I set Delta U equal to with the information I have?

6. Nov 10, 2009

### Mapes

Not quite... the energy of the tent doesn't change if it's at steady state. More like the heater puts out 1500 W, so this is the heat flux through all the layers. This leaves you with an equation with only one unknown.

Last edited: Nov 10, 2009
7. Nov 10, 2009

### charan1

What is the definition of steady state? Since that explains that the Q is 1500Watts throughout...

8. Nov 10, 2009

### Mapes

Steady state exists when none of the system's state variables (like temperature) change with time.

9. Nov 11, 2009

### charan1

For Outside Air to outside tent Surface (Which is convection) so $$\Delta$$T = tso - tao :
1500W=(10W/m2*K)(33m2)(tso - 5*C)
tso=9.595*C

For the Heat transfer between the inner and outer surface of the tent (Which is Conduction):
1500W= -(.04W/m*k)(33m2)(tsi - 9.595*C) / .005m
tsi=3.913*C

For Heat transfer between the inside air and inner tent surface (which is convection) so $$\Delta$$T = tsi - tai :
1500W=(7W/m2*k)(33m2)(3.913*C-tai)
tai=-2.58*C

these are the numbers I got, I'm not sure they are correct because it doesn't really make sense to me.....

I think the problem is something with the $$\Delta$$T's.....The way I subtracted them....I was unsure about which value actually goes first in the delta....for instance the surface temp or the air temp...

10. Nov 11, 2009

### Mapes

Well, let's look at it this way: wouldn't you expect the temperature to get colder as you transition from the heater to the outside? Does that help get the signs right?

11. Nov 11, 2009

### charan1

For Outside Air to outside tent Surface (Which is convection) so $$\Delta$$T = tso - tao :
1500W=(10W/m2*K)(33m2)(tso - 5*C)
tso=9.595*C

For the heat transfer between the inner and outer surface of the tent (Which is Conduction):
1500W= -(.04W/m*k)(33m2)(9.595*C - tsi) / .005m
tsi=15.27*C

For Heat transfer between the inside air and inner tent surface (which is convection) so $$\Delta$$ T = tsi - tai :
1500W=(7W/m2*k)(33m2)(tai - 15.27*C)
tai=21.77*C

This made more sense...what do you think?

12. Nov 11, 2009

### Mapes

Much more sense!

13. Nov 11, 2009

Thank you!