# Homework Help: Thermodynamics - Thermal Conductivity + Heat Flow

1. May 11, 2012

### daleklama

1. The problem statement, all variables and given/known data

The brick wall of a building has dimensions of 4m by 10m, it is 15cm thick with a coefficient of thermal conductivity of 0.8 (Wm^-1C^-1).

(a) How much heat flows through the wall in a 12-hour period when the average inside temperature is 20 C and the average outside temperature is 5 C?

(b) Will a 2kW heater operating in the room compensate for this?

(c) A layer of insulating material, thickness 5cm, k = 0.04 (Wm^-1C^-1) is added to the inside of the wall. Calculate the rate of heat loss after the addition of this layer.

2. Relevant equations

The Law of Heat Conduction:

H = (dQ/dt) = - k A (dT/dx)

Another equation I found is the following, but I'm not sure if it's relevant here:

H = A(T2 – T1)/(summation Ri), where the summation is taken from i = 1 to i = n,
and Ri = (Li/ki).

3. The attempt at a solution

I've attemped (a) using the Law of Heat Conduction, but I think my answer seems very wrong!

I said the area A = 4m x 10m = 40m^2
k = 0.8
dx = 15cm = 0.15m.

Using equation above:

(dQ/dt) = - (0.8) (40) (15/0.15)
(dQ/dt) = -3200

12 hours is 720 mins or 43200 seconds.

dQ = -3200 (43200)
dQ = - 138240000 W, or 138240 kW.

Fairly sure it's wrong, and even if it was right, I'm not sure how to do part (b). Do I just compare my answer to (a) with the 2kw given in (b)?

And c is beyond me at the moment, sadly! :(

Any help would be greatly appreciated, thanks.

2. May 11, 2012

### Staff: Mentor

Your numbers are okay but the final units are not. Heat energy is measured in Joules (or calories). Watts are a rate of energy movement or change, J/s. Since you've accounted for the time interval the result should just be joules.
You can compare wither the total heats for the same 12hr time period or compare the heat rates (J/s) for each.
[/quote]
And c is beyond me at the moment, sadly! :(
[/QUOTE]

You'll have to read up on how a layer of insulation affects the heat flow.

3. May 11, 2012

### LawrenceC

For part c, there are a number of ways to work the problem. All the heat that passes through one wall must pass through the other in the same amount. So equate the heat fluxes with the unknown being the interface temperature. Once you solve the equation for that temperature, just evaluate the heat flow through one portion.

Or you can use the formula you have which will be written:

H = A(T2-T1)/(dX1/k1 + dX2/k2)

where the dX's are the thicknesses of each region.

Another way is to integrate the differential equation and evaluate boundary conditions.

4. May 11, 2012

### daleklama

Thank you both a lot for your help!

gneill: I'm not completely sure I understand. So the answer I calculated ( - 138,240,000) is in Joules, okay. Or - 138,240 kJ, I assume.

But as for part (b), I'm still stumped. I want to find out if a 2kJ heater is enough to compensate for this heat loss.
I don't think I understand how to do what you said (compare the total heats for the 12 hour period, or compare the J/s rate for each).

As for part (c), I have it sorted now I think, thanks LawrenceC.
I calculated 417.4, and I think the unit is Watts, because it's a rate of heat loss, and I didn't take a time interval into account?
If I wanted to know if the 2kW heater is now sufficient to cover heat loss... I'm not sure what to do here either I'm afraid.

5. May 11, 2012

### Staff: Mentor

You could even go up another prefix and use MegaJoules (MJ).
The heater has a rating of 2 kW, not kJ. So it's proving heat at a rate of 2000 Joules per second. How much heat does it deliver in the 12 hour timeframe? Compare with the total heat lost through the wall which you calculated. Or, what is the rate of heat loss through the wall per second? Compare with the heaters rating.
Yes, that is correct.
Is 417W greater or less than 2kW?

6. May 11, 2012

### daleklama

Aha! It's all becoming clear :D

(a) We lose 138,240,000 J through the wall in 12 hours.

(b) The 2kW heater gives us 86,400,000 J in 12 hours. (2000 x 43200)
This is not enough to compensate for our heat loss through the wall.

(c) With the insulating material, the rate of heat loss is now 417.4 Watts.
If we compare this to 2,000 Watts from the heater, the heater gives us much more heat than we are losing.
Thus, the heater is now sufficient to compensate for heat loss.

Thank you very much!

7. May 12, 2012

### daleklama

Er, I'm attempting a question very similar to this one, I'm a tiny bit stuck.

"Wall consists of 10cm layer of brick lined with 5cm layer of insulating material.
(a) Calculate rate of heat flow through the wall PER UNIT AREA if Temperature on one side is 20 degrees C and on the other side is 5 degrees C.
k(brick) = 0.8 W m^-1 K^-1
k(insulator) = 0.05 W m^-1 K^-1

(b) The wall has dimensions of 12 m x 5 m. Calculate total heat flowing in 12 hours."

It's very very similar, but I have a few Qs:

For (a), it doesn't give an area, but asks for 'per unit area.' All formulas I have contain 'A' so I use 1?

The formula H = (dQ/dt) = - k A (dT/dx) doesn't seem to apply here, because I have two k's and two thicknesses.

So I instead used LawrenceC's formula:

H = A(T2-T1)/(dX1/k1 + dX2/k2)

where A = 1, T2-T1 = 15, dx1/k1 is 0.1/08 and dx2/k2 is 0.05/0.05.

I got 13.333 Watts as the rate of heat loss. I'm not sure if I did this right.

But then on part (b), I'm given an area of 60m AND a time.

Do I re-do what I did above again, but subbing in the new area? And then also multiply in the time? Doesn't seem to make sense :(

Thank you.

8. May 12, 2012

### LawrenceC

"For (a), it doesn't give an area, but asks for 'per unit area.' All formulas I have contain 'A' so I use 1?"

Correct

"The formula H = (dQ/dt) = - k A (dT/dx) doesn't seem to apply here, because I have two k's and two thicknesses."

Above will work if you equate two fluxes and solve for interface temperature, then evaluate flux.

k1(Ti-Tout1)/thick1 = k2(Tout2-Ti)/thick2, where Tout's are respective wall surface temperatures. Solve for Ti, interface temperature, then evaluate either side of equation. You will get same answer as formula for H gives.

"Do I re-do what I did above again, but subbing in the new area? And then also multiply in the time? Doesn't seem to make sense"

Yes, sub in new area in place of 1 m^2. That gives rate of heat loss per hour. Multiply by time to get total energy loss in specified time.