# Thermal conductivity through a concrete wall

1. ### Xyius

500
Hey everyone!
Got an interesting problem here, I am thinking I did it correctly but I wanted to see your opinions. Here is the problem..

A 0.2-m-thick plane wall is constructed of concrete. At steady state, the energy transfer by conduction through the wall is 0.15 kW/m2. The inside temperature of the wall is 24oC.
a. If the temperature distribution through the wall is linear, what is the temperature on the outside surface of the wall?
b. A layer of insulation is added to the inside of the wall. Assuming the same temperatures at the surfaces, determine the thickness of the insulation that will reduce the energy transfer through the composite wall by a factor of 2.

The insulation has a thermal conductivity of 0.08 W/m.K. For concrete , k = 1.4 W/m.K.

My Solution
Since it is stated that we can assume a linear relationship, the heat transfer through the wall will be proportional to the temperature differential between the inside and outside of the wall and the thickness of the wall. (Which will be denoted L) Therefore..

$$\frac{dQ}{dt} = kL(T_{outside}-T_{inside})$$

Plugging in the values we obtain..(Using Kelvin for temperature)

$$\frac{dQ}{dt} = kL(T_{outside}-T_{inside}) \rightarrow 0.15 = (1.4)(0.2)(T_{outside}-299.15) \rightarrow T_{outside} = 299.7K$$

For the second part, if we look at the wall on a cross section (Going to do my best to make a wall out of text here...)

---L-2x
| |---| |
| |---| |
| |---| |
| |---| |
x ----- x

Therefore the above equation for heat transfer turns into..

$$\frac{dQ}{dt} = 2k_{concrete}L(T_{outside}-T_{inside})+k_{insulation}(L-2x)(T_{outside}-T_{inside})$$

Plugging in all the values I get an x value of 4.6cm which means the thickness of the insulation is L-2x or 10.9cm.

This seems reasonable to me, do you guys think my logic is correct?

Thanks!

2. ### Xyius

500
I actually believe this is wrong looking at it for a second time, ill post my second attemp in a bit

EDIT:

I am trying it again with..

dq/dt=k(To-Ti)/L Since it matches the units for the heat transfer, but it still isn't working out for me..

Last edited: Feb 2, 2011
3. ### Xyius

500
Okay I seriously need help with this. None of the numbers I am getting make sense. After talking to my professor, I have gathered these facts.

The heat conductivity by conduction though a medium is..

$$Q=\frac{kA \Delta T}{\Delta x}$$

And the conductivity by convection is.. (don't know if I need this)

$$Q=hA(T_{surroundings}-T_{surface})$$

The heat transfer rate I was given is in W/m^2 so it looks like "A" is divided out in the above expressions.

According to my professor, heat transfer analysis can be related to electical voltage analysis. So the equation...
$$I = \frac{\Delta V}{R}$$
is analogous to...
$$Q = \frac{\Delta T}{R_{t}}$$

Where,

$$R_{t} = \frac{\Delta x}{k}$$ (Just the formula for heat transfer though conduction above.)

The next part of the problem is asking what would the thickness of an insulation with a thermal conductivity of k=0.08 to make the heat transfer through the composite wall reduced by a factor of 2.

I seriously need help on this one, I haven't been this stuck on a problem in a long time. :(

### Staff: Mentor

You're pretty close to solving it. The electrical circuit analogy is a good one.

First revisit part (a). You don't need to convert temperatures to absolute here, since you're really dealing with temperature differences. Suppose I is the "heat current" per square meter of wall, I = 0.15kW/m2. You've already determined that the equivalent heat resistance of the concrete wall is given by Rc = ∆x/kc. So the "potential difference" (temperature difference) across the wall is

$$\Delta T = I R_c$$

Now, the problem statement didn't specify the direction that the heat was passing through the wall. So assume that its moving inward and that the outside temperature is higher than inside (you can go the other way by a change of sign of the current, if you wish). What's the outside temperature?

5. ### Xyius

500
I am getting 45.4 degrees C. I got this before but I didn't think it was right. That is pretty hot. I am stuck on the second part.

It is saying that a layer of insulation is being installed inside the wall and is asking how thick the insulation has to be to decrease the heat flow by a factor of 2 through the WHOLE wall.

My guess if I have to somehow include not only the insulation but the thickness of the concrete slabs on either side of the insulation. But nothing I do is working out. :\

### Staff: Mentor

If you assume that the heat is flowing outward, then the temp outside would be 21.4 degrees colder than inside. Maybe that's what was intended, maybe it wasn't. They should have specified. Anyways, it won't affect the results.

Just assume that a layer of insulation is being applied to one side of the wall. You've got the thermal resistance for the wall, now what should the thermal resistance be for the insulation in order to achieve the goal?