Fourier’s Law Of Heat Conduction (Window Pane)

[jisbon]

Homework Statement

The room has double glazing windows. Each window has a dimension of 100cmx30cm and is constructed by sandwiching a 3mm air gap between the two glass panels. Calculate the power required to maintain the temperature of the room at 27 degrees.

Relevant Equations

q = Q/A = -kdT/dx

In this case with a presence of the airgap, what should I do with the equation that is provided to be? Must the temperature gradient be caculated spearately (glass+air+glass) ? My tutor provided the hint that the heat flux should be constant for the windows, so in this case should I just omit the air? If so, why?
Thanks.

 

[scottdave]

No, the air plays a role. If you are familiar with electric circuits, think of it as the 2 window panes as medium-sized resistors and the air gap is a very large resistor. These are all in series. The same current flows through all 3 components. So the heat flow from the inside of the room to the outside world will be the same as the heat flow through the first pane, which is the same as the flow through the air and the outer pane, once in steady-state.

 

[jisbon]

So if I count in the air too, how would I exactly calculate kdT/dx for the 3 mediums? Do I calculate one by one?
For example,
Glass 1- (1.3)27/3mm?
Air- (1.3)27/3mm?
Glass 2- (1.3)27/3mm?

 

[scottdave]

What can you say about the temperature differential in the first glass compared to the 2nd glass, and why?

And which constant should you be using for air? Also, look at the dimensions (units) in the constant.

 

[scottdave]

Also, what is the 27 degree difference across? How much does each element have across it?

 

[scottdave]

See if you can sketch a graph of the temperature from left to right as you cross each medium.

 

[nasu]

You will need to know the temperature outside. If you have 27 degrees outside as well the power is zero. So the answer depends on outside temperature.

 

[Chestermiller]

You will need to know the temperature outside. If you have 27 degrees outside as well the power is zero. So the answer depends on outside temperature.

The diagram shows 0 C outside.

 

[jisbon]

Also, what is the 27 degree difference across? How much does each element have across it?

Is it okay to say each has 9 degrees ?

What can you say about the temperature differential in the first glass compared to the 2nd glass, and why?

And which constant should you be using for air? Also, look at the dimensions (units) in the constant.

The temperature difference is 18 degrees? Not exactly sure. I should be using the constant given in the table for air, 0.025?

 

[Chestermiller]

No. Using 9 degrees is incorrect. Let the two unknown interface temperatures be T1 and T2. In terms of these temperatures, please write down the heat conduction equation for each of the three layers separately.

 

[jisbon]

Alright, let me try this:
For:
Glass 1- (1.3)27-T1/3mm?
Air- (0.025)T1-T2/3mm?
Glass 2- (1.3)T2-0/3mm?

 

[scottdave]

Alright, let me try this:
For:
Glass 1- (1.3)27-T1/3mm?
Air- (0.025)T1-T2/3mm?
Glass 2- (1.3)T2-0/3mm?

Make sure you are using the right units. And of course mind the parentheses.

 

[Chestermiller]

Alright, let me try this:
For:
Glass 1- (1.3)27-T1/3mm?
Air- (0.025)T1-T2/3mm?
Glass 2- (1.3)T2-0/3mm?

This is good. I'm going to re-write your equations a little better. First of all, they're not equations. They are just expressions. Each expression is equal to the heat flux Q through the layers. So here is my version of what you wrote:

$$Q=\frac{k_1}{\delta_1}(27-T_1)$$
$$Q=\frac{k_2}{\delta_2}(T_1-T_2)$$
$$Q=\frac{k_3}{\delta_3}(T_2-0)$$

If this is OK with you, then please next solve each of these equations algebraically for each of the temperature differences in terms of Q, k, and $\delta$ for each layer. What do you get?

 

[jisbon]

Hi there, sorry for the late reply.
I've solved these equations and got T2 to be 50/9K,which is theoretically impossible. Any ideas why?

 

[Chestermiller]

Hi there, sorry for the late reply.
I've solved these equations and got T2 to be 50/9K,which is theoretically impossible. Any ideas why?

Show us what you did.

 

[jisbon]

Show us what you did.

$Q=1.3\left( \dfrac {300-T_{1}}{3\times 10^{-3}}\right) = 0.025\left( \dfrac {T_{1}-T_{2}}{3\times 10^{-3}}\right) = 1.3\left( \dfrac {T_{2}}{3\times 10^{-3}}\right)$
Solving these equations,
$T_{1}=\dfrac {T_{2}+15600}{53}$
$T_{1}=50/9K$

 

[Chestermiller]

$Q=1.3\left( \dfrac {300-T_{1}}{3\times 10^{-3}}\right) = 0.025\left( \dfrac {T_{1}-T_{2}}{3\times 10^{-3}}\right) = 1.3\left( \dfrac {T_{2}}{3\times 10^{-3}}\right)$

The third equality should be T2-273, or the first equality should be 27-T!

 

[jisbon]

The third equality should be T2-273, or the first equality should be 27-T!

But seriously, thanks for the help. Careless :/
Found T2 to be 0.5 degrees celsius and T1 to be 26.5 degrees