Skip to 0 minutes and 3 seconds So this is worked example 3.5. It’s about a double-glazed window. We’re told that we have a double-glazed window for a roof light which has dimensions 1.5 by 1.2 metres, and consists of two layers of glass 5 millimetres thick, with a thermal conductivity of 0.78 watts per metre Kelvin, separated by a 20-millimetre thick space, filled with argon gas of thermal conductivity 0.016 watts per metre Kelvin. And we’re asked to determine the steady rate of heat transfer through this window when the inside air temperature is 18 degrees C and the outside temperature is 0 degrees C. And we’re told the convection heat transfer coefficients.
Skip to 1 minute and 3 seconds And we’re then asked to go on and find the temperature of the inner surface of the window, and what happens if we take one of the layers of glass away. So to start this problem, we need to draw a little picture of what’s happening. And so I’m going to represent the double-glazed window just as a series of parallel lines here. OK, so this is my argon-filled atmosphere in the middle. This is one sheet of glass, and this is the other sheet of glass. And this is going to be inside my room. And over here, this is going to be outside in the environment. And I’m going to assume that outside here, we’re going to be at temperature T1.
Skip to 1 minute and 57 seconds We’ve got at the interface here, temperature T2, at this interface T3, and at this interface T4. T5 at this interface, and then inside my room I’m going to be at temperature T6. And I’m going to mark my thicknesses on here. So I’m just going to call it tg for the moment for the glass, ta for the thickness of the argon layer, and tg for this thickness of glass here. And we’re given a heat transfer coefficient for the inside and for the outside heat transfer, and some thermal conductivities for the glass. And, perhaps unusually for the argon, you might expect that the heat transfer through here would be by convection.
Skip to 2 minutes and 51 seconds But the argon’s trapped in between the two sheets of glass, and there’ll be no movement, and so the heat transfer across here will be as a consequence of conduction. And so we end up with a thermal conductivity being relevant here. And so what we’re expecting to happen is that heat will move in this direction. So we’ll have a heat movement with temperature moving across here as we move from the warm inside, down the temperature gradient, to the outside world. And we can tackle this problem by applying the first law of thermodynamics to this unit. So if we write the first law of thermodynamics, and we apply it for unit area of glass.
Skip to 3 minutes and 46 seconds So we’re going to apply to a unit area of glass window.
Skip to 4 minutes and 2 seconds And so we can use the usual form of it. So the heat transfer in, minus the heat transfer out, plus the work terms - so the work in, minus the work out - plus the energy flow in as a consequence of any mass flow, minus the energy flow out of the consequence of any mass flow out. And all those energy flows across the system boundary will be equal to the change in the internal energy in the system, plus the change in the kinetic energy of the system, plus the change in the potential energy of the system. And because this thing is stationary, then there’ll be no change in potential energy, so this term will go to 0.
Skip to 4 minutes and 53 seconds There’ll be no change in kinetic energy, so this term here will also go to 0. The heat flows at a steady rate through here, and so we would not expect there’ll be any change in internal energy. It’s a steady state system. There’s no mass moving here at all, so this term will also go to 0. And we’re not doing any work in this system, and so this term here will also go to 0. And so what we’re left with is the heat transfer in is going to be equal to the heat transfer out, because this is the only term that we’re left with in the first law of thermodynamics.
Skip to 5 minutes and 40 seconds And so we can write down that the rate of heat transfer will be constant.
Skip to 5 minutes and 58 seconds And so we could say we can express that little q-dot will be equal to big Q dot over A. So this is the rate of heat transfer divided by the area of the window. So when we use little q-dot, this is heat transfer per unit area. And what we’re saying is that is equal to a constant. And so now we can consider the types of heat transfer that will occur at each step here. So we’re going to have a convection heat transfer here, then a conduction, another conduction heat transfer process, conduction again across this glass, and then convection out into the outside world. And so we can use the standard expressions.
Skip to 6 minutes and 44 seconds So by convection, we would expect that Q-dot over A will be equal to h naught for this process here, and it’s happening between T6 and T5, and so we’re going to have T6 minus T5. So that’s a standard expression for heat transfer by conduction. We said it’s going to be constant through this process, so we’d expect the same rate to occur at this side from this surface of the glass to the outside world. So we’re going to have another term here that’s going to be h naught into T2 minus T1. And the alert amongst you will notice that I’ve used the wrong h here. This should be the inside one for this one here, and this should be hi.
Skip to 7 minutes and 38 seconds And then by conduction, we are going to have a heat flow across the glass, the argon, and then the glass again. And so we’re going to have a standard expression for the heat transfer by conduction. So this is going to be k. Let’s do the glass first. K for g divided by tg, and the first process here’s going to be from 4 to 5, and so this is applying the standard expression for heat transfer by conduction to what’s happening across this glass here. And then we’re going to have the equivalent amount of heat moving across the argon, and so we can write a similar expression, and it’s going to go from T4 here to T3.
Skip to 8 minutes and 34 seconds And then we can do the same again for the last piece of conduction from this surface of glass, to this surface of the second sheet of glass. So now we’ll have kg divided by tg. It’s going to be T3 minus T2.
Skip to 8 minutes and 54 seconds And so I want to combine these expressions now, so I’m going to clear the board. OK, so we can move on now and make use of the fact that we’ve got a continuous temperature distribution across this system here. And so we can write that delta T, the total change in temperature from the outside to the inside– so that’s going to be T6 minus T1– is actually made up of a number of little steps here. And so we can say that this is equal to T6 minus T5, plus T5 minus T4, plus T4 minus T3, plus T3 minus T2 , plus T2 minus T1. So these are the individual temperature gaps across the system.
Skip to 9 minutes and 56 seconds And so these terms also appear in the heat conduction and convection equations that I wrote down a moment ago. And so we can write that hence, Q-dot equals - I’m going to take the A that’s normally underneath it and put it on the top over here - the total temperature difference divided by the terms from each of the heat convection equations. So we’re going to end up with two terms here for the glass because we’ve got two pieces of glass, a conduction term here for the argon, and then finally, a convection term here for the heat transfer that’s occurring at the inside of the glass here. And so now we can substitute the values we have into this expression.
Skip to 10 minutes and 52 seconds So we can write here, substituting appropriate values. And so we’re going to have that at Q-dot is equal to– so we have an area that we were told at the beginning of the question was 1.2 times 1.5, so that’s our area term, times the total temperature gradient across here, which we were told in the question was 18 degrees inside and 0 outside. And then we can divide that by the terms with the coefficients for the heat transfer.
Skip to 11 minutes and 38 seconds And so 10 for the heat transfer occurring at the outside, plus 2 times the coefficient - I’m sorry, the thickness of the glass, so that’s 10 to the minus 3, to put it into metres, divided by the coefficient for the glass, which was a 0.78. Plus the thickness of the argon, so that was 20 times 10 to the minus 3, to express it in metres, divided by the coefficient for the argon, which is 0.016. Plus, finally, 1 over h, which is 40. And so if you get your calculator out and do that calculation, then Q-dot comes in at 23.4 watts as the heat transfer rate through our double-glazed window.
Skip to 12 minutes and 33 seconds And so we’re asked in the question, then, to carry on and find the temperature of the inner surface of the window, i.e., T5 here. So I’m just going to clear a little bit of space to do that part of the problem. The second part the question, we’re asked to find the temperature on the inner surface of the window. And so we can do that– we can say– now for convection at the inner surface, and we can say Q-dot over A. And for convection here, it’s going to be at this surface here, so it’s hi. And it’s going from T6 to T5, like that. And we now know what Q-dot is, we found it in the previous part the question.
Skip to 13 minutes and 35 seconds We know what the area of our window is - we’re told hi - and we know what T6 is, that’s given to us in the question, so we can find T5. So we can rearrange this and say that T5 will be equal to T6 minus Q-dot over Ahi. And so, if we substitute some values into here, we’ll now have to do this in degrees Kelvin. So this is 291 Kelvin minus the Q-dot we found, which was 23.4 divided by the area of the window, which is 1.2 times 1.5. Times the hi value that we have, which is 10. And if you get your calculator out and do that calculation, it should come out at 289.7 Kelvin.
Skip to 14 minutes and 35 seconds And for the final part of the question, we’re asked to consider what happens if we take away one of these panes of glass, and, of course, the argon disappears, as well. So we’re just left with one pane of glass. So we can say– and for a single pane of glass. So now we’re going to have Q-dot single, and we could use the same form of equation we had for the double-glazed situation. So area of the glass times the temperature gradient, divided by– and we’ll have the heat convection terms. But now we’re only going to have one conduction term here, which will be tg over kg, and another convection term for the heat transfer to the outside world.
Skip to 15 minutes and 38 seconds So this is now simplified compared to double-glazed case. And we can substitute into here, and so we’ll have Q-dot single equal to 1.2 times 1.5 for the area. Same temperature gradient again. And then divide through and substitute values for the heat transfer coefficient, the thickness of the glass, which is 5 times 10 to the minus 3, divided by the kg value for the glass, which is 0.78, plus the heat transfer coefficient to the outside world, which is 40. And if you get your calculator out and do that, it comes out at 246.5 watts. And that’s quite dramatically different to what we got with the double-glazed system shown in the picture. It’s an order of magnitude more.
Skip to 16 minutes and 38 seconds So the effectiveness of double-glazing is huge in reducing the amount of heat transfer from 246 watts for a single pane of glass down to about 23 watts for our double-glazed unit. And so that’s why it’s so advantageous to use a double-glazed system in your house.
Double glazed window (worked example)
A double-glazed window for a roof-light is 1.5m by 1.2m and consists of two layers of glass 5mm thick of thermal conductivity 0.78 W/m.K separated by a 20mm thick space filled with Argon gas of thermal conductivity 0.016 W/m.K. Determine the steady rate of heat transfer through this window when the inside air temperature is 18°C and the outside temperature is 0°C. The convection heat transfer coefficients of the inner and outer surfaces of the window are hi =10 W/m2.K and ho =40 W/m2.K respectively.
What is the temperature of the inner surface of the window?
The double-glazed unit replaced a window made from a single layer of glass 5mm thick, what was the reduction in heat loss under corresponding conditions?
Have a go at this question and then watch the video or check your answer with the solution by reading through the attached solution at the bottom of this page.
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