Question on heat transfer
Hi
I am stuck on these questions, I know the answers but can't seem to get to them.
Help please?
I am stuck on these questions, I know the answers but can't seem to get to them.
Help please?
Original post by SomeGuy21
Hi
I am stuck on these questions, I know the answers but can't seem to get to them.
Help please?
I am stuck on these questions, I know the answers but can't seem to get to them.
Help please?
Are you sure that the answer you have for the second one is correct? It doesn't even have the right units (given in meters, while the question is about power). The first answers are fine though, so if you want I can explain them.
Original post by lerjj
Are you sure that the answer you have for the second one is correct? It doesn't even have the right units (given in meters, while the question is about power). The first answers are fine though, so if you want I can explain them.
If you wouldn't mind.
EDIT:
Sorry about delayed reply, was kinda busy.
(edited 9 years ago)
Original post by SomeGuy21
If you wouldn't mind.
EDIT:
Sorry about delayed reply, was kinda busy.
EDIT:
Sorry about delayed reply, was kinda busy.
Ok, if you know the formula then you just need to plug it in- otherwise think about what four things will affect the rate of transfer of heat. One of them is the u-value, or the coefficient in the question. What are the other three and how do they affect it?
Original post by lerjj
Ok, if you know the formula then you just need to plug it in- otherwise think about what four things will affect the rate of transfer of heat. One of them is the u-value, or the coefficient in the question. What are the other three and how do they affect it?
Just realised I know the answer for 1, didn't realise when I did that.
Was meant to be the answer for 2, couldn't answer the plane question...
Original post by lerjj
Ok, if you know the formula then you just need to plug it in- otherwise think about what four things will affect the rate of transfer of heat. One of them is the u-value, or the coefficient in the question. What are the other three and how do they affect it?
Also, I keep getting 60,000 for the plane because I keep using HcA(T1-T2) formula. =l
Original post by SomeGuy21
Just realised I know the answer for 1, didn't realise when I did that.
Was meant to be the answer for 2, couldn't answer the plane question...
Was meant to be the answer for 2, couldn't answer the plane question...
Ok, so since you did question 1 I'm going to assume that you used an equation that looked something like:
With the 2nd one, you should just have to multiply the co-efficient you're given by the contact area and by the temperature gradient.
The reason for the first equation (hopefully you know this, but anyway) is that more heat is transferred if the conductor is thicker (similar to how resistance is inversely proportional to area in electronics) and that it will take longer if the heat transfer is taking place across a large area, while it will be faster with a steep temperature gradient. (In fact, very similar to how to maximise diffusion)
The second one removes the distance part as they are in contact already, and just replaces cross-sectional area with contact area.
You ought to be able to guess at the formulas from the units for the co-efficients. If you read the first one as "1 watt per meter, per degree Kelvin" and the second one as "1 Watt per meter squared, per degree Kelvin" then it becomes clearer what the equation ought to be.
Original post by SomeGuy21
Also, I keep getting 60,000 for the plane because I keep using HcA(T1-T2) formula. =l
I think that's correct. The answer given certainly appears wrong as it hasn't got units for power.
Someone more knowledgeable than myself may need to tell you whether there is a trick here, just because it seems relatively simple. But I'm fairly certain 60 kW is the right answer.
Your equations should be much simpler if you use dimensional analysis. Think about what you have to do to cancel out the extra units from your equations to get W on it's own for power, and it should be simple.
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