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# Heat Transfer Through Material

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## Heat Transfer Through Material

(OP)
Hi All,

Trying to get my head around some heat transfer calculations at the moment so I'm after a sanity check please on the daily heat loss in kW experienced through the following:

Material area: 28m2
U Value: 3.1
R Value: 0.32
Temp Dif: 15C

The info I have doesn't give the units, but given the U Value and that the material is sold in metric I'm guessing it's W/m2/K - BTU/ft2/F seems like it would be incredibly excessive. Heat transfer is from water to air.

I'm coming up with around 31kW/day.

The second part I'm struggling to get around even more is if there is a second layer (U=6.2, R=0.16) and if there is any difference between the order of the layers...

The scenario is a swimming pool cover.

Materials and heat transfer isn't really my thing so any help explaining it would be really appreciated!

### RE: Heat Transfer Through Material

Your units don't make sense. You specified a constant temperature difference, so all you should have is a constant power. The 24-hr quantity should in joules, i.e., watts * time = joules

What does your calculation look like? I don't get what you got

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

### RE: Heat Transfer Through Material

(OP)
I used Ht = U * A * dt and multiplied by 24 hours trying to calculate the heat input required to account for the losses.

So working towards "this much reduction in heat loss = this much reduction in heating cost".

As for the dt I was expecting that it could average out over time. I'm not expecting 100% accuracy particularly due to other factors that can't be accounted for.

### RE: Heat Transfer Through Material

Again, your units are not correct. What was your raw Ht? You have to multiply by 86400 seconds to get joules of energy. But since you should have already calculated Ht in watts, this extra calculation doesn't make sense

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

### RE: Heat Transfer Through Material

(OP)
Ht was 1300 with the formula used.

### RE: Heat Transfer Through Material

OK, so you already have the heat loss that you need to make up. However, R-value is typically used for things like home insulation, so there's typically air spaces on either side of the insulation, which requires calculation of natural or forced convection heat flow. Your calculated heat loss is only appropriate if both sides of the insulation are held to maintain a 15C delta

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

### RE: Heat Transfer Through Material

(OP)
So heat energy required over 24 hours to cover the losses based on what's given IS 31kW? 1300 * 24 = 31,000

I make it that the 1300 is W/hr - or is that wrong?

This would go into heating cost calculation of 31kW*(£/kW / 0.8) = daily cost to cover losses ... The .8 accounting for efficiency losses within the boiler and heat exchanger. I would then take this and compare to the existing heat retention solution's losses to give an approximate saving (which would forego the need for assumptions and calculations about losses through other surfaces and methods of loss not being accounted for in an attempt to work out the ACTUAL heating cost).

Checking other dt values it looks like it averages out if there is a temperature swing over time, so picking an approximate average should get close, right? Not looking for accuracy to the level of building regs or architecture or whatever.

### RE: Heat Transfer Through Material

NO. 1300 W PERIOD. W/hr is nonsensical, and an unnecessary calculation. You started with 1300 W in your first expression, then you multiplied 24 hr.

ENERGY has units of JOULES.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

### RE: Heat Transfer Through Material

(OP)
Either way, 31.2 is the correct value to plug into the cost calculation considering gas usage is measured in kW, is it not? You said at the top that you "don't get what I get", but after that haven't debated this figure and it's still the only one I get:

i.e. 1300W = 1300J/s
1300 * 86400 = 112,320,000J/day
Convert to kWh (1.3) and * 24 is 31.2/day

or more simply

1300W * 24 = 31.2kWh/day (so although not a unit, W/hr is the reality of the Ht value in this calculation)

Therefore, assuming 80% efficiency of the heating system and a kWh price of 5p, the daily cost of replacing the heat lost through this particular cover is:

(£0.05 / 0.8) * 31.2 = £1.95

If I'm way off base and still completely missing something then please just lay it out, because I'm sorry but in that case I'm missing the broader context of your statements. If I'm right, then I feel this has been more frustrating for both of us than it needed to be. However, if I am right, then your replies have helped me understand the "underneath" of the calculation so thank you.

### RE: Heat Transfer Through Material

(OP)
Yay for me, but is the easy answer "yes, 31.2kWh/day is correct"?

### RE: Heat Transfer Through Material

Please think about what you are saying. What could kW/day possibly mean to you? A kW is an unit of instantaneous power. Power is energy used per unit of time. One Watt is a unit of power equal to one joule per second, where a Joule is equal to the work done by one newton of force moving one meter.

If you do not understand the basic conceptual difference between power and energy, you will not progress further in your understanding of science. It is apparent that your mind is focusing on the numerical part of the answer without understanding the concepts. The units(Watts, Joules, meters) are part of the numbers and they go through the same operations as the numbers. This is called dimensional analysis, and it is a technique that is essential to avoiding the errors that you have made in trying to solve this problem.

### RE: Heat Transfer Through Material

(OP)
I hold my hands up that I screwed the units early on and would like to point out that I'd already made clear that I was (now) aware of the relationship W=J/s - note also that my final calculations included kWh/day.

We've done that bit.

So NOW I'm asking if the numbers are correct, because initially I was told my result was wrong but I haven't been told the same since, despite still sitting on the same figure having gone through everything else (and, indeed, come to understand it better)...

As for "What could kW/day possibly mean to you?" I would use common sense and contextual reasoning to infer kWh, while issuing polite caution to be more careful not to omit the "h". I'm only focussed on the values so much, because I've now been taught my lesson regarding the units and done more research about those relationships and concepts and would like to further my understanding of this particular scenario and formula by knowing if the numbers are right. As I mentioned, this is a new area for me (dealing more often with fluid flow, volume, velocity, filtration rate, and chemicals) so it is of importance to know that I'm getting the numbers right as well as the units and calculations.

I think I am because my initial method bears the same result as working it out with Joules, but it hasn't been explicitly confirmed and without that I can't be confident. Likewise, if there are still errors they aren't being pointed out.

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