# Heat Transfer Issue

1. Jul 28, 2011

### math111

I am getting crazy values so I need help.

I want to find out how many watts or even Kw I need to heat up a space.
Here are the givens:
Sizes: Width:1.1176m, Length:1.6764m, Height:1.8288m
Temp-outside: -9.44C
Temp-inside:12.78C
Material: Steel[thinckness(.00635m), thermal K=43]

I just want to calculate 1 face(Front--?widthxheight) of the wall because that will help me on the other faces

I did Q=(k*Area*Temp Difference)/thk
Q=(43*2.04*22.22)/.00635
Q=306950.5watts or 307kW.

Does this mean I will need at the minimum at 307kW heater just based on one wall and that mean it will be even greater by alot!!!!!!That seems too high or is this correct and I am not giving the proper information to solve the problem at hand??

Steel has a very high K so it might make sense. I am concerned about why making the mat'l thinnner would need more watts to make a higher temperature difference ?

Last edited: Jul 28, 2011
2. Jul 28, 2011

### edgepflow

You need to include the natural convection heat transfer coefficients on the inside and outside of the walls. These are much more significant than the conduction term you used.

3. Jul 28, 2011

### timthereaper

Well, that's not really that high considering your problem. You've got a big, 6mm thin steel box that's -10 Celsius on the outside and 12 Celsius inside. There's not that much thermal mass and steel is a fairly good conductor of heat. If you thickened up the walls with an insulator, you would severely decrease the loss of heat. However, in any case, I do agree with edgepflow that convection (natural and even forced if there's wind) is a major source of heat loss, so reducing that is a primary concern.

4. Jul 28, 2011

### Zeekar

5. Jul 28, 2011

### Zeekar

You have to imagine it with heat resistivity. The thicker the wall/plate is the higher resistivity is and less heat will pass trough.

6. Jul 28, 2011

### AlephZero

Post #2 is right. The basic mistake in your calculations is assuming the temperature of the inside and outside face of the METAL. are at 12.78C and -9.44C. The correct assumption is that the inside and outside AIR temperatures are 12.78C and -9.44C.

300kW can't possibly be right. Find out what is the heat output of your home heating system. Unless you live in a very large house, it will be way smaller than that, to heat the entire house not just a box smaller than one room.

You can made the heat input as small as you like by insulating the box. If you covered it in some polystyrene sheet, or even plywood, you would probably need more like 30W of heating, not 300 kW.

7. Aug 3, 2011

### math111

I red-did the problem and got around the 500-600 watts range. So I think I did something right. I am learnig this stuff at a snails pace but don't mind.

8. Aug 3, 2011

### math111

Right now I trying to figure out how to connect power with temperature.

1. Does Watts, Kw, Btu/Hr have any relation to temperature of let's say a heater??

EXAMPLE:if I have a 1kw heater compared to a 2kw heater what is the difference in temperature it will output?

I know the 2KW will output more heat but at a higher temp? or is that not true and just outputs more heat of the same temperature as the 1kw heater.