# Thermal conductivity and resistance

1. Jul 16, 2011

### justinlj

1. The problem statement, all variables and given/known data
i need to find the thickness of the insulating material that is needed to block out the heat that is coming into the room.

the area of the affected wall is 3.3m2 the temp in first room is 18 degree celsius while the 2nd room is 45 degree celsius.

the thermal conductivity of the material is 0.0698 W/mK

2. Relevant equations

thermal conductivity, lambda= W/mK

thermal transmittance, U= W/m2K

thermal resistance, R=1/U=thickness/lambda

3. The attempt at a solution
1.find K, the temp difference between the 2 rooms
2.square both sides of equation 1, since im only given area. Find the value of W.
3. subsitute the value of W into equation 2 to find U.
4. substitute the value of U into the third equation to obtain the thickness.

my teacher said the way i dealt with the area was incorrect. Is there any other way to obtain the amount of heat transfer W when im only given the temp difference and the area of the wall?

2. Jul 16, 2011

### gsal

I am an electrical kind of guy and typically solve thermal problems like this as an analogy to electrical network...I use Ohm's Law V=IR and turn it into T=QR

I typically refer to conductivity with the greek letter kappa, I will use ' k ', here.

I typically refer to resistivity with the greek letter rho, I will use ' r ' here. And so r = 1/k

I refer to a total resistance value of a given part with dimensions with the letter R, and so

R = r L/ A = L / kA

L is the length (thickness of the wall)
A is the cross-sectional area (of the wall between rooms)

In your case, these condcutivities and resistivities and resistance are thermal ones.

If Q is the amount of watts flowing through the wall...we can simply apply ' Ohm's Law ' (remember, I am electrical)

T2 - T1 = Q R
T2 - T1 = Q ( L / kA )

so

Q = (T2 - T1) k A / L

And so, the larger the L the less the heat...but I doubt very much that you will ever block all the heat...

so, when
L = (T2 - T1) k A
you have Q=1 watts going through

when L is 10 times larger, you will have Q = 0.1 watts

how small a value for Q does your teacher desire to consider " blocking " the heat?

3. Jul 16, 2011

### justinlj

hmm.. so i would actually need to know the maximum amount of heat allowed to flow through? i dont really have that data though. without that i cant solve the question rite?

what if i assume an ideal case where all the heat is blocked and Q=0? can i still solve this question?

4. Jul 16, 2011

### gsal

You mean, like L=infinity?

I think there is something missing in the statement of the problem.

You see, if the temperatures of 45C and 18C are not allowed to change for the purposes of this problem, but heat is allowed to flow from one room to the other one, then, clearly, the 45C room is being continuously heated somehow for it to remain at 45C...and the one at 18C is being cooled, somehow, for it to no heat up when the heat comes through the wall

On the other hand, if both rooms where totally insulated from everything else except between each other...then, eventually, the temperatures would equalized and only then would the flow of heat from one room to the other would stop.

5. Jul 16, 2011

### justinlj

in that case i would be juggling with too many unknown variables right?

how do u suggest i make the question solvable? will it be ok if i set these conditions?
1. the max Q allowed is 20W
2. there is a constant heat source in the room at 45C
3. the max allowed temp rise in 18C is around 2-5C

6. Jul 16, 2011

### gsal

I say Option 1

Meaning, suppose the 45C second " room " is actually the outdoors being heated by the sun and the 18C room is an actual room with an A/C in it that keeps cooling it down...in this case, the maximum heat (Q) allowed would be whatever the A/C can handle...anymore and the room would start warming up

For example, let's supposed that the 18C room is a room that is being cooled down by a small window A/C equipment that consumes 500 watts...assuming that the A/C has 100% efficiency, it means that all the energy the A/C uses is used to suck 500 watts of heat out of the air in the room...and, hence, the room can keep taking in 500 watts through the wall ...

so
L = (T2-T1) k A / Q
L = (45 - 18) x 0.0698 x 3.3 / 500
L = 0.012
and so, a 1.2 centimeter insulation in the wall would do...

Last edited: Jul 16, 2011