# Homework Help: Thermodynamics Insulation Problem.

1. Apr 15, 2007

### tsMore

1. The problem statement, all variables and given/known data

Hi guys, thanks in advance! I am trying to wrap my head around some insulation problem I have been given. I think I got it but just want to be sure, this stuff still confuses me. Did do so well on the midterm!

The problem about a joe's heated outdoor steam house, which has to be at 60 degrees celcius, in the middle of the winter where it is -10 degrees outside. The house has a surface of 10 meters2, and has some insulation (0.05 meters thick thermal conductivity 0.040 W/mC). First we have to compute the rate of heat loss.

Then we know that Joe's heating bill is too high, and he wants to cut it in 1/2, but applying another layer of insulation on top of the existing one, but he doens't know how thick of a layer he needs. The new insulation has a thermal conductity of 0.01 W/mC.

2. Relevant equations

I hope I only need that Q/t = A x K (T_hot - T_cold) / thickness

or Q/t = A x K (T_hot - T_cold) / Sum(R)

and R = thickness/K

where K = thermal conductivity.

3. The attempt at a solution

Ok, with the single insulation layer

Q/t = A x K (T_hot - T_cold) / thickness
Q/t = 10m2 x 0.040W/mC x (60 - -10)C / 0.050m

= 560

and with the multiple we need to cut 560 in 1/2 to get 280.

Rold = 0.050m / 0.040W/mC
Rnew = X / 0.010W/mC

Q/t = A x K (T_hot - T_cold) / Sum(R)
260 = 10m2 x (60 - -10)C / (0.050m / 0.040W/mC + X / 0.010W/mC)

325 + 26000X = 700
X = 0.0144m

about 1.5 cm of insulation I guess.

Thanks everyone!

tb

2. Apr 15, 2007

### Chi Meson

You mean 280 here, right? You do (with this correction) get the right answer of .0125m, but you took the long way to get there.

If you solved the problem with only variables (not plugging in the numbers right away, just use "A" and "delta T" (which cancel out), and set (Q/t)2=0.5(Q/t)1, you see that

X2=(k2X1)/k1