Thermodynamics Insulation Problem.

[tsMore]

Homework Statement

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.

Homework 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.

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

 

[Chi Meson]

260 = 10m2 x (60 - -10)C / (0.050m / 0.040W/mC + X / 0.010W/mC)

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

 

[m2003]

h, I am not entirely sure what you are asking for, but I'll try to provide some guidance.

Firstly, it seems like you have correctly calculated the rate of heat loss for the single layer of insulation. This is an important step in understanding the problem, as it tells us how much heat is being lost per unit of time.

Next, you mention that Joe wants to cut his heating bill in half by adding another layer of insulation. This is a good idea, as adding more insulation will decrease the rate of heat loss, thus reducing the amount of energy needed to maintain the desired temperature inside the steam house.

To determine the thickness of the new insulation layer, you can use the equation:

Rnew = X / Knew

Where Rnew is the new thermal resistance (thickness divided by thermal conductivity), X is the thickness of the new insulation layer, and Knew is the thermal conductivity of the new insulation.

You can then use this value of Rnew in the equation:

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

Where Q/t is the desired rate of heat loss (in this case, half of the original rate), A is the surface area of the steam house, K is the thermal conductivity of the original insulation, T_hot is the desired temperature (60 degrees Celsius), T_cold is the outside temperature (-10 degrees Celsius), and Sum(R) is the sum of the thermal resistances of both insulation layers (Rold + Rnew).

Solving for X will give you the desired thickness of the new insulation layer.

I hope this helps! Remember to always double check your equations and units to ensure accuracy. Good luck!