Register to reply 
Thickness of concrete wall to have the same insulating value of 3.5 in. fiberglass 
Share this thread: 
#1
Jan1611, 04:43 PM

P: 13

1. The problem statement, all variables and given/known data
How thick a concrete wall would be needed to give the same insulating value as 3.5 inches of fiberglass? 2. Relevant equations Q = A·k·ΔT/Δx k = Q·Δx/A·ΔT 3. The attempt at a solution So here, k is equal to the insulating value, and since I want the insulating value to be the same for the concrete wall and fiberglass, I would have the equation Q·3.5/A·ΔT = Q·Δx/A·ΔT Since it seems that area and change in temp are the same in both cases, I got Q·3.5 = Q·Δx Which is where I got stuck. Would the Q (heat flow) cancel? 


#2
Jan1611, 05:12 PM

HW Helper
P: 6,202




#3
Jan1611, 06:32 PM

P: 13

Okay, so I did try setting the heat flow equal..
A·k(concrete)·ΔT/Δx = A·k(fiberglass)·ΔT/Δx = Q So again, A and ΔT are the same k(fiberglass) / 3.5 = k(concrete) / Δx So I found that k(fiberglass) = .04 W/mk k(concrete) = 1.7 W/mk .04/3.5 = 1.7/Δx Δx = 1.7*3.5/.04 = 148.75 in. However, this answer was incorrect. *Edit. the heat conductivity of both objects were measured at 25 degrees celsius, so I am not sure of what k value to use for each. 


#4
Jan1611, 06:43 PM

HW Helper
P: 6,202

Thickness of concrete wall to have the same insulating value of 3.5 in. fiberglass
Do you happen to know the correct answer?



#5
Jan1611, 06:50 PM

P: 13

I don't know the correct answer, but I got "Try Again" when I entered it.
On another note, k(fiberglass) was rounded to 2 decimal places and I've seen it rounded to .043 or even .045 (depending on temperature), which makes a very large difference. 


#6
Jan1611, 06:54 PM

HW Helper
P: 6,202




#7
Jan1611, 07:01 PM

P: 13

The problem never tells us the heat conductivity of the two materials, so I can only try to find it online. I'm not sure which to use.
Was my setup correct when I set the heat flow? And was the final answer correct (given the correct k values)? 


#8
Jan1611, 08:19 PM

HW Helper
P: 6,202




#9
Jan1911, 02:33 AM

P: 1

Using k(concrete)=1 W/mK & k(fiberglass)=0.042 W/mK , you can use the Heat Flow equation: H=kA(ΔT/Δx)
It may help to make up imaginary values for A and ΔT Also note that 3.5 inches = 0.0889 meters Solve for H(fiberglass) by using the above equation and values. Set any values for A and ΔT (i.e., A=0.5m^2 & ΔT=65°C) Now solve for Δx(concrete) by setting the equation equal to the value you just obtained for H(fiberglass). Remember to use the same A and ΔT values. If done correctly Δx should be about 83 inches (after converting your answer from meters to inches) Hope that helps. 


#10
Jan1911, 03:29 AM

P: 13

Thanks DG0628, I used your k values and equation and it worked out fine.
I had the k value for concrete incorrect. I also didn't convert from inches to meters. 


Register to reply 
Related Discussions  
Breaking through a concrete wall easily  General Physics  13  
Axial loading and the change of wall thickness in a cylinder  Classical Physics  0  
Strength of test tube with respect to wall thickness  General Engineering  0  
Formula for allowable wall thickness  Materials & Chemical Engineering  2  
A car crashes into a concrete wall  Introductory Physics Homework  1 