# Thickness of concrete wall to have the same insulating value of 3.5 in. fiberglass

by preimmortal
Tags: concrete, fiberglass, insulating, thickness, wall
 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?
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P: 6,189
 Quote by preimmortal 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?
I think it would make more sense if the insulating value would be a measure of the heat flow. As k for concrete would not be the same as k for fibre-glass.
 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.
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P: 6,189

## Thickness of concrete wall to have the same insulating value of 3.5 in. fiberglass

Do you happen to know the correct answer?
 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.
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P: 6,189
 Quote by preimmortal 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.
Try using the most accurate sets of values you can find.
 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)?
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P: 6,189
 Quote by preimmortal 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)?
The method is correct. However, if your textbook or the question does not give you the values for k, then the answer will vary as different places will have different values.
 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.
 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.

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