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

In summary: Oops.In summary, to find the thickness of a concrete wall needed to have the same insulating value as 3.5 inches of fiberglass, you can use the heat flow equation Q = kA(ΔT/Δx) and set the values for k and A to those of concrete and fiberglass, respectively. By setting the heat flow values equal to each other and solving for Δx, you can find that the thickness of the concrete wall would need to be approximately 83 inches.
  • #1

Homework Statement


How thick a concrete wall would be needed to give the same insulating value as 3.5 inches of fiberglass?


Homework Equations


Q = -A·k·ΔT/Δx
k = -Q·Δx/A·ΔT


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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  • #2


preimmortal said:

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.
 
  • #3


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.
 
Last edited:
  • #4


Do you happen to know the correct answer?
 
  • #5


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


preimmortal said:
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.
 
  • #7


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


preimmortal said:
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.
 
  • #9


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


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.
 

1. How thick does a concrete wall need to be to have the same insulating value as 3.5 in. of fiberglass?

The thickness of a concrete wall needed to match the insulating value of 3.5 in. of fiberglass will vary depending on the type and density of the concrete. However, on average, a concrete wall would need to be at least 8 inches thick to provide the same level of insulation as 3.5 inches of fiberglass.

2. Can a concrete wall provide the same level of insulation as fiberglass?

Yes, a concrete wall can provide the same level of insulation as fiberglass if it is thick enough. Concrete has a lower insulation value than fiberglass, but it can make up for this by being thicker and denser.

3. How does the density of concrete affect its insulating value?

The denser the concrete, the higher its insulating value will be. This is because the air pockets within the concrete act as insulators. The more air pockets there are, the better the insulation.

4. Is there a specific type of concrete that is best for insulation purposes?

There are certain types of concrete that are better for insulation purposes, such as lightweight concrete or aerated concrete. These types of concrete have a higher number of air pockets and therefore provide better insulation than traditional concrete.

5. Are there any other factors besides thickness that affect the insulating value of a concrete wall?

Yes, besides thickness, factors such as the quality of the concrete, the temperature and humidity, and the presence of any cracks or gaps can also affect the insulating value of a concrete wall. It is important to ensure that the concrete is of high quality and properly sealed to achieve the best insulation.

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