Table 2.3 - thermal insulation regulations

maximum u-values of elements/Wm-2 K-1

dwellings - (floors) - 0.45, (walls) - 0.45, (roofs) - 0.25
offices,shops - (floors) - 0.45, (walls) - 0.45, (roofs) - 0.45
industrial,storage - (floors) - 0.45, (walls) - 0.45, (roofs) - 0.45

maximum single-glazed areas

dwellings - (windows) - 15%, (rooflights) - 15%
offices,shops - (windows) - 35%, (rooflights) - 20%
industrial,storage - (windows) - 15%, (rooflights) - 20%

1. a house has a living room with an external wall of area 9.2m2. The single galzed window (U=5.3 W m-2 K-1) is 15% of this area. Calculate the rate of heat loss through this wall on a day when the outside temperature is 10 degrees Celsius less than the inside temperature.

2. a warehouse has the maximum permitted area of single-glazed window and rooflight. It has a height of 4m, one length is 8m, and the other length is 10m.

a) estimate the power requirements of a heating system on a day when the temperature difference in 10 degrees Celsius.

b) how much energy would be saved in a day if the windows and rooflights were double glazed (U=3.0 W m-2 K-1)?

I looked it up:

Ht = U A dt (1)
where

Ht = heat loss (Btu/hr, W)
U = "U-value" (Btu/hr ft2 oF, W/m2K)
A = wall area (ft2, m2)
dt = temperature difference (oF, K)

1. You were given the total area and from that you can calculate the window area. Subtract the window area from the total area to get the wall area. DT is 10 degrees F. You are given U for the varoius areas. Do the calculation (equation 1) for both areas. That is the total heat loss.

2. a. Figure out the area of the area of the warehouse. Assume each wall is a rectangle and I assume the roof is also a rectangle. Multiply the total area by the maximum permitted area (a fraction you read from your "maximum single-glazed areas" for each of the areas. Subtract each area from the total. The remainder of the area is wall. Perform the heat loss calculation on each of the areas separately. Add all of the heat losses. That will give you the power requirement.

b. Do the same calculation, except use the given U value for the window and rooflights.

This is very straight-forward. Don't make it harder than it really is.

Quote:

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Originally Posted by Perito

I looked it up:

Ht = U A dt (1)
where

Ht = heat loss (Btu/hr, W)
U = "U-value" (Btu/hr ft2 oF, W/m2K)
A = wall area (ft2, m2)
dt = temperature difference (oF, K)

1. You were given the total area and from that you can calculate the window area. Subtract the window area from the total area to get the wall area. dT is 10 degrees F. You are given U for the varoius areas. Do the calculation (equation 1) for both areas. That is the total heat loss.

2. a. Figure out the area of the area of the warehouse. Assume each wall is a rectangle and I assume the roof is also a rectangle. Multiply the total area by the maximum permitted area (a fraction you read from your "maximum single-glazed areas" for each of the areas. Subtract each area from the total. The remainder of the area is wall. Perform the heat loss calculation on each of the areas separately. Add all of the heat losses. That will give you the power requirement.

b. Do the same calculation, except use the given U value for the window and rooflights.

This is very straight-forward. Don't make it harder than it really is.

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I'm sorry but I am getting kind of confused... can you show me how to work it out please with the working out, thank you

Ok, I never did that, but I understand what's being explained. You really have a problem Nargis... :(

1.
Window area = 9.2 x 15% = 1.38 m^2.

Wall area = 9.2 - 1.38 = 7.82 m^2

(Could also use 85% of total area to find wall area, giving 85% x 9.2 = 7.82 m^2)

Ok, using formula, Ht = (5.3)(7.82)(10) = 414.46 J/s