Overall heat transfer loss from buildings - transmission, ventilation and infiltration
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The overall heat loss from a building can be calculated as
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H = Ht + Hv + Hi (1)
where
H = overall heat loss (W)
Ht = heat loss due to transmission through walls, windows, doors, floors and more (W)
Hv = heat loss caused by ventilation (W)
Hi = heat loss caused by infiltration (W)
1. Heat loss through walls, windows, doors, ceilings, floors, etc.>
The heat loss, or norm-heating load, through walls, windows, doors, ceilings, floors etc. can be calculated as
Ht = A U (ti - to) (2)
where
Ht = transmission heat loss (W)
A = area of exposed surface (m2)
U = overall heat transmission coefficient (W/m2K)
ti = inside air temperature (oC)
to= outside air temperature (oC)
Heat loss through roofs should be added 15% extra because of radiation to space. (2) can be modified to:
H = 1.15 A U (ti - to) (2b)
For walls and floors against earth (2) should be modified with the earth temperature:
H = A U (ti - te) (2c)
where
te= earth temperature (oC)
Overall Heat Transmission Coefficient
The overall of heat transmission coefficient - U - can be calculated as
U = 1 / (1 / Ci + x1 / k1 + x2 / k2 + x3 / k3 + .. + 1 / Co) (3)
where
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Ci = surface conductance for inside wall (W/m2K)
x = thickness of material (m)
k = thermal conductivity of material (W/mK)
Co= surface conductance for outside wall (W/m2K)
The conductance of a building element can be expressed as:
C = k / x (4)
where
C = conductance, heat flow through unit area in unit time (W/m2K)
Thermal resistivity of a building element is the inverse of the conductance and can be expressed as:
R = x / k = 1 / C (5)
where
R = thermal resistivity (m2K/W)
With (4) and (5), (3) can be modified to
1 / U = Ri + R1 + R2 + R3 + .. + Ro (6)
where
Ri = thermal resistivity surface inside wall (m2K/W)
R1.. = thermal resistivity in the separate wall/construction layers (m2K/W)
Ro = thermal resistivity surface outside wall (m2K/W)
For walls and floors against earth (6) - can be modified to
1 / U = Ri + R1 + R2 + R3 + .. + Ro + Re (6b)
where
Re = thermal resistivity of earth (m2K/W)
2. Heat loss by ventilation
The heat loss due to ventilation without heat recovery can be expressed as:
Hv = cp ρ qv (ti - to) (7)
where
Hv = ventilation heat loss (W)
cp = specific heat air (J/kg K)
ρ = density of air (kg/m3)
qv = air volume flow (m3/s)
ti = inside air temperature (oC)
to = outside air temperature (oC)
The heat loss due to ventilation with heat recovery can be expressed as:
Hv = (1 - β/100) cp ρ qv (ti - to) (8)
where
β = heat recovery efficiency (%)
An heat recovery efficiency of approximately 50% is common for a normal cross flow heat exchanger. For a rotating heat exchanger the efficiency may exceed 80%.
3. Heat loss by infiltration
Due to leakages in the building construction, opening and closing of windows, etc. the air in the building shifts. As a rule of thumb the number of air shifts is often set to 0.5 per hour. The value is hard to predict and depend of several variables - wind speed, difference between outside and inside temperatures, the quality of the building construction etc.
The heat loss caused by infiltration can be calculated as
Hi = cp ρ n V (ti - to) (9)
where
Hi = heat loss infiltration (W)
cp = specific heat air (J/kg/K)
ρ = density of air (kg/m3)
n = number of air shifts, how many times the air is replaced in the room per second (1/s) (0.5 1/hr = 1.4 10-4 1/s as a rule of thumb)
V = volume of room (m3)
ti = inside air temperature (oC)
to = outside air temperature (oC)
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- Heating - Heating systems - capacity and design of boilers, pipelines, heat exchangers, expansion systems and more
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Heat can move, transfer, through material by conduction. In most cases this is considered a loss. The following calculator and examples show how to compute the amount of heat transfer/loss.
Heat transfer/loss calculators
How heat transfer is calculated
(Examples for solar air heaters, houses and underground tanks are given below.)
where:
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- heat_transferred is the amount of heat transferred, either lost or gained, through the material separating both sides (BTU / hour)
- surface_area is the surface area of the material separating the two areas that are at different temperatures, it's the material that the heat is being transferred through (Examples: wall, window) (ft2),
- temperature_A is the larger of the two temperatures on both sides of the separator (F),
- temperature_B is the smaller of the two temperatures on both sides of the separator (F),
- R_value is the R-value of the separating material (ft2 x F x hours / BTU).
The R-value is the resistance to the movement of heat through the separating material, a house wall in the diagrams on the right. So the formula is saying, get the difference in the temperatures on the two sides of the wall and divide it by the resistance to heat flow through the wall. Once you've done that, multiply it by the area of the wall since the heat is being transferred everywhere on the wall, not just at one location.
Example 1: Heat loss calculations through a house wall in winter.
In the second example on the right we've significantly increased the wall's R-value, the resistance to heat transfer, from 14 to 28. The effect was to slow down heat loss from 228 BTU per hour to 114 BTU per hour, a major improvement. BTU is a measure of energy. Example 2: Heat loss calculations through a house wall in winter.
Note that the direction of heat flow is from the higher temperature side to the lower temperature side. Since the formula is subtracting the two temperatures, it's clearer if you use the higher temperature for temperature_A otherwise you'll end up with a negative result, which may be confusing to some.
Examples of heat transfer/loss calculations
Solar heater glazing
The interesting thing when looking at these, especially in the solar heater case, is the difference the two temparatures make. On the left below is a screen solar air heater where the input air moves between the absorber and the glazing before being heated as it passes through the absorber and reenters the house. That means the air between the absorber and the glazing starts out at room temperature and consists of a constant inflow of room temperature air, though it does get heated either by contacting the absorber but not going through it or by radiation from the absorber.
On the right below is a backpass solar air heater such as a downspout one. In that case the air between the absorber and the glazing is never refreshed and increases in temperature from contact with the absorber. This means, theoretically it reaches higher temperatures and so has greater heat loss through the glazing.
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More about the different types of solar air heaters can be found on solar air heater types page.
Theoretical comparison of heat loss through the glazing between a screen solar air heater and a backpass solar air heater.
Note that the above conclusions regarding what happens in the two types of collectors are theoretical and do not take into account the many other variables. The intention is only to highlight the affects of a larger temperature difference.
House window with solar air heater
An interesting example of heat loss through a house window is found when people put solar air heaters inside their house directly behind a window as in the ilustration below. In both cases the same amount of solar energy enters the window. Without a solar air heater, the solar energy is absorber/converted to heat by the furniture. With a solar air heater the solar energy is still converted to heat but results in higher temperature air next to the window. This cause a greater rate of heat loss at the window.
Assuming a window R-value of 2, and a window area of 6 square feet, the heat loss with the solar air heater is 321 BTU per hour ([6 x (130 - 23)] / 2). For the window without the solar air heater the heat loss is 150 BTU per hour ([6 x (73 - 23)] / 2).
Heat loss by putting a solar air heater indoors in a window.
You can find out more about this, most often incorrect, use of a solar air heater on this page about window covering solar air heaters.
Normal wall
The examples given at the top of this page are for a wall. Keep in mind that a house wall is actually a little more complicated since it has windows, wooden studs, electrical boxes, ...
Assuming that none of these things overlap each other you simply take each object's surface area and divide it by the R-value for that object. This gives you the thermal conductance for each individual object. You then add up all the conductances and divide that total by the temperature difference.
For example, here are calculations for a simple 20 ft x 10 ft house wall with two windows.
Window R-value = 2
Window 1 area = 5 ft x 3 ft = 15 sq ft
Window 1 conductance = 15 / 2 = 7.5 BTU/hour x F
Window 2 area = 4 ft x 3 ft = 12 sq ft
Window 1 conductance = 12 / 2 = 6 BTU/hour x F
Wall R-vallue = 14
Wall area = (20 ft x 10 ft) - 15 sq ft - 12 sq ft = 173 sq ft
Notice that to get the area of just the wall material we subtracted the area taken up by the windows.
Wall conductance = 173 / 14 = 12.4
Outdoor temperature = 50F, Indoor temperature = 70F
Heat loss = (6 + 12.4) x (70 - 50) = 368 BTU per hour
Window R-value = 2
Window 1 area = 5 ft x 3 ft = 15 sq ft
Window 1 conductance = 15 / 2 = 7.5 BTU/hour x F
Window 2 area = 4 ft x 3 ft = 12 sq ft
Window 1 conductance = 12 / 2 = 6 BTU/hour x F
Wall R-vallue = 14
Wall area = (20 ft x 10 ft) - 15 sq ft - 12 sq ft = 173 sq ft
Notice that to get the area of just the wall material we subtracted the area taken up by the windows.
Wall conductance = 173 / 14 = 12.4
Outdoor temperature = 50F, Indoor temperature = 70F
Heat loss = (6 + 12.4) x (70 - 50) = 368 BTU per hour
Underground tank
The temperature around 5 feet underground (or below the frost line) stays relatively constant throughout the year. As such it provides a constant temperature for a heat storage tank. In North America in many areas the temperature at that depth is around 40F to 50F, though it varies widely. The following are some sample calculations for a cylindrical container.
Heat loss calculations for an underground heat storage tank