A temperature difference between the outside and inside air will create a "natural draft" forcing air to flow through the building.
The direction of the flow depends on the temperatures. If inside temperature is higher than outside temperature, inside air density is less than outside air density, and inside air will flow up and out of the upper parts of the building. Cold outside air will flow into the lower parts of the building.
If outside temperature is higher than inside air temperature - the air flow will be in the opposite direction.
The natural draft is caused by the difference in outside and inside air density. The natural draft head can therefore be expressed as
dpmmH2O = (ρo - ρr) h (1)
dpmmH2O = head in millimeter water column (mm H2O)
ρo = density outside air (kg/m3)
ρr = density inside air (kg/m3)
h = height between outlet and inlet air (m)
Equation (1) can be modified to SI pressure units like
dp = g (ρo - ρr) h (1b)
dp = pressure (Pa, N/m2)
g = acceleration of gravity - 9.81 (m/s2)
With air density of 1.293 kg/m3 at 0oC, the air density at any temperature can be expressed as
ρ = (1.293 kg/m3) (273 K) / (273 K + t) (2)
ρ = 353 / (273 + t) (2b)
ρ = density of air (kg/m3)
t = the actual temperature (oC)
Equation (1) above can easily be modified by replacing the densities with equation (2).
The calculator below can be used to calculate the natural draft pressure generated by the inside and outside temperature difference.
The natural draft force will be balanced to the major and minor loss in ducts, inlets and outlets. The major and minor loss in the system can be expressed as
dp = λ (l / dh) (ρr v2 / 2) + Σξ 1/2 ρr v2 (3)
dp = pressure loss (Pa, N/m2, lbf/ft2)
l = length of duct or pipe (m, ft)
dh = hydraulic diameter (m, ft)
Σ ξ = minor loss coefficient (summarized)
Equation (1) and (3) can be combined to express the air velocity through the duct
v = [ (2 g (ρo - ρr) h ) / ( λ l ρr / dh + Σξ ρr ) ]1/2 (4)
Equation (4) can also be modified to express the air flow volume through the duct
q = π dh2 /4 [ (2 g (ρo - ρr) h ) / ( λ l ρr / dh + Σξ ρr ) ]1/2 (5)
q = air volume (m3/s)
The calculator below can be used to calculate the air flow volume and velocity in a duct similar to the drawing above. The friction coefficient used is 0.019 which is appropriate for normal galvanized steel ducts.
Calculate the air flow caused by natural draft in a normal family house with two floors. The height of the hot air column from ground floor to outlet air duct above roof is approximately 8 m. The outside temperature is -10 oC, the inside temperature is 20 oC.
A duct of diameter 0.2 m goes from 1. floor to the outlet above the roof. The length of the duct is 3.5 m. Air leakages through the building are neglected. The minor coefficients are summarized to 1.
The density of the outside air can be calculated like
ρo = (1.293 kg/m3) (273 K) / ((273 K) + (-10 oC))
= 1.342 kg/m3
The density of the inside air can be calculated like
ρr = (1.293 kg/m3) ( 273 K) / ((273 K) + (20 oC))
= 1.205 kg/m3
The velocity through the duct can be calculated like
v = [ (2 (9.81m/s2) ((1.342 kg/m3) - (1.205 kg/m3)) (8 m) ) / ( 0.019 (3.5m)(1.205 kg/m3)/(0.2 m) + 1(1.205 kg/m3) ) ]1/2
= 3.7 m/s
The air flow can be calculated like
q = (3.7 m/s) 3.14 (0.2 m)2 / 4
= 0.12 m3/s
that these equations can be used for dry air, not for mass flow and energy loss calculations where air humidity may have vast effects.