# How do I calculate the overall vapour flow?

• Engineering
okandrea
Homework Statement:
A cavity wall consists of a 100mm brick exterior with 50mm cavity and 10mm gypsum board. The inside temperature is 20°C and RH of 40%. The outside temperature is -10°C with RH of 85%. Calculate overall vapour flow over one day.
Relevant Equations:
R = 1/M, RH = Pw/Pws x 100%, Qv=A(µ/l)(pw,1 - pw,2)
Step 1:
Values are from textbook 'Building for a Cold Climate'
Mbrick = 46 ng/s*Pa*m^2
Mgypsum = 2870 ng/s*Pa*m^2 (for 9.5mm)
took the above value and used ratios to determine permeance for 10mm Gypsum board (2870/9.5 = X/10)
Mgypsum (new value) = 3021.05 ng/s*Pa*m^2

Step 2:
Values are from textbook 'Building for a Cold Climate' (pressure over ice)
T in = 20°C --> Pws = 2.337 kPa = 2337 Pa
T out = -10°C --> Pws = 259.7 Pa

Step 3:
Resistance --> R = 1/M
R brick (common) = 1/46 = 0.0217
R gypsum = 1/3021 = 0.000331

Step 4:
Used relative humidity (and saturated vapour pressure) to find vapour pressure
RH = Pw/Pws x 100%
(1) 20°C Temp (in), 40% RH --> 0.40 = Pw/2337
0.40 x 2337 = Pw
Pw (in) = 934.8 Pa
(2) -10°C Temp (out), 85% RH --> 0.85 = Pw/259.7
0.85 x 259.7 = Pw
Pw (out) = 220.75 Pa

Step 5:
I understand that normally I would add all M-values (permeances) to get total permeance and then use the vapour flow (Qv) rate equation:
Qv = A(µ/l)(pw,1 - pw,2) or Qv = A*M*(pw,1 - pw,2)
With the values that I have - and am sure of - the only part of the calculation I've figured out was the change in pressure, where I subtract the high vapour pressure with the low vapour pressure:
pw,1 - pw,2 = Δp
934.8 - 220.75 = 714.05 Pa

I'm confused as to how the cavity part comes in ("50mm cavity") and the value for area since the givens only consist of thicknesses. I know that the textbook example uses 1sqm so I feel that I should assume the A-value as well. Are there any more assumptions I am missing? Are the steps that I've done so far correct?