Estimating the rate of leakage of air into a closed room
I have a digital thermometer-cum-hygrometer. If I measure both T ( degrees Celsius) and relative humidity RH (in %) at night when I go to sleep, and again in the morning, when I wake up, I can estimate the leak rate. The logic is that the temperature will dip towards the morning, and the RH may also change - but the absolute humidity (AH) will remain the same. Unless, there is an inleak of air. So the percentage change in the AH and the volume of the room can give an estimate of the average leak rate (knowing the time beween the two measurements).
Note: the AH is related to the RH & T by the Antoine equation. (I have assumed the values are correct without getting into whether this set of values is the most reliable set.)
I put this to my colleague Haldar and he did not disagree with the above argument but pointed out that if I was sleeping in the room, then the amount of water vapour contributed by exhalation and perspiration would change the AH. I did the calculations and I do not think it makes a large difference - at least in winter! (which is what it is now) - and I also checked it in an unoccupied guest bedroom, and the results are similar. However, the idea of measuring the leakage rate in this way has limited utility - as is explained in detail below...
The bedroom doors (and windows) are closed at night and so
the difference in absolute humidity (AH) between night & morning should
give the leak rate:
Leakage rate:
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
22:45
|
20.0
|
74
|
12.957
|
|
06:30
|
18.9
|
76
|
12.46
|
|
06:30
|
18.8
|
75
|
12.18
|
The AH is calculated from T and RH by the Antoine equation:
Log10
(P) = 8.07131 – [1730.63/(233.426 +T)]
Where p is in Torr, and T is in °C.
In the morning, average AH = 12.32
Percent change: (12.957-12.32)/12.957 = 4.9 % - over 7hrs 45 mins.
Percent change per hr: 0.0063 /hr.
Total volume of room: 11.25 x 15 x 9 = 1519 ft3 = 42.3 m3
Total volume leak rate per hr: 0.27 m3/hr = 4.5
L/min.
However, a small reality check: the AH in the room went down
from 22:45 to 06:30 – but that can only if the AH outside the room went down in
this time period.
This did happen, as can be seen from this data from Greater Kailash-I
from www.bluebus.com:
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
21:30
|
14.9
|
90
|
11.66
|
|
07:15
|
10.3
|
98
|
9.18
|
That is, AH outside decreased by about 21%.
Next day:
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
23:00 8-12-16
|
19.4
|
76%
|
12.82
|
|
07:00 9-12-16
|
18.8
|
76%
|
12.35
|
Percent change: (12.82-12.35)/12.82 = 3.7 % - over 8 hrs
Percent change per hr: 0.0046 /hr.
Total volume of room: 11.25 x 15 x 9 = 1519 ft3 = 42.3 m3
Total volume leak rate per hr: 0.20 m3/hr = 3.3
L/min.
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
23:30 9-12-16
|
19.4
|
76
|
12.82
|
|
06:30 10-12-16
|
18.4
|
77
|
12.20
|
|
07:45 10-12-16
|
18.2
|
77
|
12.05
|
Percent change: (12.82-12.2)/12.82 = 4.8% - over 7 hrs
Percent change per hr: 0.0069 /hr.
Total volume of room: 11.25 x 15 x 9 = 1519 ft3 = 42.3 m3
Total volume leak rate per hr: 0.29 m3/hr = 4.9
L/min.
Change in 1.25 hrs: 1.2% i.e. 0.0098 /hr i.e. 7 L/min
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
23:00 10-12-16
|
18.3
|
77%
|
12.125
|
|
07:15 11-12-16
|
16.6
|
81%
|
11.453
|
Percent change: (12.125-11.453)/12.125 = 5.54% - over 8.25 hrs
Percent change per hr: 0.0067 /hr.
Total volume of room: 11.25 x 15 x 9 = 1519 ft3 = 42.3 m3
Total volume leak rate per hr: 0.284 m3/hr = 4.74
L/min.
Readings for
8th & 9th Jan.17:
a a) Guest bedroom: empty
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
21:30 8-1-17
|
17.9
|
74%
|
11.363
|
|
08:00 9-1-17
|
16.9
|
74%
|
10.665
|
AH decreases by: 6.14%.
b b) Main bedroom: 1 sleeper
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
21:30
8-1-17
|
17.7
|
80%
|
12.13
|
|
08:00 9-1-17
|
17.1
|
79%
|
11.531
|
AH decreases by: 4.94%.
c c) Outside temperatures:
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
21:00
8-1-17
|
13.8
|
90%
|
10.62
|
|
08:00
9-1-17
|
10.3
|
96%
|
8.99
|
AH decreases by: 15.35%.
The AH decreases by a larger percentage in the unoccupied room
than in the occupied room – and by the largest percentage in the outside of the
house.
Reading on 10th Jan17:
a d) Main bedroom: 2 sleepers
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
21:30
9-1-17
|
16.1
|
71%
|
9.724
|
|
08:00
10-1-17
|
14.9
|
73%
|
9.255
|
AH decreases by: 4.82%.
b e) Outside temperatures:
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
21:00
9-1-17
|
11.0
|
65%
|
6.38
|
|
08:00
10-1-17
|
5.8
|
82%
|
5.648
|
AH decreases by: 11.47%.
Breathing
contribution:
inhaled water vapour: 4.5 ml per breath (total
500 ml of air per breath).
exhaled water vapor: 31 ml per breath. (total
volume exhaled 500 ml per breath).
Total: about 17,000 breaths per day.
Assume 15 breaths per minute, and 10.5 hrs is
630 mins. Total water vapour exhaled:
(0.0265)(630) = 16.7 Litres.
Assuming a 27,000 Litre room, the percentage
change in AH due to one sleeper is: 16.7/27000 = 0.062%.
Three sites below add to the confusion – but
they are all smaller than
what is calculated above:
1)
For the average 70 kg man, it has been measured
to be 400 ml due to respiration and 400 ml due to perspiration per 24 hour
period of time.
The quora site suggests that an equal amount is
lost due to breathing and perspiration.
Note: the quora value for breathing is much smaller than the one calculated
above, which is of the order of 38 litres per day!
2)
The average human exhales
0.35 liters of water each day.
Every day, we breathe in about 14000L of air.
Assuming that the humidity of exhaled air is 100% and inhaled air is 20%, and
using the carrying capacity of 1kg of air to be 20g of water vapour, we can
calculate how much water is lost simply by breathing. This estimate gives 400ml
of water lost per day too.
3)
During physical exercise
amount of exhaled H2O is linear, but not proportional to heart rate.
And so at the heart rate of 140 bpm amount of exhaled water is approximately
four times higher than during the rest and equals about 60-70 ml/h.
The site below gives a formula and analysis that is most
credible, but the difference with the earlier estimate (of mb-soft i.e. 31
ml per 500 ml) is 43%:
According to this blog, the air exhaled by a
human is at 35 °C and 95% RH.
The partial pressure is calculated from:
P = RH [510.8 exp(17.2694T/(T + 238.3))]
Where T is in °C and p is in Pa.
Plugging in this gives a partial pressure Pw
= 4430 Pa
This gives a value of 0.0437, or about 4.4%.
For comparison: the value 31.5/500 = 0.063, or
6.3%.
The agreement is not good (43% off!) – but at
least the order of magnitude is right!
Sweating contribution:
“Chemistry of
the Textiles Industry”
edited by C. Carr (1995) p.236
“Sweat rate is 15-30 ml/hr
when sleeping in comfort zone temperatures.”
The quora.com site estimates
400 ml/day for a 70 kg man i.e. 17 ml/hr on average, which is within the
range of the above values.
LSB error:
However, the problem with the measurements above is the
least significant bit (LSB) in the digital display of the meter that I have.
The humidity is accurate to 1% and the temperature to 0.1 °C.
Suppose one takes the worst case (error in both RH and T to
minimize the difference in AH), on the measurements above for the guest bedroom a) above:
f)
|
Time
|
Temp (°C)
|
RH (%)
|
AH
|
|
21:30 8-1-17
|
17.8
|
73%
|
11.363
|
|
08:00 9-1-17
|
17.0
|
75%
|
10.878
|
In this case, the AH decreases by 2.33% - as compared with
6.14% above! That means that unless the measurements are done with significantly
better accuracies the estimated inleak rate could be off by a factor of 2-3X!
Further, one can only use this idea on two measurements done
overnight, to get the maximum difference in AH. Also, to avoid door opening which
would happen many times in the day!
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