Estimating the leakage of air into a closed bedroom

Estimating the rate of leakage of air into a closed room

This entire exercise is connected with my air purifier which is located in my bedroom and which is turned on after we go to sleep. Unfortunately, despite the fact that all the windows and doors are closed, air from outside the bedroom does get in. I wanted to estimate the rate at which that happens.
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 8^th & 9^th 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 10^th 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:

http://mb-soft.com/public3/weight7.html

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)

https://www.quora.com/How-much-water-does-a-person-lose-in-a-day-through-breathing

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)

https://www.reference.com/science/many-liters-water-humans-lose-day-breathing-b3c4caad4688336e

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)

https://www.ncbi.nlm.nih.gov/pubmed/22714078

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%:

http://www.sciencebits.com/exhalecondense

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 P_w = 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!

How much energy is used in pumping down to high vacuum?

How much energy is used in pumping down to high vacuum?

This seems to be an extremely simple question and I was quite surprised when I couldn't find an answer on the Internet - since the time required to pump down to a given vacuum is well-known, and addressed in all elementary texts on vacuum technology.
I had the feeling that the power consumed should increase as the vacuum level goes further down from the bar to the millibar to the microbar range... but it turns out not to be so...
Since I couldn't find an answer I asked my colleague Dr.Shiv Kumar who promptly sent me a thread on researchgate started by a researcher Xuezhong He of the Norway Institute of Science and Technology who asked the same question, along with some responses - to which I added my two cent's worth, as reproduced below:

https://www.researchgate.net/post/How_can_I_estimate_vacuum_pump_power_consumption

"If we want to use a vacuum pump to suck a gas stream from 1 bar to 100mbar, how to estimate the power consumption for this vacuum pump? Is it identical to use a compressor from 100mbar to 1bar?
How can I estimate vacuum pump power consumption?" 

Christian Binek (Univ. of Nebraska at Lincoln) responded:

I would argue that the work Integral[PdV] is just a minuscule and most likely negligible fraction of the internal power consumption of the pump, e.g., due to friction losses and acceleration of mechanical parts. 
Pavel Soucek of Masaryk University suggested that we just use a wattmeter, and Bhupendra Desai of IIT Roorkee suggested a clip-on wattmeter.

Prashant Thankey of the Institute of Plasma Research added:

"The actual power consumed by a vacuum pump in its pumping action is actually the throughput (Q = PS) of the vacuum system. As pumping continues, Q generally decreases, and so the power consumption decreases. However, the total power consumption of the pump is the sum of the power consumed in the pumping action plus power consumed in overcoming the friction, heat losses etc."

However, let's just try to do the PdV thing:

                                                         Q = SP

Where s = speed of pump (litre/sec), P = pressure in torr and flow rate Q is in torr-litre/sec.

The standard equation is:

                                                                                                                                                                                             P = P₀ exp(- St/V)

Since PV has units of energy, the total energy is:

                                                 E = P₀ òexp(- St/V) dt between P₁ and P₂.

If S is independent of pressure and time:

                                                 E =P₀V[exp(-St₂/V) - exp(-St₁/V)]

But P = P₀ at t = 0, so:

                                                 E =P₀V [1 – exp(-St/V)]

This means that the rate of energy consumption is not constant with time because:

                                                                                                                                                                                            dE/dt = P₀S exp(-St/V)

But the actual pump-down time from P₀ to P_f is given as:

http://www.engineeringtoolbox.com/vacuum-evacuation-time-d_844.html

                                                                                                                                                                                             t = (V/S) ln(P₀/P_f)

Numerically, if V= 1 m³ and S = 0.1 m³/sec, P₀ = 1 bar, and P_f = 10 millibar: t = 46 sec.

The energy consumed in this process is: 

                                            E = (1.013x10⁵)(1)[1 – exp(-4.6)] = 1.003x10⁵ J

Suppose p_f =1 mbar, then t = 69 secs.

As long as speed s remains constant, each decade reduction in p_f would only add 23 secs.

The power consumption on going from 1 bar to 1 millibar is = 1x10⁵/69 = 1450 W.

And there is literally no extra energy consumed in this process… because the maximum energy consumed can only be 1.013x10⁵ J!

So this is just what Prashant Thankey also said...with equations, and some numbers.

However, the reason extra energy is consumed is because any pump consumes power by being on, overcoming internal friction etc.

Also, pump speed s will not remain constant. As you approach the ultimate pressure, the speed will go to zero – and each vacuum pump, has an ultimate pressure, whose

value depends upon the operating principle. For example, cryopumps do not have mechanical moving parts. But they do have an ultimate vacuum limit. In addition, one would have to incorporate the effects of outgassing, diffusion and permeation that act to limit the highest attainable vacuum level.

The power consumption of most diffusion pumps is also of the order of 1500 W.

Thus, the power consumed by pump-down is not a large fraction of the total power consumed – but it is not ‘minuscule’ as Christian Binek would suggest.

However, the pump-down power will indeed become a small fraction if we are talking about ultrahigh vacuum since it takes hours (if not days!) to reach these vacuum levels.

Realistic calculations would take into account actual geometry of a vacuum vessel including the vacuum duct used and the effective speed at the mouth of the vessel.

However, that is a rather more detailed analysis than I plan to do!

As Soucek & Desai argued, it is best to just use a wattmeter and do the actual measurements. This is obviously the correct approach... but it is complicated a bit by the fact that that the total time required to reach a fairly high vacuum will depend upon the degree of outgassing, diffusion & permeation in the material of the vessel, assuming that there are no microleaks.

Endnote: Clearly, the power consumption is not going to increase as you go to higher vacuum levels (more rarefied), but you could argue that since the time required to attain ultrahigh vacuum is much more, it is the total energy consumption (in Joules) that will increase as the level of vacuum increases. However, the same problem occurs as in the previous para: it depends more on the details of the vessel (outgassing, diffusion, permeation etc), than on the pump...