# How to calculate the amount of water needed to keep relative humidity 95% in a 20'\*6.5'\*8' room? **URL:** \ **Category:** Factual Questions\ **Created:** [August 12, 2017, 6:05pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453 "2017-08-12T18:05:13Z")\ **Posts on this page:** 19\ **Page:** 1
**Author:** ![oparthiv](https://avatars.discourse-cdn.com/v4/letter/o/aca169/32.png) [@oparthiv](https://boards.straightdope.com/u/oparthiv)\ **Post date:** [August 12, 2017, 6:05pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/1 "2017-08-12T18:05:13Z")
How to calculate the amount of water needed to keep relative humidity 95% in a 20’\*6.5’\*8’ room? The room temperature is 60 degree celcius. ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [August 12, 2017, 6:51pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/2 "2017-08-12T18:51:42Z")
I went ahead and converted the 1040 ft[sup]3[/sup] to 29.4 m[sup]3[/sup] for your volume … near as I can figure, density at 1 atm and 60ºC is about 1.5 kg/m[sup]3[/sup] {[Cite](http://www.engineeringtoolbox.com/air-temperature-pressure-density-d_771.html)} … so we have about 45 kg of air … The absolute humidity of saturated air at 60ºC is 150g/kg {[Cite](http://www.engineeringtoolbox.com/moisture-holding-capacity-air-d_281.html)} … so that gives 6-3/4 kg of water vapor for 100% Relative Humidity … or about **6.4 kg of water** in your room at 95% RH … A little nitpick … once we have 95% RH in a 60ºC room … we don’t need to add any water to “_keep_” the room as such … what I’ve given is the total mass of water in this room … hope this is helpful … ---
**Author:** ![Common\_Tater](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/common_tater/32/13846_2.png) [@Common\_Tater](https://boards.straightdope.com/u/Common_Tater)\ **Post date:** [August 13, 2017, 1:17am UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/3 "2017-08-13T01:17:19Z")
This room ain’t for human habitation I take it? ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [August 13, 2017, 1:52am UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/4 "2017-08-13T01:52:38Z")
> [@Common\_Tater](#): > > This room ain’t for human habitation I take it? The information was a little harder to find … the better graphs all stop at 35ºC … and better keep the heat turned on, things will get real wet if the temperature drops … ---
**Author:** ![oparthiv](https://avatars.discourse-cdn.com/v4/letter/o/aca169/32.png) [@oparthiv](https://boards.straightdope.com/u/oparthiv)\ **Post date:** [August 13, 2017, 2:06pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/5 "2017-08-13T14:06:33Z")
> [@Common\_Tater](#): > > This room ain’t for human habitation I take it? Nope.This is a battery curing chamber. I need to calculate the amount of water needed to keep the rh 95% ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [August 13, 2017, 2:17pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/6 "2017-08-13T14:17:37Z")
> [@oparthiv](#): > > Nope.This is a battery curing chamber. I need to calculate the amount of water needed to keep the rh 95% It might be easiest to just to make more water available than needed … keep a 5 gallon tray of water in the room and just let the air take what she needs … ---
**Author:** ![oparthiv](https://avatars.discourse-cdn.com/v4/letter/o/aca169/32.png) [@oparthiv](https://boards.straightdope.com/u/oparthiv)\ **Post date:** [August 13, 2017, 2:20pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/7 "2017-08-13T14:20:07Z")
Thanks for your help. It would be much helpful if you could give the mathematical calculation with reference ---
**Author:** ![oparthiv](https://avatars.discourse-cdn.com/v4/letter/o/aca169/32.png) [@oparthiv](https://boards.straightdope.com/u/oparthiv)\ **Post date:** [August 13, 2017, 2:22pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/8 "2017-08-13T14:22:26Z")
> [@watchwolf49](#): > > It might be easiest to just to make more water available than needed … keep a 5 gallon tray of water in the room and just let the air take what she needs … I just need the calculation process. Can you please help me on this? ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [August 13, 2017, 3:03pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/9 "2017-08-13T15:03:56Z")
We have a known volume and temperature, and I’m assuming this is at standard atmospheric pressure … we can look up the air’s density using charts available at a variety of websites, I’ve posted a link to the “Engineer’s Toolbox” site above, and either convert all your measures to metric or to Standard American … Then — Density times Volume equals Mass From another chart, which I linked to above, we can find the ratio of water to air (by mass) for saturated air (which means 100% RH) … this is called “Absolute Humidity”, and although technically a dimensionless ratio, it’s typically presented as either grams of water per kilograms of air or kilograms of water per kilogram of air … Then — Mass of Air times Absolute Humidity equals Mass of Water Vapor for 100% RH Finally — Mass of Water (100% RH) times 0.95 equals Mass of Water (95% RH) The biggest problem is trying to cipher out these charts in the links I’ve given, they are not too exacting at 60ºC … perhaps you can find better information at another web site … the difficulty is that the conditions in your specifications are rather uncommon … all the best charts and graphs stop at around 35ºC … ---
**Author:** ![Ruken](https://avatars.discourse-cdn.com/v4/letter/r/f475e1/32.png) [@Ruken](https://boards.straightdope.com/u/Ruken)\ **Post date:** [August 13, 2017, 3:14pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/10 "2017-08-13T15:14:35Z")
OP should search for: humidity salt bath I’ve never actually used one but they’re used to maintain humidity. ---
**Author:** ![oparthiv](https://avatars.discourse-cdn.com/v4/letter/o/aca169/32.png) [@oparthiv](https://boards.straightdope.com/u/oparthiv)\ **Post date:** [August 13, 2017, 3:24pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/11 "2017-08-13T15:24:49Z")
> [@watchwolf49](#): > > We have a known volume and temperature, and I’m assuming this is at standard atmospheric pressure … we can look up the air’s density using charts available at a variety of websites, I’ve posted a link to the “Engineer’s Toolbox” site above, and either convert all your measures to metric or to Standard American … > > Then — Density times Volume equals Mass > > From another chart, which I linked to above, we can find the ratio of water to air (by mass) for saturated air (which means 100% RH) … this is called “Absolute Humidity”, and although technically a dimensionless ratio, it’s typically presented as either grams of water per kilograms of air or kilograms of water per kilogram of air … > > Then — Mass of Air times Absolute Humidity equals Mass of Water Vapor for 100% RH > > Finally — Mass of Water (100% RH) times 0.95 equals Mass of Water (95% RH) > > The biggest problem is trying to cipher out these charts in the links I’ve given, they are not too exacting at 60ºC … perhaps you can find better information at another web site … the difficulty is that the conditions in your specifications are rather uncommon … all the best charts and graphs stop at around 35ºC … So, according to your calculation: 1. 29.44 cubic feet\*1.5 kg/m3=44.16 kg of air. 2. 44.16\*.15 kg/kg (150g/kg)=6.624 kg of water vapor 3. 6.624\*.95=6.29 kg of water Is this the process? ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [August 13, 2017, 3:31pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/12 "2017-08-13T15:31:30Z")
> [@oparthiv](#): > > So, according to your calculation: > > 1. 29.44 cubic feet\*1.5 kg/m3=44.16 kg of air. > 2. 44.16\*.15 kg/kg (150g/kg)=6.624 kg of water vapor > 3. 6.624\*.95=6.29 kg of water > Is this the process? Yes … except that’s 29.44 cubic meters of air … and in (1) the 1.5 kg/m[sup]3[/sup] is _very_ approximate … could be +/- 0.2 kg/m[sup]3[/sup] … you’ll need to make a more exhaustive search on the internet to get a more exact number … ---
**Author:** ![oparthiv](https://avatars.discourse-cdn.com/v4/letter/o/aca169/32.png) [@oparthiv](https://boards.straightdope.com/u/oparthiv)\ **Post date:** [August 13, 2017, 3:39pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/13 "2017-08-13T15:39:35Z")
> [@watchwolf49](#): > > Yes … except that’s 29.44 cubic meters of air … and in (1) the 1.5 kg/m[sup]3[/sup] is _very_ approximate … could be +/- 0.2 kg/m[sup]3[/sup] … you’ll need to make a more exhaustive search on the internet to get a more exact number … Thanks a lot ---
**Author:** ![Riemann](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/riemann/32/3133_2.png) [@Riemann](https://boards.straightdope.com/u/Riemann)\ **Post date:** [August 13, 2017, 6:03pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/14 "2017-08-13T18:03:41Z")
**watchwolf49** is calculating the total amount of water vapor present in that volume of air at 95%RH, which doesn’t seem particularly relevant, since increasing a sealed volume of air from zero to 95%RH is not a likely practical objective. You are asking for the amount of water required (presumably per unit time) to _maintain_ humidity at a certain level. This can’t be calculated from fundamental physical principles. It depends on how tightly your room is sealed from the outside environment, and the conditions in the outside environment (which will vary). ---
**Author:** ![oparthiv](https://avatars.discourse-cdn.com/v4/letter/o/aca169/32.png) [@oparthiv](https://boards.straightdope.com/u/oparthiv)\ **Post date:** [August 13, 2017, 6:22pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/15 "2017-08-13T18:22:35Z")
> [@Riemann](#): > > **watchwolf49** is calculating the total amount of water vapor present in that volume of air at 95%RH, which doesn’t seem particularly relevant, since increasing a sealed volume of air from zero to 95%RH is not a likely practical objective. > > You are asking for the amount of water required (presumably per unit time) to _maintain_ humidity at a certain level. This can’t be calculated from fundamental physical principles. It depends on how tightly your room is sealed from the outside environment, and the conditions in the outside environment (which will vary). I need to find the amount of water to be vaporized for 95% RH at room Temp 60 degree Celsius at a sealed 39.64 cubic meter room. can you help me on this? ---
**Author:** ![Riemann](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/riemann/32/3133_2.png) [@Riemann](https://boards.straightdope.com/u/Riemann)\ **Post date:** [August 13, 2017, 6:52pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/16 "2017-08-13T18:52:24Z")
> [@oparthiv](#): > > I need to find the amount of water to be vaporized for 95% RH at room Temp 60 degree Celsius at a sealed 39.64 cubic meter room. can you help me on this? Then watchwolf’s approach is correct, although I haven’t checked his arithmetic. You will, of course, need to subtract out the humidity of the air at the start, which is unlikely to be zero. ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [August 13, 2017, 8:28pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/17 "2017-08-13T20:28:57Z")
> [@Riemann](#): > > Then watchwolf’s approach is correct, although I haven’t checked his arithmetic. You will, of course, need to subtract out the humidity of the air at the start, which is unlikely to be zero. Yeah … I’m not sure what the application here is exactly … nor why the 95% RH is critical … however, I did some quick checking and for air at 60% RH and 25ºC we have an absolute humidity of about 12 g/kg … Be advised, I’m seeing some websites that define “absolute humidity” as the density of water vapor (in g/m[sup]3[/sup]) … and then using “specific humidity” for the mass ratio (in g/kg) … so just check and make sure the dimensions match if you find some value … Yes … the OP is advised to check **watchwolf’s** arithmetic … we’re not optimizing for maximum tax returns here … [wolfish grin] … ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [August 13, 2017, 10:17pm UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/18 "2017-08-13T22:17:42Z")
Check with hatchery operations, They maintain about 98% RH but also have some fresh air coming in. To be exact will require sensors fans and electronic controls over introduced moisture. 1 gallon of water would certainly be enough to get it up there but you will not be able to calculate this, you have to control it. ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [August 14, 2017, 1:59am UTC](https://boards.straightdope.com/t/how-to-calculate-the-amount-of-water-needed-to-keep-relative-humidity-95-in-a-20-6-5-8-room/793453/19 "2017-08-14T01:59:24Z")
Another way, and this assumes an air tight volume, is to heat the room up to 58ºC and provide plenty of liquid water so that the RH rises to 100% … seal the liquid water off and raise the temperature to 60ºC … now your RH is 95% …