Increasing Relative Humidity

1. Dec 2, 2013

Avis

I have to design a system to increase the relative humidity in a bunker. I would like to use a spray nozzle and service water. How could I determine how much the relative humidity would raiser per volume of water sprayed into the bunker.

Known:
- Volume of Bunker
- Temperature
- Initial Relative Humidity
- Air Flow (negligable)

2. Dec 2, 2013

Staff: Mentor

Check the vapor pressure of water at that temperature. This gives you how much water the air needs (per volume) to get from 0% to 100% relative humidity. If you want smaller changes in humidity, just scale down. This assume that all the air in the bunker is mixed properly. If that is not the case, it gets more complicated.

3. Dec 3, 2013

Avis

Not really following you. So if say at 20°C the vapor pressure is 2.3 kPa how much water per volume do i need?

4. Dec 3, 2013

Staff: Mentor

Look up "psychrometrics". From a psychometric chart or table you can find the water content of air at different conditions and thus calculate exactly how much water you need to evaporate to achieve the desired humidity.

5. Dec 5, 2013

Avis

Ok What If i were to have the following conditions.

Volume of space: 2200m^3
Temperature: 20°C
Initial Relative Humidity: 18%
Desired Relative Humidity: 65%
Air exhauted from space at: 0.25m^3/s

How much water would I have to spray in per hour or second to acheive that humidity level?

6. Dec 6, 2013

Staff: Mentor

Ok, typically we would want you to do more of it yourself, but there is enough SWAG in this that I'll have to help some anyway...

So here's an SI psych chart: http://www.uigi.com/UIGI_SI.PDF

1. How much moisture is gained/lost through the walls?
2. How much moisture is lost through exhaust/infiltration?

I'm assuming by "bunker", you mean concrete, underground, with no insulation or vapor barrier. Typically such bunkers are very humid because of moisture infiltrating through the concrete (like a basement). But if you are exhausting from it, you'll also be pulling in outside air, potentially at roughly zero humidity in winter. The infiltration rate you gave is small relative to the room volme (about 0.4 air changes an hour), but not negligible.

So my assumption will be that all of the added moisture is used to counteract the dry air coming in from outside.

20C @ 65% RH is about 9.5 grams of water per kg of air (from the chart). 0.25 m^3/sec is 0.3 kg of air per sec, so that's about 10 kg of water per hour that needs to be evaporated to counter-act the dehumidification of the entering dry air.

Of course, all bets are off if my assumption about infiltration through the walls is wrong. Indeed, for best results, you want a push-pull ventilation system with a positive airflow balance (supply airflow larger than the exhaust airflow) to eliminate other sources of infiltration.

7. Dec 6, 2013

Avis

We call it a bunker on site but a better term would be a silo. It is a metal container which is located inside. It is a safe assumtion that the intitial inside conditions are the same as the incoming air (air is pulled through filtered grates)

This is what I've done so far: (leas note I was way off on my estimation of the volume and have refined it)

Temperature: 20°C
Initial Relative Humidity: 18%
Desired Relative Humidity: 65%
Air exhauted from space at: 0.25m^3/s

@ initital conditions the enthalpy is 25 KJ/kg
@ final conditions the enthalpy is 40 kJ/kg
Change in enthalpy is 15 kJ/kg

Density of air at 18°C is 1.21246 kg/m^3
Mass of air in bunker is 1016kg

Therefore 15240 KJ of energy must be added.

Not sure where to go from there.

8. Dec 9, 2013

Staff: Mentor

Ok, sounds good - so no infiltration. And no heat loss then either?
That's an odd choice of words. Air is pulled in through filtered grates by the exhaust fan negatively pressurizing the room? Not even with fans to push through the filters? Does this air come in from outside? Or from adjacent rooms that are similarly conditioned? It would have to be from already similarly conditioned rooms to have the same conditions.
That calculation looks right, but if the air in the room is completely isolated from outside (of the room), then the air coming in through the filtered grates minus the air leaving through the exhaust is the only source of lost moisture/enthalpy. So while you correctly calculated the energy required to humidify the space, you didn't calculate the energy (power) required to keep it humidified. The calculations I made previously were for dry entering air, but we're in the same order of magnitude of each other.

9. Dec 11, 2013

Avis

Here is a rough drawing of what I'm dealing with.

I would assume No heat loss as bunker inside temperature is equal to bunker exterior temperature.

This air comes in from the adjacent area which has the same intital conditons and remain at thoes conditions. This areas humidity would not increase as the humidity in the bunker increased.

It would stay at:

Temperature: 20°C
Initial Relative Humidity: 18%

So to get from 18-65% relative humidity I would need to add roughly 7g of water/kg of dry air.

Thus the exhaust would take away 2.107g of water every second.

The whole volume of air would need 7.06kg of water to increase the humidity.

Not sure how to do the balance here.

10. Dec 11, 2013

Staff: Mentor

Does the "product" absorb water vapor?
Agreed.
Er, no - that's 2.1g * 3600s. That's kg/hr, not kg/ air change. But that's ok; that's what you want.
7 kg/hr is the answer you are looking for, though I'd throw in a healthy safety factor of perhaps 50%.