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equation to size the room de-humidifier 2

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moideen

Mechanical
May 9, 2006
359
I am seeking for the engineering equation that will tell me what size dehumidifier will be most suited for the room in question. I am appreciative of the reaction.

The problem with the world is that intelligent people are full of doubts, while stupid ones are full of confidence.
-Charles Bukowski-
 
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Buy one. If it's not enough buy another, bigger one. Their efficiency is unlikely to vary much with size, so running in parallel costs as much as one that has the same total capacity. In addition, if multiple are used then they can be dispersed to where the humidity is highest or needs the most control.
 
Due to the fact that a website provided the dehumidifier calculation formula (attached) for a 3000 m3 warehouse, which is 328 kg per hour for ambient 30°C with 70% RH for the needed 50% RH, I entered the same figure into their online calculator; nevertheless, it provided me with a different response. Confusing..

Link




The problem with the world is that intelligent people are full of doubts, while stupid ones are full of confidence.
-Charles Bukowski-
 
 https://files.engineering.com/getfile.aspx?folder=bf518345-5043-4d8d-b17d-6c9ecf26ae8a&file=how-to-select-dehumidifier.pdf
It looks like the example is wrong - air at 30C 70%RH should be about 21g/m^3 water content, not 138g/m^3.
 
@3DDave
Exactly. I had noticed this before, but I was unsure of my understanding. I calculated and found a result of 24 kg/hr based on this calculation, but the same website gave a different value like 12 kg/hr.
Space volume 2000m3
Air flow 4000 m3/hr
Airchange 4000/2000=2
Current humidity 70% at 30c(21 g/m3)
Required humidity 50%(15g/m3)
Capacity of dehumidifier = (21-15)*2000*2/1000=24kg/hr


The problem with the world is that intelligent people are full of doubts, while stupid ones are full of confidence.
-Charles Bukowski-
 
Hi moideen, this Condensate Rate Calculator comes up with the same solution you calculated, 24 kg/hr.

condensate_rate_kqxksl.png


Adrienne Gould-Choquette, P.E.
 
@AdrienneGC -thank you , calculator, an instrument that is helpful

The problem with the world is that intelligent people are full of doubts, while stupid ones are full of confidence.
-Charles Bukowski-
 
Forget the equations that you requested instead make a copy of a blank psychrometric graph to determine the amount of water vapor to be removed by the dehumidifier. If you took thermodynamics in the past your text book will have such chart. The only thing that I see from your data is the exit temperature from the dehumidifier, however,you could estimate that value if you knew the bulk air temperature exiting the cooling coil of the dehumidifier; that air temperature is somewhat dry and will be slightly higher than the cooling coil fins so some judgement will be required. You can plot the dry air temperatures and relative humidity of the incoming and exiting air such psychrometry graph sheet to calculate the difference of mass of water per mass of dry air to size the unit. A recommended reference which I suggest you read before hand is Perry Chemical Engineering handbook on the subject of psychrometry as there is a section which deals on determining the air temperature exiting cooling coils.
 
Doing your problem long hand while using English units and then reverting back to metric for an answer I got 53.6 Lbm/Hr (=24.3Kg/hr) of water vapor to be removed as condensate by assuming that the exit temperature from the dehumidifier to be the same as the inlet temperature of 86dF (=30dC). Note that the exit temperature will be a little higher then 86dF (30dC) since dehumidifiers incorporate an evaporator (cooling coil) and a condensor (heating coil)behind the cooling coil. A psychrometic graph will will give a close approximation to your calculator and long hand calculation.
 
Are you double counting here?

Unless you're pulling in fresh air into this unit, then you only have 2000m3 of air available to reduce the humidity.

So yes, right at the start you might be pulling 0.4kg/min, but after 30 mins you've theoretically pulled all the water out of the room no?

Think you need to see over what sort of timescale you're talking to reduce the humidity and also how many air changes are happening with the outside more humid air.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Ll, there is no re circulation of room air in the calculations. If re circulation was to be accounted then there are several factors such as occupant load, moisture generating equipment, time element to reach the 50% relative humidity, weather conditions affecting ambient air leakage into the room which all must be accounted. At that point a more rigorous analysis involving the time element and a spread sheet program would be needed other than hand calculations.
 
My point exactly...

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
LI, the 24.3kg/hr would be a valid number if the OP intended to have re circulation of room air maintained at a relative humidity of 70% by dehumidifing to a relative humidity of 50%. For lack of exit temperature from the dehumidifier, the 24.3kg /hr was based on the dry bulb temperature of 30dC (=86 dF for my long calculations). Essentially I based the OP's request as a steady state problem instead of considering a time element to dehumidify the room which would involve a first order differential equation.
 
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