That would depend upon the sensible heat factor...the lower that factor higher the water output or moisture removal per BTU or kJ. The latent heat removal rate should be divided by latent heat of water.

Hi D.B.,

For convenience your question is rewritten here.

“How much condensed water can I obtain from each BTU of an air-conditioner equipment in a building?” 

If the relative humidity is zero, no water vapor would condense. So, what is being sought is an estimate of the maximum amount of water that can be condensed using 1 BTU with a non-zero relative humidity.  Implicit in the question is the assumption the energy available for condensation is used only for condensation; no energy is going into cooling or freezing of the condensate.  

It is further assumed, to estimate the maximum amount of water that can be condensed using 1BTU, that the maximum air temperature the air conditioner works with  is 120 F, which is approximately 322 K, and  that the relative humidity is 100%

So let us calculate the maximum amount of condensed water for each BTU, with these assumptions.

If Q is to be the maximum mass of the condensed water then:

Qx(heat of condensation) =1BTU=250.164 calorie

 Q= (250.164 calorie)/(Heat of condensation)

The heat of condensation is the same as the heat of vaporization. Please see:  https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

At 320 K, the heat of condensation is estimated to be about   43000 joules/mole

So with some rounding,  Q is found by first getting somewhat consistant units, and then dividing.

43000 (joules/mole) x 1 cal/4.184joules = 10277.246 cal/mole

10227.246( cal/mole)  1 (mole/ 18 gms) =571 cal/gram

43000 joules/mole = 571 cal/gram

Q= (250.164 cal)/(571 [cal/gram])

Q= .44 grams

So, .44 grams appears to be the maximum amount of water that can be condensed using 1 BTU.

Best regards,

Jack

thanks for the valuable answer.