Dehumidifying vs. Cooling
Dehumidifying vs. Cooling
(OP)
I had a question regarding the amount of power required to dehumidify a space vs. the quanity of power to cool the same space. Is there a thumbrule for the power consumption to maintain a space at a low humidity level vs. maintaining the same space at a low humidity and temperature level? For example, does it require 20%(?) less power to only dehumidify and not cool?
Also, in a typically insulated space does humidity cycle back to exterior conditions at the same rate as temperature cycles?
Thanks.
Also, in a typically insulated space does humidity cycle back to exterior conditions at the same rate as temperature cycles?
Thanks.





RE: Dehumidifying vs. Cooling
Often the cooling coil is also used to dehumidify the space, and the costs to dehumidify are related to the outdoor design temperatures, and the desired indoor humidity levels.
Humidity tends to cycle with occupancy and ventilation, as people are the major internal producers of latent energy (unless you have a pool, or a group of showers, you get the idea).
More information about what you are trying to do, and where would result in more specific answers.
RE: Dehumidifying vs. Cooling
QS = m cp dT sensible (cool only)
QL = m (dW) ifg latent (dehumidify only)
QT = m (di) total (both)
Q = energy used
m = mass flow rate
cp = specific heat of air at constant pressure
dT = difference in temperature during cooling process
dW = change in humidity ratio (grains water/grains air)
ifg = change in enthalpy between states (latent heat)
di = difference in enthalpy
The power required will also be affected by the efficiency of the "machine" that accomplishes the objective.
RE: Dehumidifying vs. Cooling
Quick quick example:
Entering Air: 90 FDB, 70% RH
Supply Air condition you want: 55 F, 40%
For both 55 F and 40% you will need to cool and dehumidify (say using a typical cooling coil), running along the dew point until you reach your desired 50% RH - which happens to be at ~49 F. Now you will need to use reheat to bring the air up to the 55 F you desire, now using heating energy during a summer condition.
To go for just temperature you would be riding the saturation curve down to 55 F, and would deliver at a higher RH.
Aiming for RH alone, you could use a desiccant system to remove water, though comparing a cost vs. a cooling coil would require more detail on the amount of air, running time per year, and available equipment.
RE: Dehumidifying vs. Cooling
The rule of thumb is that it takes~ 970 btu's of energy to condense 1 lb of water(7000) grams (latent heat)and it takes only takes 1 btu to change 1-lb of water/air 1 degree F (sensible heat)
Also, in a typically insulated space does humidity cycle back to exterior conditions at the same rate as temperature cycles?
Ans. yes and no, it would depend on the amount of difference on the humidity level between inside and outside
RE: Dehumidifying vs. Cooling
First of all I am looking at the cost difference for cooling vs. dehumidification for a large, insulated building (e.g., open metal building 100 x 80 x 14-ft). The building will be located south of Houston. Assume outside conditions 85degreesF, 90% humidity.
The interior conditions to compare are:
a) cooling (and maintaining temperature) at 70 degreesF and not regulating humidity directly OR
b) dehumidifying (and maintaining humidity) at 40% humidity and not regulating temperature directly.
1. What is the best form of dehumidification for scenario "b"? I am unfamiliar with desiccant systems. Can they be used for large facilities?
2. In scenario "a", without using reheating, approximately what will the humidity level be maintained at?
3. In scenario "b", will the dehumidification process affect air temperature?
4. What is the relative power consumption for scenario "a" compared to "b"?
RE: Dehumidifying vs. Cooling
This is what I get
1. You will need about a 30 ton unit that can do 6 air changes/hr
2. This gives you about 12,000 CFM supply air
3. If 2400 CFM OSA air at 85*f and 90 % RH and 11,000 Return Air of 71*F and 47% RH has a dew point of 58*F so if you can keep the OSA at 2400 CFM or less you will be doing only a small amount of latent heat removal and mostly sensible cooling which requires less energy and cheaper to operate, however most air conditioning applications for comfort require about 15/20 CFM/person and I don’t know what the code requirements are for your application.
If you devide 12,00 CFM by the specific volume of 13.687 cuft/lb at the mixed air conditions you will get 877 lbs of air/min then x 5.92 btu/lb = 5,200 btu/lb min
4.. You would be condensing about 11 grains /lb of air x 877 lbs = ~ 1.38 lbs/min x 60 min/hr =83 lbs./8.33lbs/gal = ~ 10 gal hr of condensate. This is one scenario, others may have better ideas.
RE: Dehumidifying vs. Cooling
The return air should be only 9600 CFM I didn't make the correction above.
RE: Dehumidifying vs. Cooling
Most assuredly
"b) dehumidifying (and maintaining humidity) at 40% humidity and not regulating temperature directly"
aiming for a similar target myself in similar OA conditions in a bindery (paper is very sensitive to RH) we must reheat the dehumidified air for human comfort, something you may need consider.