Dehumidifier Runtime Calculator
Enter the room size, humidity readings, outdoor air, leakage, and dehumidifier capacity to estimate the drawdown time and whether the target can be held.
Estimated time to target RH2 hr 12 min
- Target achievable
- Yes
- Effective removal capacity (pints/day)
- 24.5
- Water initially removed from room air (pints)
- 0.89
- Continuous moisture load at target (pints/day)
- 18.06
- Estimated duty cycle after reaching target
- 73.69%
- Minimum nameplate capacity to hold target (pints/day)
- 25.79
Assumes one well-mixed air volume, sea-level pressure, and steady temperature, outdoor air, leakage, and dehumidifier capacity.
The initial water removed is airborne vapor only; damp materials or standing water can add much more moisture.
How to use this calculator
- Measure the starting indoor RH and enter the target RH you want the dehumidifier to reach.
- Enter the connected floor area, ceiling height, indoor temperature, and outdoor conditions.
- Use the nameplate pints per day and reduce it with the capacity factor if the room is cool or the unit is aging.
- Set air changes per hour and any added moisture sources, then read the runtime, steady load, and minimum capacity.
How the runtime is calculated
This calculator converts each relative-humidity reading into an actual humidity ratio, which is pounds or kilograms of water vapor per unit of dry air. It uses the Buck saturation-vapor-pressure equation over water:
pws = 0.61121 × exp[(18.678 − T ÷ 234.5) × T ÷ (257.14 + T)]
Here T is temperature in °C and pws is in kPa. Vapor pressure is pv = RH × pws, and the humidity ratio is W = 0.621945 × pv ÷ (101.325 − pv). The tool calculates W0 for the starting room air, Wt for the target, and Wo for the outdoor air.
The room air mass comes from floor area, ceiling height, and dry-air density. The dehumidifier rating is converted from pints per day to kg per hour and multiplied by the capacity factor. That derating matters because a unit rated at warm, humid test conditions often removes less water in a cool basement or at lower RH.
For a leaky or ventilated room, the model is a mixed-air moisture balance:
Weq = Wo + (Q − C) ÷ (ACH × Mda)
t = −ln[(Wt − Weq) ÷ (W0 − Weq)] ÷ ACH
C is effective dehumidifier removal, Q is internal moisture added per hour, ACH is air changes per hour, and Mda is dry-air mass. If ACH is zero, the calculator uses the sealed-room form t = Mda × (W0 − Wt) ÷ (C − Q).
What changes the answer most
Target feasibility is usually controlled by outdoor moisture, air leakage, and effective capacity. A large room can hold more airborne water, but the water stored in the air is often modest compared with moisture stored in concrete, wood, drywall, carpet, and furnishings. That is why a real damp basement can take much longer than the initial-air number suggests.
The continuous load at the target is max(0, Q + ACH × Mda × (Wo − Wt)). If outdoor air is drier than the target, ventilation can help and the mechanical load can fall to zero. If the load is greater than the derated capacity, the duty-cycle row reports the shortfall instead of implying that continuous operation can hold the target.
What this leaves out
The model assumes the room is well mixed and that temperature, outdoor conditions, air leakage, and removal capacity stay constant. It does not diagnose leaks, flooding, mold, drainage failures, or a failing dehumidifier. It also does not model moisture slowly released from building materials, so use the result as a planning estimate rather than a promise.
Worked example
For a 500 ft² area with an 8 ft ceiling at 70°F, starting at 70% RH and targeting 50% RH, the room contains about 294 lb of dry air. With outdoor air at 80°F and 70% RH, 0.35 ACH, a 35 pint/day unit derated to 70%, and no added internal moisture, the effective capacity is 24.5 pints/day.
The airborne vapor difference from 70% to 50% RH is about 0.9 pint. Outdoor leakage adds about 18.1 pints/day at the target, so the unit is modeled to reach the target in about 2 hr 12 min and then run at about 74% duty cycle. The minimum nameplate rating to hold that target at the same derating is about 25.8 pints/day.
Common questions
How long should a dehumidifier run before humidity starts dropping?
In a well-mixed room, the air reading can start moving within minutes, but the calculated time is the time to reach the target RH. If wet materials are releasing moisture, the hygrometer may fall slowly or plateau even when the airborne water calculation is small.
Why does room temperature change the drying time?
Relative humidity is a percentage of the water vapor the air can hold at that temperature. Warm air at 50% RH contains more actual moisture than cool air at 50% RH, so the same RH change can mean a different amount of water removed.
Why is a 35-pint unit unlikely to remove 35 pints every day in my basement?
Nameplate capacity is measured under standardized conditions, not necessarily your room conditions. Cooler air, lower RH, defrost cycling, clogged filters, poor airflow, and an older coil can all reduce real removal, which is why the calculator includes a capacity factor.
Can a dehumidifier reach 50% RH when it is humid outside?
Yes, if its effective removal rate is larger than the outdoor-air moisture load and any internal moisture sources. If leakage brings in moisture faster than the unit can remove it, the calculator reports the target as unreachable with the entered capacity.
What ACH value should I use for a basement or room?
Many closed residential spaces are roughly in the 0.2 to 0.5 ACH range, but actual leakage varies widely. Use a higher value for drafty rooms, open stairwells, frequent door opening, exhaust fans, or HVAC operation that exchanges air with other spaces.
Should I run the dehumidifier continuously or use its humidistat?
Use the drawdown time to estimate the initial run. After the target is reached, the duty cycle estimates how much of the day the unit may need to run to hold that RH under the entered conditions.