3.6 Humidification & Dehumidification Equipment: Steam, Adiabatic, Desiccant & Low-Temperature Processes

Key Takeaways

  • Humidifier capacity at sea level is m_w (lb/hr) = 4.5 x CFM x (W2 - W1), where 4.5 comes from 60 min/hr multiplied by the 0.075 lbm/ft3 standard air density.
  • At 5,000 ft the standard air density falls to about 0.0625 lbm/ft3, so the humidification constant drops from 4.5 to roughly 3.75 and a sea-level sizing overstates required capacity by about 17%.
  • Isothermal (steam) humidification moves the air state almost vertically at nearly constant dry-bulb temperature, while adiabatic (evaporative) humidification follows a line of constant wet-bulb temperature and depresses dry-bulb temperature.
  • Cooling-coil dehumidification cannot produce a leaving dew point below the coil apparatus dew point, so deep drying below roughly 40 F dew point requires a desiccant wheel or a liquid desiccant contactor.
  • Desiccant regeneration air is typically heated to 180 F to 280 F, and the process air leaves the wheel hot and dry, requiring downstream sensible cooling that must be added to the system load.
Last updated: August 2026

3.6 Humidification & Dehumidification Equipment: Steam, Adiabatic, Desiccant & Low-Temperature Processes

The NCEES specification lists sub-topic 1C, Humidification/Dehumidification Processes (sea level, 5,000-ft elevation, low temperature) as its own line item, and the parenthetical is the giveaway: you are expected to size moisture equipment away from sea level and below freezing, not only at standard conditions. Sections 3.3 and 3.5 developed the psychrometric processes. This section supplies the equipment, the sizing arithmetic, and the selection logic that turns a state-point change into a specified humidifier or dryer.


1. Sizing the Humidification Load

Humidification is a mass balance on water, not an energy balance. Air entering at humidity ratio W1 must leave at W2, and the difference is supplied as liquid or vapor:

m˙w=60ρQ(W2W1)\dot{m}_w = 60 \, \rho \, Q \, (W_2 - W_1)

where the mass flow is in lb/hr, Q is volumetric airflow in cfm, and the density is that of the air being humidified. At sea level with standard air at 0.075 lbm/ft3, the constant collapses to the familiar 4.5:

m˙w=4.5Q(W2W1)\dot{m}_w = 4.5 \, Q \, (W_2 - W_1)

Worked Example - Winter Humidification at Sea Level

A 100% outdoor-air unit delivers 10,000 cfm. Outdoor air at 10 F and 60% relative humidity carries about W1 = 0.0011 lb/lb. After preheat, the space requires supply air at W2 = 0.0055 lb/lb.

  • Moisture added per pound of dry air: 0.0055 - 0.0011 = 0.0044 lb/lb
  • Humidifier capacity: 4.5 x 10,000 x 0.0044 = 198 lb/hr of steam
  • Latent energy carried in: roughly 198 lb/hr x 1,100 Btu/lb = 218,000 Btu/hr

Cross-check the latent load directly with the 4,840 constant, which is simply 4.5 multiplied by the approximate latent heat of vaporization of 1,076 Btu/lb: 4,840 x 10,000 x 0.0044 = 213,000 Btu/hr. The two agree within chart-reading precision.

The Same Load at 5,000 ft

Density is what changes, and nothing else in the equation does. Standard air at 5,000 ft is about 0.0625 lbm/ft3, roughly 83% of sea-level density.

  • Corrected constant: 60 x 0.0625 = 3.75
  • Humidifier capacity: 3.75 x 10,000 x 0.0044 = 165 lb/hr

Sizing that same unit with the sea-level constant would specify 198 lb/hr against a true requirement of 165 lb/hr - a 20% oversize that shows up as short-cycling and poor control, not as extra capacity you can use. Every "1.08 / 4.5 / 4,840" constant in HVAC is a density constant in disguise, and every one of them must be scaled by the actual-to-standard density ratio when a question names an elevation.


2. Isothermal vs. Adiabatic Humidification

AttributeIsothermal (Steam)Adiabatic (Evaporative)
Water phase suppliedVapor, already boiledLiquid, evaporated in the airstream
Psychrometric pathNearly vertical; dry-bulb rises 1 F to 3 F from steam superheatFollows a line of constant wet-bulb temperature
Effect on dry-bulbEssentially unchangedFalls; the latent heat of vaporization is drawn from the air
Energy sourceBoiler, electrode, resistive, or gas-fired steam generatorFan energy plus the sensible heat already in the air
Typical useWinter humidification, hospitals, museums, cleanroomsData centers, dry climates, free cooling and pre-cooling
Water qualityTolerant; minerals stay in the boiler and are blown downDemands treated or reverse-osmosis water to avoid dust

The exam distinction to internalize: an evaporative humidifier is also an evaporative cooler. Adding 0.0044 lb/lb adiabatically to 10,000 cfm removes roughly 213,000 Btu/hr of sensible heat from the air, dropping dry-bulb temperature by about 20 F. In winter that heat must be replaced by the preheat coil, so adiabatic humidification does not avoid the energy - it relocates it from a steam plant to a heating coil.

Absorption Distance

Both types require an absorption distance downstream of the dispersion manifold before the moisture is fully mixed. Wetting a downstream filter, coil, duct liner, or humidity sensor is the classic field failure. Practical rules: keep the high-limit duct humidistat well downstream of the manifold, and never place a humidifier immediately upstream of a filter bank or a turning vane.


3. Dehumidification and the Apparatus Dew Point Floor

There are only two ways to remove moisture from air: cool it below its dew point and condense the water out, or transfer the water into a sorbent.

Cooling-Coil Dehumidification

The coil is analyzed with the apparatus dew point and bypass factor developed in Section 3.5. The consequence that matters here is a hard physical limit: the leaving air dew point can never be lower than the coil apparatus dew point. With chilled water at 42 F to 45 F supply, a realistic ADP is 48 F to 52 F, which places the practical floor for a chilled-water coil at roughly a 50 F leaving dew point - about 0.0076 lb/lb.

That is adequate for comfort conditioning and for most 62.1 ventilation air. It is not adequate for:

  • Ice rinks and refrigerated warehouse anterooms, where a 20 F to 35 F dew point is needed to stop fog and frost
  • Pharmaceutical and lithium battery manufacturing, at dew points down to -40 F
  • Supermarket cases, where the store dew point drives case-coil frosting and defrost energy

Pushing a chilled-water coil colder is not the answer: to hit a 40 F dew point you would need roughly 33 F to 35 F ADP, which freezes the coil and requires glycol, and the deep reheat penalty grows with every degree.

Desiccant Dehumidification

A solid desiccant wheel rotates slowly (typically 6 to 20 rph) between the process airstream and a heated regeneration airstream. Silica gel, molecular sieve, and lithium-chloride-impregnated media adsorb water vapor because the vapor pressure at the desiccant surface is below the vapor pressure of the process air.

Two effects that surprise candidates:

  1. Process air leaves hot. Adsorption releases the heat of vaporization plus a heat of wetting, so process air can leave the wheel 30 F to 60 F warmer than it entered. That sensible gain lands on the downstream cooling coil and must be counted in the system load.
  2. Regeneration air must be heated to 180 F to 280 F, typically by gas burner, steam coil, or recovered condenser heat. Regeneration energy, not the wheel, dominates the operating cost.

Liquid desiccant systems spray a lithium chloride or glycol solution over a contactor. They dehumidify at a lower regeneration temperature, tolerate deeper drying, and can remove particulates and bioaerosols, but they carry the risk of desiccant carryover into the airstream.


4. Low-Temperature Behavior

Below freezing, three things change and each has appeared in different forms on mechanical exams:

  • Humidity ratios become very small. Saturated air at 0 F holds about 0.0008 lb/lb; at 70 F it holds about 0.0158 lb/lb. Outdoor air in winter is, in absolute terms, almost perfectly dry, which is exactly why winter humidification loads are dominated by ventilation rate rather than by envelope leakage.
  • Frost, not condensate. A coil or wheel operating below 32 F builds frost rather than draining condensate, which blocks airflow and destroys capacity. Refrigerated coils need a defrost strategy; energy recovery wheels need preheat, wheel-speed modulation, or bypass to hold the exhaust-side surface above frosting.
  • The psychrometric chart runs out. Standard sea-level charts stop near 32 F. Low-temperature work uses either a dedicated low-temperature chart or the humidity ratio equation directly:

W=0.622pwppwW = 0.622 \, \frac{p_w}{p - p_w}

with the saturation pressure taken over ice rather than over liquid water below 32 F.


5. Selection Logic

Required Leaving Dew PointPractical Equipment Choice
Above 50 FChilled-water cooling coil alone
45 F to 50 FDeep-row coil with low chilled-water supply temperature, or coil plus modest reheat
35 F to 45 FDirect-expansion coil at low suction, or coil plus desiccant polishing
Below 35 FDesiccant wheel or liquid desiccant, with precooling to reduce the desiccant load

The economical arrangement for deep drying is almost always hybrid: use a cooling coil to knock the bulk moisture out at high efficiency, and use the desiccant only for the last increment of drying, where its regeneration cost buys something a coil cannot deliver at all.

Test Your Knowledge

A 100% outdoor-air handling unit at an elevation of 5,000 ft delivers 8,000 cfm and must raise the humidity ratio from 0.0012 lb/lb to 0.0052 lb/lb. Standard air density at that elevation is approximately 0.0625 lbm/ft3. What steam humidifier capacity is required?

A
B
C
D
Test Your Knowledge

An evaporative (adiabatic) humidifier is added to a 12,000 cfm supply duct carrying 70 F air, and it raises the humidity ratio by 0.0030 lb/lb. Neglecting fan heat, what happens to the dry-bulb temperature of the air leaving the humidifier?

A
B
C
D
Test Your Knowledge

A process requires supply air at a 30 F dew point. The available chilled water plant supplies 44 F water to a deep-row coil with an apparatus dew point of 50 F. What is the correct engineering conclusion?

A
B
C
D
Test Your Knowledge

Air leaves a solid desiccant wheel considerably warmer than it entered. What is the primary reason, and what design consequence follows?

A
B
C
D