L5 · HVAC Loads and Psychrometrics

Foundation lesson · 25 minute study session

Track moisture removal before adding reheat

A wet cooling coil changes enthalpy and humidity ratio. Reheat then raises temperature without restoring the moisture removed by the coil.

By the end of this lesson

Calculate condensate using a dry-air mass balance.
Distinguish coil duty from reheat duty.
Explain direct evaporation and steam humidification.

Condensation requires a wet process

Cooling alone does not guarantee dehumidification. Moisture condenses when air contacts sufficiently cold surfaces and leaves at a lower humidity ratio. Calculate condensate from the dry-air mass flow multiplied by the change in humidity ratio. Reheat increases $h$ while leaving $W$ unchanged in a dry coil.

Reheat changes temperature

After dehumidification, sensible reheat raises dry-bulb temperature at essentially constant humidity ratio. The resulting lower RH reflects warmer air, not additional water removal. Coil leaving condition and final supply condition must be kept separate.

Choose the humidification model

Direct evaporative cooling generally lowers dry bulb while increasing humidity ratio, with nearly constant enthalpy under common approximations. Steam injection adds both moisture and energy. Use the complete energy balance when water or steam enthalpy is supplied. For a wet coil, condensate carries some energy out; neglect it only when the problem permits that approximation.

A decision map for this lesson

Cooling removes moisture; dry reheat changes temperature

Track the drain and the enthalpy change separately. Dry reheat does not undo the moisture removal.

A realistic air-handler cutaway reveals a cooling coil and condensate pan followed by a separate reheat coil and supply fan. — **Follow the drain before adding reheat**Connect the equipment to the calculation

Cooling changes enthalpy from 36 to 22 and humidity ratio from 0.014 to 0.008; reheat changes enthalpy to 25 while moisture ratio stays 0.008. At 9000 pounds dry air per hour, condensate is 54 pounds per hour. — **Follow the drain before adding reheat**Connect the equipment to the calculation

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Cooling coil with a falling W	Find condensate from $\dot m_{cond}=\dot m_{da}(W_i-W_o)$ .	Using the drop in RH as the water-removal rate.
Coil thermal duty	Use the air enthalpy decrease; include condensate enthalpy when the specified model requires it.	Omitting an outlet energy stream when the question has not authorized that approximation.
Dry reheat after the coil	Increase h while keeping W fixed when no moisture transfer occurs.	Assuming that a lower leaving RH means more water was removed.
Net cooling across coil plus reheat	Subtract the heat added by reheat from the cooling-coil removal.	Calling gross coil duty the net duty across both devices.

On a narrow screen, scroll across the table to compare the decision and the error.

Relationships and their limits

Condensate flow

\dot m_w=\dot m_{da}(W_{in}-W_{out})

Dry-air mass flow and humidity ratios must have matching time and mass units.

Cooling duty with condensate

\dot Q=\dot m_{da}[(h_{in}-h_{out})-(W_{in}-W_{out})h_w]

Use a compatible enthalpy datum. A problem may authorize neglecting the condensate enthalpy term.

Sensible reheat

\dot Q_{reheat}=\dot m_{da}(h_{supply}-h_{coil})

Humidity ratio is unchanged across a dry reheat coil.

Symbols and unit basis
Symbol	Meaning	Units or basis
$\dot m_w$	Condensate mass flow	lb water/h
$\dot m_{da}$	Dry-air mass flow	lb dry air/h
$W$	Humidity ratio	lb water/lb dry air
$h$	Moist-air enthalpy	Btu/lb dry air
$h_w$	Condensate enthalpy on compatible datum	Btu/lb water

One decision, fully worked

Apply the method.

Air flows at 9,000 lb dry air/h through a cooling coil. Inlet h = 36 Btu/lb and W = 0.014; leaving h = 22 Btu/lb and W = 0.008. Neglect condensate enthalpy. Reheat increases enthalpy to 25 Btu/lb without adding moisture.

Calculate water removal from the humidity-ratio difference.
Use inlet and coil-leaving enthalpy for cooling duty.
Use coil-leaving and final-supply enthalpy for reheat duty.
Keep the final humidity ratio at the coil-leaving value.

\dot m_w=9{,}000(0.014-0.008)=54\ \mathrm{lb/h}

\dot Q_{coil}=9{,}000(36-22)=126{,}000\ \mathrm{Btu/h}

\dot Q_{reheat}=9{,}000(25-22)=27{,}000\ \mathrm{Btu/h}

Decision

The coil removes 54 lb/h of water and 126,000 Btu/h. Reheat adds 27,000 Btu/h; final W remains 0.008.

Independent check

Net air-stream energy removal is 99,000 Btu/h, matching 9,000 × (36 − 25).

Specialist extension

Separate cooling, moisture removal and reheat

Cooling below the entering dew point can remove moisture. A subsequent dry reheat coil changes enthalpy and dry-bulb temperature without changing humidity ratio. For a wet-coil energy balance, distinguish a simple air-enthalpy approximation from a full balance that includes liquid condensate energy. Apply the explicitly supplied model and boundary.

A new applied case

Air at 6,000 lb dry air/h leaves a cooling coil at h = 22 Btu/lb dry air and W = 0.008. Dry reheat adds 24,000 Btu/h.

h_s=22+24{,}000/6000=26\ \mathrm{Btu/lb_{da}}

\dot m_v=6000(0.008)=48\ \mathrm{lb\ water/h}

Decision. Supply enthalpy is 26; water-vapor flow remains 48 lb/h.

Cross-check. A dry reheat step cannot by itself solve an excessive supply humidity ratio.

Unit basis. h is Btu/lb dry air; W is lb water/lb dry air; mass rate is lb dry air/h; heat input is Btu/h.

NCEES PE Mechanical Reference Handbook v2.1; see the module reference path.

Open targeted practice in the study studio →

Transfer the method

HVAC Loads and Psychrometrics · HV-LC-L5

Try a new decision.

Select one

A cooling coil processes 12,000 lb dry air/h. Inlet h = 38 Btu/lb dry air and W = 0.015; outlet h = 24 and W = 0.009. Neglect condensate enthalpy as authorized here. The total cooling duty and condensate rate, respectively are most nearly:

A168,000 Btu/h; 0 lb water/hB456,000 Btu/h; 180 lb water/hC72 Btu/h; 168,000 lb water/hD168,000 Btu/h; 72 lb water/h

A lesson check supports practice, not a readiness score. Your response stays in this page session and clears on reload.

Catch the common errors

Using final supply h to calculate the gross cooling-coil duty.
Assuming reheat adds moisture because dry bulb rises.
Applying an enthalpy approximation without checking the stated condensate assumption.

Handbook navigation drill

Find it, then explain it.

Search: Moist-Air Cooling and Dehumidification

Find the water and energy balances, then trace cooling and reheat as two separate steps.

Check your navigation

You can state which quantities change and which remain constant in each step.

Reference path: §9.2.2–9.2.6, pp. 480–482; §9.3.10–9.3.11, pp. 525–526. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Theory sources

NCEES handbook v2.1, §§9.2.2 and 9.2.4.
Grondzik and Kwok, Mechanical and Electrical Equipment for Buildings, 12th ed., ch. 12.
Lechner, Heating, Cooling, Lighting, 4th ed., §§10.12 and 10.15.

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