E6 · HVAC Equipment and Components

Foundation lesson · 18 minute study session

Treat a coil as an air process and a heat exchanger.

Connect air-side enthalpy, moisture removal and a supplied bypass model without confusing sensible and total duty.

By the end of this lesson

Determine coil duty on a dry-air mass basis.
Explain apparatus dew point and an explicitly supplied bypass-factor model.
Check temperature and humidity results against their endpoint states.

Follow both heat and moisture.

A dry heating or cooling coil changes sensible energy without changing humidity ratio. A wet cooling coil can remove water and reduce enthalpy. Total air-side duty uses dry-air mass flow times enthalpy change; the stated model may neglect condensate enthalpy, or may require it explicitly.

A bypass model describes an idealized outlet.

In the supplied model, a bypassed fraction remains at the inlet state and the remaining fraction reaches the apparatus-dew-point state. Humidity ratio and enthalpy follow mass-weighted mixing. A temperature-based bypass factor is a commonly used approximation; it is not proof that a physical fraction of air avoids all contact with the coil.

Match the second side and the selection basis.

Water or refrigerant must carry the same heat across the chosen coil boundary, subject to stated losses. Coil area, water flow, face velocity, pressure drop and fouling are separate selection constraints. Do not use a room sensible load alone as a wet-coil total-capacity requirement.

A decision map for this lesson

A wet coil needs both a thermal and a moisture check

Use the explicitly supplied bypass model and state data. Apparatus dew point is not automatically the leaving dry bulb.

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Temperature bypass factor is specified	Use $BF=(T_o-T_{adp})/(T_i-T_{adp})$ within that model.	Using room dry bulb in place of the coil inlet or apparatus state.
Humidity mixing model is supplied	Use $W_o=BF\,W_i+(1-BF)W_{adp}$ .	Assuming a temperature result alone establishes latent performance.
Face velocity or water flow is limited	Check geometry and fluid-side duty as separate sizing constraints.	Using nominal tonnage to bypass a stated face-velocity limit.
Catalog and design entering states differ	Obtain performance at the actual specified air and fluid conditions.	Selecting the most favorable catalog capacity regardless of the design state.

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

Relationships and their limits

Air-side total duty

\dot Q\approx\dot m_{da}(h_{in}-h_{out});\quad\dot m_w=\dot m_{da}(W_{in}-W_{out})

Cooling positive in this convention. Heat relation neglects condensate enthalpy only when authorized.

Supplied temperature bypass approximation

BF\approx\frac{T_{out}-T_{adp}}{T_{in}-T_{adp}}

A stated idealized model, not a guaranteed handbook lookup. Inlet and outlet states must fit the process.

Idealized humidity mixing

W_{out}=BF\,W_{in}+(1-BF)W_{adp}

Treat BF as dry-air mass fraction in this simplified mixture model; no additional moisture source.

Symbols and unit basis
Symbol	Meaning	Units or basis
$BF$	Supplied bypass fraction/model parameter	dimensionless
$T_{adp}$	Apparatus-dew-point model temperature	°F
$W$	Humidity ratio	lb water/lb dry air
$h$	Moist-air enthalpy	Btu/lb dry air
$\dot m_{da}$	Dry-air mass flow	lb dry air/h
$T_{in},\ T_{out}$	Air dry-bulb inlet and outlet temperatures	°F
$\dot m_w$	Water removal rate	lb water/h
$\dot Q$	Coil cooling rate	Btu/h

One decision, fully worked

Apply the method.

For a simplified coil model, inlet air is 85°F and outlet air is 55°F. The apparatus dew point is 45°F. Supplied inlet and apparatus-dew-point humidity ratios are 0.012 and 0.0063 lb water/lb dry air. Use the stated temperature bypass approximation and humidity mixing model.

Measure both temperature differences from the apparatus dew point.
Find the bypass parameter from the supplied model.
Use the parameter as the dry-air mixing fraction for humidity ratio.

BF=\frac{55-45}{85-45}=0.25

W_{out}=0.25(0.012)+0.75(0.0063)=0.007725\ \mathrm{lb_w/lb_{da}}

Decision

The simplified bypass parameter is 0.25 and outlet humidity ratio is 0.007725 lb water/lb dry air. Coil duty still requires dry-air mass flow and an enthalpy balance; bypass factor alone is not a capacity rating.

Independent check

Both modeled outlet properties lie between their inlet and apparatus-dew-point endpoints. Outlet humidity is closer to the apparatus value because the treated fraction is 75%.

Specialist extension

Account for condensate energy when the case requires it

A wet-coil total-duty shortcut uses dry-air mass flow times air-enthalpy change. A full steady balance also accounts for the energy carried out by condensate on a compatible datum. The liquid term is often small, but omitting it when expressly requested changes the requested boundary. Never subtract liquid energy on a datum inconsistent with the air properties.

A new applied case

Air flow is 5,000 lb dry air/h. Inlet h = 35 and outlet h = 23 Btu/lb dry air. W falls from 0.014 to 0.008. Condensate exits at supplied liquid enthalpy 15 Btu/lb on a compatible datum.

\dot m_{cond}=5000(0.014-0.008)=30\ \mathrm{lb/h}

\dot Q_{removed}=5000(35-23)-30(15)=59{,}550\ \mathrm{Btu/h}

Decision. The full stated cooling balance gives 59,550 Btu/h.

Cross-check. The air-only approximation gives 60,000 Btu/h. The 450 Btu/h liquid outlet energy explains the difference.

Unit basis. h is Btu/lb dry air; W is lb water/lb dry air; liquid h is Btu/lb water on a compatible datum; rates are lb/h.

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

Open targeted practice in the study studio →

Transfer the method

HVAC Equipment and Components · HV-LC-E6

Try a new decision.

Select one

A supplied idealized coil model uses BF = ( $T_{out}$ − $T_{adp}$ )/( $T_{in}$ − $T_{adp}$ ) and $W_{out}$ = BF $W_{in}$ + (1 − BF) $W_{adp}$ . Given $T_{in}$ = 90°F, $T_{out}$ = 60°F, $T_{adp}$ = 50°F, $W_{in}$ = 0.014 and $W_{adp}$ = 0.0076 lb water/lb dry air, the outlet humidity ratio is most nearly:

A0.0092 lb water/lb dry airB0.0108 lb water/lb dry airC0.0035 lb water/lb dry airD0.0124 lb water/lb dry air

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 relative humidity in place of humidity ratio in the moisture balance.
Claiming the temperature bypass equation is an exact physical mass split for every coil.
Computing total wet-coil duty from temperature change alone.

Handbook navigation drill

Find it, then explain it.

Search: Cooling and Dehumidification

Find the handbook air-side energy and moisture relationships. Explain which additional properties are needed to turn this model state into a coil-duty calculation.

Check your navigation

Section 9.2.2 supports the energy and moisture balances. This lesson explicitly supplies the bypass model; it does not promise that model is printed in the handbook.

Reference path: §9.2.1–9.2.2, pp. 479–480; §5.5, pp. 312–314. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Theory sources

NCEES PE Mechanical Reference Handbook v2.1: §9.2.1–9.2.2, pp. 479–480; §5.5, pp. 312–314
Mechanical and Electrical Equipment for Buildings, 12th ed.: mechanical systems and energy principles. Supporting theory; not an examination reference.

Worked examples and lesson prose are original ReasonSpan^™ material. Obtain authorized publications separately; this preview does not reproduce the supplied books or handbook.