L1 · HVAC Loads and Psychrometrics

Foundation lesson · 25 minute study session

Build the thermal path before calculating a load

A correct load begins with a physical boundary and the right thermal path. More insulation, shading and thermal mass affect a building in different ways.

By the end of this lesson

Distinguish heat gain from cooling load.
Combine series resistances and parallel heat-flow paths.
Use the correct area and design temperature difference.

Choose the boundary

Heat gain is energy entering the space. Cooling load is the rate that must be removed to hold the desired conditions at a particular time. Radiation absorbed by room surfaces may become a cooling load later. Equipment load can also include outdoor-air treatment, duct gains and fan heat. Identify which quantity the problem requests before adding terms. Define $U$ on the full assembly area and verify that $\sum f_i=1$ .

Series and parallel paths

For a uniform layer, area-normalized thermal resistance is thickness divided by conductivity. Add layers in series, including surface films when applicable. Framing and insulation paths act in parallel: calculate each path U-factor and area-weight those U-factors. Averaging their R-values gives the wrong assembly performance.

Solar gain and storage

Opaque conduction, glazing conduction and solar gain are separate contributions. If a problem supplies incident solar irradiance and SHGC, multiply them by the applicable glazing area. Thermal mass shifts heat flow in time; it does not substitute for insulation. Passive strategies help explain load reduction, but the examination calculation must follow the stated design method and data.

A decision map for this lesson

Choose the thermal path before the coefficient

Separate series layers from parallel area paths before calculating the assembly heat rate.

A realistic wall cutaway reveals insulation bays, wood framing and continuous sheathing. — **Series layers and parallel wall paths**Connect the equipment to the calculation

An assembly with 85 percent insulated bays at R-20 and 15 percent framing at R-6; conductances are weighted by area to give U equal to 0.0675. — **Series layers and parallel wall paths**Connect the equipment to the calculation

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Layers along one heat path	Add area-normalized resistances: $R_{path}=\sum R_i$ . Then $U_{path}=1/R_{path}$ .	Adding conductances for layers that are in series.
Framing and insulated bays side by side	Weight path conductances by area: $U_{assembly}=\sum f_iU_i$ .	Averaging the path R-values instead of their conductances.
A supplied U or R already includes films	Use the stated assembly basis and count each physical resistance once.	Adding inside and outside films again.
An instantaneous space peak	Check timing, thermal storage and simultaneous zone loads.	Treating steady conduction or separate zone maxima as the complete coincident peak.

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

Relationships and their limits

Area-normalized resistance

R_A=L/k;\qquad U=1/\sum R_A

Keep thickness and conductivity units consistent. U includes all resistances included in the specified assembly.

Parallel paths

U_{assembly}=\sum_i f_iU_i;\qquad \dot Q=UA\Delta T

Area fractions sum to one. This is a steady-state conduction relationship, not a complete transient cooling-load method.

Symbols and unit basis
Symbol	Meaning	Units or basis
$R_A$	Area-normalized thermal resistance	h·ft²·°F/Btu
$U$	Overall heat-transfer coefficient	Btu/(h·ft²·°F)
$f_i$	Area fraction for path i	dimensionless
$A$	Assembly area	ft²
$\dot Q$	Heat-transfer rate	Btu/h
$\Delta T$	Driving temperature difference	°F difference
$L$	Layer thickness	ft
$k$	Layer conductivity	Btu/(h·ft·°F)

One decision, fully worked

Apply the method.

An opaque wall has 800 ft² of net area. Insulated bays occupy 85% with R-20; framing occupies 15% with R-6. The stated path R-values already include films. The design temperature difference is 35°F. Find steady-state heat transfer.

Use the net opaque wall area; no window area is included.
Invert each path resistance before weighting.
Combine U-factors using the stated fractions.
Multiply assembly U by area and temperature difference.

U=0.85/20+0.15/6=0.0675\ \mathrm{Btu/(h\,ft^2\,^{\circ}F)}

\dot Q=0.0675(800)(35)=1{,}890\ \mathrm{Btu/h}

Decision

The wall transfers 1,890 Btu/h under the stated steady-state conditions.

Independent check

The assembly R-value is about 14.8, lower than R-20 because framing bypasses some insulation.

Specialist extension

Turn an envelope limit into a design requirement

A design constraint is often given as an allowable heat rate rather than a required R-value. Solve for total resistance on the stated area basis, then subtract the layers already present. For parallel paths, solve the conductance-weighted area fractions instead. A steady conduction target does not by itself establish the hourly cooling peak of a massive enclosure.

A new applied case

An 800 ft² roof has existing series resistance 6 h·ft²·°F/Btu. At a 45°F difference, conductive gain must not exceed 1,800 Btu/h. Ignore thermal bridges.

R_{total}=A\Delta T/\dot Q_{max}=800(45)/1800=20

R_{add}=20-6=14\ \mathrm{h\,ft^2\,^{\circ}F/Btu}

Decision. Add at least 14 on this resistance and area basis.

Cross-check. Substitution gives 800 × 45 / (6 + 14) = 1,800 Btu/h.

Unit basis. A is area in ft²; ΔT is °F; heat rate is Btu/h; area-normalized R is h·ft²·°F/Btu.

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-L1

Try a new decision.

Select one

A wall has 600 ft² of area. Insulated paths cover 80% with R = 16 h·ft²·°F/Btu; framing paths cover 20% with R = 4 on the same basis. Ignore other resistances. For a 30°F temperature difference, the steady conduction rate is most nearly:

A4,500 Btu/hB1,800 Btu/hC1,324 Btu/hD1,125 Btu/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

Averaging R-values across parallel paths.
Adding film resistance again when already included.
Calling instantaneous heat gain a complete equipment cooling load.

Handbook navigation drill

Find it, then explain it.

Search: Thermal Resistance

Find the composite-wall relationship and identify whether the resistance is for the whole object or per unit area.

Check your navigation

Your units reduce to Btu/h after multiplying U, area and temperature difference.

Reference path: §5.1–5.2, pp. 303–308; §9.1, pp. 446–478. 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, §§5.1–5.2 and 9.1.9–9.1.13.
Grondzik and Kwok, Mechanical and Electrical Equipment for Buildings, 12th ed., ch. 7, Heat Flow.
Lechner, Heating, Cooling, Lighting, 4th ed., chs. 3 and 9.
Brown and DeKay, Sun, Wind & Light, 3rd ed., Energy Programming and climate-responsive load reduction.

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