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Heating & Cooling Loads

10% of exam

Envelope conduction, solar and internal gains, infiltration and ventilation loads, supply-air flow from room sensible heat, and reheat.

5 concepts

A. Heating/Cooling Loads

Envelope Conduction Loads: U-Factor, R-Value and CLTD

Build the overall U-factor from series and parallel thermal resistances, then drive heating loads with UAΔT and cooling loads with the CLTD method.

Every heating and cooling load calculation begins at the envelope, and the envelope begins with one number: the overall heat-transmission coefficient

U

. Get

U

wrong and the wall, roof, and slab loads that often dominate a winter heating bill are wrong by the same factor. The handbook gives you the raw material — the surface film and plane air-space resistances in §9.1.11 "Thermal Resistance Properties," plus the conduction equations — in the NCEES PE Mechanical Reference Handbook — §9 Heating, Ventilation, and Air Conditioning. What it does not spell out is the workflow that turns a stack of materials into a defensible load, and the seasonal split between the simple steady-state heating method and the time-lag CLTD cooling method. That workflow is exactly where exam points are won and lost.

Resistance in series: the R-value chain

Heat crossing a wall passes through each layer in turn — outdoor air film, cladding, sheathing, insulation, gypsum, indoor air film — so the thermal resistances add like resistors in series. The unit resistance

R

of a homogeneous layer is its thickness over its conductivity,

R = L/k

, and the overall coefficient is simply the reciprocal of the total:

U = 1/R_{\text{total}}

. Always include both air films; the handbook lists

R_i \approx 0.68

for a still indoor vertical surface and

R_o \approx 0.17

for a 15 mph winter wind (use

0.25

for the 7.5 mph summer film). Forgetting the films is the single most common envelope error.

R_{\text{total}} = R_o + \sum_j \frac{L_j}{k_j} + R_i, \qquad U = \frac{1}{R_{\text{total}}}

Resistance in parallel: framing and the U-average

Real walls are not one-dimensional: a stud and the insulated cavity beside it are two parallel heat paths with very different resistances. The parallel-path (area-weighted) method computes a separate

U

for each path and combines them by area fraction

f

. A nominal

2\times4

wood-frame wall is roughly

25\%

framing once plates and headers are counted, and the wood stud short-circuits the insulation — this "thermal bridging" can raise the assembly

U

by a third or more above the clear-cavity value. The isothermal-planes method (combining the bridged-layer resistances in parallel before adding the common layers in series) allows lateral heat flow that short-circuits the insulation, so it gives a slightly higher

U

(lower

R

). The true assembly value lies between the two: parallel-path is the lower-

U

/upper-

R

bound and isothermal-planes the higher-

U

/lower-

R

bound. Either is acceptable on the exam unless one is specified.

U_{\text{avg}} = \sum_i f_i\,U_i = f_{\text{cav}}U_{\text{cav}} + f_{\text{frame}}U_{\text{frame}}, \qquad \sum_i f_i = 1

Steady-state heating: UAΔT

Winter heating loads are computed at the steady-state design condition with no solar credit and no internal-gain credit (the plant must hold setpoint at 6 a.m. with the lights off). For each surface the conduction load is the product of the transmittance, the area, and the indoor-to-outdoor temperature difference. Sum over all surfaces and add the ground, glazing, and infiltration pieces. Because there is no thermal storage in the steady model, the same

UA\,\Delta T

form governs walls, roof, doors, and the partition gains between conditioned and unconditioned spaces.

q_{\text{heat}} = U A\,(t_i - t_o) = U A\,\Delta T

Cooling is not steady: the CLTD method

On a sunny afternoon the outdoor air temperature, the solar irradiation, and the wall's own thermal mass all matter, and the peak heat entering the room lags the peak outdoor condition by hours. The Cooling Load Temperature Difference (CLTD) method folds all of that — the sol-air temperature, the daily swing, and the mass time lag — into a single equivalent temperature difference tabulated by surface type, orientation, and hour. The cooling load is then the familiar

U A \times \text{CLTD}

. CLTD is not in the handbook, so know it cold: it is a published ASHRAE simplified method, and the table value must be corrected for your actual indoor and outdoor design temperatures.

q_{\text{cool}} = U A \,(\text{CLTD})_{\text{corrected}}

Correcting the table CLTD

Tabulated CLTD values assume a

78^\circ\text{F}

indoor temperature and a

95^\circ\text{F}

outdoor design with a

21^\circ\text{F}

daily range (mean

85^\circ\text{F}

). For any other design you shift the table value by the indoor and the mean-outdoor differences. The mean outdoor temperature is the design dry bulb minus half the daily range. Latitude/month and color corrections (

LM

,

K

) refine it further, but the temperature correction below is the one examiners test most.

(\text{CLTD})_{\text{corr}} = \text{CLTD}_{\text{table}} + (78 - t_i) + (t_{o,\text{mean}} - 85)

Opaque conduction vs the CLF for internal/solar gains

Keep two cousins distinct. Opaque envelope conduction (walls, roof) uses CLTD directly because the table already contains the storage lag. Heat gains that originate as radiation inside the room — sunlight through glass, lights, people, equipment — are instead converted to cooling load with a separate Cooling Load Factor (CLF) or radiant time series, because their stored fraction reappears later. Conflating the two double-counts or omits storage. This concept owns the opaque-conduction half; fenestration and internal gains carry their own CLF treatment in their concepts.

q_{\text{cool,load}} = (\text{instantaneous radiant + convective gain}) \times \text{CLF}

Exam strategy

Build

R_{\text{total}}

from the outside in and never drop the two air films — write them first so you cannot forget them. Decide season before you pick an equation: heating is steady

UA\,\Delta T

with no solar/internal credit; cooling is

UA\times\text{CLTD}

for opaque surfaces. For framed assemblies, state whether you used parallel-path or isothermal-planes and area-weight the

U

values, not the

R

values. Always correct a table CLTD for your real

t_i

and

t_{o,\text{mean}}

, and remember

t_{o,\text{mean}} = t_{o,\text{design}} - \tfrac{1}{2}(\text{daily range})

. Carry

U

to three significant figures; small

U

differences scale the whole envelope load.

Key equations

Layer resistance (conduction)

R = \dfrac{L}{k}

Unit thermal resistance of a solid layer; $L$ = thickness (ft or in), $k$ = conductivity (Btu·in/(h·ft²·°F) or Btu/(h·ft·°F)). Units of $R$ are h·ft²·°F/Btu.

Series resistance and U

R_{\text{total}} = R_o + \textstyle\sum_j \dfrac{L_j}{k_j} + R_i, \quad U = \dfrac{1}{R_{\text{total}}}

Parallel-path (area-weighted) U

U_{\text{avg}} = \textstyle\sum_i f_i\,U_i

Steady-state heating conduction

q_{\text{heat}} = U A\,(t_i - t_o)

Heat gain through interior partitions

q = U A\,(t_b - t_i)

Gain from an adjacent unconditioned space at temperature $t_b$

CLTD cooling load (opaque)

q_{\text{cool}} = U A\,(\text{CLTD})_{\text{corr}}

Cooling load through a sunlit wall or roof, Btu/h. CLTD (°F) embeds sol-air temperature and mass time lag.

CLTD temperature correction

(\text{CLTD})_{\text{corr}} = \text{CLTD}_{\text{table}} + (78 - t_i) + (t_{o,\text{mean}} - 85)

Worked examples

Overall U-factor of a cavity wall

Problem. A wall is built (outside to inside) of a 7.5 mph summer air film, $4\,\text{in}$ face brick ( $R = 0.44$ ), exterior sheathing ( $R = 1.32$ ), an R-13 insulated cavity, $1/2\,\text{in}$ gypsum board ( $R = 0.45$ ), and a still indoor air film ( $R = 0.68$ ). The handbook lists the winter outdoor film as $R_o = 0.17$ and the indoor film as $R_i = 0.68$ . Using the winter outdoor film, find the clear-cavity $U$ -factor and the steady-state heating load through $200\,\text{ft}^2$ when indoors is $70^\circ\text{F}$ and outdoors is $5^\circ\text{F}$ .

Solution. Add resistances in series:

R_{\text{total}} = 0.17 + 0.44 + 1.32 + 13.0 + 0.45 + 0.68 = 16.06\ \tfrac{\text{h·ft}^2\text{·}^\circ\text{F}}{\text{Btu}}.

Thermal bridging by wood framing

Problem. The same wall has its cavity interrupted by nominal $2\times4$ wood studs that occupy $25\%$ of the wall area (the cavity is the other $75\%$ ). The stud path replaces the R-13 cavity with $3.5\,\text{in}$ of wood at $R = 4.38$

Sunlit-wall cooling load with corrected CLTD

Problem. A west-facing masonry wall has $U = 0.090\ \text{Btu/(h·ft}^2\text{·}^\circ\text{F})$ and area $320\,\text{ft}^2$

Common pitfalls

•Omitting the air-film resistances. The indoor still-air film ( $R_i \approx 0.68$ ) and outdoor wind film ( $R_o \approx 0.17$ winter, $0.25$ summer) are real resistances — leaving them out overstates $U$ and the load.
•Averaging R-values for parallel paths. Combine framed assemblies by area-weighting the U-values ( $U_{\text{avg}} = \sum f_i U_i$ ), not the resistances — averaging R underestimates the thermal-bridge penalty.
•Using $UA\,\Delta T$ for a sunlit cooling load. Steady-state conduction ignores solar and mass lag; cooling through opaque surfaces needs CLTD, which embeds the sol-air temperature and storage delay.
•Forgetting to correct the table CLTD. Tables assume 78 °F indoors and an 85 °F outdoor mean; shift by $(78 - t_i) + (t_{o,\text{mean}} - 85)$ before using it, with $t_{o,\text{mean}} = t_{o,\text{design}} - \tfrac12(\text{range})$
•Applying a Cooling Load Factor (CLF) to opaque conduction. CLF/RTS converts internal and solar radiant gains; CLTD already carries the opaque-wall storage lag — using both double-counts storage.
•Crediting solar or internal gains in the heating design. Winter loads are computed cold and dark at the worst hour; only conduction, ground, and infiltration count.
•Mixing conductivity units. $k$ for building materials is usually Btu·in/(h·ft²·°F), so $R = L_{\text{in}}/k$ with $L$ in inches; using $L$ in feet with that $k$

References

NCEES PE Mechanical Reference Handbook — §9 Heating, Ventilation, and Air Conditioning (§9.1.9 interior surfaces, §9.1.11 thermal resistance, §9.1.13 fenestration U-factors)
ASHRAE Handbook—Fundamentals, Ch. 18 (Nonresidential Cooling and Heating Load Calculations) and Ch. 26 (Material Properties) — Source of the CLTD/CLF method, film resistances, and framing factors.
ASHRAE Standard 90.1 — Energy Standard for Buildings — Assembly U-factor and insulation requirements by climate zone.

Internal Gains: People, Lights and Equipment

Quantify sensible and latent gains from occupants, lighting (use and ballast factors), and motors/appliances, then split them radiant/convective and diversify.

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Supply Airflow from Room Sensible Heat and Reheat

Size supply CFM from the room sensible load and supply-to-room ΔT using the 1.10 equation, find the leaving-air temperature, and add reheat at part load.

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Ventilation and Infiltration Loads (Sensible + Latent)

Compute outdoor-air sensible (1.10·CFM·Δt) and latent (4,840·CFM·ΔW) loads, find infiltration by air-change and leakage methods, and add OA load to the coil.

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Fenestration and Solar Heat Gain (SHGC, Shading, Glazing)

Compute solar gain through glazing with SHGC and irradiance, relate it to the shading coefficient and external shading, and convert to cooling load with the CLF.

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Add layer and film resistances in series; $U$ is the reciprocal. $R_i \approx 0.68$ (still indoor air), $R_o \approx 0.17$ winter / $0.25$ summer.

For framing and other parallel paths; $f_i$ = area fraction, $U_i$ = transmittance of path $i$ . Weight U-values, never R-values.

Design heating load through a surface, Btu/h. No solar or internal credit; $A$ in ft², $\Delta T$ in °F.

to the conditioned space at

t_i

(handbook §9.1.9).

Adjusts a table value (78 °F indoor, 85 °F mean outdoor basis) to actual design temperatures; $t_{o,\text{mean}} = t_{o,\text{design}} - \tfrac12(\text{daily range})$ .

Overall coefficient:

U = 1/16.06 = 0.0623\ \text{Btu/(h·ft}^2\text{·}^\circ\text{F})

. Heating load:

\Delta T = 70 - 5 = 65^\circ\text{F}

, so

q = U A\,\Delta T = 0.0623\,(200)(65) = 809\ \text{Btu/h}.

Sanity:

R\approx16

is a respectable insulated wall, so

U

near

0.06

is expected;

809\,\text{Btu/h}

over

200\,\text{ft}^2

is about

4\,\text{Btu/h·ft}^2

, reasonable for a

65^\circ\text{F}

winter

\Delta T

. Answer:

U = 0.0623

,

q = 809\ \text{Btu/h}

.

; all other layers are unchanged. Find the area-weighted assembly

U

and compare with the clear-cavity value.

Solution. Common layers (both paths):

0.17 + 0.44 + 1.32 + 0.45 + 0.68 = 3.06

. Cavity path:

R_{\text{cav}} = 3.06 + 13.0 = 16.06 \Rightarrow U_{\text{cav}} = 0.0623

. Stud path:

R_{\text{stud}} = 3.06 + 4.38 = 7.44 \Rightarrow U_{\text{stud}} = 0.1344

. Area-weight the U-values:

U_{\text{avg}} = 0.75(0.0623) + 0.25(0.1344) = 0.0467 + 0.0336 = 0.0803\ \tfrac{\text{Btu}}{\text{h·ft}^2\text{·}^\circ\text{F}}.

Effective

R = 1/0.0803 = 12.5

, versus

16.1

for the clear cavity. Sanity: framing raises

U

by

0.0803/0.0623 - 1 = 29\%

— the classic one-quarter-third penalty for

2\times4

thermal bridging. Answer:

U_{\text{avg}} = 0.0803\ \text{Btu/(h·ft}^2\text{·}^\circ\text{F})

.

. The CLTD table gives

30^\circ\text{F}

at the design hour. The space is kept at

75^\circ\text{F}

; the outdoor design dry bulb is

99^\circ\text{F}

with a

22^\circ\text{F}

daily range. Find the wall cooling load.

Solution. Mean outdoor temperature:

t_{o,\text{mean}} = 99 - \tfrac12(22) = 88^\circ\text{F}

. Correct the table CLTD:

(\text{CLTD})_{\text{corr}} = 30 + (78 - 75) + (88 - 85) = 30 + 3 + 3 = 36^\circ\text{F}.

Cooling load:

q = U A\,(\text{CLTD})_{\text{corr}} = 0.090\,(320)(36) = 1{,}037\ \text{Btu/h}.

Sanity: the corrections add

6^\circ\text{F}

because this building runs cooler and hotter than the

78/85

table basis;

1{,}040\,\text{Btu/h}

for a heavy sunlit wall this size is reasonable (about

0.086\,\text{ton}

). Answer:

q = 1{,}040\ \text{Btu/h}

(3 s.f.).

.

gives

R

twelve times too small.