6.2 Cooling & Heating Load Calculations (CLTD/SCL/CLF & Heat Balance)

Cooling & Heating Load Calculations (CLTD/SCL/CLF & Heat Balance)

Accurate building load calculations are critical for proper HVAC equipment selection, preventing inefficient short-cycling from oversized chillers and uncomfortable temperature/humidity drift from undersized systems. Heating load analysis is modeled under steady-state worst-case winter conditions (no solar or internal heat credits), whereas cooling load analysis requires dynamic thermal modeling to account for solar radiation, internal heat generation, and thermal mass storage lag.

---

1. Outdoor Design Conditions & ASHRAE Weather Criteria

Design weather conditions are established statistically by ASHRAE based on multi-decade meteorological observations:

+-----------------------------------------------------------------------------------------+
|                         ASHRAE DESIGN WEATHER CRITERIA                                  |
|                                                                                         |
|   HEATING DESIGN CONDITIONS:                                                            |
|   - 99.6% Value: Temperature exceeded 99.6% of annual hours (Coldest, standard design)  |
|   - 99.0% Value: Temperature exceeded 99.0% of annual hours                             |
|   - Zero credit for solar radiation or internal occupant/equipment heat gain.           |
|                                                                                         |
|   COOLING DESIGN CONDITIONS:                                                            |
|   - 0.4% Dry-Bulb / MCWB: Peak dry-bulb exceeded 0.4% of annual hours (35 hrs/year)     |
|   - 1.0% Dry-Bulb / MCWB: Dry-bulb exceeded 1.0% of annual hours (88 hrs/year)          |
|   - Peak Dew-Point / MCDB: Governs high-latent cooling coil design.                     |
+-----------------------------------------------------------------------------------------+

---

2. Thermal Envelope Transmission Fundamentals

Heat transmission across multi-layered building assemblies (walls, roofs, fenestrations) is calculated using the overall heat transfer coefficient ($U$-factor):

$U = \frac{1}{R_{\text{total}}}$

Where $R_{\text{total}}$ is the summation of individual material thermal resistances, enclosed air space resistances, and internal/external surface air film resistances:

$R_{\text{total}} = R_i + \sum_{k=1}^n \frac{x_k}{k_k} + R_{\text{air space}} + R_o$

+-----------------------------------------------------------------------------------------+
|                        COMPOSITE WALL THERMAL RESISTANCE PROFILE                        |
|                                                                                         |
|   OUTDOOR AIR (T_o)                                            INDOOR AIR (T_i)         |
|      |                                                                 |                |
|      v  R_o (Film)    R_brick      R_sheath    R_insul     R_gypsum    v  R_i (Film)    |
|   ---+------------+------------+------------+------------+------------+------------+--  |
|      |  Outside   | 4" Face    | 1/2" Ply-  | 3.5" R-13  | 1/2" Dry-  | Inside     |    |
|      |  Air Film  | Brick      | wood       | Fiberglass | wall       | Air Film   |    |
|      | (0.17/0.25)| (R = 0.80) | (R = 0.62) | (R = 13.0) | (R = 0.45) | (R = 0.68) |    |
|   ---+------------+------------+------------+------------+------------+------------+--  |
+-----------------------------------------------------------------------------------------+

Standard Film Resistances ($R_i$ and $R_o$):

• Inside Still Air Film ($R_i$): $R_i = 0.68\text{ hr}\cdot\text{ft}^2\cdot^\circ\text{F/BTU}$ ($0.12\text{ m}^2\cdot\text{K/W}$) for vertical surfaces with non-reflective emissivity ($\epsilon = 0.90$).
• Outside Moving Air Film ($R_o$):
  • Winter ($15\text{ mph}$ wind): $R_o = 0.17\text{ hr}\cdot\text{ft}^2\cdot^\circ\text{F/BTU}$ ($0.030\text{ m}^2\cdot\text{K/W}$).
  • Summer ($7.5\text{ mph}$ wind): $R_o = 0.25\text{ hr}\cdot\text{ft}^2\cdot^\circ\text{F/BTU}$ ($0.044\text{ m}^2\cdot\text{K/W}$).

Thermal Bridging & Framing Factors

In wood and steel stud assemblies, heat flows simultaneously through the insulated cavity and through the solid structural studs. The effective assembly $U$-factor ($U_{\text{eff}}$) is calculated via parallel path area-weighting:

$U_{\text{eff}} = f_{\text{framing}} U_{\text{framing}} + (1 - f_{\text{framing}}) U_{\text{cavity}}$

Where $f_{\text{framing}}$ is typically $0.15$ to $0.25$ for $16\text{ in.}$ on-center wall framing.

---

3. Heating Load Calculation (Steady-State Framework)

Because peak heating loads typically occur at night or during overcast winter mornings, solar radiation and internal occupant/equipment loads are neglected (safety conservatism).

+-----------------------------------------------------------------------------------------+
|                        HEATING LOAD COMPONENT EQUATIONS                                 |
|                                                                                         |
|   1. Opaque Conduction:      \dot{q}_{cond} = U \cdot A \cdot (T_i - T_o)                |
|                                                                                         |
|   2. Slab-on-Grade:          \dot{q}_{slab} = F_p \cdot P \cdot (T_i - T_o)              |
|                                                                                         |
|   3. Basement Wall Loss:     \dot{q}_{bw} = U_{avg} \cdot A_{bw} \cdot (T_i - T_g)       |
|                                                                                         |
|   4. Infiltration Sensible:  \dot{q}_{inf,s} = 1.08 \cdot CFM_{inf} \cdot (T_i - T_o)   |
|                                                                                         |
|   5. Infiltration Latent:    \dot{q}_{inf,L} = 4840 \cdot CFM_{inf} \cdot (\omega_i - \omega_o)|
+-----------------------------------------------------------------------------------------+

• Slab-on-Grade Perimeter Heat Loss: Heat loss through unheated or heated slabs occurs primarily at the exposed perimeter edge, not uniformly across the floor slab area: $\dot{q}_{slab} = F_p \times P \times (T_i - T_o)$ Where $F_p$ is the perimeter heat loss coefficient [$\text{BTU/hr}\cdot\text{ft}\cdot^\circ\text{F}$] and $P$ is the linear perimeter length [$\text{ft}$].
• Infiltration Airflow Rate ($CFM_{\text{inf}}$): Calculated using the Air Change Method: $CFM_{\text{inf}} = \frac{ACH \times \text{Room Volume (ft}^3)}{60}$
• Outdoor Ventilation Air Requirement (ASHRAE 62.1): $V_{oz} = R_p P_z + R_a A_z$ Where $R_p$ is outdoor airflow rate per person [$\text{CFM/person}$], $P_z$ is zone population, $R_a$ is outdoor airflow rate per unit area [$\text{CFM/ft}^2$], and $A_z$ is zone floor area [$\text{ft}^2$].

---

4. Cooling Load Calculation (Dynamic Thermal Dynamics)

Heat Gain vs. Cooling Load

A fundamental concept tested on the PE exam is the physical difference between Instantaneous Heat Gain and Cooling Load.

+-----------------------------------------------------------------------------------------+
|                        HEAT GAIN VS. COOLING LOAD MECHANICS                             |
|                                                                                         |
|   INSTANTANEOUS HEAT GAINS                                                              |
|   |                                                                                     |
|   +---> Convective Gain (Lights, Air, Convection) ---> [ INSTANTANEOUS COOLING LOAD ]   |
|   |                                                              ^                      |
|   +---> Radiant Gain (Solar, Radiation, High Mass)               |                      |
|             |                                                    | (Convective release  |
|             v                                                    |  with time lag)      |
|         [ THERMAL STORAGE IN SLAB, WALLS & FURNITURE ] ----------+                      |
+-----------------------------------------------------------------------------------------+

1. Convective Gains: Directly transfer heat to room air by convection (e.g., infiltration air, equipment convection) and become instantaneous cooling load immediately.
2. Radiant Gains: Transfer heat by radiation (e.g., solar beam radiation, lighting radiation, occupant radiant dissipation) to the room floor, walls, and furniture. This energy is absorbed, stored in the structural thermal mass, and subsequently released convectively into the space with a time lag of several hours.

Sol-Air Temperature ($T_{\text{sol-air}}$)

The fictitious outdoor air temperature that, in the absence of all radiation exchanges, would produce the same rate of heat transfer through an exterior surface as the actual combined outdoor air temperature, incident solar radiation, and longwave sky radiation:

$T_{\text{sol-air}} = T_o + \frac{\alpha I_t}{h_o} - \frac{\epsilon \delta R}{h_o}$

Where:

• $T_o = \text{outdoor ambient dry-bulb temperature } [^\circ\text{F}]$
• $\alpha = \text{solar absorptance of the surface (0.90 for dark surfaces, 0.45 for light surfaces)}$
• $I_t = \text{total solar irradiance incident on surface } [\text{BTU/hr}\cdot\text{ft}^2]$
• $h_o = \text{outdoor surface heat transfer coefficient } [\approx 3.0\text{ BTU/hr}\cdot\text{ft}^2\cdot^\circ\text{F}]$
• $\epsilon = \text{surface hemispherical emittance } [\approx 0.90]$
• $\delta R = \text{radiation correction factor: } 7^\circ\text{F (horizontal roofs)}, , 0^\circ\text{F (vertical walls)}$

Fenestration Solar Heat Gain

$\dot{q}_{fen} = A \times SHGC \times IAC \times E_t$ Where:

• $A = \text{fenestration gross area } [\text{ft}^2]$
• $SHGC = \text{Solar Heat Gain Coefficient of the glazing assembly (dimensionless, 0 to 1)}$
• $IAC = \text{Interior Attenuation Coefficient (1.0 for unshaded, 0.5 to 0.7 with blinds/drapes)}$
• $E_t = \text{incident peak solar irradiance } [\text{BTU/hr}\cdot\text{ft}^2]$

CLTD / SCL / CLF Method

1. Opaque Walls and Roofs: $\dot{q} = U \times A \times CLTD_{\text{corrected}}$ $CLTD_{\text{corr}} = CLTD + (78 - T_{\text{room}}) + (T_{\text{mean, outdoor}} - 85)$
2. Fenestration Conduction: $\dot{q}_{cond,fen} = U \times A \times (T_o - T_i)$
3. Fenestration Solar Load: $\dot{q}_{solar,fen} = A \times SC \times SCL$ (where $SC$ is Shading Coefficient, $SCL$ is Solar Cooling Load).

---

5. Internal Heat Gains & Space Load Summary

+-----------------------------------------------------------------------------------------+
|                        INTERNAL HEAT GAIN FORMULATIONS                                  |
|                                                                                         |
|   1. Lighting Load:      \dot{q}_{light} = Total Watts \cdot 3.412 \cdot F_{ul} \cdot F_{sa} |
|      - F_ul = Lighting Use Factor (typically 1.0 for occupied commercial)               |
|      - F_sa = Ballast Allowance Factor (1.0 for LED/electronic, 1.15-1.25 for magnetic) |
|                                                                                         |
|   2. People Sensible:    \dot{q}_{occ,s} = N_{people} \cdot q_{s,person} \cdot CLF         |
|   3. People Latent:      \dot{q}_{occ,L} = N_{people} \cdot q_{L,person}                    |
|      - Office Sitting: q_s = 245 BTU/hr, q_L = 205 BTU/hr (Total = 450 BTU/hr)          |
|      - Heavy Exercise: q_s = 525 BTU/hr, q_L = 1075 BTU/hr (Total = 1600 BTU/hr)        |
|                                                                                         |
|   4. Equipment / Power:  \dot{q}_{equip} = Total Operating Watts \cdot 3.412            |
|      - Electric Motors:  \dot{q}_{motor} = \frac{HP \cdot 2545}{\eta_{motor}} (in space)  |
+-----------------------------------------------------------------------------------------+

---

6. Step-by-Step Worked Engineering Problem

Problem Statement

A commercial corner office ($30\text{ ft} \times 40\text{ ft} \times 10\text{ ft}$ ceiling height) located in Chicago requires peak cooling load sizing at $3:00\text{ PM}$.

• South Exterior Wall: Gross area $= 400\text{ ft}^2$, containing $100\text{ ft}^2$ of double-pane tinted glass ($U_{\text{glass}} = 0.55\text{ BTU/hr}\cdot\text{ft}^2\cdot^\circ\text{F}$, $SHGC = 0.40$, $IAC = 0.80$, $E_t = 120\text{ BTU/hr}\cdot\text{ft}^2$). Opaque wall net area $= 300\text{ ft}^2$ ($U_{\text{wall}} = 0.080\text{ BTU/hr}\cdot\text{ft}^2\cdot^\circ\text{F}$, $CLTD_{\text{corr}} = 25.0^\circ\text{F}$).
• East Exterior Wall: Gross area $= 300\text{ ft}^2$, opaque, no glass ($U_{\text{wall}} = 0.080\text{ BTU/hr}\cdot\text{ft}^2\cdot^\circ\text{F}$, $CLTD_{\text{corr}} = 15.0^\circ\text{F}$).
• Interior Conditions: $T_i = 75.0^\circ\text{F}$. Outdoor ambient temperature: $T_o = 95.0^\circ\text{F}$.
• Internal Loads:
  • Occupancy: 8 persons doing moderately active office work ($q_s = 250\text{ BTU/hr}\cdot\text{person}$, $q_L = 200\text{ BTU/hr}\cdot\text{person}$, $CLF = 1.0$).
  • Lighting: $1,200\text{ Watts}$ LED ($F_{sa} = 1.0, F_{ul} = 1.0, CLF = 1.0$).
  • Computers: $1,500\text{ Watts}$ total operating load.
• Infiltration: $0.25\text{ ACH}$.

Calculate the total Space Sensible Cooling Load and Space Latent Cooling Load.

+-----------------------------------------------------------------------------------------+
|                          OFFICE COOLING LOAD BREAKDOWN                                  |
|                                                                                         |
|   1. South Opaque Wall Conduction:  q = 0.080 * 300 * 25.0         =    600 BTU/hr      |
|   2. East Opaque Wall Conduction:   q = 0.080 * 300 * 15.0         =    360 BTU/hr      |
|   3. South Glass Conduction:        q = 0.55 * 100 * (95.0 - 75.0) =  1,100 BTU/hr      |
|   4. South Glass Solar Gain:        q = 100 * 0.40 * 0.80 * 120    =  3,840 BTU/hr      |
|   5. Lighting Sensible Load:        q = 1,200 * 3.412              =  4,094 BTU/hr      |
|   6. Computer Power Load:           q = 1,500 * 3.412              =  5,118 BTU/hr      |
|   7. Occupant Sensible Load:        q = 8 * 250 * 1.0              =  2,000 BTU/hr      |
|   8. Infiltration Sensible Load:    CFM = 0.25 * 12,000 / 60 = 50                       |
|                                     q = 1.08 * 50 * (95.0 - 75.0)  =  1,080 BTU/hr      |
|   -----------------------------------------------------------------------------------   |
|   TOTAL SPACE SENSIBLE COOLING LOAD:                               = 18,192 BTU/hr      |
|                                                                                         |
|   9. Occupant Latent Load:          q_L = 8 * 200                  =  1,600 BTU/hr      |
|   TOTAL SPACE LATENT COOLING LOAD:                                 =  1,600 BTU/hr      |
+-----------------------------------------------------------------------------------------+

Summary Calculations:

• $\text{Total Space Sensible Load} = 600 + 360 + 1,100 + 3,840 + 4,094 + 5,118 + 2,000 + 1,080 = 18,192\text{ BTU/hr}$
• $\text{Total Space Latent Load} = 1,600\text{ BTU/hr}$
• $\text{Total Space Cooling Load} = 18,192 + 1,600 = 19,792\text{ BTU/hr} = 1.65\text{ TR}$
• $\text{Space SHR} = \frac{18,192}{19,792} = 0.919$

---

7. Common Exam Traps & PE Pro-Tips

• Trap 1 — Confusing Space Cooling Load with Central Coil Load: Ventilation outdoor air introduced at the central air handling unit is a coil load, not a space heat gain. Infiltration air (air leaking through cracks, envelope seams, and doors directly into the conditioned envelope) is a space load.
• Trap 2 — Slab Heat Loss Calculation: Never multiply slab floor surface area by a $U$-factor! Slab heat loss is strictly a perimeter phenomenon governed by $F_p \times P \times \Delta T$.
• Trap 3 — Forgetting Ballast Allowance on Old Fluorescent Systems: Magnetic ballast fixtures consume $15\text{--}25%$ more electric power than rated lamp tube wattage ($F_{sa} = 1.15\text{--}1.25$). Modern LEDs operate with $F_{sa} \approx 1.0$.