6.4 Internal Heat Gains: Lighting, People, Equipment & Diversity Factors

6.4 Internal Heat Gains: Lighting, People, Equipment & Diversity Factors

In modern energy-efficient commercial buildings with highly insulated building envelopes and high-performance glazing, internal heat gains frequently dominate the total cooling load. Internal heat gains originate from four primary sources: building occupants, electric lighting fixtures, office and process plug load equipment, and electric motors/machinery. Unlike external weather loads that reverse direction between summer and winter, internal heat gains generate continuous cooling loads during all occupied hours, reducing winter heating demand while increasing year-round cooling coil loads.

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1. Occupant Heat Gain (People)

Human beings generate heat through biological metabolism. Heat is released to the surrounding environment through four simultaneous thermodynamic pathways:

1. Convection: Heat transfer from skin and clothing to ambient air (Sensible).
2. Radiation: Longwave radiant heat transfer from outer surface clothing/skin to cooler surrounding room surfaces (Sensible).
3. Evaporation (Sweating & Respiration): Moisture vapor release from skin and breathing exhalation (Latent phase change: $h_{fg} \approx 1050\text{ Btu/lb}$).
4. Conduction: Direct physical contact with furniture/flooring (Sensible, negligible).

+-----------------------------------------------------------------------------------------+
| HUMAN THERMAL DISSIPATION PATHWAYS                                                      |
+-----------------------------------------------------------------------------------------+
| Total Metabolic Rate (M) = Sensible Heat (Convection + Radiation) + Latent Heat (Evap)  |
|                                                                                         |
| Space DB Effect: At 70°F -> 70% Sensible / 30% Latent                                   |
|                  At 75°F -> 50% Sensible / 50% Latent (Standard Office Design)          |
|                  At 80°F -> 30% Sensible / 70% Latent (Sweating Dominates)              |
+-----------------------------------------------------------------------------------------+

ASHRAE Standard Occupant Heat Output Table (at $75^\circ\text{F}$ Indoor Dry-Bulb)

Occupant Cooling Load Calculation Formulation

$q_{\text{people, sensible}} = N \cdot q_{s,\text{person}} \cdot F_D \cdot \text{CLF}_{\text{people}}$

$q_{\text{people, latent}} = N \cdot q_{l,\text{person}} \cdot F_D$

Where:

• $N = \text{Number of occupants in the space}$
• $q_{s,\text{person}}, q_{l,\text{person}} = \text{Sensible and latent heat dissipation rates per person } (\text{Btu/hr})$
• $F_D = \text{Occupant Diversity Factor } (0.7\text{ to }1.0)$
• $\text{CLF}_{\text{people}} = \text{Cooling Load Factor for people (accounts for radiant thermal storage, } \approx 1.0\text{ for 24-hr operation)}$

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2. Electric Lighting Heat Gains

All electrical energy supplied to interior lighting fixtures is ultimately converted entirely into thermal heat energy ($100%$ conservation of energy). Light photons emitted by the fixtures strike walls, floors, and furniture where they are absorbed and degraded into thermal heat.

General Lighting Heat Gain Formulation

$q_{\text{light}} = W_{\text{installed}} \times 3.412\left(\frac{\text{Btu/hr}}{\text{Watt}}\right) \times F_u \times F_{sa} \times \text{CLF}_{\text{light}}$

Where:

• $W_{\text{installed}} = \text{Total connected lighting electrical power (Watts) or Area } (\text{ft}^2) \times \text{LPD } (\text{W/ft}^2)$
• $3.412 = \text{Conversion constant } (1\text{ Watt} = 3.41214\text{ Btu/hr})$
• $F_u = \text{Lighting Use Factor (fraction of installed lights energized during peak load, typically } 0.85\text{ to }1.0)$
• $F_{sa} = \text{Special Allowance Factor (Ballast or LED driver power multiplier: } 1.0\text{ for direct LED, } 1.15\text{ for fluorescent/HID)}$
• $\text{CLF}_{\text{light}} = \text{Cooling Load Factor for lighting (accounting for thermal mass storage delay)}$

+-----------------------------------------------------------------------------------------+
| LIGHTING HEAT SPLIT: ROOM SPACE VS. RETURN AIR PLENUM                                   |
+-----------------------------------------------------------------------------------------+
| Recessed Fixture in Dropped Ceiling:                                                    |
|                                                                                         |
|   +---------------------------------------+  <-- Floor Slab Above                       |
|   | RETURN AIR PLENUM (T_plenum > T_room) |                                             |
|   |   q_plenum = (1 - F_space) * q_light  |  --> Heat goes to central return air        |
|   +================[ FIXTURE ]============+  <-- Acoustic Ceiling Tile                  |
|   |                                       |                                             |
|   | CONDITIONED ROOM SPACE (T_room = 75°F)|                                             |
|   |   q_space  = F_space * q_light        |  --> Heat goes directly into zone CFM       |
|   +---------------------------------------+                                             |
+-----------------------------------------------------------------------------------------+

Space vs. Return Air Plenum Heat Fraction ($F_{\text{space}}$)

When lighting fixtures are recessed into a suspended return air plenum:

• Heat to Room Space ($q_{\text{room}}$): $F_{\text{space}} \approx 0.70\text{ to }0.80$ enters the conditioned space directly.
• Heat to Return Air Plenum ($q_{\text{plenum}}$): $1 - F_{\text{space}} \approx 0.20\text{ to }0.30$ is carried directly by return air back to the air handling unit. This fraction increases the mixed air temperature ($T_{\text{mix}}$) at the central cooling coil but does not increase the required zone supply airflow ($\text{CFM}_{\text{supply}}$).

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3. Plug Loads & Office Equipment

Plug loads include computers, monitors, printers, copiers, servers, and laboratory equipment. Using the manufacturer's maximum nameplate power rating severely overestimates cooling loads because equipment operates at idle or partial load states for the majority of the working day.

Sensible Equipment Heat Gain Equation

$q_{\text{equip}} = P_{\text{operating}} \times 3.412\left(\frac{\text{Btu/hr}}{\text{Watt}}\right) = W_{\text{nameplate}} \times F_{\text{use}} \times 3.412$

Representative Commercial Equipment Heat Gains

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4. Electric Motors & Machinery Heat Gains

Electric motors drive fans, pumps, compressors, and industrial machines. Sizing motor heat gains requires evaluating whether the motor itself, the driven machine, or both are physically located inside the conditioned space.

+-----------------------------------------------------------------------------------------+
| THE THREE CANONICAL MOTOR HEAT DISSIPATION CONFIGURATIONS                               |
+-----------------------------------------------------------------------------------------+
| Case 1: Motor IN Space, Driven Machine IN Space                                         |
|         q = (hp * 2,545) / eta                                                          |
|                                                                                         |
| Case 2: Motor IN Space, Driven Machine OUTSIDE Space                                    |
|         q = hp * 2,545 * (1 - eta) / eta  = Motor Inefficiency Heat                     |
|                                                                                         |
| Case 3: Motor OUTSIDE Space, Driven Machine IN Space                                    |
|         q = hp * 2,545                    = Mechanical Work Converted to Heat           |
+-----------------------------------------------------------------------------------------+

Mathematical Formulation of the Three Motor Cases

1. Case 1: Motor IN Space, Machine IN Space (e.g., AHU supply fan inside conditioned air stream, indoor machine tool): All electrical power input to the motor ($P_{\text{in}} = P_{\text{brake}} / \eta$) degrades into heat within the space: $q_{\text{motor}} = \frac{P_{\text{hp}} \times 2544.43}{\eta_{\text{motor}}} = \frac{P_{\text{hp}} \times 2545}{\eta_{\text{motor}}}\quad (\text{Btu/hr})$
2. Case 2: Motor IN Space, Machine OUTSIDE Space (e.g., exhaust fan motor inside conditioned room driving rooftop impeller via shaft): Only the motor electrical and magnetic losses ($P_{\text{in}} - P_{\text{shaft}}$) remain in the space: $q_{\text{motor}} = P_{\text{hp}} \times 2545 \times \left(\frac{1 - \eta_{\text{motor}}}{\eta_{\text{motor}}}\right)\quad (\text{Btu/hr})$
3. Case 3: Motor OUTSIDE Space, Machine IN Space (e.g., outdoor motor driving an indoor chilled water pump): The motor losses are rejected to the outdoors, but the useful shaft work delivered to the indoor fluid/machine degrades into friction/viscous heat: $q_{\text{motor}} = P_{\text{hp}} \times 2545\quad (\text{Btu/hr})$

Where:

• $P_{\text{hp}} = \text{Brake horsepower delivered to the driven load } (\text{hp})$
• $\eta_{\text{motor}} = \text{Full-load electric motor efficiency (decimal, } 0.80\text{ to }0.95)$
• $2545 = \text{Conversion constant } (1\text{ hp} = 2544.43\text{ Btu/hr} = 745.7\text{ Watts})$

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5. Diversity Factors & Block Load Synthesis

In multi-room facilities, not all internal loads peak simultaneously. Occupants move between offices and conference rooms; lights are switched off in vacant areas; and computers cycle power.

Space Peak Load vs. Block (Coincident) Peak Load

• Space (Zone) Peak Load: Sized using a diversity factor $F_D = 1.0$ (worst-case individual peak) to ensure terminal VAV boxes, ductwork, and zone coils can meet individual room maximums.
• Central Block Load: Sized using the Coincident Peak Diversity Factor ($D < 1.0$) to size the central air handling units, chillers, and cooling towers.

+-----------------------------------------------------------------------------------------+
| DIVERSITY FACTOR (D) = Coincident Building Peak Load / Sum of Individual Zone Peaks     |
+-----------------------------------------------------------------------------------------+
| Office Buildings:     D_people = 0.70 to 0.85  |  D_lights = 0.85 to 0.95               |
| Hotels / Apartments:  D_people = 0.50 to 0.70  |  D_plug   = 0.50 to 0.75               |
+-----------------------------------------------------------------------------------------+

$Q_{\text{chiller, design}} = D_{\text{envelope}} Q_{\text{envelope, sum}} + D_{\text{people}} Q_{\text{people, sum}} + D_{\text{lights}} Q_{\text{lights, sum}} + D_{\text{plug}} Q_{\text{plug, sum}} + Q_{\text{ventilation}}$

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6. NCEES Reference Handbook Navigation Tactics

• Occupant Heat Gains: Search "Rates of Heat Gain from Occupants" or "Metabolic Rate" in Section 7 (HVAC & Refrigeration Applications) to find the table of sensible and latent heat outputs for seated, office, walking, and athletic activities.
• Motor Heat Gain Formulas: Search "Heat Gain from Electric Motors" to find the exact three-case equations relating brake horsepower ($P_{\text{hp}}$), motor efficiency ($\eta$), and heat output.
• Lighting Heat Multipliers: Search "Lighting Heat Gain" to locate the $3.412$ conversion constant and special ballast allowance factor ($F_{sa}$).

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7. Worked Computational Examples

Example 1: Large Training Facility Internal Cooling Load

A corporate training room ($2,500\text{ ft}^2$) has an occupancy of 60 seated students engaged in light computer training. The following design parameters are specified:

• Occupant heat dissipation: $q_s = 240\text{ Btu/hr}$, $q_l = 210\text{ Btu/hr}$ per person.
• Lighting: Recessed LED fixtures with total installed power of $1,800\text{ Watts}$. 25% of lighting heat is extracted directly into the return air plenum.
• Plug loads: 60 laptops at $40\text{ Watts}$ actual operating power each, plus one instructor workstation and projector totaling $450\text{ Watts}$.
• Diversity factor for training occupancy: $F_D = 0.90$.

Calculate: (a) Space sensible internal cooling load, (b) Space latent internal cooling load, (c) Total space internal cooling load, and (d) Heat rejected to the return air plenum.

Solution:

1. Calculate occupant heat gains: $q_{\text{people, sensible}} = 60 \times 240\text{ Btu/hr} \times 0.90 = 12,960\text{ Btu/hr}$ $q_{\text{people, latent}} = 60 \times 210\text{ Btu/hr} \times 0.90 = 11,340\text{ Btu/hr}$
2. Calculate lighting heat gains: $q_{\text{light, total}} = 1,800\text{ W} \times 3.412\text{ Btu/(hr}\cdot\text{W)} = 6,141.6\text{ Btu/hr}$ $q_{\text{light, room}} = 6,141.6 \times (1 - 0.25) = 6,141.6 \times 0.75 = 4,606.2\text{ Btu/hr}$ $q_{\text{light, plenum}} = 6,141.6 \times 0.25 = 1,535.4\text{ Btu/hr}$
3. Calculate plug load equipment heat gains (100% sensible): $W_{\text{plug}} = (60 \times 40\text{ W}) + 450\text{ W} = 2,400 + 450 = 2,850\text{ Watts}$ $q_{\text{plug}} = 2,850\text{ W} \times 3.412\text{ Btu/(hr}\cdot\text{W)} = 9,724.2\text{ Btu/hr}$
4. Synthesize space loads: $q_{\text{space, sensible}} = 12,960 + 4,606.2 + 9,724.2 = 27,290.4\text{ Btu/hr}$ $q_{\text{space, latent}} = 11,340.0\text{ Btu/hr}$ $q_{\text{space, total}} = 27,290.4 + 11,340.0 = 38,630.4\text{ Btu/hr} = 3.219\text{ Tons}$
5. Heat to return air plenum: $q_{\text{plenum}} = 1,535.4\text{ Btu/hr}$

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Example 2: Mechanical Room Motor Heat Dissipation (Case 2)

A commercial mechanical equipment room houses a $25\text{ hp}$ continuous exhaust fan motor operating at full load with an efficiency of $\eta = 91.0%$. The motor is mounted inside the conditioned mechanical room, but its drive shaft passes through an exterior partition wall to rotate the fan impeller located outside in the ambient air stream.

Calculate: The sensible heat dissipation rate into the mechanical room from the motor.

Solution:

1. Identify the appropriate motor case: The motor is located inside the room, while the driven machine (fan impeller and moving air stream) is outside the room (Case 2).
2. Apply the Case 2 heat dissipation formula: $q = P_{\text{hp}} \times 2545 \times \left(\frac{1 - \eta_{\text{motor}}}{\eta_{\text{motor}}}\right)$ $q = 25\text{ hp} \times 2545\left(\frac{\text{Btu/hr}}{\text{hp}}\right) \times \left(\frac{1 - 0.910}{0.910}\right)$ $q = 63,625 \times \left(\frac{0.090}{0.910}\right) = 63,625 \times 0.09890 = 6,292.6\text{ Btu/hr}$

(Note: If the fan impeller were also located inside the room (Case 1), the heat dissipation would be $q = 63,625 / 0.910 = 69,917.6\text{ Btu/hr}$—over 11 times higher!)