Cleanroom heat load calculation: worked example with fan filter unit, process, and recirculation fan heat
A worked cleanroom heat load for a 1,200 square foot ISO 7 room: fan filter unit heat, process and occupant loads, envelope, makeup air latent load, and airflow checks.
What this means
A cleanroom heat load adds the usual process, occupant, lighting, and envelope gains to two loads the air system creates itself: the full electrical input of recirculation fans or fan filter units, which becomes heat in the room, and the makeup air load, which carries most of the moisture. In the worked 1,200 square foot ISO 7 room below, 16 fan filter units drawing 250 watts each add 13,650 BTU per hour, 34 percent of a 39,940 BTU per hour room sensible load, and the makeup air unit carries ten times the room's own latent load.
Equipment and model context
- A 1,200 square foot ISO 7 non-unidirectional cleanroom with a fan filter unit ceiling, a dry recirculation coil, and a makeup air unit
- Equipment power, fan input, outdoor design, and envelope values are stated assumptions; use measured and manufacturer data for a real room
This works one cleanroom heat load from start to finish to show the method and where cleanroom loads differ from comfort cooling. Every input is a stated assumption. A real calculation uses measured process power, the manufacturer's fan input at the operating point, and local design weather.
What this covers
- Every line of a cleanroom sensible load with its basis, for one worked ISO 7 room.
- How much heat fan filter units and recirculation fans add, and why it rises with air changes.
- Why the makeup air unit carries more latent load than the room produces.
- How to check whether cooling or the air change rate sets the airflow, and what happens at night.
What changes the result
- The air change rate, which fixes fan filter unit count and therefore fan heat before any load is known.
- Process equipment entered at operating power rather than nameplate, and the share of its heat lost to exhaust.
- Outdoor design moisture and the makeup airflow needed for exhaust and pressurization.
- Occupancy and gowning level, which set the room's moisture release.
| Load | Basis in the example | BTU per hour |
|---|---|---|
| Fan filter units | 16 units at 250 W input each, all of it released in the room | 13,650 |
| Process equipment | 5.0 kW operating power from tool data, none rejected to exhaust | 17,060 |
| Lighting | 1.0 W per square foot over 1,200 square feet | 4,090 |
| Occupants, sensible | 8 people at 275 BTU per hour, light bench work | 2,200 |
| Walls | 1,400 square feet, U-factor 0.30, 7°F difference to surrounding space | 2,940 |
| Room sensible total | Sum of the lines above | 39,940 |
| Makeup air credit | 700 cfm delivered at 42°F into a 68°F room | minus 19,660 |
| Recirculation coil sensible duty | Dry coil in the recirculation path | 20,280 |
Where a cleanroom load departs from a comfort cooling load
Three things separate the calculation from an office load. Airflow is set by an air change rate before the load is known, so fan heat is a consequence of the cleanliness class rather than of the cooling. Makeup air runs continuously to replace process exhaust and pressurization leakage, so outdoor moisture arrives every hour the room operates. And the process load can outweigh everything else put together.
The order of work follows from that. Fix the airflow from the air change rate and the room's particle and regulatory needs, count the fan heat that airflow brings, add the room gains, and only then split the latent load between the room and the makeup air unit. The cleanroom HVAC design guide places this step within the full design sequence.
The example room and its basis of design
The room is 30 by 40 feet with a 10 foot ceiling, 12,000 cubic feet, classified ISO 7 and held at 68°F (20°C) and 45 percent relative humidity. The basis of design sets 50 air changes per hour, 10,000 cfm, delivered by 16 fan filter units at 625 cfm each. Eight people work at benches, process tools draw 5.0 kW in operation, and 700 cfm of makeup air replaces 400 cfm of process exhaust and 300 cfm of pressurization leakage.
Outdoor summer design is taken as 92°F dry bulb with 120 grains of moisture per pound of dry air, a humid location. The room sits inside a building held at 75°F, so its walls gain heat.
Fan filter unit and recirculation fan heat
A fan filter unit sits in the room's air stream with its motor, so its entire electrical input ends up as heat in the room: motor losses warm the air directly, and the work the fan does on the air turns into heat as friction through the filter and the room. Sixteen units at 250 watts each draw 4,000 watts, and 4,000 × 3.412 = 13,650 BTU per hour.
The 250 watts is the assumption to check. Input power depends on the unit, its speed, and how loaded its filter is, and a unit held at constant airflow draws more as its filter loads. Take the manufacturer's input watts at the design airflow with the filter at its planned replacement pressure drop. A flat 1,000 BTU per hour per unit, a figure quoted on some cleanroom pages, corresponds to about 293 watts and fits some units and not others.
A central recirculation fan follows the same rule with one distinction. When its motor sits in the air stream, the heat added is 2,545 × brake horsepower ÷ motor efficiency, in BTU per hour. When the motor sits outside the air stream, only the shaft power reaches the air, 2,545 × brake horsepower, and the motor losses go to the mechanical room instead. The AHU static pressure budget shows how to find brake horsepower from airflow and pressure.
Process, lighting, occupants, and walls
Enter process tools at operating power from the tool maker's facilities data or from measurement, never at nameplate, which describes the largest current a circuit must carry. Where a tool rejects part of its heat into process exhaust, subtract that part. The example assumes 5.0 kW released entirely to the room, 17,060 BTU per hour.
Lighting at 1.0 watt per square foot over 1,200 square feet adds 1,200 watts, 4,090 BTU per hour. For occupants, the ASHRAE Handbook, Fundamentals occupant table gives light bench work as 275 BTU per hour sensible and 475 BTU per hour latent per person, so eight people add 2,200 sensible and 3,800 latent.
The walls are 1,400 square feet of cleanroom panel with an assumed U-factor of 0.30 BTU per hour per square foot per °F, facing a 75°F building across a 7°F difference: 1,400 × 0.30 × 7 = 2,940 BTU per hour. The room sensible total is 39,940 BTU per hour, and fan filter unit heat is 34 percent of it.
Makeup air carries the moisture
At 68°F and 45 percent relative humidity the room holds about 45.7 grains of moisture per pound of dry air, a dew point of about 46°F. To absorb 3,800 BTU per hour of occupant moisture, the 700 cfm of makeup air has to arrive drier than the room by 3,800 ÷ (0.68 × 700) = 8.0 grains, at about 37.7 grains per pound, which is a dew point of about 41°F. The factor 0.68 combines 60 minutes, standard air density of 0.075 pounds per cubic foot, and the latent heat of water, divided by 7,000 grains per pound.
Drying outdoor air from 120 to 37.7 grains removes 0.68 × 700 × 82.3 = 39,200 BTU per hour of latent heat at the makeup air unit coil, and cooling it from 92°F to 42°F removes 1.08 × 700 × 50 = 37,800 BTU per hour of sensible heat. The makeup air unit therefore carries about 77,000 BTU per hour, while the room's own latent load is 3,800. The factor 1.08 is 60 minutes times 0.075 pounds per cubic foot times the specific heat of air, 0.24 BTU per pound per °F.
The makeup air leaves its coil at 42°F and goes into the recirculation path without reheat. Reheating it to 55°F first would spend 9,830 BTU per hour of heat only for the recirculation coil to remove it again. Delivered cold, it offsets 1.08 × 700 × (68 − 42) = 19,660 BTU per hour of room sensible load, leaving 20,280 BTU per hour for the dry recirculation coil. Ducts carrying air that cold need insulation, because their surfaces sit below the dew point of the spaces they pass through. The sensible and latent load split article covers the general relationship, and the psychrometric chart shows each state point.
Checking airflow against the load
With 10,000 cfm supplied and a 39,940 BTU per hour room sensible load, supply air needs to be only 39,940 ÷ (1.08 × 10,000) = 3.7°F below the room. Cooling alone, at a 13°F supply difference, would need 39,940 ÷ (1.08 × 13) = 2,845 cfm, about 14 air changes per hour. The 50 air change basis of design governs the airflow, which is the result the air changes by ISO class page predicts for a room of this kind.
The dry recirculation coil has to keep its surface above the room's 46°F dew point, or it starts condensing moisture the makeup air unit is already controlling. That points to chilled water supplied warmer than the makeup air coil receives, or a separate coil loop, which the cleanroom humidity control guide covers in detail.
Part load: the room at night
Fan filter units run whenever the room must stay classified, and makeup air runs whenever pressurization must hold. At night, with process tools idle, lights off, and nobody inside, the room gains only fan heat and wall gain, 13,650 + 2,940 = 16,590 BTU per hour, while 42°F makeup air still offsets 19,660. The room would overcool by about 3,070 BTU per hour.
The controls need an answer for that before handover: reset the makeup air leaving temperature upward when humidity allows, reduce makeup airflow to what pressurization needs with process exhaust off, or add a small reheat on the makeup air duct. A load calculation that stops at the design day misses the condition the room spends the most hours in.
| Item | Recirculation coil and room | Makeup air unit |
|---|---|---|
| Moisture source | 3,800 BTU per hour latent from 8 occupants | Outdoor air at 120 grains per pound in summer design |
| How moisture is removed | Absorbed by supply air drier than the room | Condensed on the coil, 39,200 BTU per hour latent |
| Sensible cooling | 20,280 BTU per hour on a dry coil | 37,800 BTU per hour cooling 700 cfm from 92°F to 42°F |
| Coil surface condition | Dry, kept above the 46°F room dew point | Wet, leaving air near a 41°F dew point |
Fan filter unit heat for the worked room at air change rates from 20 to 60, holding each unit at the same operating point and adding units in proportion to airflow, beside the room's other sensible gains, which do not change with airflow.
- Fan filter unit heat
- All other room sensible gains
- Fan filter unit heat is 17 percent of the room sensible load at 20 air changes per hour, 29 percent at 40, and 38 percent at 60.
- Process, lighting, occupant, and wall gains stay at 26,290 BTU per hour whatever the airflow, so every added air change raises the cooling load without any change in the room's activity.
- Real units draw more power as their filters load if they are controlled to constant airflow, so the operating point used for the load should be the loaded filter condition.
Questions people ask about this
How much heat does a fan filter unit add to a cleanroom?
All of its electrical input, because the motor losses and the work the fan does on the air both end as heat inside the room. Take the input watts at the unit's operating point from the manufacturer and multiply by 3.412 for BTU per hour, so a unit drawing 250 watts adds about 850 BTU per hour.
Should process exhaust be subtracted from the room heat load?
Subtract only the heat the exhaust actually carries away. A tool in a ducted enclosure that rejects its heat into the exhaust stream adds less to the room, while a tool that merely has a fume connection still releases its electrical heat into the room. The tool maker's heat rejection split, or a measurement, decides which applies.
Why is the makeup air unit larger than the recirculation coil?
Outdoor air on a humid design day carries far more moisture than the room produces, and removing it means cooling that air to a low dew point. In the example, 700 cfm of makeup air needs about 77,000 BTU per hour of cooling, almost four times the 20,280 BTU per hour dry recirculation coil duty.
Does a cleanroom heat load need a safety factor?
Add margin where an input is uncertain, such as future process tools, rather than a blanket percentage on every line. Airflow is already fixed by the air change rate, so a blanket factor only oversizes the coils and valves, and an oversized chilled water valve controls poorly at the light loads the room sees overnight.
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