Air horsepower, abbreviated as AHP, is the useful power that a fan transfers to the air.
It represents the theoretical power required to move a specified airflow against a specified pressure. Air horsepower does not include efficiency losses in the fan wheel, bearings, drive system, or motor.
For this reason, air horsepower is lower than the actual shaft power and electrical input power required by a fan system.
How Is Air Horsepower Calculated?
In imperial units, air horsepower is commonly calculated using:
AHP = Q × P / 6,356
Where:
- AHP = air horsepower
- Q = airflow, CFM
- P = pressure, inches of water gauge
- 6,356 = unit conversion constant
This formula calculates the minimum theoretical power required to move the specified airflow against the specified pressure at 100% efficiency.
In SI units, air power can be calculated using:
Air Power (kW) = Airflow (m³/s) × Pressure (Pa) / 1,000
When airflow is expressed in cubic meters per hour:
Air Power (kW) = Airflow (m³/h) × Pressure (Pa) / 3,600,000
The calculated value is expressed in kilowatts. It can be converted to horsepower using:
1 hp ≈ 0.746 kW
Static Air Horsepower vs. Total Air Horsepower
Air horsepower can be calculated using either fan static pressure or fan total pressure.
Static Air Horsepower
Static air horsepower is calculated using fan static pressure:
Static AHP = Q × Ps / 6,356
Where Ps is the fan static pressure.
Static air horsepower represents the useful power associated with overcoming resistance from ducts, filters, dust collectors, dampers, elbows, hoods, and other system components.
Total Air Horsepower
Total air horsepower is calculated using fan total pressure:
Total AHP = Q × Pt / 6,356
Where Pt is the fan total pressure.
Fan total pressure includes both static pressure and velocity pressure. Total air horsepower is therefore normally higher than static air horsepower at the same airflow.
When comparing fan efficiencies, the pressure type must remain consistent. Static air horsepower should be compared with static efficiency, while total air horsepower should be compared with total efficiency.
Air Horsepower vs. Brake Horsepower
Air horsepower is the useful power delivered to the airflow. Brake horsepower, abbreviated as BHP, is the mechanical power required at the fan shaft.
Because no fan operates at 100% efficiency, brake horsepower is higher than air horsepower.
The relationship can be expressed as:
Fan Efficiency = Air Horsepower / Brake Horsepower × 100%
Brake horsepower can therefore be estimated using:
Brake Horsepower = Air Horsepower / Fan Efficiency
For example, if the required air horsepower is 10 hp and the fan efficiency is 70%, the estimated brake horsepower is:
10 / 0.70 = 14.29 hp
A more efficient fan requires less shaft power to produce the same airflow and pressure.
Air Horsepower vs. Motor Horsepower
Air horsepower should not be used directly as the required motor rating.
Power passes through several stages in a fan system:
Electrical input power → Motor output power → Fan shaft power → Air power
Energy losses may occur in:
- The electric motor
- Belts, couplings, or other transmission components
- Bearings and mechanical components
- The fan wheel
- Airflow through the fan inlet and outlet
As a result, the electrical input power is normally higher than the brake horsepower, and the brake horsepower is higher than the air horsepower.
Motor selection should be based on the fan manufacturer’s performance curve, the brake horsepower at the operating point, transmission efficiency, expected operating range, and an appropriate design margin.
Air Horsepower Calculation Example
Consider an industrial dust collection system with the following requirements:
- Airflow: 10,000 m³/h
- System static pressure: 2,000 Pa
First, convert the airflow to cubic meters per second:
10,000 / 3,600 = 2.78 m³/s
The theoretical air power is:
2.78 × 2,000 / 1,000 = 5.56 kW
Converted to horsepower:
5.56 / 0.746 = 7.45 hp
This means that approximately 5.56 kW, or 7.45 hp, of useful air power is theoretically required to move 10,000 m³/h of air against 2,000 Pa of static pressure.
If the fan static efficiency is 70%, the estimated shaft power is:
5.56 / 0.70 = 7.94 kW
This calculation does not mean that a 7.94 kW motor can be selected directly. The final motor rating must be confirmed using the fan curve, actual air density, drive arrangement, operating range, and manufacturer’s selection data.
Why Is Air Horsepower Important in Dust Collection?
Air horsepower helps explain the relationship between airflow, system pressure, and fan power.
When pressure increases while airflow remains constant, the required air horsepower increases proportionally.
When airflow increases while pressure remains constant, the required air horsepower also increases proportionally.
Reducing unnecessary pressure loss can therefore reduce both theoretical air power and actual energy consumption. This may involve:
- Selecting appropriate duct diameters
- Reducing unnecessary duct length
- Avoiding sharp elbows and poor transitions
- Maintaining clean filter media
- Controlling excessive filter pressure drop
- Designing efficient hoods and branch connections
In an industrial dust collection system, air horsepower can be used to:
- Estimate the theoretical fan power requirement
- Compare different airflow and pressure conditions
- Calculate fan static or total efficiency
- Evaluate the effect of pressure loss on energy demand
- Check whether a proposed fan or motor selection appears reasonable
Can Air Horsepower Be Used for Fan Selection?
Air horsepower is a theoretical calculation and cannot replace formal fan selection.
Actual fan power depends on:
- Fan type and wheel design
- Fan operating efficiency
- The system operating point
- Air density and temperature
- Fan speed
- Drive efficiency
- Inlet and outlet system effects
- Dust collector pressure drop
- Changes in filter resistance during operation
Final fan selection should be based on the brake horsepower or shaft power shown on the manufacturer’s fan curve at the required operating point.
The selected motor must also be checked across the expected operating range to ensure that it will not become overloaded.