Engineering task and calculation objective
The VENT module calculates the input power of fans: from the inlet temperature, inlet pressure and inlet volume flow of the gas, the required pressure increase and the fan efficiency, it determines the necessary shaft power. A distinction is made between the arrangement – induced draft (fan downstream of the equipment, drawing from the sub-atmospheric side) or forced draft (fan upstream of the equipment) – because the density and volume flow at the fan inlet depend on where the fan is installed in the system.
The need to calculate the power consumption of a fan arises in almost every air handling system in plant engineering: combustion air and induced draft fans on furnaces, cooling air fans on air coolers and cooling towers, process exhaust, drying plants, or the ventilation of buildings and enclosures. The pressure increase follows from the pressure drop calculation of the connected system (ducts, equipment, silencers, dampers), the volume flow from the process balance.
The result – the power required at the fan shaft – is the basis for motor selection, for assessing energy costs and for comparing control concepts. Since fans often run continuously, even a few percentage points of efficiency decide over substantial operating costs.
Calculation workflow
- Define the arrangement and suction conditions: First, the fan arrangement is chosen (induced or forced draft). This determines the thermodynamic state at the fan inlet: for an induced draft fan, inlet pressure and density are below ambient, and the volume flow drawn in is correspondingly larger.
- Define the operating point: Inlet temperature, inlet pressure and inlet volume flow are entered; they determine the gas density at the inlet and thus the conversion between mass flow and volume flow.
- Apply the pressure increase: The required total pressure increase of the fan is taken over from the pressure drop calculation of the connected system – including all ducts, equipment, dampers and exit losses at the design volume flow.
- Calculate the input power: From volume flow and pressure increase follows the theoretical air power; dividing by the fan efficiency yields the power required at the shaft. For motor selection, gearbox/belt and motor losses plus a design margin are added.
Input quantities
| Quantity | Symbol | Unit |
|---|---|---|
| Fan type 1 = sucking / 2 = blowing | < > | - |
| Inlet temperature | ϑE | °C |
| Inlet pressure | pE | Pa |
| Inlet volume flow | VE | m³/s |
| Outlet temperature | ϑA | °C |
| Outlet pressure | pA | Pa |
| Outlet volume flow | VA | m³/s |
| Pressure increase | dp | Pa |
| Fan efficiency | n | - |
| Manufacturer | Lüfterhersteller | - |
| Type | Lüftertyp | - |
| Noise level distance | in | m |
| Noise level distance | in | dB(A) |
| Maximum noise level | m | dB(A) |
| Number of fans | Lüfter | - |
| Connection (D=delta/S=star) | (D=Dreieck/S=Stern) | - |
| Number of poles | Polanzahl | - |
| Data per motor | ( ) | - |
| (Nennwert) | (Nennwert) | W |
| (Nennwerte) | (Nennwerte) | A |
| (Betriebspunkt) | (Betriebspunkt) | 1/min |
| (Betriebspunkt) | (Betriebspunkt) | W |
| (Betriebspunkt) | (Betriebspunkt) | A |
| Operating temperature | Betriebstemperatur | °C |
Calculated results
| Quantity | Symbol | Unit |
|---|---|---|
| Input power | P | W |
Worked example
A forced draft fan is to deliver 20,000 m³/h of air (drawn in at 20 °C, 1 bar) against a total pressure increase of 2,500 Pa. The fan efficiency is 72%. This worked example determines the power required at the fan shaft.
Given values
| Inlet volume flow V̇ | 20,000 m³/h = 5.556 m³/s |
| Pressure increase Δp | 2,500 Pa |
| Fan efficiency η | 0.72 |
Solution
Theoretical air power
Pair = V̇ · Δp = 5.556 · 2,500 ≈ 13,890 W ≈ 13.9 kW
Power required at the shaft
Pshaft = Pair / η = 13,890 / 0.72 ≈ 19,290 W ≈ 19.3 kW
For motor selection, transmission and motor losses plus a design margin (typically 10 to 15%) must additionally be considered – here one would typically select a 22 kW standard motor.
Result
| Air power | ≈ 13.9 kW |
| Power required at the shaft | ≈ 19.3 kW |
| Motor proposal | 22 kW standard motor |
All values are illustrative. The applicable standard and project-specific boundary conditions remain authoritative.
Frequently asked questions
Why must induced draft and forced draft arrangements be distinguished?
Because a fan always handles the volume flow at its inlet. If it sits on the suction side, downstream of equipment with a high pressure drop, the inlet pressure is lower and the density smaller – the same mass flow target then means a larger volume flow and thus a higher power requirement. The temperature at the inlet (e.g. downstream of a hot process) acts in the same way via the density.
What efficiency is realistic to assume?
That depends on the fan type and the operating point: centrifugal fans with backward-curved blades achieve roughly 75 to 85% at their best efficiency point, axial fans similar values, simple forward-curved (squirrel cage) impellers considerably less (50 to 65%). What matters is that the design point lies close to the efficiency optimum of the fan curve – an oversized fan operated with throttling wastes energy permanently.
Should the pressure increase be entered as static or total pressure?
For the power requirement, the total pressure increase governs, i.e. including the dynamic pressure components. Manufacturer curves are quoted partly as static and partly as total pressure increase – when comparing offers and applying the system curve, the definition must match, otherwise the fan will be sized incorrectly.
Does the compressibility of the air have to be considered?
At the pressure increases typical for fans, up to about 10,000 to 30,000 Pa, the density change is small and the incompressible calculation P = V̇·Δp/η is sufficiently accurate. At higher pressure ratios, one speaks of blowers or compressors; polytropic or isentropic compression calculations with temperature rise are then required.