【2026 Transmission Equipment】Key points for selecting motors for different types of fans
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This post was last edited by The one on 2026-5-28 15:20. ▶ One-sentence conclusion: First calculate the shaft power → then multiply by a safety factor to determine the motor power → and finally select the appropriate speed. The key to selecting a fan motor is to convert \"air volume × total pressure\" into \"motor power\".
I. Why are mistakes common when selecting fan motors? Fans are one of the most common types of loads in industrial settings, but there are many cases of incorrect motor selection: ●❌ Insufficient motor power → difficult starting, overheating during operation, and frequent tripping. ●❌ Choosing an overly high motor power → wasted investment, low operational efficiency, and using overpowered equipment for minor tasks. ●❌ Mismatched rotation speed → Deviation in fan performance, high noise, and high energy consumption. ●❌ Load characteristics were ignored → A centrifugal fan motor was used for an axial flow fan → Startup failed.
💡The essence of proper selection: Using mathematical formulas to convert the fan’s \"air volume and total pressure\" into the motor’s \"power, speed, and torque\".
II. Analysis of Core Parameters 2.1 Parameters on the Fan Side • Air volume Q (m³/h): The volume of air delivered by the fan per hour. • Total pressure P (Pa): The pressure difference between the outlet and inlet of the fan (including dynamic pressure + static pressure). • Efficiency η (—): The efficiency with which the fan converts shaft power into wind pressure (usually 0.5~0.85).
2.2 Motor-side parameters • Shaft power Pₛ (kW): The power required by the fan shaft. • Motor power Pₘ (kW): The rated power of the selected motor. • Speed n (rpm): The rated speed of the motor.
III. Three core calculation formulas Formula 1: Calculation of shaft power Pₛ = (Q × P) / (3600 × 1000 × η) Parameter explanations: Q (m³/h), P (Pa), η (efficiency; values range from 0.6 to 0.8 as specified in the manual).
Example: Q=10000 m³/h, P=2000 Pa, η=0.7 → Pₛ = (10000×2000)/(3600×1000×0.7) = 7.94 kW
Formula 2: Motor power (× safety factor) Pₘ = Pₛ × K (safety factor). • For a steady load (fan): K = 1.1~1.2. • Moderate fluctuation (exhaust fan): K = 1.2~1.3. • Large fluctuations (dust extraction fan): K = 1.3~1.5.
Continuing with the example: for the ventilator, K=1.2 → Pₘ = 7.94×1.2 = 9.53 kW → Therefore, a 11 kW motor should be selected
Formula 3: Relationship between rotational speed and fan performance – Similarity law: Q₁/Q₂ = n₁/n₂； P₁/P₂ = (n₁/n₂)² ； Pₛ₁/Pₛ₂ = (n₁/n₂)³
Key conclusion: A 20% reduction in rotational speed results in a roughly 50% decrease in power (this is the principle behind energy savings through variable frequency control!) ）