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PE Architectural Engineering · Complete public lesson

Fan affinity laws: check airflow, pressure and motor power

Work a fan speed change using one consistent ratio, test the cubic shaft-power demand against a motor limit, and recognize when the installed system needs a new curve intersection.

Suggested 20 minutes · No account required

Fan affinity laws: airflow, pressure and the motor-power limit

Follow one speed ratio through the first, second and third powers, then check the shaft-power limit and a changed speed.

3 minutes 8 seconds · English captions and transcript

This supplementary SI example teaches the governing relationships; the October 2026 Architectural exam specification lists USCS units. The narrated example uses the same givens as the complete lesson below. The Your Turn prompt precedes the solution. Pause the video to work at your own pace.

Read the video transcript

ReasonSpan™. P E Architectural Engineering. Fan affinity laws.

A fan speed increase can meet an airflow target and still exceed the motor's available output. In this example, we will keep one speed ratio throughout, apply the first, second and third powers, and check the decision before accepting the change.

The fan originally runs at eight hundred revolutions per minute. Airflow is four cubic meters per second, pressure rise is six hundred pascals, and required shaft power is five kilowatts. The proposed speed is nine hundred and twenty revolutions per minute. The available motor shaft-output limit is seven point five kilowatts. Assume the same diameter, density and efficiency, dynamically similar operation, and unchanged system resistance proportional to airflow squared, without a fixed pressure offset.

Divide new speed by original speed: nine hundred and twenty divided by eight hundred is one point one five. Airflow uses the first power. Four times one point one five gives four point six cubic meters per second. Pressure rise uses the second power. Six hundred times one point one five squared gives seven hundred and ninety-three point five pascals. Keep the same ratio direction for both calculations.

Shaft power uses the third power. Five times one point one five cubed gives approximately seven point six zero kilowatts. A fifteen percent speed increase has produced about a fifty-two percent power increase. The unrounded requirement exceeds the supplied seven point five kilowatt limit. Airflow alone cannot establish that this proposed change is acceptable.

Under this limited model, what speed would put required shaft power exactly at seven point five kilowatts?

Set five times the ratio cubed equal to seven point five. The limiting ratio is the cube root of one point five, approximately one point one four four seven. Multiplying by eight hundred gives about nine hundred and fifteen point seven seven revolutions per minute. This is a theoretical equality with no margin, not a recommended setpoint.

Now reduce the original speed to seven hundred and twenty revolutions per minute. The ratio becomes zero point nine. Airflow is three point six cubic meters per second, pressure rise is four hundred and eighty-six pascals, and shaft power is three point six four five kilowatts. The quantities fall by ten, nineteen and twenty-seven point one percent. Then check whether the reduced airflow still meets the actual requirement.

The useful sequence is ratio, exponent, limit, and system check. If a filter, damper or duct resistance changes, obtain the new fan curve and intersect it with the actual system curve. Confirm manufacturer limits, motor output, drive capability and controls. Open the complete ReasonSpan lesson and its application check, then try eight different free Architectural questions. The links are in the description.

The situation and givens

A fixed-diameter fan operates at 800 rpm, delivering 4.00 m³/s at a pressure rise of 600 Pa and requiring 5.00 kW of shaft power. A proposed speed increase to 920 rpm is intended to increase airflow. The motor's available shaft-output limit is 7.50 kW. Determine whether the proposal passes that power comparison.

  • Original speed N₁ = 800 rpm; proposed speed N₂ = 920 rpm.
  • Original airflow Q₁ = 4.00 m³/s; pressure rise Δp₁ = 600 Pa; shaft power W₁ = 5.00 kW.
  • Same fan diameter, air density and efficiency; dynamically similar operation.
  • The unchanged system resistance is proportional to Q², with no fixed pressure offset. The motor's available shaft-output limit is 7.50 kW.
Conceptual illustration of a centrifugal ventilation fan, motor and ductwork in a commercial mechanical room.
Conceptual equipment illustration. Use the stated numerical givens for the calculation.

Complete worked method

1. Form the new-to-original speed ratio

Keep subscript 1 for the original condition and 2 for the proposed condition. The ratio is 920/800 = 1.15. A 15% increase means 1.15, not 0.15. Use this same direction throughout.

r=N2N1=920800=1.15r=\frac{N_2}{N_1}=\frac{920}{800}=1.15

2. Predict airflow and pressure rise

Airflow uses the first power: 4.00 × 1.15 = 4.60 m³/s. Pressure rise uses the second power: 600 × 1.15² = 793.5 Pa. These predictions depend on the stated similarity and unchanged-system assumptions.

Q2=4.00(1.15)=4.60 m3/sΔp2=600(1.15)2=793.5 PaQ_2=4.00(1.15)=4.60\ \mathrm{m^3/s}\qquad \Delta p_2=600(1.15)^2=793.5\ \mathrm{Pa}

3. Calculate shaft power before accepting the change

Required shaft power uses the third power. The result is 5.00 × 1.15³ = 7.604375 kW, approximately 7.60 kW. That is a 52.09% increase in power for a 15% speed increase. Compare unrounded demand with the limit: it exceeds 7.50 kW by about 0.104 kW. The proposed change fails this supplied shaft-power comparison.

W2=5.00(1.15)3=7.604375 kW>7.50 kWW_2=5.00(1.15)^3=7.604375\ \mathrm{kW}>7.50\ \mathrm{kW}

4. Find the theoretical power-limited speed

For this limited model, set 5.00r³ = 7.50. The limiting ratio is the cube root of 1.50, approximately 1.144714. The corresponding speed is 915.77 rpm and airflow is 4.57886 m³/s. This equality is a theoretical limit with no margin; it is not a recommended operating setpoint.

rlimit=7.50/5.003=1.144714Nlimit=800rlimit=915.77 rpmr_{limit}=\sqrt[3]{7.50/5.00}=1.144714\qquad N_{limit}=800r_{limit}=915.77\ \mathrm{rpm}

5. Return to the actual equipment and system

Verify the manufacturer's speed and performance limits, motor output, drive capability, controls and stable operating region. If a filter, damper, duct or fixed pressure requirement changes, obtain or construct the new fan curve and intersect it with the actual system curve. The original airflow ratio cannot establish the changed installed operating point.

Common errors and why they fail

  • Using 800/920 reverses the change and predicts reductions when speed increases.
  • Scaling shaft power linearly gives 5.75 kW and hides the power-limit failure.
  • Comparing required shaft power with electrical input treats different quantities as interchangeable.
  • Keeping the original curve intersection after resistance changes ignores the installed system.

Apply the method

Under the stated assumptions, what is the most defensible conclusion about the 920-rpm proposal?

Choose one response

Change one condition

Instead reduce speed from 800 to 720 rpm. With the same similarity assumptions, calculate airflow, pressure rise and shaft power before revealing the answer.

Reveal the worked transfer

The ratio is 0.90. Airflow becomes 3.60 m³/s, pressure rise becomes 486 Pa, and shaft power becomes 3.645 kW. The reductions are 10%, 19% and 27.1%, respectively. Meeting the motor limit does not prove that reduced airflow meets the ventilation or process requirement.

Put the method to work

Try different questions to check whether you can identify the governing condition and explain the next action. The free sampler includes answers and worked explanations.

Try 8 free questions →See program and price

Original learning example, separate from exam forms. These affinity-law predictions are conditional; actual equipment selection and electrical input require additional manufacturer and system evidence.

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