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Fan Performance and Damper Losses

A fan does not know there is a damper in front of it. It makes the pressure its curve says it will make, and anything the duct does not need is burnt across the damper as heat and noise — you pay for every kilowatt of it. This works out how efficient your fan really is, how much of its power the damper is throwing away, and what a speed drive would give you back in a year. Press YOUR FAN on the board to type your numbers in.

Why this matters now

Fan power, false air and unreliable gas measurements quietly increase fuel and electricity cost while weakening every heat and mass balance in the plant.

What this tool helps decide

See whether the fan is operating where it should and quantify pressure, shaft power and energy lost across the damper before changing the fan or drive.

What subscription adds

Keep the gas-side audit together: save readings by fan, duct or machine, revisit the same equipment later and produce a report with findings, cost signals and priorities.

Connected to current industry priorities around energy efficiency, emissions measurement, waste heat and combustion optimisation reported in World Cement, CemNet and Global Cement.See process-tool plans

Where your fan is running

The gas

This only sets the composition below. Change any of it afterwards.
Degrees C at the fan.
Per cent by volume.
% dry
% dry
% dry
Pa at sea level.
Metres.
Pa. Negative if the fan sucks.
Grammes per m³.

The fan as it runs today

Revolutions per minute.
Per hour.
Pascals, inlet to outlet. 1 mbar is 100 Pa, 1 mmH₂O is 9.81 Pa.
Pascals. This is what the damper is throwing away. Put 0 if the damper is wide open or there is not one. If you have never measured it, read the pressure either side of the damper — it is the single most valuable reading on this page.
Kilowatts. If you only know the motor, use motor kW times motor efficiency.

The fan curve from the maker

Points off the manufacturer's curve, at the speed below. Flow in the same units you chose above. Leave rows blank if you have fewer points. Without a curve the tool still gives you efficiency and the damper loss, but it cannot work out the speed drive saving.
Flow per hourPressure, Pa

What power costs you

In your own money.
Running hours.
THE GAS AND THE FAN
HOW HARD IT IS THROTTLED
WHAT A SPEED DRIVE WOULD DO
Gas density here
—
Actual gas flow
—
Fan efficiency
—
Power per 1000 m³/h
—
Power it should draw
—
Damper is burning
—
Share of the pressure
—
Flow with damper open
—
Duct resistance, k
—
Throttled back to
—
Speed it would need
—
Power at that speed
—
You would save
—
Worth a year
—
Power saved a year
—

The fan curve, point by point

What to do about it

How the numbers are worked out

Every formula here was checked against your own TCEA fan spreadsheet before this page was written.

Gas density. ρN = MW / 22.414, then ρ = ρN × 273.15/(273.15+T) × Pabs/101325, where the absolute pressure is the barometric pressure plus the static pressure at the fan — which is negative when the fan sucks. Checked: your sheet gives 0.834656 kg/m³ at 150 °C and 0.616217 at 300 °C, and this reproduces both to six figures.

The fan laws. Q₂/Q₁ = N₂/N₁, ΔH₂/ΔH₁ = (N₂/N₁)² × (ρ₂/ρ₁), P₂/P₁ = (N₂/N₁)³ × (ρ₂/ρ₁). Your sheet lists those ratios in its own rows and this matches every point of its curve — 2962.2 Pa at 6.944 m³/s becomes 2125.11 Pa at 5.882 m³/s for a speed ratio of 0.847, and 0.847² is exactly 0.717409.

Note the density term. Gas that gets hotter gets thinner, so a fan makes less pressure and draws less power on hot gas. That is why a fan pulls more amps on a cold start than when the system is up to temperature. The site's older assets/js/fan-curve.js has this term the wrong way up; your spreadsheet has it right, and this page follows the spreadsheet.

Efficiency is the definition, not a model: η = Q [m³/s] × ΔH [Pa] ÷ shaft power [W]. A cement plant fan in good order runs 70–85 %. Below about 60 % something is wrong — a worn or coated impeller, the wrong wheel, or a fan running far off its best point.

The damper loss. The fan makes the pressure its curve says it will make. The duct only needs k × Q². Everything else is burnt across the damper. Measure the pressure either side of the damper and this becomes exact: kduct = (ΔHfan − Δpdamper) / Q².

The speed drive saving. With the damper opened and the speed dropped instead, the fan meets the duct at the same flow. The speed it needs is Q ÷ Qfree, where Qfree is where the fan curve crosses the duct resistance. Power then falls with the cube of speed. The cube law assumes the efficiency does not change with speed, which is not quite true, so the saving is an estimate — a good one, and it is always conservative in practice because a drive also removes the throttling turbulence.

What this cannot tell you. It does not know whether your fan can be fitted with a drive, what the motor will tolerate, or the critical speeds of the shaft. It tells you whether the saving is big enough to be worth asking those questions.