INTRODUCTION

Air movement is one of the biggest areas of energy use in buildings. One of the primary measures to save energy these days is to reduce the volume if air that is delivered to heat and cool. The TABB technician knows that reducing air flow will also reduce the fan power required to move the air. Energy savings is one of the primary reasons that variable air volume systems are very popular design choices today: with these systems, delivered air volumes are reduced whenever operating loads are less than design loads.

The technician, however, may not know how to calculate the fan power, energy and operating cost savings that result from a reduction in air flow. This article provides a simple method of estimating the power, energy and cost savings that result from a reduction in air flow through a fan.

THE FAN LAWS

The fan laws tell us that for a given fan system and condition of air, the air flow in cfm is directly proportional to rotational speed of the fan.

cfm2/cfm1 = rpm2/rpm1 (Equation 1)

The fan laws also tell us that fan power (brake horsepower or bhp) is proportional to the cube of the fan speed.

bhp2/bhp1 = (rpm2/rpm1)3 (Equation 2)

Combining the two equations and substituting from equation 1,

bhp2/bhp1 = (cfm2/cfm1)3 (Equation 3)

By rearranging this equation , it is possible to calculate the new fan horsepower when the air flow is reduced:

bhp2 = bhp 1 x (cfm2/cfm1)3 (Equation 3)

EXAMPLE CALCULATION

For instance, say a fan is operating at 10 bhp at 10,000 cfm and the air flow is reduced 10% (to 9,000 cfm) by reducing the fan speed by 10%. The equation becomes:

bhp2 = 10 bph x (9,000 cfm/10,000 cfm)3

bhp2 = 7.2 hp (72% of the original 10 bph)

Fan power can be converted to motor input power by multiplying the brake horsepower by the horsepower to kilowatt conversion factor (1 bhp = 0.746 kW) and dividing by the fan efficiency. For the example above, assuming that the motor has an efficiency of 90% at 100% and 75% of full load, (for most motors, the efficiency curve is fairly flat between 75% and 100% of load) the difference in input power will be:

At the original load the motor input power is:
10 bhp x 0.746 kw/bhp/0.90 (motor efficiency) = 8.3 kW

At the reduced air flow the motor input is:
7.2 bhp x 0.746 kW/bhp / 0.90 = 6.0 kW

The savings is therefore 8.3 kW Ð 6.0 kW = 2.3 kW

If the fan operates 3000 hours per year, the annual kilowatt hours saved is

3,000 hours x 2.3 kW = 6,900 kilowatt-hours.

ENERGY COST SAVINGS

A typical average cost per kilowatt-hour is about 10 cents ($0.10) per kWh, (however the cost may be up to twice this under some electric rates in some areas) the annual cost savings resulting from the reduction in fan energy is:

6,900 kWh x $0.10/kWh = $690 per year.

METHODS OTHER THAN FAN SPEED TO REDUCE AIR VOLUME DO NOT RESULT IN AS GREAT POWER/ENERGY REDUCTION

Note that this amount of reduction is only possible if the reduction in air flow is achieved by reducing the fan speed. Permanent reductions in fan speed and volume can be made by changing the fan and/or motor sheave diameter. A popular option chosen in recent years is to install a variable-speed drive to change the motor speed so that the fan speed can change dynamically in response to the demand for cooling or heating.

Other methods that reduce delivered air flow include discharge dampers, inlet vanes, fan cycling or diverting air. Options to reduce air flow that do not change the fan speed will not result in as much energy reduction. Discharge dampers and inlet vanes trade pressure drop for the reduction in air flow. Although there are some savings they are not as great as with fan speed control. With fan cycling there is no change in fan power, the air flow and fan energy are reduced in proportion to the amount of time that the fan is cycled off. With bypassing there is no savings at all because the fan delivers the same flow at the same pressure, and the by-passed air is essentially wasted. In the next month, we will compare several methods of volume reduction.