Heat gain arising from fan power

The flow of air along a duct results in the airstream suffering a loss of energy, and for the

Flow to be maintained a fan must make good the energy loss (see section 15.4). The energy

Dissipated through the ducting system is apparent as a change in the total pressure of the airstream and the energy input by the fan is indicated by the fan total pressure.

Ultimately, all energy losses appear as heat (although, on the way to this, some are evident as noise, in duct systems). So an energy balance equation can be formed involving the energy supplied by the fan and the energy lost in the airstream. That is to say, the loss of pressure suffered by the airstream as it flows through the ducting system and past the items of plant (which offer a resistance to airflow) constitutes an adiabatic expansion which must be offset by an adiabatic compression at the fan.

So, all the power supplied by the fan is regarded as being converted to heat and causing an increase in the temperature of the air handled, At, as it flows through the fan.

A heat balance equation can be written, accepting an expression for air power derived later in section 15.4.

Air power = fan total pressure (N rrf2) x volumetric flow rate (m3 s-1)

The rate of heat gain corresponding to this is the volumetric flow rate xpxcxAt, where p and c are the density and specific heat of air, respectively. Hence

_ fan total pressure (N irf2)

TOC o "1-5" h z 1 ~ P c

The air quantities have cancelled, indicating that the rise in air temperature is independent of the amount of air handled, and using p = 1.2 kg m-3 and c = 1026 J kg-1 K-1 we get

, fan total pressure (N m“2 ) . iri>

At =————— 1231 ^ ( ^

Thus, the air suffers a temperature rise of 0.000 812 K for each N irf2 of fan total pressure.

The energy the fan receives is in excess of what it delivers to the airstream, since
frictional and other losses occur as the fan impeller rotates the airstream. The power input to the fan shaft is termed the fan power (see section 15.4) and the ratio of the air power to the fan power is termed the total fan efficiency and is denoted by T|. Not all the losses occur within the fan casing. Some take place in the bearings external to the fan, for example. Hence, for the case where the fan motor is not in the airstream, full allowance should not be made. It is suggested that a compromise be adopted.

If an assumption of 70 per cent is made for the fan total efficiency and if it is assumed that, instead of 30 per cent, only 15 per cent of the losses are absorbed by the airstream (since some are lost from the fan casing and the bearings) equation (6.10) becomes

A, _ fan total pressure (Nm’2)

1 ~ 1231 x 0.85

. fan total pressure (N m“2) „

* =———————- 1045————————- (6I1)

Thus almost one thousandth of a degree rise in temperature for each N itT2 of fan total pressure results from the energy input at the fan. In other words, a degree rise occurs for each kPa of fan total pressure.

When the fan and motor are within the airstream, as is the case with many air handling units, all the power absorbed by the driving motor is liberated into the airstream. Full account must then be taken of the motor inefficiency as well as all the fan inefficiency. Assuming a total fan efficiency of 70 per cent and a motor efficiency of 90 per cent the temperature rise of the airstream is

^ _ fan total pressure (N m“2)

1231 x 0.7 x 0.9 fan total pressure (Nm"2)

116 This represents a temperature rise of about 1.3 K for each kPa of fan total pressure.

(6.12)

Posted in Engineering Fifth Edition