# Variable Frequency Drives - Generated Heat and Required Ventilation

## A variable frequency drive develops heat during work and to avoid overheating ventilation is required

Variable frequency drives are common for controlling the electric motor speed in applications with fans, pumps, compressors, elevators, extruders etc.

### Heat Loss from a Variable Frequency Drive

An amount of the power transferred through a variable frequency drive to the motor is lost as heat. The heat loss from a drive can be expressed as

H_{loss}= P_{t}(1 -η_{d})(1)

where

H_{loss}= heat loss to the variable-frequency drive surroundings (kW)

P_{t}= electrical power through the variable-frequency drive (kW)

η_{d}= variable-frequency drive efficiency

The heat loss expressed in imperial units

H_{loss}= P_{t}3412 (1 -η_{d})(1b)

where

H_{loss}= heat loss to the variable-frequency drive surroundings (btu/h)

P_{t}= power in to the frequency drive (kW)

η_{d}= variable-frequency drive efficiency

To calculate maximum heat loss - the maximum power transmission through the variable-frequency drive must be used.

It is common that the heat loss from a frequency drive is in the range* 2 - 6%* of the KVA rating.

### Necessary Ventilation for Cooling a Variable-Frequency Drive

Maximum ambient temperature for a frequency-drive is approximately *40 ^{o}C (104^{o}F)*. Since frequency-drives often are physical protected in small cabinets or small rooms, ventilation - or even cooling - may be needed to avoid overheating.

The mass flow of air needed for transporting heat from the variable-frequency drive can be expressed as

m_{air}= H_{loss}/ c_{p}(t_{out}- t_{in})(2)

where

m_{air}= mass flow of air (kg/s)

H_{loss}= heat loss to the frequency-drive surroundings (W)

c_{p}= specific heat of air (kJ/kg^{ o}C) (1.005 kJ/kg^{ o}C standard air)

t_{out}= temperature of air out (^{o}C)

t_{in}= temperature of air in (^{o}C)

Combined with (1), the mass flow (2) can be expressed as:

m_{air}=P_{t}(1 -η_{d}) / c_{p}(t_{out}- t_{in})(2b)

The volume flow can be calculated by multiplying (2b) with the specific volume or inverted density:

q_{air}= (1 / ρ_{air})P_{t}(1 -η_{d}) / c_{p}(t_{out}- t_{in})(2c)

where

ρ_{air}= density of air at the actual temperature (1.205 kg/m^{3}standard air)

#### Example - Ventilation and Cooling of a Variable-Frequency Drive

The amount of air for cooling a cabinet mounted variable-frequency drive with maximum power of *100 kW* and efficiency of *0.95*, maximum ambient operating temperature *40 ^{ o}C* and outside temperature

*20*, can be expressed as (2b):

^{ o}C

m_{air}= (100 kW)(1 - 0.95) / (1.005 kJ/kg.^{o}C) ((40^{ o}C) - (20^{ o}C))=

0.25 kg/s

The volume and density of air depends on the temperature of the air. The density of air at *20 ^{o}C* is

*1.205 kg/m*and

^{3}*1.127 kg/m*at

^{3}*40*.

^{ o}CThe volume flow at the inlet (*20 ^{ o}C*):

q_{air}= (1 / (1.205 kg/m^{3})) (0.25 kg/s)

= 0.208 m^{3}/s

= 749 m^{3}/h

The volume flow at the outlet (*40 ^{ o}C*):

q_{air}= (1 / (1.127 kg/m^{3})) (0.25 kg/s)

= 0.222 m^{3}/s

= 799 m^{3}/h

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