Cooling time of an Induction motor
cooling motor

Cooling status is referred to the condition defined by a motor temperature reduction caused by a decreasing current from running condition to a less than the Full Load Amps (FLA) condition known as minimum running condition. The cooling status is also initiated by a stopped motor following a running condition.

Motor thermal status, also referred as motor thermal capacity is a parameter used to evaluate the thermal condition of the motor.

Thermal status is measured in percentage of the motor thermal limit. When a motor is stopped for a long time, the thermal status is 0% which means that zero percent of its capacity is used. At motor starting, the thermal capacity used can reach let's said 40%. If the motor has two cold starts, then the thermal capacity after these actions would be 80%. Exceeding the thermal capacity of the motor produces a loss of insulation life due to deterioration by overheating the windings. 

During the cooling time, the thermal status decreases from a higher running thermal status to a lower one.

The equation of the cooling motor is given by:

θactual=θfinal+θinitial-θfinale-t/τ

where:

θinitial: Thermal status at the start of the cooling period (t= 0) in percent

θfinal   : Thermal status at the end of the cooling period (t → ꝏ) in percent

θactual: Thermal status after time t in percent

t           : Time in seconds

τ           : Cooling time constant in seconds

Thermal status is proportional to I squared and the equation can be expressed in the following form: 

I2(t)=I2f+I20-I2fe-t/τ

where:

I20: Motor current squared at the start of the cooling period (t= 0) in per unit (pu)

I2f : Motor current squared at the end of the cooling period (t → ꝏ) in pu

I2(t): Motor current squared after time t in pu

t      : Time in seconds

τ      : Cooling time constant in seconds

 

Example

A motor with a cooling time constant of 3.5 hours is stopped with a thermal status of 90%. Calculate the time required to reach 40% motor status.

 

Replacing the data in the first equation gives:

40=0+90-0e-t/12600

t=-12600 ln4090=10217 s

The required time is 2 hours 50 minutes.

The curve below shows the motor thermal status variation with time for this example.  

Ernesto Hidalgo-Tupia, Eng., M. A. Sc. 
Electrical Engineer

 

 

 

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