What happens if connecting a 50hz induction motor to 60hz power supply
What happens if connecting a 50hz induction motor to 60hz power supply
To answer this question will require a little background on induction motor and some basic maths.
All commonwealth countries has 50hz power supply, where as USA and Europe have 60hz. It is not a matter of which is better.
Synchronous speed, Ns = 120 * f/p
where
f = frequency
p = the number of poles in the motor
If it is a 4 poles motor, and when the power supply is 50hz, the Ns will be 120 x 50/4 = 1500 rpm
what if it is a 4 poles motor operating at 60hz, the the Ns will be 120 x 60/4 = 1800 rpm
this represents a 20% operating speed increase. [ie (1800-1500)/1500* 100% = 20%]
Here are the problems when operating a motor beyond it’s design operating specifications.
1. if a motor is designed for 1500rpm beyond its specification%, i.e. you operate it at 1800rpm, you will increase heat generated, vibration, wear and tear on it’s bearing and shafts hence reduce operating life of the motor.
2. The increase speed will affect the flux density (B) of the motor.
Flux density = B = Ø/Area ∝ (Volt/frequency)
where
Ø = flux
V = voltage
f = frequency
Assuming the voltage (V) is the same (constant) when running the motor at 50hz and 60hz,= we can rewrite the above flux density formula to the below
B ∝ 1/frequency, flux density is inversely proportion with frequency, so when frequency increases the flux density decreases and vice versa. And as flux density decreases, flux Ø decreases. Reduce flux will reduce the torque capability of the motor.
3. The third problem operating a 50hz induction motor at 60hz, will increase the rotor reactance that will cause the power factor to reduce.
Let’s take a look at he maths
Reactance X = 2 π fr L,
where
fr = the frequency of the rotor
L = inductance of the rotor
if fr increases X increases
Power factor, pf = R/Z
where
R = resistance
Z = impedance
When Z increases while R remained constant, pf will reduce.
and P = I2R; increase current (I) will increase hence the motor will generate more heat.
So what is the solution?
We recall that B ∝ V/f
therefore
if V1 = 400V, f1 = 50hz and f2 = 60hz,=
then
V1/f1 = V2/f2
hence
V2 = V1/f1 * f2 = 400V x 60hz/50hz,= we will need to increase the supply voltage to 480V which is not always practical. Therefore it is always best to run the induction motor within its design specification.
