Can We Operate a 60Hz Transformer on 50Hz Supply Source and Vice Versa?

Electrical Technology

What Happens When a 50Hz Transformer Operates on 60Hz and Vice Versa?

Let’s suppose we have a transformer with same rating and parameters when operated at 60Hz and 50Hz supply frequency then. Lets see what happens.

Transformer Rating & Parameters

Suppose, a transformer rating as follow where we will connect it first to the 60Hz supply frequency and 50Hz later.

V = Voltage = 5kV
R = Resistance = 6.1Ω
L = Inductance = 0.3 Henry
f = Frequency = 60Hz

Transformer Operated on 60Hz

To find the current in the primary of transformer by using I = V / Z, so we will first find the impedance which depends on inductive reactance.

Inductive Reactance = X_L = 2πfL

X_L = 2πfL = 2 x 3.1415 x 60 x 0.3 = 113.1 Ω

And

Impedance Z = √ (R²+X_L²)

Z = √ (6.1²+113.1²)

Z = 113.26 Ω

Now the value of current

I = V / Z

I = 5kV / 113.26 Ω

I = 44.14 A

Related Post: Which Transformer is More Efficient When Operates on 50Hz or 60Hz?

Transformer Operated on 50Hz

The same transformer is now connected to the 50Hz frequency. We will do the same calculation as above for 60Hz transformer.

Inductive Reactance = X_L = 2πfL

X_L = 2πfL = 2 x 3.1415 x 50 x 0.3 = 94.245 Ω

And

Impedance = Z = √ (R²+ X_L²)

Z = √ (6.1²+ 94.245 ²)

Z = 94.44 Ω

Now the value of current

I = V / Z

I = 5kV / 94.44 Ω

I = 52.94 A

Related Post: Is it Possible to Operate a 50Hz Transformer on 5Hz or 500Hz Frequency?

We see that the value of primary current in case of 50Hz transformer is more than in case of 60Hz transformer.

Now the more current, the more problems i.e.

X_L ∝ f
Z ∝ XL

But

X_L ∝ 1/I

And

I ∝ I/Z

In simple words:

When frequency (f) increases, Inductive reactance (X_L) increases.
When Inductive reactance (X_L) increases, Impedance (Z) increases.
When Impedance Increase, Current decreases.

So we can say that:

When we operate a 60Hz transformer on 60Hz supply source, the current were 44.14 A.

But when operated a 60Hz transformer on 50Hz supply source, the current were 52.94 A.

The extra current in this case may causes copper loss (P = I²R) and produces heat.

We may use an additional current limiter device (Inductor or resistor) in series with the primary winding of a transformer by which we can increase the overall impedance to reduce the extra current. This way, we can use a 60Hz transformer on 50Hz AC Supply.

If we still able to operate a 50Hz transformer on 60Hz, we may face the following results.

The applied voltage to the primary of transformer windings

V(t) = V_M Sin ωt … Volts

is applied to a transformer’s primary winding, the flux of the transformer is given by

If the applied voltage V is increased by 10%, the resulting maximum flux in the clore also increases by 10%.

The 10% increase in flux requires an increase in magnetization current much larger than 10% (see fig 1). As the voltage increase, the high magnetization current soon becomes unacceptable. The maximum applied voltage and rated voltage is set by the maximum acceptable magnetization current in the core. Keep in mind that voltage and frequency are inversely proportional (i.e. V 1/f) when the value of maximum flux is constant.

Φ_Max = – V_M / ωN_P

Where

Φ_Max = Maximum flux
V_M = Max voltage
N_P = Number of turns in primary
ω = 2πf ( where f = frequency)

Related Post: Why Flux in Primary and Secondary Winding is Always Equal?

Therefore, if a 60Hz transformer is to be operated on 50Hz, its applied voltage must also be reduced by 1/6 or the peak flux in the core will be too high/. This reduction in applied voltage with frequency is known as derating.

Similarly, a 50Hz transformer may be operated at a 20% higher voltage on 60Hz if this action does not cause insulation problems.