Electrical Machines · Transformers

Transformer Tests

Dr. Mithun Mondal BITS Pilani, Hyderabad Campus Electrical Machines

Demonstrative Video

Trn Tests
trn tests
Trn Tests

  • The performance of a transformer can be calculated on the basis of its equivalent circuit which contains four parameters (as referred to primary)

    1. the equivalent resistance \(R_{01}\)

    2. the equivalent leakage reactance \(X_{01}\)

    3. the core loss conductance \(G_0\) or resistance \(R_0\)

    4. the magnetizing susceptance \(B_0\) or reactance \(X_0\)

  • These constants or parameters can be determined by two tests:

    1. open-circuit tests

    2. short-circuit tests

  • Tests also determine the voltage regulation and efficiency.

  • Tests are very economical and convenient, because they furnish the required information without actually loading the transformer

Open Circuit Test
open circuit
Open Circuit

  • The shunt branch parameters are determined by performing this test

  • The core loss and the magnetizing current depend on applied voltage only and are practically unaltered by the load current

  • Hence to get these parameters, Rated voltage is supplied generally to LV winding keeping HV open

  • \(I_0\) is very small as compared to \(I_{FL}\) the loss in \(R_1\) is neglected

  • Thus \(W_0\) drawn from source is dissipated as heat in the core

\[\begin{aligned} W_{0} & =V_{1}I_{0}cos\Phi_{0} & \Rightarrow cos\Phi_{0} & =\dfrac{W_{0}}{V_{1}I_{0}}\\ \Longrightarrow I_{w} & =I_{0}cos\Phi_{0} & \Longrightarrow I_{m} & =I_{0}sin\Phi_{0}\\ \Rightarrow R_{0} & =V_{1}/I_{w} & \Rightarrow X_{m} & =V_{1}/I_{m} \end{aligned}\]
and to evaluate constants but with same Carry out two experiments using two different
Short Circuit Test
short circuit
Short Circuit

  • Input voltage is reduced to a small fraction of rated value and secondary terminals are short-circuited.

  • A current will circulate in the secondary winding.

  • Since a small fraction of rated voltage is applied to the primary winding, the flux in the core and hence the core loss is very small.

  • Hence, the power input on short circuit is dissipated as heat in the winding

  • Since, the applied voltage is very small (may be of the order of 5-8%), the magnetizing branch can now be eliminated from the equivalent circuit.

\[\begin{aligned} Z_{sc} & =\dfrac{V_{sc}}{I_{sc}} & \Rightarrow R_{T}=\dfrac{W_{sc}}{I_{sc}^{2}}\\ \Rightarrow & X_{T}=\sqrt{Z_{sc}^{2}-R_{T}^{2}} \end{aligned}\]
Why Transformer Rating is given in KVA and not in KW?

  • Copper losses \((I^2R)\) depends on current which passing through transformer winding

  • Iron losses or core losses or Insulation losses depends on Voltage.

  • Total losses depends on voltage (V) and current (I) which expressed in Volt ampere (VA) and not on the load power factor (p.f).

  • Hence, the transformer rating may be expressed in VA or kVA, not in W or kW

  • Moreover manufactures design a transformer with no idea which kind of load will be connected to the transformer

  • The load may be resistive (R), inductive (L), capacitve (C) or mixed load (R, L and C).

  • Its mean, there would be different power factor (p.f) at the secondary (load) side on different kind of connected loads depends on R, L and C.

Voltage Regulation of a transformer
  • \[\mbox{voltage regulation}(\%)=\dfrac{E_{2}-V_{2}}{V_{2}}\times100\]
    The percentage of voltage difference between no load and full load voltages of a transformer with respect to its full load voltage

  • At no load, the secondary terminal voltage is \(E_2\)

  • At full load, rated current \(I_2\) flows through the secondary circuit and voltage drop comes into picture.

  • At this situation, primary winding will also draw equivalent full load current from source.

  • The voltage drop due to the secondary impedance is \(I_2Z_2\)

  • \[\mbox{voltage regulation}(\%) = \dfrac{I_2R_2\cos\theta_2 \pm I_2X_2\sin\theta_2 }{V_2}\]
    \(-\)\(+\)In general, voltage regulation is given by
    voltage regulation
    Voltage Regulation