Transformer 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

• 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

$$\begin{aligned} W_{0} & =V_{1}I_{0}cos\Phi_{0} & \Rightarrow cos\Phi_{0} & =\dfrac{W_{0}}{V_{1}I_{0}}\\ \Longrightarrow I_{w} & =I_{0}cos\Phi_{0} & \Longrightarrow I_{m} & =I_{0}sin\Phi_{0}\\ \Rightarrow R_{0} & =V_{1}/I_{w} & \Rightarrow X_{m} & =V_{1}/I_{m} \end{aligned}$$ Separation of Core losses $$\begin{aligned} W_{i} & =W_{h}+W_{e}\\ & =PB_{m}^{1.6}f+QB_{m}^{2}f^{2} \end{aligned}$$ Carry out two experiments using two different $f$ but with same $B_m$ to evaluate constants $P$ and $Q$

Short Circuit Test

• 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

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

• 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 $I_2Z_2$

• In general, voltage regulation is given by $$\mbox{voltage regulation}(\%) = \dfrac{I_2R_2\cos\theta_2 \pm I_2X_2\sin\theta_2 }{V_2}$$ $+$ for lagging and $-$ for leading load