An open-circuit test was conducted on the transformer given in Example 6. The following results were obtained when the low-voltage winding was excited. (In some cases, such as in a repair shop, a 69 kV voltage may not be available and the opencircuit test has to be done by exciting the LV

Question

An open-circuit test was conducted on the transformer given in Example 6. The following results were obtained when the low-voltage winding was excited. (In some cases, such as in a repair shop, a 69 kV voltage may not be available and the opencircuit test has to be done by exciting the LV winding at its rated voltage.)

[latex]E_{ s }=4160 V \quad I_{ o }=2 A \quad P_{ m }=5000 W[/latex]

Using this information and the transformer characteristics found in Example 6, calculate:

a. the values of [latex]X_{ m } ext { and } R_{ m }[/latex] on the primary side (Fig. 21)

b. the efficiency of the transforamer when it supplies a load of 250 kVA, whose power factor is 80% (lagging)

Accepted Answer

a. Applying Eq. 1 to the secondary side

[latex]R_{ m }=E_{ p }^{2} / P_{ m }[/latex] (Equation 1)

[latex]\begin{aligned}R_{ m } &=E_{ s }^{2} / P_{ m } \&=4160^{2} / 5000=3461 \Omega\end{aligned}[/latex]

The apparent power [latex]S_{ m }[/latex] is

[latex]\begin{aligned}S_{ m } &=E_{ s } I_{0}=4160 imes 2=8320 VA \Q_{ m } &=\sqrt{S_{ m }^{2}-P_{ m }^{2}} \&=\sqrt{8320^{2}-5000^{2}}=6650 \Omega \X_{ m } &=E_{ s }^{2} / Q_{ m } \&=4160^{2} / 6650=2602 \Omega\end{aligned}[/latex]

The values of [latex]R_{ m } ext { and } X_{ m }[/latex] referred the primary side will be (69 000/4160)² = 275 times greater. The values on the primary side are therefore:

[latex]\begin{aligned}&X_{ m }=275 imes 2602 \Omega=715 imes 10^{3} \Omega=715 k \Omega \&R_{ m }=275 imes 3461 \Omega=952 imes 10^{3} \Omega=952 k \Omega\end{aligned}[/latex]

These are the values that would have been found if the primary had been excited at 69 kV.

b. Industrial loads and voltages fluctuate all the time. Thus, when we state that a load is 250 kVA, with [latex]\cos heta=0.8[/latex], it is understood that the load is about 250 kVA and that the power factor is about 0.8. Furthermore, the primary voltage is about 69 kV.

Consequently, in calculating efficiency, there is no point in arriving at a precise mathematical answer, even if we were able to give it. Knowing this, we can make certain assumptions that make it much easier to arrive at a solution.

Question 10.7: An open-circuit test was conducted on the transformer given ...

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