Demystifying Autotransformers: Taming Power for Efficiency

Problem-1

An auto-transformer supplies a load of 3 kW at 115 volts at a unity power factor. If the applied primary voltage is 230 volts, calculate the power transferred to the load

inductively
conductively

Solution-1

\[\begin{aligned} K & = 115/230 = 1/2 \\ \text{Power transferred inductively} & = P_i\left(1-K\right) \\ & = 3\left(1-1/2\right) = 1.5~\mathrm{kW} \\ \text{Conductively transferred power} & = K\times P_i \\ &= (1/2)\times 3 = 1.5~\mathrm{kW} \end{aligned}\]

Problem-2

The primary and secondary voltages of an auto-transformer are 500 V and 400 V respectively. Show with the aid of diagram, the current distribution in the winding when the secondary current is 100 A and calculate the economy of Cu in this particular case.

Solution-2

\[\begin{aligned} K & = V_2/V_1 = 400/500=0.8\\ I_1 & = KI_2 = 0.8\times 100= 80~\mathrm{A}\\ \text{saving}& = KW_0 = 0.8\times 100 = 80\% \end{aligned}\]

Problem-3

A 5-kVA, 110/110 V single phase, 50 -Hz transformer has full-load efficiency of 95% and an iron loss of 50 W. The transformer is now connected as an auto-transformer to a 220-V supply. If it delivers a 5-kW load at UPF to a 110-V circuit, calculate the efficiency of the operation and the current drawn by the high-voltage side.

Solution-3

2-windings of auto-TF are connected in series, voltage across each is 110 V.
The iron loss would remain the same in both connections.
Auto-TF windings will carry half the current as compared to the conventional 2-winding TF.
Hence, the Cu loss will be 1/4th of the previous.

\[\begin{aligned} 0.95 & = \dfrac{\text{output}}{\text{output + losses}} \\ & = \dfrac{5000}{5000+50+\text{Cu-loss}}\\ \text{Cu-loss} & = 212~\mathrm{W} \end{aligned}\]
2-winding TF:
\[\begin{aligned} \text{Cu-loss} & = 212/4 = 53~\mathrm{W} \\ \text{Iron loss} & = 50~\mathrm{W} \\ \eta & = \dfrac{5000}{5000+53+50} = 0.9797 = 97.97\% \\ \text{current on HV side} & = 5103/220 = 23.2~\mathrm{A} \end{aligned}\]
Auto-TF:

Problem-4

A 2-winding TF is rated at 2400/240 V, 50 kVA. It is re-connected as a step-up auto-transformer with 2400 V input. Calculate the rating of the auto-TF and the inductively and conductively transferred powers while delivering the rated output at UPF.

Solution-4

With 50 kVA rating, the rated currents on two sides are 20.8 A (2400-V side) and 208 A (240-V side)
2400-V side will work as common winding

\[\begin{aligned} \text{Output} & = 2640\times 208 =550~ \mathrm{kVA}\\ \text{input current} & = 208 \times 2640/2400 = 229~\mathrm{A} \\ K & = 2640/2400 = 1.1 \end{aligned}\]

The rating of auto-TF then 550 kVA
At UPF rated load = 550 kW
\[\begin{aligned} &= \text{Power handled by the common winding}\\ & = 2400~\mathrm{V} \times 20.8~\mathrm{A} \times 10^{-3} =50~\mathrm{kW} \\ & = \text{rated output as 2-winding TF} \end{aligned}\]
Inductively transferred power
Remaining power = 550-50=500 kW = conductively transferred