Question C.4: In Example C.3, the transformers have the same percentage im......
In Example C.3, the transformers have the same percentage impedances and the same X/R ratios; the secondary voltage of the 10 MVA transformer is 4 kV and that of the 5 MVA transformer is 4.16 kV. Calculate the circulating current at no load.
We will work on per-phase basis. The 10 MVA transformer impedance referred to 4 kV secondary is 0.008 + j0.08 ohm, and the 5 MVA transformer impedance at 4.16 kV secondary is 0.0208 + j0.1384 ohm; Z_{1} + Z_{2} = 0.0288 + j0.2184. Assume that the load voltage is 4 kV; then, on a per-phase basis, the load is 5 MVA at 0.8 power factor and the load impedance is 0.853 + j0.64 ohm.
\frac{E_{1}}{Z_{1}} + \frac{E_{2}}{Z_{2}}= \frac{4000}{\sqrt{3}(0.008 \ + \ j0.08)} + \frac{4160}{\sqrt{3}(0.0208 \ + \ j0.1384)}= 5409 – j45.550 \ kAAlso,
(\frac{1}{Z_{1}}+ \frac{1}{Z_{2}}+ \frac{1}{Z_{L}})=3.048 – j20.003From Equation C.18,
V (\frac{1}{Z_{1}}+ \frac{1}{Z_{2}} + \frac{1}{Z_{L}}) = (\frac{E_{1}}{Z_{1}}+ \frac{E_{2}}{Z_{2}}) (C.18)
the load voltage is 2260 – j74.84 volts phase to neutral. From Equation C.16,
I_{1} = \frac{ E_{1} \ – \ V}{Z_{1}} \ \ \ I_{2} = \frac{E_{2} \ – \ V}{Z_{2}} (C.16)
the 10 MVA transformer load current is 980.04 – j445.347 and that of the 5 MVA transformer is 672.17 – j874.42. The total load current is 1652.2 – j1319.75 and the singlephase load MVA is 3.645 MW and 3.112 Mvar. This is much different from the desired loading of 4 MW and 3 Mvar. This is due to assumption of the load voltage. The calculation can be repeated with a lower estimate of load voltage and recalculation of load impedance.