A 3ϕ 240 kV :15 kV transformer supplies a 13.8kV 60 MVA pf = 0.8 lagging load, and is connected to a 230 kV source on the high-voltage (HV) side, as shown in Figure 1.6. (a) Determine all base values on both sides for S3ϕbase = 100 MVA. At the low-voltage (LV) bus,

Question

A 3ϕ 240 kV :15 kV transformer supplies a 13.8kV 60 MVA pf = 0.8 lagging load, and is connected to a 230 kV source on the high-voltage (HV) side, as shown in Figure 1.6.

(a) Determine all base values on both sides for [latex]S_{3ϕbase} [/latex] = 100 MVA. At the low-voltage (LV) bus, [latex]V_{Lbase} [/latex] = 13.8kV.

(b) Draw the positive sequence circuit in per-unit, modeling the transformer as ideal.

(c) Determine all currents and voltages in SI and per-unit.

Accepted Answer

(a) Base values on the LV side are the same as in Example 1.3. The turns ratio may be derived from the voltage ratings ratios:

[latex]\frac{N_{1} }{N_{2} } =\frac{240\sqrt{3} }{15\sqrt{3} } =16[/latex]

∴[latex](V_{base} )_{HVside} =\frac{N_{1} }{N_{2} } (V_{base} )_{LVside}=16.00(7.967)=127.5 kV [/latex]

[latex](I_{base} )_{HVside} =\frac{S_{base} }{(V_{base} )_{HVside} }=\frac{33.33}{0.1275} =261.5 A[/latex]

Results are presented in the following chart.

	[latex]S_{3ϕbase} [/latex]	[latex]V_{Lbase} [/latex]	[latex]S_{base} [/latex]	[latex]I_{base} [/latex]	[latex]V_{base} [/latex]	[latex]Z_{base} [/latex]
Bus	MVA	kV	MVA	kA	kV	Ohm
LV	100	13.8	33.33	4.184	7.967	1.904
HV	100	220.8	33.33	o.2615	127.5	487.5

(b)

[latex]V_{LV} =\frac{7.967\angle 0°}{7.967} =1\angle 0°pu[/latex]

[latex]S_{1Φ} =\frac{60}{3} = 20 MVA[/latex]

[latex]S_{1Φ} =\frac{20}{33.33} = 0.6 pu[/latex]

The positive sequence circuit is shown as Figure 1.7.

(c) All values determined in pu are valid on both sides of the transformer! To determine SI values on the HV side, use HV bases. For example:

[latex]V_{an} =(1\angle 0°)127.5=127.5\angle 0° kV[/latex]

[latex]V_{ab} =(1.732\angle3 0°)(127.5)=220.8\angle 30° kV[/latex]

[latex]I_{a} =(0.6\angle-36.9°)(261.5)=156.9\angle- 36.9° A[/latex]

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