Three 1-phase 20-kVA, 2000/200-V transformers identical with that of Ex. 3.3 are connected in Y/ Δ in a 3-phase, 60 kVA bank to step-down the voltage at the load end of a feeder having impedance of 0.13 + j 0.95 Ω/phase. The line voltage at the sending-end of the feeder is 3464 V. The transformers

Question

Three 1-phase [latex] 20-kVA[/latex], [latex] 2000/200-V[/latex] transformers identical with that of Ex. 3.3 are connected in [latex] Y/ \Delta [/latex] in a 3-phase, [latex] 60 kVA[/latex] bank to step-down the voltage at the load end of a feeder having impedance of [latex] 0.13 + j 0.95 \Omega /phase[/latex]. The line voltage at the sending-end of the feeder is 3464 V. The transformers supply a balanced 3-phase load through a feeder whose impedance is [latex] 0.0004 + j 0.0015 \Omega/phase[/latex]. Find the load voltage (line-to-line) when it draws rated current from transformers at 0.8 lagging power factor.

Accepted Answer

Figure 3.58 gives the circuit diagram of the system. The computations will be carried out on per phase-Y basis by referring all quantities to the HV (Y-connected) side of the transformer bank.

LV feeder impedance referred to the HV side is

[latex] \left(\frac{2000\sqrt{3} }{200} ight) ^{2} (0.0004+j0.0015) =0.12+ j0.45 \Omega /phase[/latex]

The total series impedance of the HV and LV feeders referred to the HV side is

[latex] Z_{F}=(0.13+j0.95) +(0.12+ j0.45)[/latex]

[latex] = 0.25 + j 1.4 \Omega /phase[/latex]

From Ex. 3.5, the equivalent impedance of the transformer bank is referred to the HV side

[latex] Z_{T}= 0.82 + j 1.02 \Omega /phase Y[/latex]

Sending-end feeder voltage [latex] =\frac{3464}{\sqrt{3} } =2000V/phase Y[/latex]

Load current on the HV side = rated current of transformer

[latex] = 10 A/phase Y[/latex]

It is now seen that the equivalent circuit for one phase referred to the Y-connected HV side is exactly the same as in Ex. 3.6, Fig. 3.20. Thus the load voltage referred to the HV side is 197.692 V to neutral. The actual load voltage is 197.688 V, line-to-line (since the secondaries are [latex] \Delta [/latex]-connected).

PU method In such problems it is convenient to use the pu method. We shall choose the following base values:

[latex] (MVA)_{B}=\frac{3 imes 20}{1000} =0.06[/latex]

Voltage base on HV side [latex] =2\sqrt{3} kV[/latex] (line-to-line)

Voltage base on LV side [latex] =0.2 kV[/latex] (line-to-line)

Note The voltage base values are in the ratio of line-to-line voltages (same as phase voltages on equivalent star basis)

[latex] \bar{Z}_{1} (LV line)(pu) = (0.0004 + j 0.0015) imes \frac{0.06}{(0.2)^{2}} = (0.06 + j 0.225) imes 10^{-2}[/latex]

[latex] \bar{Z}_{2} (HV line)(pu) = (0.13 + j 0.95) imes \frac{0.06} {(2\sqrt{3}) ^{2}} = (0.065 + j 0.475) imes 10^{-2} [/latex]

[latex] \bar{Z}_{T} (star side)(pu) = (0.82 + j 1.02) imes \frac{0.06}{(2\sqrt{3} )^{2}} = (0.41 + j 0.51) imes 10^{-2} [/latex]

Note Suppose the transformer impedance was given on the delta side

[latex] \bar{Z}_{T} (delta side)(pu) = \left(\frac{200}{2000} ight)^{2} imes (0.82+ j1.02)[/latex]

[latex] (0.82 + j 1.02) imes 10^{–2} \Omega (delta phase)[/latex]

Equivalent star impedance [latex] =\frac{1}{3} (0.82 + j 1.02) imes 10^{-2}\Omega[/latex]

[latex] \bar{Z}_{T} (pu) = \frac{0.06}{(0.2)^{2}} imes \frac{1}{3} (0.82 + j 1.02) imes 10^{–2}[/latex]

[latex] = (0.41 + j 0.51) imes 10^{–3}[/latex] (same as calculated above)

[latex] \bar{Z}(total) (pu) = 0.06 + j 0.225[/latex]

0.065 + j 0.475

0.41 + j 0.51

[latex] (0.535 +j1.21) imes 10^{2}[/latex]

[latex] V_{1}(sending–end voltage)= 2\sqrt{3} kV[/latex] (line) or 1 pu [/latex]

[latex] I_{1} (= rated current)= 1 pu; pf= 0.8 lag[/latex]

[latex] V_{1}(load voltage) = 1 – 1 (0.535 imes 0.8 + 1.2 imes 0.6) imes 10^{–2}[/latex]

[latex] = 0.98846 pu[/latex]

[latex] = 0.98846 imes 200 = 197.692 V (line)[/latex]

Question 3.20: Three 1-phase 20-kVA, 2000/200-V transformers identical with...

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