Question 1 of 6: Per-unit analysis of a plant distribution system

Question 1: Per-unit analysis of a plant distribution system (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the high-voltage base current and base impedance; (b) the transformer and motor per-unit impedances on the chosen base; (c) the actual primary current in amperes with L1 out of service; and (d) the largest unity-power-factor load that L1 could carry before T1 reaches its 300 kVA nameplate.

[Figure not reproduced: Figure Q1 — single-line diagram of the plant distribution system, redrawn from the examination paper. The delta winding of T1 faces the utility; the wye winding feeds the 600 V plant bus. See the official exam paper.]

Approach. Fix the base set from the low-voltage side, transfer the bases through the transformer turns ratio, convert each nameplate impedance to the common base, then work the load flow in per unit and convert the single answer that is wanted in amperes at the end.

1. Part (a) — carry the base voltage through the transformer ratio. Bases transform exactly as rated quantities do, so the high-voltage base voltage is the low-voltage base multiplied by the nameplate ratio. Because 600 V was chosen as the low-voltage base and the transformer is rated 12.5 kV/600 V, the ratio is unity times the nameplate: $$V_{base,HV}=V_{base,LV}\times\frac{12.5\ \text{kV}}{600\ \text{V}}=600\times\frac{12500}{600} =12\,500\ \text{V}$$ The base power is the same at every voltage level in a per-unit study.
2. Compute the high-voltage base current and base impedance. For a three-phase base defined on the total power and the line-to-line voltage, $$I_{base}=\frac{S_{base}}{\sqrt{3}\,V_{base}}=\frac{500\,000}{\sqrt{3}\times 12\,500} =23.094\ \text{A}, \qquad Z_{base}=\frac{V_{base}^{2}}{S_{base}}=\frac{(12\,500)^{2}}{500\,000}=312.5\ \Omega$$ so that $$\boxed{I_{base,HV}=23.09\ \text{A}\ ,\qquad Z_{base,HV}=312.5\ \Omega}$$ The corresponding low-voltage figures, needed later as a cross-check, are $I_{base,LV}=500\,000/(\sqrt{3}\times 600)=481.13$ A and $Z_{base,LV}=600^{2}/500\,000=0.72\ \Omega$. Note that the ratio $I_{base,LV}/I_{base,HV}=481.13/23.094=20.83$ is exactly the turns ratio, as it must be.
3. Part (b) — move the transformer reactance onto the study base. A per-unit impedance changes base by the ratio of the powers and the inverse square of the voltages: $$X_{new}=X_{old}\left(\frac{S_{base,new}}{S_{base,old}}\right) \left(\frac{V_{base,old}}{V_{base,new}}\right)^{2} =0.10\times\frac{500}{300}\times\left(\frac{600}{600}\right)^{2}$$ The voltage bases coincide with the nameplate voltages, so only the power ratio acts: $$\boxed{X_{T1}=j0.1667\ \text{pu on }500\ \text{kVA}}$$ In ohms referred to the 600 V side this is $0.1667\times 0.72=0.120\ \Omega$, and referred to the 12.5 kV side, $0.1667\times 312.5=52.08\ \Omega$.
4. Convert the motor nameplate into an electrical loading. The horsepower rating is mechanical output; the electrical input is larger by the efficiency, and the apparent power is larger again by the power factor: $$P_{out}=125\times 746=93\,250\ \text{W},\qquad P_{in}=\frac{P_{out}}{\eta}=\frac{93\,250}{0.854}=109\,192\ \text{W}$$ $$S_{M1}=\frac{P_{in}}{\mathrm{pf}}=\frac{109\,192}{0.8}=136\,490\ \text{VA} =136.49\ \text{kVA at }\cos^{-1}0.8=36.87^\circ\ \text{lagging}$$ so the motor draws $Q_{M1}=136.49\times 0.6=81.89$ kVAr.
5. Express the motor as a per-unit impedance at rated bus voltage. A constant-power load seen at a known voltage is equivalent to the impedance that would draw the same complex power, $Z=V^{2}/S^{*}$. In per unit, with $V=1.0\angle 0^\circ$ and $S_{M1}=136.49/500=0.27298\angle 36.87^\circ$ pu, $$Z_{M1}=\frac{|V|^{2}}{S_{M1}^{*}} =\frac{1.0}{0.27298\angle-36.87^\circ}=3.6633\angle 36.87^\circ\ \text{pu}$$ $$\boxed{Z_{M1}=2.931+j2.198\ \text{pu}\quad(3.663\angle 36.87^\circ\ \text{pu})}$$ Multiplying by $Z_{base,LV}=0.72\ \Omega$ gives $2.110+j1.583\ \Omega$ per phase, wye-equivalent, at 600 V. For completeness the heating load L2, at $150/500=0.30$ pu and unity power factor, is the pure resistance $Z_{L2}=1/0.30=3.333$ pu.
6. Part (c) — add the connected loads in complex power. With L1 open the transformer carries only M1 and L2. Real and reactive powers add arithmetically at a common bus: $$P_{tot}=109.19+150.00=259.19\ \text{kW},\qquad Q_{tot}=81.89+0=81.89\ \text{kVAr}$$ $$S_{tot}=\sqrt{P_{tot}^{2}+Q_{tot}^{2}}=\sqrt{259.19^{2}+81.89^{2}}=271.82\ \text{kVA}$$ at a plant power factor of $259.19/271.82=0.9535$ lagging, i.e. an angle of $17.53^\circ$.
7. Convert to per unit and then to primary amperes. With the bus held at rated voltage the per-unit current equals the per-unit apparent power, and the per-unit current is the same on both sides of the transformer — that is the whole convenience of the method: $$I_{pu}=\frac{S_{tot}}{S_{base}}\Big/|V_{pu}|=\frac{271.82}{500}=0.5436\ \text{pu} \ \angle-17.53^\circ$$ Multiplying by the high-voltage base current found in step 2, $$I_{HV}=0.5436\times 23.094=\boxed{12.56\ \text{A at }0.9535\ \text{pf lagging}}$$ The same per-unit current on the secondary is $0.5436\times 481.13=261.6$ A, and $261.6/12.56=20.83$ recovers the turns ratio, which is the arithmetic check on the answer.
8. Part (d) — load the spare feeder up to the nameplate, not up to the base. The limit is the transformer's own 300 kVA rating, not the 500 kVA study base. Adding a unity-power-factor block $P_{L1}$ leaves the reactive demand unchanged, so $$\left(P_{tot}+P_{L1}\right)^{2}+Q_{tot}^{2}=S_{T1,rated}^{2} \;\Longrightarrow\; P_{L1}=\sqrt{300^{2}-81.89^{2}}-259.19$$ Evaluating, $\sqrt{90\,000-6\,706.6}=288.61$ kW, hence $$\boxed{P_{L1,max}=29.41\ \text{kW at unity power factor}}$$ At that loading the transformer sits at exactly 300 kVA and $0.962$ power factor lagging. The result is modest because the reactive demand of the motor and the 150 kW heater already consume $271.8/300=90.6\ \%$ of the nameplate.