Line current of a mixed three-phase load: 22-Elec-B8 Power Electronics and Drives, December 2019, Q2

Question 2: Line current of a mixed three-phase load (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.

Quantity	Symbol	Value
Supply, three-phase three-wire (line-to-line)	$V_{LL}$	$208\ \text{V}$
Heating unit, unity power factor	$P_h$	$1\,500\ \text{W}$
Motor rated shaft output	$P_{\text{out}}$	$5\ \text{hp}$
Motor full-load efficiency	$\eta$	$87\%$
Motor full-load power factor	$\cos\phi_m$	0.85 lagging

Find. The total line current drawn from the 208 V supply with the motor delivering its rated 5 hp.

Approach. Reduce each load to a real and a reactive power, add them separately, and recover the line current from the total apparent power and the three-phase relation $S = \sqrt{3}\,V_{LL} I_L$.

Convert the motor's shaft rating into electrical input power. Efficiency is defined on real power, so $$P_m = \frac{P_{\text{out}}}{\eta} = \frac{5 \times 745.7}{0.87} = \frac{3\,728.5}{0.87} = 4\,285.6\ \text{W}$$ The 556.9 W difference is the motor's own copper, iron, friction and windage loss.
Attach the motor's reactive demand. At $\cos\phi_m = 0.85$ the phase angle is $\phi_m = 31.79^\circ$, so $$Q_m = P_m \tan\phi_m = 4\,285.6 \times 0.6197 = 2\,656.0\ \text{var}$$ The heater, being resistive, contributes $P_h = 1\,500\ \text{W}$ and $Q_h = 0$.
Add real and reactive powers separately. Powers superpose; currents and power factors do not. $$P_T = P_m + P_h = 4\,285.6 + 1\,500 = 5\,785.6\ \text{W}$$ $$Q_T = Q_m + Q_h = 2\,656.0 + 0 = 2\,656.0\ \text{var}$$
Combine into the total apparent power. $$S_T = \sqrt{P_T^2 + Q_T^2} = \sqrt{5\,785.6^2 + 2\,656.0^2} = 6\,366.1\ \text{VA}$$ $$\boxed{\,\cos\phi_T = \frac{P_T}{S_T} = \frac{5\,785.6}{6\,366.1} = 0.909\ \text{lagging}\,}$$ Adding the resistive heater has pulled the combined power factor up from the motor's 0.85.
Recover the line current. For any balanced three-phase load the total apparent power is $S_T = \sqrt{3}\,V_{LL} I_L$, so $$I_L = \frac{S_T}{\sqrt{3}\,V_{LL}} = \frac{6\,366.1}{\sqrt{3} \times 208} = \frac{6\,366.1}{360.3}$$ $$\boxed{\,I_L = 17.67\ \text{A}\,}$$
Check against the individual branch currents. Taken alone the motor draws $4\,285.6/(\sqrt{3}\cdot 208\cdot 0.85) = 13.99\ \text{A}$ and the heater $1\,500/(\sqrt{3}\cdot 208) = 4.16\ \text{A}$. Their arithmetic sum is 18.15 A, which correctly exceeds the 17.67 A computed above: the two currents are $31.79^\circ$ apart, so they add as phasors, not as numbers. A result larger than 18.15 A, or equal to it, would signal an error.

Figure 2.1 — The two loads stacked on one power triangle. The heater adds length along the real axis only, so the resultant angle — and therefore the power factor — improves even though the current rises.

Final results.

Quantity	Result
Motor electrical input power	$4\,285.6\ \text{W}$
Motor reactive power	$2\,656.0\ \text{var}$
Total real power	$5\,785.6\ \text{W}$
Total apparent power	$6\,366.1\ \text{VA}$
Combined power factor	0.909 lagging
Line current	$\mathbf{17.67\ \text{A}}$

Topic. Combining unlike three-phase loads through the power triangle.

Key relations. Efficiency $\eta = P_{\text{out}}/P_{\text{in}}$ (real power only); $Q = P\tan\phi$; superposition of complex power $S_T = \sum P_i + j\sum Q_i$; and the balanced three-phase relation $S = \sqrt{3}\,V_{LL} I_L$ with $1\ \text{hp} = 745.7\ \text{W}$.

Why this works. Complex power is conserved and additive at a common bus, which currents and power factors are not: two loads on the same bus share a voltage, so their currents can be added only as phasors, whereas their $P$ and $Q$ components add as plain numbers regardless of angle. Reducing every load to a $(P, Q)$ pair therefore removes all the trigonometry from the middle of the problem and leaves exactly one right-triangle at the end. Efficiency belongs to the real-power axis alone because the losses it accounts for — copper, core, friction — are dissipative; the magnetising demand that sets the power factor is a separate, non-dissipative quantity, which is why the efficiency and the power factor must be applied in that order and never multiplied together.

Common pitfall. Using the 5 hp shaft output as the electrical input, which understates the current by 13 per cent; and dividing by the power factor and the efficiency in one step to get an apparent power for the motor of $3\,728.5/(0.87 \times 0.85)$ — that value, $5\,042$ VA, is correct for the motor alone but cannot then be added to the heater's 1500 VA arithmetically. A third trap is dropping the $\sqrt{3}$ or applying it to the phase rather than the line quantity.

Source. Grainger & Stevenson, Power System Analysis, §1.6–1.9 (complex power, balanced three-phase relations); Rashid, Power Electronics, 4th ed., §15.2 for the motor-loading conventions.

Question 2 of 5: Line current of a mixed three-phase load

Question 2: Line current of a mixed three-phase load (20 marks)