--- title: "Current, pf, power and efficiency of 3.3 kV motor" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:motor-3-3kv-efficiency" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-3-3kv-efficiency site: BCT Engineering --- # Current, pf, power and efficiency of 3.3 kV motor Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2074 Chaitra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-3-3kv-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question A 3.3 kV, 50 Hz star connected synchronous motor has a synchronous impedance of (0.8 + j55) Ω. It is synchronized to 3.3 kV main from which it is drawing 750 kW at an excitation emf of 4.27 kV (line). Determine the armature current, power factor and power angle. Also find the mechanical power developed. If the stray load loss is 30 kW, find the efficiency. ## Short answer With Zs = (0.8+j5.5) Ω per phase, Vph=1905.26 V, Ef=2465.29 V, armature current = 163.0 A, power factor = 0.805 leading, power angle = 18.91°. Mechanical power developed = 686.2 kW, efficiency = 87.5%. ## Answer **Assumption:** the printed $X_s = 55\ \Omega$ cannot be right: with 55 Ω the largest power the motor could take is about $3VE/X_s \approx 0.26$ MW, less than 750 kW. The standard version of this problem has $Z_s = (0.8 + j5.5)\ \Omega$ per phase, which is used here. **Given:** $V = 3300/\sqrt3 = 1905.26$ V, $E_f = 4270/\sqrt3 = 2465.29$ V per phase, input $750$ kW, i.e. $250$ kW per phase. $$ Z_s = 0.8 + j5.5 = 5.5579\angle 81.72^\circ\ \Omega $$ #### Power angle Power input per phase of a motor: $$ P = \frac{V^2}{Z_s}\cos\theta - \frac{VE_f}{Z_s}\cos(\theta + \delta) $$ $$ \begin{aligned} \frac{V^2}{Z_s}\cos\theta &= \frac{1905.26^2}{5.5579} \times \cos81.72^\circ = 94011\ \text{W} \\ \frac{VE_f}{Z_s} &= \frac{1905.26 \times 2465.29}{5.5579} = 845107\ \text{W} \\ 250000 &= 94011 - 845107\cos(\theta + \delta) \\ \cos(\theta+\delta) &= -0.1846 \Rightarrow \theta + \delta = 100.64^\circ \\ \delta &= 100.64^\circ - 81.72^\circ = 18.91^\circ \end{aligned} $$ #### Armature current and power factor $$ \begin{aligned} \vec E_r &= \vec V - \vec E_f = 1905.26 - 2465.29\angle -18.91^\circ = -426.94 + j799.06 \\ &= 906.0\angle 118.12^\circ\ \text{V} \\ \vec I_a &= \frac{\vec E_r}{Z_s} = 163.01\angle 36.39^\circ\ \text{A} \end{aligned} $$ Power factor $= \cos36.39^\circ = 0.8050$ **leading** (over-excited, $E_f > V$). Check: $\sqrt3 \times 3300 \times 163.01 \times 0.8050 = 750$ kW. #### Mechanical power and efficiency $$ \begin{aligned} P_{cu} &= 3I_a^2R_a = 3 \times 163.01^2 \times 0.8 = 63.77\ \text{kW} \\ P_{mech} &= 750 - 63.77 = 686.23\ \text{kW} \\ P_{out} &= 686.23 - 30 = 656.23\ \text{kW} \\ \eta &= \frac{656.23}{750} \times 100 = 87.50\ \% \end{aligned} $$ **Answer:** $I_a = 163.0$ A, pf $= 0.805$ leading, power angle $= 18.91^\circ$, mechanical power developed $= 686.2$ kW, efficiency $= 87.5\ \%$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the printed Xs = 55 ohm replaced by 5.5 ohm? A. With 55 ohm the motor could take at most about 0.26 MW, less than the 750 kW given B. Because 55 ohm would make the motor generate C. Because Ra must always be ten times smaller than Xs D. Because the frequency is 50 Hz Answer: A. The maximum power 3VE/Xs would be only about 0.26 MW with 55 ohm, so the problem as printed is impossible. (Key point: Data correction) 2. Which relation is used to find the power angle for a motor with Ra not negligible? A. P = V^2 cos(theta)/Zs - (V Ef/Zs) cos(theta + delta) B. P = 3 V Ef sin(delta)/Xs only C. P = V I sin(phi) D. P = Ef^2/Zs Answer: A. With the motor input per phase, 250000 = 94011 - 845107 cos(theta + delta), giving delta = 18.91 degrees. (Key point: Power angle method) 3. Why is the power factor 0.805 leading? A. Because the motor is under-excited B. Because Ef (4.27 kV) is greater than V (3.3 kV), so it is over-excited C. Because the load is purely resistive D. Because Ra is larger than Xs Answer: B. Over-excitation (Ef > V) makes Ia lead V. (Key point: Leading power factor) 4. How are the mechanical power and efficiency obtained? A. Pmech = input + copper loss; efficiency = Pmech/input B. Pmech = input - armature copper loss; output = Pmech - stray loss; efficiency = output/input C. Pmech = input - stray loss only; efficiency = 100 percent D. Pmech = 3 V I; efficiency = cos(phi) Answer: B. Pcu = 63.77 kW, so Pmech = 686.2 kW; output = 656.2 kW, efficiency = 656.2/750 = 87.5 percent. 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