Question 5 of 5: Plant power-factor correction with a synchronous motor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Professional Engineers Ontario — 16-Elec-A6 Power Systems and Machines, May 2017. Closed-book; formula sheet supplied. Five questions of equal value; all five constitute a complete paper, and all are solved here. All ac quantities are rms; three-phase quantities are line-to-line and total real power unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers ch. 2, DC machines ch. 8, synchronous machines ch. 4–5, induction machines ch. 6); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis, ch. 2).
Question 5: Plant power-factor correction with a synchronous motor (20 marks)
Given. A 460 V, 60 Hz plant with the three loads below; the synchronous motor has $X_s=1.45\,\Omega$/phase and draws 80 kW throughout.
Given data
Load
Data
Induction motor
50 kVA, 0.84 pf lagging
Synchronous motor
125 hp, 6-pole, Y, $X_s$=1.45 Ω/ph, 80 kW
Heating
30 kW (unity pf)
Source
460 V (L-L), 60 Hz
Find. The synchronous motor P and Q for unity-pf operation, its armature current and excitation EMF, then its power factor, current and torque angle at 0.80 leading plant pf, and finally the vars it handles when its excitation is reduced 10%.
Over-excited synchronous motor phasor diagram: E_a = V - jX_s I_a with I_a leading V.
Approach. Resolve every load into P and Q, impose the plant reactive requirement to size the synchronous motor vars, then use $E_a=V-jX_sI_a$ (motor convention, $R_a$ neglected) for current, EMF and angle.
Resolve the fixed loads. Induction motor: $P=50(0.84)=42$ kW, $Q=50\sin(\cos^{-1}0.84)=+27.13$ kvar (lagging). Heating: $30$ kW, $0$ kvar. Synchronous motor: $P=80$ kW.
(a) Synchronous-motor P and Q for unity plant pf. Unity pf requires total $Q=0$, so the motor must supply $Q_{sm}=-(27.13+0)=-27.13$ kvar (i.e. it delivers 27.13 kvar), while its active power stays at $P_{sm}=80$ kW.
(b) Armature current. $|S_{sm}|=\sqrt{80^2+27.13^2}=84.48$ kVA, so $I_a=\dfrac{|S_{sm}|}{\sqrt3\,V_L}=\dfrac{84{,}480}{\sqrt3(460)}=106.0$ A (pf $=80/84.48=0.947$ leading).
(c) Excitation EMF. With $V=460/\sqrt3=265.6$ V/phase and $I_a=106.0\angle{+18.7^\circ}$ A (leading), $E_a=V-jX_sI_a$ gives $$\boxed{\,E_a=347.0\ \text{V/phase},\quad \delta=-24.8^\circ\,}$$ The motor is over-excited ($E_a\gt V$), confirming that it supplies vars.
(d) Plant at 0.80 leading pf. Total $P=42+80+30=152$ kW, so total $Q=-152\tan(\cos^{-1}0.80)=-114$ kvar. The motor must then handle $Q_{sm}=-114-27.13=-141.13$ kvar with $P_{sm}=80$ kW. Hence $|S_{sm}|=\sqrt{80^2+141.13^2}=162.2$ kVA and: (i) new pf $=80/162.2=0.493$ leading; (ii) $I_a=162{,}230/(\sqrt3\cdot460)=203.6$ A; (iii) with $I_a=203.6\angle{+60.5^\circ}$ A, $E_a=V-jX_sI_a$ gives torque angle $\delta=-15.6^\circ$.
(e) Excitation reduced by 10%. Starting from the part-(a) value, $E_a=0.9(347.0)=312.3$ V. Holding $P=80$ kW, $\sin\delta=P X_s/(3VE_a)=0.466\Rightarrow\delta=27.8^\circ$, and $Q=\dfrac{3V(E_a\cos\delta-V)}{X_s}=+5.87$ kvar — the motor still supplies about 5.9 kvar, but far less than before because it is now only slightly over-excited. (The question does not say which operating point is reduced; part (a)'s design excitation is taken as the base. If instead the part-(d) excitation, $E_a=542.3$ V, is reduced 10% to 488.1 V with $P$ still 80 kW, then $\delta=17.4^\circ$ and the motor supplies about 110.1 kvar — in either case it remains over-excited and supplies vars.)
