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Leading kVAR supplied by synchronous motor

Short answer

Factory: 240 kW at 0.8 lag => 180 kVAR lag. Combined 300 kW at 0.9 lag => 145.3 kVAR lag. Motor supplies 34.7 kVAR leading. Motor kVA = √(60²+34.7²)=69.31 kVA, pf = 0.866 leading.

A synchronous motor absorbing 60 kW is connected in parallel with a factory load of 240 kW having a lagging p.f. of 0.8. If the connected combined load has a p.f. of 0.9, what is the value of the leading kVAR supplied by the motor and at what p.f. is it working?

The synchronous motor is over-excited so it supplies leading kVAR, which cancels part of the lagging kVAR of the factory load.

Factory load

P1=240 kW,cos⁡ϕ1=0.8 lag,tan⁡ϕ1=0.75Q1=240×0.75=180 kVAR (lagging)\begin{aligned} P_1 &= 240\ \text{kW},\quad \cos\phi_1 = 0.8\ \text{lag},\quad \tan\phi_1 = 0.75 \\ Q_1 &= 240 \times 0.75 = 180\ \text{kVAR (lagging)} \end{aligned}

Combined load

P=240+60=300 kWcos⁡ϕ=0.9 ⇒ tan⁡ϕ=tan⁡(25.84∘)=0.4843Q=300×0.4843=145.30 kVAR (lagging)\begin{aligned} P &= 240 + 60 = 300\ \text{kW} \\ \cos\phi &= 0.9 \ \Rightarrow\ \tan\phi = \tan(25.84^\circ) = 0.4843 \\ Q &= 300 \times 0.4843 = 145.30\ \text{kVAR (lagging)} \end{aligned}

kVAR from the motor

Qm=Q1−Q=180−145.30=34.70 kVAR (leading)Sm=602+34.702=69.31 kVAcos⁡ϕm=6069.31=0.866 leading\begin{aligned} Q_m &= Q_1 - Q = 180 - 145.30 = 34.70\ \text{kVAR (leading)} \\ S_m &= \sqrt{60^2 + 34.70^2} = 69.31\ \text{kVA} \\ \cos\phi_m &= \frac{60}{69.31} = 0.866\ \text{leading} \end{aligned} 
     kVAR (lag +)
      |
 180  +  factory load
      |
145.3 +  combined (pf 0.9)
      |   motor gives -34.7 kVAR
      +--------------------------- kW
         60   240   300

Answer: The motor supplies 34.70 kVAR leading and works at a power factor of 0.866 leading (input 69.31 kVA).

✓ Answer checked by Claude Sonnet 5.5

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