Adding an additional motor to a load

Question

You have a 500kVA load with a power factor of 0.6 lag. You add an additional synchronous motor with an input of 80kW and max power factor of 0.8 leading. Determine the reactive power and the overall power factor.

Can anyone help me with this question? I don't even know where to start. I don't think my teacher even knows how to do it.

Thanks.

---

Is this correct?

Existing installation

$$P=S\cdot pf = 500\cdot 0.6 = 300\text{ kW}$$

$$Q=\sqrt{500^2-300^2} = 400\text{ KVAR}$$

$$S=500\text{ kVA}$$

New synchronous motor

$$P=80\text{ kW}$$

$$Q=\sqrt{100^2 - 80^2} = 60\text{ KVAR}$$

$$S=\frac{80}{0.8} = 100\text{ kVA}$$

Added together

$$P= 80\text{ kW} + 300\text{ kW} = 380\text{ kW}$$

$$Q= 400\text{ KVAR} - 60\text{ KVAR} = 340\text{ KVAR}$$

$$S= \sqrt{380^2+340^2} = 509.9\text{ kVA}$$

Overall power factor = \$\frac{P}{S} = \frac{380}{509.9} = 0.75\$ lagging.

So reactive power = 340 kVAR and pf = 0.75 lagging.

Answer 1

Is this correct?

Yes.

---

For any future readers, the key formulae here are:

$$ P_{kW} = S_{kVA} * pf $$ $$ S_{kVA}^2 = P_{kW}^2 + Q_{kvar}^2 \quad\text{therefore}\quad Q_{kvar} = \sqrt{S_{kVA}^2 - P_{kW}^2}$$

Which can be derived from an example of the power triangle below and the fact that \$ pf = \cos\theta \$.