Question

Calculate the resulting power factor if a synchronous motor rated at 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor.

Assume constant voltage (1 hp = 0.746 kW).

Answer 1

The question is incomplete. Below is the complete question

"An industrial plant consisting primarily of induction motor loads absorbs 500 kW at 0.6 power factor lagging. (a) Compute the required kVA rating of a shunt capacitor to improve the power factor to 0.9 lagging. (b) Calculate the resulting power factor if a synchronous motor rated 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor. Assume constant voltage. (1 hp = 0.746 kW)"

Answer:

a. 424.5KVA

b. 0.808 lagging.

Step-by-step explanation:

Let's first determine the real power, reactive and the apparent power delivered.

Ql= Ptan(the real power angle)

Ql=500*tan53.13°

Ql=666.7 Kvar

The reactive angle is

Arccos(0.9)=25.84°

Now we calculate the reactive power

Qs=Ptan25. 84°

Qs=242.4KVAR

Hence the apparent power is

Qc=Ql-Qs

Qc=666.7-242.2

Qc=424.5KVAR

The required KVA rating of the shunt capacitor is 424.5KVA

b. To calculate the resulting power factor we first determine the power absorbs by the synchronous motor.

Pm=(500*0.746)/0.9

Pm=414.4KW

The total reactive power is

Ps=P+Pm

Ps=414.4+500=914.4KW

Hence we compute the source power factor as

PF=cos[arctan(Qs/Ps)]

PF=cos[arctan(666.7/914.4)

From careful calculation, we arrive at

PF=0.808.

Note this power factor will be lagging.