Question

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power factor of 0.75 lagging from the substation bus. The supply voltage is 12.47 kV. The power factor can be improved by connected a capacitor in parallel with the supply or by using a synchronous motor, which generates reactive power. Analyze both of these cases independently:(a) Find the required kVAR rating of a capacitor connected across the load to raise the power factor to 0.95 lagging.

(b) Assuming a synchronous motor rated at 250 hp, with an 80% efficiency is operated from the same bus at rated conditions and a power factor of 0.85 leading, calculate the resultant supply power factor.

Answer 1

(a)
`Q=332`
`kvar` and
`C=5.66`
`uF`

(b)
`pf=0.90` lagging

Step-by-step explanation:

Given Data:

`P=600kW`

`V=12.47kV`

`f=60Hz`

`pf_(old) =0.75`

`pf_(new) =0.95`

(a) Find the required kVAR rating of a capacitor

`\alpha _(old)=cos^(-1)(0.75) =41.41`°

`\alpha _(new)=cos^(-1)(0.95) =18.19`°

The required compensation reactive power can be found by

`Q=P(tan(\alpha_(old)) - tan(\alpha_(new)))`

`Q=600(tan(41.41) - tan(18.19))`

`Q=332`
`kvar`

The corresponding capacitor value can be found by

`C=Q/2\pi fV^(2)`

`C=332/2*\pi *60*12.47^(2)`

`C=5.66`
`uF`

(b) calculate the resultant supply power factor

First convert the hp into kW

`P_(mech) =250*746=186.5`
`kW`

Find the electrical power (real power) of the motor

`P_(elec) =P_(mech)/n`

where
`n` is the efficiency of the motor

`P_(elec) =186.5/0.80=233.125`
`kW`

The current in the motor is

`I_(m) =(P/\*V*pf)<cos^(-1)(pf)`

The pf of motor is 0.85 Leading

Note that
`<` represents the angle in complex notation (polar form)

`I_(m) =(233.125/12.47*0.85)<cos^(-1)(0.85)`

`I_(m)=18.694+11.586`
`j`
`A`

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

`I_(load) =(600/12.47*0.75)<-cos^(-1)(0.75)`

`I_(load) =48.115-42.433`
`j`
`A`

Now the supply current is the current flowing in the load plus the current flowing in the motor

`I_(supply) =I_(m) + I_(load)`

`I_(supply)= (18.694+11.586)+(48.115-42.433)`

`I_(supply) =66.809-30.847j`
`A`

or in polar form

`I_(supply) =73.58<-24.78`°

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

`pf=cos(24.78)=0.90` lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90