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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.

Respuesta :

Answer:

(a) [tex]Q=332[/tex] [tex]kvar[/tex] and [tex]C=5.66[/tex] [tex]uF[/tex]

(b) [tex]pf=0.90[/tex] lagging

Explanation:

Given Data:

[tex]P=600kW[/tex]

[tex]V=12.47kV[/tex]

[tex]f=60Hz[/tex]

[tex]pf_{old} =0.75[/tex]

[tex]pf_{new} =0.95[/tex]

(a) Find the required kVAR rating of a capacitor

[tex]\alpha _{old}=cos^{-1}(0.75) =41.41[/tex]°

[tex]\alpha _{new}=cos^{-1}(0.95) =18.19[/tex]°

The required compensation reactive power can be found by

[tex]Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))[/tex]

[tex]Q=600(tan(41.41) - tan(18.19))[/tex]

[tex]Q=332[/tex] [tex]kvar[/tex]

The corresponding capacitor value can be found by

[tex]C=Q/2\pi fV^{2}[/tex]

[tex]C=332/2*\pi *60*12.47^{2}[/tex]

[tex]C=5.66[/tex] [tex]uF[/tex]

(b) calculate the resultant supply power factor

First convert the hp into kW

[tex]P_{mech} =250*746=186.5[/tex] [tex]kW[/tex]

Find the electrical power (real power) of the motor

[tex]P_{elec} =P_{mech}/n[/tex]

where [tex]n[/tex] is the efficiency of the motor

[tex]P_{elec} =186.5/0.80=233.125[/tex] [tex]kW[/tex]

The current in the motor is

[tex]I_{m} =(P/\*V*pf)<cos^{-1}(pf)[/tex]

The pf of motor is 0.85 Leading

Note that [tex]<[/tex] represents the angle in complex notation (polar form)

[tex]I_{m} =(233.125/12.47*0.85)<cos^{-1}(0.85)[/tex]

[tex]I_{m}=18.694+11.586[/tex][tex]j[/tex] [tex]A[/tex]

Now find the Load current

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

[tex]I_{load} =(600/12.47*0.75)<-cos^{-1}(0.75)[/tex]

[tex]I_{load} =48.115-42.433[/tex][tex]j[/tex] [tex]A[/tex]

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

[tex]I_{supply} =I_{m} + I_{load}[/tex]

[tex]I_{supply}= (18.694+11.586)+(48.115-42.433)[/tex]

[tex]I_{supply} =66.809-30.847j[/tex] [tex]A[/tex]

or in polar form

[tex]I_{supply} =73.58<-24.78[/tex]°

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

therefore, the supply power factor is

[tex]pf=cos(24.78)=0.90[/tex] 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

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