Question

A factory has an electrical load of 1600 kW at a lagging power factor of 0.8. An additional variable power factor load is to be added to the factory. The new load will add 320 kW to the factory. The power factor of the variable power factor load is to be adjusted so that the overall power factor of the factory is 0.96 lagging.a) What is the reactive power associated with the added load?b) Does that load absorb or deliver magnetising vars (reactive power is also sometime referred to as magnetising vars. Inductors absorb magnetising vars and capacitors deliver magnetising vars)c) What is the power factor of the adjustable power factor load?d) If the voltage at the input is 2400 V(rms), what is the magnitude of the current into the factory before the adjustable load?e) What is the magnitude of the current with the adjustable load?

Answer 1

Step-by-step explanation:

a)

Apparent power|S0| of the original load:

|S0|=P0/0.8=1600000/0.8

|S0|=2000kvA

P0 is the average power of the original load.

To determine the relative power Q0 of the original load :

Q0=√|S0|^2-P0^2

=√(2000000^2)-(1600000)^2=1200kVAR

Complex power of the original load S0:

S0=P0+jQ0

=1600+j1200kvA

After the additional power factor load, the final power Pf is :

Pf=1600000+320000=1920kw

Therefore the final apparent power |Sf| of the load will be:

|Sf |=Pf/0.96

|Sf|=1920000/0.96=2000kvA

Hence, the reactive power (Qf) of the composite Load:

Qf=√|Sf|^2-P^2

Qf=√(2000000)^2-(1920000)^2=560kVAR

Now, the final complex power(Sf) associated with the composite load is:

Q(add)=Qf-Q0

Q(add)=560000-1200000

=-640kVAR

b) Yes, since the reactive power value of the added load is negative, the added load delivers magnetizing vars:

c) Complex power S(add) of the added load:

S(add)=P(add) + jQ(add)

S(add)=320000-j640000

S(add)=715541.753<-63.435

The phase difference between the added load voltage and current is:

Qv-Qi=-63.435

The power factor Pf of the added load :

Pf=cos(-63.435)=0.447

Therefore, the added load current leads the voltage, the power of the additional load or adjustable power factor is leading by:

Pf=0.447leading

d) the current into the factory (I) before adding the power factor load is:

I=S0/2400

I=1600000-j1200000/2400

I=666.667-j500

I=833.33<-36.87A(rms)

rms magnitude of the current into the factory| I| before adding the power factor load is given as:

|I|=833.33A(rms)

e) current into the factory ,I, after adding the power factor load :

I=Sf*/2400

I=1920000-j560000/2400

I=833.33<-16.25•(rms)

Therefore, rms magnitude of the current into the factory |I| after adding the power factor load is :

|I|=833.33A(rms)

Finally ,adding the capacitor has not changed the current rms magnitude but it's phase only.