Electrical Engineering Power (Ac)

Question

A small industrial plant has a 10-𝑘𝑊 heating load and a 20kVA inductive load due to a bank of induction motors. The heating elements are considered purely resistive ( 𝑝𝑓=1 ), and the induction motors have a lagging power factor of 70%. If the supply is 1000𝑉𝑟𝑚𝑠  at 60𝐻𝑧, determine the capacitive element required to raise the power factor to 95%

Thanks in advance!

Answer 1

To determine the capacitive element required to raise the power factor to 95%, we need to calculate the reactive power before and after the power factor correction and then find the difference, which represents the capacitive reactive power required.

Given data:

1. Heating load: 10 kW (purely resistive load)
2. Inductive load: 20 kVA (lagging power factor of 70%)
3. Supply voltage: 1000 Vrms
4. Frequency: 60 Hz

1. Calculate the reactive power before power factor correction:

For the inductive load, we'll find the reactive power 𝑄before​.

𝑄before=𝑆×sin⁡𝜃

Given that the power factor PF=0.70, we can find the angle 𝜃 using the power factor:

cos⁡𝜃=PF

𝜃=arccos⁡(PF)

𝜃=arccos⁡(0.70)

𝜃≈45.57

Now, we can calculate the reactive power:

𝑄before=20×sin⁡(45.57∘)

𝑄before≈14.14 kVAR

2. Calculate the reactive power after power factor correction:

For the desired power factor of 95%, we need to find the angle 𝜃new

PFnew=0.95

cos⁡𝜃new=0.95

𝜃new=arccos⁡(0.95)

𝜃new≈18.19∘

Now, we can calculate the reactive power after power factor correction:

𝑄after=𝑆×sin⁡𝜃new

𝑄after=20×sin⁡(18.19∘)

𝑄after≈6.58 kVAR

3. Calculate the capacitive reactive power required:

The difference between the reactive power before and after power factor correction gives the capacitive reactive power required.

𝑄capacitive=𝑄before−𝑄after

𝑄capacitive=14.14−6.58

𝑄capacitive≈7.56 kVAR



So, the capacitive element required to raise the power factor to 95% is approximately 7.56 kVAR.