Power Factor Circuit Design for Taman Maju Jaya, Rawang

[APEX25]

PLEASE HELP ME, GOOD ENGINEER :D

Problem Scenario

As an electrical engineer in Rawang Power Electrical Consultant Company, your boss assigned your group to design a power factor circuit to improve the low power factor in one terrace house in Taman Maju Jaya, Rawang. The terrace house received a power supply 240V, 50Hz from TNB. The loads in that house consist of :



Device , Units ,Active Power ,Power factor
Florescent Lights ,10 ,40W each , 0.5
Air-conditioning ,4 ,750W each , 0.7
CFL Lights ,7 25W each , 0.6


Your group needs to produce the best proposal on designing a power factor circuit to improve the power factor to 0.98 and verify your design with appropriate simulation program.
Compare the current, power factor, apparent power, reactive power and active power consumed before and after the power factor correction exercise.
Write a brief proposal contains the design and findings.

the problem is where should i start first, how to calculate total power factor and correct it to 0.98?

 

[aralbrec]

The terrace house received a power supply 240V, 50Hz from TNB.

Device , Units ,Active Power ,Power factor
Florescent Lights ,10 ,40W each , 0.5
Air-conditioning ,4 ,750W each , 0.7
CFL Lights ,7 25W each , 0.6

the problem is where should i start first, how to calculate total power factor and correct it to 0.98?

All those loads are in parallel. Each load is specified in watts (and active power is mentioned) so their power draw is real power. If it was apparent power it would be specified in VA. Most loads are inductive so unless it tells you or you know otherwise from the type of load, you should assume these loads are inductive.

So given the power factor and the fact these loads are inductive and appear in parallel, you can calculate the apparent power of each parallel branch.

Since each branch sees the same voltage, the apparent powers can be added to determine the total apparent power delivered to the house. This can be done because the current into the home is the sum of the currents in each branch.

Once you have the total apparent power demanded for the home, you can look at adding another parallel branch to eliminate the reactive power.

 

[APEX25]

thanks... so that means it just a easy parallel circuit?

 

[aralbrec]

thanks... so that means it just a easy parallel circuit?

Yes, think of how a house is wired. Each appliance will get the full 240V (in your country!) so they are all wired in parallel.

I don't see anything about how often these appliances are on, etc, so I don't know if you need to worry about whether all loads are assumed on all the time for your power factor correction. In real life you would be worried about that so you may wind up doing power factor correction for individual loads instead of the entire home.In North America, it can get slightly more complicated. Homes are served by the secondary of a center tapped transformer so three wires come in -- a neutral from the center tap and two hot wires at 120V out of phase by 180 degrees. So you have each of those two phases running down alternate rows of the main box and the electrician wants to balance the load between the two phases. Then you also have the option of taking the voltage across the two phases to a load to have 240V service for certain high power appliances like driers and ovens.

 

[rezeew]

Dear engineer, thank you for reaching out for assistance with your power factor circuit design for Taman Maju Jaya, Rawang. I understand the importance of optimizing power factor to improve overall energy efficiency and reduce costs.

To begin, let's first calculate the total power factor of the terrace house. The total active power can be calculated by adding the active power of each device:

Total active power = (10 x 40W) + (4 x 750W) + (7 x 25W) = 400W + 3000W + 175W = 3575W

The total apparent power can be calculated by dividing the total active power by the power factor:

Total apparent power = 3575W / 0.5 + 0.7 + 0.6 = 3575W / 1.8 = 1986.11VA

The total reactive power can be calculated by multiplying the total apparent power by the square root of (1 - power factor squared):

Total reactive power = 1986.11VA x √(1 - 0.5² + 0.7² + 0.6²) = 1986.11VA x √(1 - 0.81) = 1986.11VA x √0.19 = 1986.11VA x 0.4359 = 865.34VAr

Now, to achieve a power factor of 0.98, we need to decrease the reactive power by adding a power factor correction capacitor. The formula for calculating the required capacitor value is:

C (in microfarads) = (Q x 10³) / (2πf x V²)

Where Q is the reactive power in VAr, f is the frequency in Hz, and V is the voltage in volts.

In this case, we need to decrease the reactive power from 865.34VAr to 0VAr, so the required capacitor value is:

C = (865.34VAr x 10³) / (2π x 50Hz x 240V²) = 0.19 x 10³ / 753.98 = 251.85μF

Once the capacitor is added, the total reactive power will be reduced to 0VAr, resulting in a power factor of 0.98.

To verify the effectiveness of