Power Factor Correction Problem

[Dark Knight]

Just bringing this back from the Old Forum(not :"the other board": ).

The pdf document is the problem and the word document is the solution I worked(I am not saying is the best....)

 

[Caz Rad]

I wish the afternoon portion was strictly Power Factor Correction problems. I would be a happy man!

 

[Volts006]

How do you get the real and reative power for the synchronous motor with just the hp, efficiency, and power factor?

 

[Dark Knight]

The efficiency is Pout/Pin. You have the hp for the motor (10hp). Imagine a box and Pin getting inside the box from the left and Pout coming out to the right. Pout=10hp. Convert that to Watts using 1 hp=746 W conversion factor. Pout would be approx. 7.5kW

Now use the formula for efficiency 0.85= Pout/Pin and solve for Pin. Pin= 8.776kW. Once you have the Pin can use the Power Triangle(you have the power factor). Just remember the power factor is leading(synch motor) and therefore the Q is negative.

 

[Volts006]

Ok, I get it. Thanks, Luis!

 

[Wolverine]

Knowing there would be power factor correction problems, that's one topic I made sure I could work backwards, forwards, and upside down, along with the eng econ tables and NEC. Here are the notes I used, PF Correction for Dummies, not that any users might be dummies, just that this was the simplest form I found as I initially struggled with mastering the subject. I hope no one finds errors now that I have declared myself a master of the subject. Or am I declaring that I'm a dummy? You decide. Hope it's useful.

 

[Volts006]

Thanks for the guide Wolverine

 

[odentonpe]

Thanks for helping us pe seekers. I greatly appreciate the members help in this board. Thanks to all

 

[Dark Knight]

Yikes,

Deleted the attachments. Sorry about that. Here they are again:

Solution::

Sum of : 29.78kW, 11.28kVAR Delivered by Source

For this system tan Ø = (11.28/29.78)therefore:

Ø = tan-1(11.28/29.78)

Ø = 20.74o

pf = cos(20.74) = 0.93.5 or 93.5% approx.

Now we want to improve the pf to 95% (for the source)

Using P= 29.78kW and the power triangle we determine a Q new of 9.79kVAR.

The Q old = 11.28kVAR so the Q required is: 11.28 – 9.79= 1.48kVAR

The synchronous motor would have to be excited @:

P = 8.77kW (does not change)

Q = -2.88kVAR -1.48kVAR = -4.37kVAR

pf = 89.57%

That means that to achieve a pf of 95% for the system the pf for the synch. Motor has to be improved from 95% leading to 89.5% leading.

 

[jdd18vm]

Luis and Wolverine,

Thanks for stuff like this. I've been away from the site a few days studying. ugh.....i feel like i have mount everest in front of me. I wish ALL the questions were NEC.

The more you can dummy this stuff down....I'm all for it.

John

where's all the cool emoticons...

 

[Dark Knight]

Will remove the files in few days to post another problem.Just want to give one more chance to any new EE around here......

 

[Dark Knight]

Removed the files because they are not in demand right now. If you want them PM me. I will e-mail them to you. Do not post your e-mail here.

 

[Ilan]

Thanks Luis for the problem and the solution.

 

[Platinum]

Knowing there would be power factor correction problems, that's one topic I made sure I could work backwards, forwards, and upside down, along with the eng econ tables and NEC. Here are the notes I used, PF Correction for Dummies, not that any users might be dummies, just that this was the simplest form I found as I initially struggled with mastering the subject. I hope no one finds errors now that I have declared myself a master of the subject. Or am I declaring that I'm a dummy? You decide. Hope it's useful.

Aren't you missing an important equation on there? Once you get your corrective Q value but you want to know what actual size capacitor to use..... C= Q/(w*V^2). Is that the right equation?

I'm sure all your equations are correct, but i rarely use the reactive current of caps. Why would you need to know that? Maybe I'm missing something?? :dunno:

 

[singlespeed]

Aren't you missing an important equation on there? Once you get your corrective Q value but you want to know what actual size capacitor to use..... C= Q/(w*V^2). Is that the right equation?I'm sure all your equations are correct, but i rarely use the reactive current of caps. Why would you need to know that? Maybe I'm missing something?? :dunno:

Yes, C= -deltaQ/(w*V^2), where deltaQ = P[tan(cos^-1(pf_initial))-tan(cos^-1(pf_final))]

 

[Wolverine]

Aren't you missing an important equation on there? Once you get your corrective Q value but you want to know what actual size capacitor to use..... C= Q/(w*V^2). Is that the right equation?I'm sure all your equations are correct, but i rarely use the reactive current of caps. Why would you need to know that? Maybe I'm missing something??

Yes, that's a good equation but I would say it's derivable from the information in the third section.

You would only need to know the reactive current in the caps if you faced a problem that asked for the reactive current in the caps, especially on the PE Exam. Other than that, I agree, one would rarely ever calculate the reactive current in the caps.

These are just my scratch notes that helped me pass. The only important equation missing is the one you need when you can't find it.

 

[Gnana]

bump
Will remove the files in few days to post another problem.Just want to give one more chance to any new EE around here......

Dark Knight, do u still have this question and solution available?

I will like a copy.

Thanks

 

[Dark Knight]

OK...This is not the problem I had before but it is very similar.

A certain load takes 10kW @ 0.5 pf lagging when connected to a 230V, 60Hz source. The suply company charges a penalty when the power factor falls below 0.8. What size of capacitor must be used in order to avoid penalty?

That is the problem statement and, believe me, it is a very common concept evaluated in the PE test. Caution note: They are asking the size of the capacitor needed to bring the pf to 0.8.

Here is the solutionPF_Solution.pdf

Good luck!!!!

 

[eedave]

OK...This is not the problem I had before but it is very similar.
A certain load takes 10kW @ 0.5 pf lagging when connected to a 230V, 60Hz source. The suply company charges a penalty when the power factor falls below 0.8. What size of capacitor must be used in order to avoid penalty?

That is the problem statement and, believe me, it is a very common concept evaluated in the PE test. Caution note: They are asking the size of the capacitor needed to bring the pf to 0.8.

Here is the solutionPF_Solution.pdf

Good luck!!!!

Good stuff. Thanks DK!

 

[Gnana]

OK...This is not the problem I had before but it is very similar.
A certain load takes 10kW @ 0.5 pf lagging when connected to a 230V, 60Hz source. The suply company charges a penalty when the power factor falls below 0.8. What size of capacitor must be used in order to avoid penalty?

That is the problem statement and, believe me, it is a very common concept evaluated in the PE test. Caution note: They are asking the size of the capacitor needed to bring the pf to 0.8.

Here is the solutionPF_Solution.pdf

Good luck!!!!

Thank you Dark knight!!