Kaplan Exam Morning Question #3

[mull982]

In the attached question and solution for problem #3 of the Kaplan morning power exam I dont understand why the solution states to use Q=(sqrt3)(Pb-Pa) to find the total reactive power.

Now I understand that with a two wattmeter configuration reactive power Q=(sqrt3)(P1-P2) however I cant seem to figure out why the solution used P2-P1 instead of P1-P2? Does it have something to do with the phase sequence?

Can anyone help?

 

[mull982]

For some reason it will not let me post the attachment. It keeps saying "upload failed file larger then avaliable space" It tells me I have already used 84.67 of 100k attachment space before I even attach anything?

 

[DK PE]

For some reason it will not let me post the attachment. It keeps saying "upload failed file larger then avaliable space" It tells me I have already used 84.67 of 100k attachment space before I even attach anything?

Not sure but you may have to go into "My controls" at top right --> then Options at lower left ---> then manage attachments and delete anything that isn't currently in use.

 

[mull982]

For some reason it will not let me post the attachment. It keeps saying "upload failed file larger then avaliable space" It tells me I have already used 84.67 of 100k attachment space before I even attach anything?

Not sure but you may have to go into "My controls" at top right --> then Options at lower left ---> then manage attachments and delete anything that isn't currently in use.

Yup that was the trick. Here is the attachment with the problem statement.

KMBT25020101018115822.pdf

 

[mull982]

For some reason it will not let me post the attachment. It keeps saying "upload failed file larger then avaliable space" It tells me I have already used 84.67 of 100k attachment space before I even attach anything?

Not sure but you may have to go into "My controls" at top right --> then Options at lower left ---> then manage attachments and delete anything that isn't currently in use.

Yup that was the trick. Here is the attachment with the problem statement.

Does anyone know why they used P2-P1 instead of P1-P2?

 

[cableguy]

I think that with this problem they assume you are looking at a source like this:

http://users.encs.concordia.ca/~lalopes/Co...rs%20method.pdf

Since that's the exact same circuit (wattmeters on A & C phase with the pressure coils connected to B phase), you can use the formulas in that example.

If they had connected the pressure coils to somewhere else, you'd have to have more information to solve the problem.

I dug in to it a little deeper, and it looks like there are 2 different wattmeter methods. I've been using a different one than presented in that PDF (and different than Kaplan).

Kaplan Method:

(assuming abc rotation, W1 = A phase, W2 = C phase, common PC on B phase)

WA = V I cos (30 + theta)

WC = V I cos (30 - theta)

WA + WC = sqrt(3) V I cos theta

WC - WA = V I sin theta

Method I've been using, same assumptions (Camara, Testmasters, etc):

WA = V I cos (30 - theta)

WC = V I cos (30 + theta)

WA + WC = sqrt(3) V I cos theta

WA - WC = V I sin theta

You don't get a real handle on power factor lead/lag until you get in to the last line there.

I *think* that the NCEES uses the second method (don't have my sample exam with me).

Once again, I think the Kaplan problem doesn't include enough information for us to make an informed decision.

Edit: I have been waiting on our IT department to give me access to a database... meanwhile, I used it to work on Wattmeters even further. I hope you guys don't mind me posting all this crap. Like my dog with a rawhide bone, I chew on things until they're completely done.

In actuality, I don't solve wattmeter problems like above. I lied a bit...

I solve them using the formula

Wattmeter Reading = V I cos (angle pressure coil - angle current)

Note that if you set your Van reference as 0, your Vab pressure coil angle is 30 degrees for the abc rotation, or -30 degrees in the cba rotation.

Then your current angle for Ia is plain and simple, reference Van as 0.

When you get to the second wattmeter, this is where things get weird.

Your Vcb angle is -(Vbc)'s angle. So if Vab is 30 degrees, Vcb angle is 90 degrees in the abc rotation. Conversely, Vcb angle is -90 degrees in the cba rotation.

Current angle theta for Ic is 120 degrees from Ia, calculated against the phase rotation.

When I use WA - WC = V I sin theta, my power factor angle does come out correct (lead or lag) using this method. Why do I use this method? Because not all wattmeter problems are presented between A&C phases, with the coil on B. I've seen them to neutral, seen A&B phases, etc. I stick with my formula because it allows me to work on any phase.

Having a drawn out phasor diagram for both rotations is the only way I can keep all this junk straight in my head, to be honest.

 

[mull982]

I think that with this problem they assume you are looking at a source like this:
http://users.encs.concordia.ca/~lalopes/Co...rs%20method.pdf

Since that's the exact same circuit (wattmeters on A & C phase with the pressure coils connected to B phase), you can use the formulas in that example.

If they had connected the pressure coils to somewhere else, you'd have to have more information to solve the problem.

I dug in to it a little deeper, and it looks like there are 2 different wattmeter methods. I've been using a different one than presented in that PDF (and different than Kaplan).

Kaplan Method:

(assuming abc rotation, W1 = A phase, W2 = C phase, common PC on B phase)

WA = V I cos (30 + theta)

WC = V I cos (30 - theta)

WA + WC = sqrt(3) V I cos theta

WC - WA = V I sin theta

Method I've been using, same assumptions (Camara, Testmasters, etc):

WA = V I cos (30 - theta)

WC = V I cos (30 + theta)

WA + WC = sqrt(3) V I cos theta

WA - WC = V I sin theta

You don't get a real handle on power factor lead/lag until you get in to the last line there.

I *think* that the NCEES uses the second method (don't have my sample exam with me).

Once again, I think the Kaplan problem doesn't include enough information for us to make an informed decision.

Edit: I have been waiting on our IT department to give me access to a database... meanwhile, I used it to work on Wattmeters even further. I hope you guys don't mind me posting all this crap. Like my dog with a rawhide bone, I chew on things until they're completely done.

In actuality, I don't solve wattmeter problems like above. I lied a bit...

I solve them using the formula

Wattmeter Reading = V I cos (angle pressure coil - angle current)

Note that if you set your Van reference as 0, your Vab pressure coil angle is 30 degrees for the abc rotation, or -30 degrees in the cba rotation.

Then your current angle for Ia is plain and simple, reference Van as 0.

When you get to the second wattmeter, this is where things get weird.

Your Vcb angle is -(Vbc)'s angle. So if Vab is 30 degrees, Vcb angle is 90 degrees in the abc rotation. Conversely, Vcb angle is -90 degrees in the cba rotation.

Current angle theta for Ic is 120 degrees from Ia, calculated against the phase rotation.

When I use WA - WC = V I sin theta, my power factor angle does come out correct (lead or lag) using this method. Why do I use this method? Because not all wattmeter problems are presented between A&C phases, with the coil on B. I've seen them to neutral, seen A&B phases, etc. I stick with my formula because it allows me to work on any phase.

Having a drawn out phasor diagram for both rotations is the only way I can keep all this junk straight in my head, to be honest.

If you would have used your method and used WA-WC then you would come up with an answer of .8 lagging as opposed to the .8 lagging they give in the solution.

 

[cableguy]

If you would have used your method and used WA-WC then you would come up with an answer of .8 lagging as opposed to the .8 lagging they give in the solution.

Agreed.

While their formulas do work, I am wondering if there is a "wattmeter convention" that needs to be applied to these generic formulas. By "generic", I mean that they slap that 30 degree in the cos(30 +/- theta) and tell you to work with that. I almost think they have their formulas set up opposite of standard convention.

I'm going to look at the NCEES convention when I get home this evening. I know I've solved their sample problems using my method with no issues. I've also solved a different Kaplan problem with no issue. I think Kaplan simply botched this problem, and I think the PDF that I referenced has things set up opposite to standard convention.

I'll stick with using V I cos (voltage angle minus current angle) for all of my wattmeters, and using the phasor diagrams to guide me. I'm not going to rely on generic "30+theta" type equations, as they'll fall to pieces if you're given W1 as an A-B wattmeter, and W2 as a C-A wattmeter, or something crazy like that.

 

[VectrenEng]

Again, I don't trust the Kaplan answer. I think it is 0.80 lagging.

In the Camara reference book - can't remember the section (42 Instrumentation?) or page number because I don't have my book in front of me - is the following equation:

tan Φ = SQRT(3)(PH - PL) / (PH + PL)

where PH is the reading of the higher power and PL is the reading of the other meter (I assume this to mean PH > PL).

 

[cableguy]

OK, after looking through the NCEES Sample exam, we were only given one wattmeter problem (509), and it wasn't even a numerical problem, it was theory. They did not give us enough information to be able to work it as a numerical problem (they did not give us phase rotation - which is critical for these problems). It was, however, connected with wattmeters on B and C phase, with the pressure coil connected to A.

Kaplan morning problem 7 is a single Wattmeter problem. Their setup and answer are consistent with 'my' method.

I don't think you can do PH-PL and get a valid answer for Φ though. Then angle will be negative for lagging power factor, and positive for leading. If you do PH-PL, you'll never get a negative angle. Keep in mind that W1-W2=VI sin Φ ..... so Φ can be positive or negative, depending on the difference.

 

[knight1fox3]

They did not give us enough information to be able to work it as a numerical problem (they did not give us phase rotation - which is critical for these problems).

Actually phase rotation is noted below the figure for problem 509. However, it is still a theoretical problem and not an in depth numerical problem.

 

[cableguy]

Actually phase rotation is noted below the figure for problem 509. However, it is still a theoretical problem and not an in depth numerical problem.

You're right, my bad.

Looking at the answers, they do give us formulas though. Which are different from any of the 3 methods I've posted above! LOL.

W1 = VI cos (Φ-30)

W2 = VI cos (Φ+30)

Now, this is likely due to the fact that W1 is on C phase, W2 on B, and since B leads C, that's why they have it written as they do. I'm going to dig in to this later today.

 

[VectrenEng]

OK, after looking through the NCEES Sample exam, we were only given one wattmeter problem (509), and it wasn't even a numerical problem, it was theory. They did not give us enough information to be able to work it as a numerical problem (they did not give us phase rotation - which is critical for these problems). It was, however, connected with wattmeters on B and C phase, with the pressure coil connected to A.
Kaplan morning problem 7 is a single Wattmeter problem. Their setup and answer are consistent with 'my' method.

I don't think you can do PH-PL and get a valid answer for Φ though. Then angle will be negative for lagging power factor, and positive for leading. If you do PH-PL, you'll never get a negative angle. Keep in mind that W1-W2=VI sin Φ ..... so Φ can be positive or negative, depending on the difference.

I had that epiphany (i.e. always get positive angle) last night. Too much studying....

 

[VectrenEng]

Sorry for the double-post, but I did some more research. I hate when I have unclear, or conflicting sources of information, especially this close to the test.

This is a good source for the two-wattmeter method.

Two-Wattmeter Method

For this problem, I changed the notation for PB to WC, since it is measuring C-phase.

Using the methods outlined in the link, for ABC rotation:

tan Φ = SQRT(3)(WC - WA) / (WC + WA) = -36.87º

pf = cos(36.87º) = 0.80 LEAD

So the Kaplan solution appears to be correct. Does anyone else agree?

 

[mull982]

Sorry for the double-post, but I did some more research. I hate when I have unclear, or conflicting sources of information, especially this close to the test.
This is a good source for the two-wattmeter method.

Two-Wattmeter Method

For this problem, I changed the notation for PB to WC, since it is measuring C-phase.

Using the methods outlined in the link, for ABC rotation:

tan Φ = SQRT(3)(WC - WA) / (WC + WA) = -36.87º

pf = cos(36.87º) = 0.80 LEAD

So the Kaplan solution appears to be correct. Does anyone else agree?

O.K. after looking at it heres the way I see it.

using tan(theta)=1.73 (Wc-Wa)/(WC+Wa) Wc is always the power meter that has the cos(30-theta associated with it) Depending on the phase rotation this will be WC for ABC rotation, but for CBA rotation this will be Wa. So ingoring the subscripts for a second alwys use the meter that has (30-theta) associated with it for Wc in the above equation and use the meter that has (30+theta) associated with it for Wa in the above equation.

 

[mull982]

Sorry for the double-post, but I did some more research. I hate when I have unclear, or conflicting sources of information, especially this close to the test.
This is a good source for the two-wattmeter method.

Two-Wattmeter Method

For this problem, I changed the notation for PB to WC, since it is measuring C-phase.

Using the methods outlined in the link, for ABC rotation:

tan Φ = SQRT(3)(WC - WA) / (WC + WA) = -36.87º

pf = cos(36.87º) = 0.80 LEAD

So the Kaplan solution appears to be correct. Does anyone else agree?

O.K. after looking at it heres the way I see it.

using tan(theta)=1.73 (Wc-Wa)/(WC+Wa) Wc is always the power meter that has the cos(30-theta associated with it) Depending on the phase rotation this will be WC for ABC rotation, but for CBA rotation this will be Wa. So ingoring the subscripts for a second alwys use the meter that has (30-theta) associated with it for Wc in the above equation and use the meter that has (30+theta) associated with it for Wa in the above equation.

To look at this even further I put some numbers to problem 509 in the NCEES exam to try to solve for the power factor.

Lets say that in problem 509 we have the same connections that are shown but we are told that W1=800W and W2=400W and we are asked to find the power factor assuming the sam ABC rotation. I come up with a p.f. of .86 leading. Does anyone else get the same?

Now if we have a CBA rotation then I come up with a p.f. of .86 lagging. Do you guys agree?

 

[VectrenEng]

Sorry for the double-post, but I did some more research. I hate when I have unclear, or conflicting sources of information, especially this close to the test.
This is a good source for the two-wattmeter method.

Two-Wattmeter Method

For this problem, I changed the notation for PB to WC, since it is measuring C-phase.

Using the methods outlined in the link, for ABC rotation:

tan Φ = SQRT(3)(WC - WA) / (WC + WA) = -36.87º

pf = cos(36.87º) = 0.80 LEAD

So the Kaplan solution appears to be correct. Does anyone else agree?

O.K. after looking at it heres the way I see it.

using tan(theta)=1.73 (Wc-Wa)/(WC+Wa) Wc is always the power meter that has the cos(30-theta associated with it) Depending on the phase rotation this will be WC for ABC rotation, but for CBA rotation this will be Wa. So ingoring the subscripts for a second alwys use the meter that has (30-theta) associated with it for Wc in the above equation and use the meter that has (30+theta) associated with it for Wa in the above equation.

To look at this even further I put some numbers to problem 509 in the NCEES exam to try to solve for the power factor.

Lets say that in problem 509 we have the same connections that are shown but we are told that W1=800W and W2=400W and we are asked to find the power factor assuming the sam ABC rotation. I come up with a p.f. of .86 leading. Does anyone else get the same?

Now if we have a CBA rotation then I come up with a p.f. of .86 lagging. Do you guys agree?

I agree - Using the methods outlined in the link, my calculations match yours.

(I let W1=WC and W2=WB)

 

[mofet]

nice thread.