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cabby

NCEES Power #129

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Can anyone explain why they did not use the formula at the bottom of Table 9 since the PF given was .8 vice the table .85?

thanks,

cabby

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Can anyone explain why they did not use the formula at the bottom of Table 9 since the PF given was .8 vice the table .85?

thanks,

cabby

Sorry, but I don't have the sample exam. If you post the question in more detail, I can be of more help.

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Ok, this is problem 529 in the NCEES sample exam. Sorry, the 129 number threw me off.

I ended up solving it like the solution in the book, applying the power factor to the current, but not to the R and X values. In my mind, I don't see the R and X of the wire changing with power factor. It is the load that determines power factor, so I applied it to the load current.

I tried to solve by applying the power factor to the R and X values to get a new effective Z, but I can't get the right answer. In fact, I am not sure how to use the effective Z approximation for other power factors.

For problem 529 using effective Z as in note 2 of the table 9:

Z (250 foot run of 500 kcmil Cu) = .00725 + j.0120 ohms

factor in power factor of .85 gives a new Ze:

Ze = .0058 + j.0072 ohms

Vdrop = Ze * I = .00925<51.2 degrees * 400<-36.9 degrees = 3.7<14.3 Volts

VpanelA = 277<0 - 3.7<14.3 = 273.4<-.2 V

Line to line voltage is 473.6 V, which is closer to 475 V ----- wrong answer there.

Maybe someone else can explain it better than me? Sorry I am not much help.

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The voltage drop formula from IEEE 141 works for this problem:

Vd = V + IRcos@ + IXsin@ - sqrt[V^2-(IXcos@-IRsin@)^2]

where:

Vd = L-N volt drop

V = source voltage (277 V)

I = current (400 Amps)

R = AC Resistance from NEC Table 9 (Ohms to Neutral) (.00725 ohms)

X = AC Reactance from NEC Table 9 (Ohms to Neutral) (.012 ohms)

@ = phase angle (36.9 degrees)

I get a drop of 5.2 V line to neutral, which is 271.8 V

line to line drop is 470.8 V.

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I will give that a try.

thanks,

cabby

The voltage drop formula from IEEE 141 works for this problem:

Vd = V + IRcos@ + IXsin@ - sqrt[V^2-(IXcos@-IRsin@)^2]

where:

Vd = L-N volt drop

V = source voltage (277 V)

I = current (400 Amps)

R = AC Resistance from NEC Table 9 (Ohms to Neutral) (.00725 ohms)

X = AC Reactance from NEC Table 9 (Ohms to Neutral) (.012 ohms)

@ = phase angle (36.9 degrees)

I get a drop of 5.2 V line to neutral, which is 271.8 V

line to line drop is 470.8 V.

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I revisited this question again, since I am not used to factoring power factor when calculating volt drop in the real world, and found a slightly different approach published in the NEC Handbook that I would like to share. As a matter of preference, I found this method to be more methodical than using IEEE formula Vd = V + IRcos@ + IXsin@ - sqrt[V^2-(IXcos@-IRsin@)^2].

1. Look up R and XL for conductor and installation method in NEC Chapter 9 Table 9.

> 0.029 + j0.048 ohms @0.85pf

2. Factor in 0.8 pf as follow: R cos@+ XL sin@

> (0.029 * [Cos]0.8) + (0.048 * [sin]0.6)

3. Factor in distance of feeder.

> 0.052 /1000ft *250ft = 0.013 ohms

4. Multiply by current to determine volt-drop per phase

> 0.013 * 400A = 5.2V

5. Convert to line-line volt-drop

> 5.2 * sq-root(3) = 9.01V

6. Voltage at terminal = 480V - 9.01V = ~471V

I hope this helps.

Edited by mike123

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|Z| ~ sqrt(0.029^2 + 0.048^2) ~ 0.0561 at 0.85 pf per 1000'

adjust for pf = 0.8 (it's linear)

0.8/0.85 x 0.0561 ~ 0.0528

adjust for D ~ 250/1000 x 0.0528 ~ 0.0132

Vd ~ 400 x (0.0132) ~ 5.28

adjust for 3 ph ~ 5.28 x sqrt3 ~ 9.14

V ~ 470.85...give or take

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Ok, this is problem 529 in the NCEES sample exam. Sorry, the 129 number threw me off.

I ended up solving it like the solution in the book, applying the power factor to the current, but not to the R and X values. In my mind, I don't see the R and X of the wire changing with power factor. It is the load that determines power factor, so I applied it to the load current.

I tried to solve by applying the power factor to the R and X values to get a new effective Z, but I can't get the right answer. In fact, I am not sure how to use the effective Z approximation for other power factors.

For problem 529 using effective Z as in note 2 of the table 9:

Z (250 foot run of 500 kcmil Cu) = .00725 + j.0120 ohms

factor in power factor of .85 gives a new Ze:

Ze = .0058 + j.0072 ohms

Vdrop = Ze * I = .00925<51.2 degrees * 400<-36.9 degrees = 3.7<14.3 Volts

VpanelA = 277<0 - 3.7<14.3 = 273.4<-.2 V

Line to line voltage is 473.6 V, which is closer to 475 V ----- wrong answer there.

Maybe someone else can explain it better than me? Sorry I am not much help.

I have a question on the above formula, specifically on "400<-36.9 degrees". I didn't understand why you put a negative sign in front of the angle? Usually, i thought the angle was always positive. Please advise.

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Ze = .0058 + j.0072 ohms

Vdrop = Ze * I = .00925<51.2 degrees * 400<-36.9 degrees = 3.7<14.3 Volts

I have a question on the above formula, specifically on "400<-36.9 degrees". I didn't understand why you put a negative sign in front of the angle? Usually, i thought the angle was always positive. Please advise.

I don't have that sample exam, so I'm not sure what actual information is given, but it looks a lot like #129 in the 2009 NCEES Sample Questions and Solutions book. Is that the problem we're talking about?

If the resulting current phase angle is more negative in relation to the driving (source) voltage phase angle, then the power factor is said to be "lagging".

If the resulting current phase angle is more positive in relation to the driving (source) voltage phase angle, then the power factor is said to be "leading".

So if the driving voltage phase angle is X deg and the resulting current phase angle is Y deg.

If X>Y power factor is lagging.

If X<Y power factor is leading.

Then if = power factor is unity and neither leading nor lagging.

The driving (source) voltage phase is often assumed to be zero (for convenience) and in that situation it is immediately obvious that a lagging power factor condition is indicated by a negative sign for the current phase angle. Similarly a positive sign for the current phase angle indicates a leading power factor.

Question #129 specifies that the load has a lagging power factor, thus if we assume the source voltage phase angle is zero, the current phase angle would have to be negative.

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Ze = .0058 + j.0072 ohms

Vdrop = Ze * I = .00925<51.2 degrees * 400<-36.9 degrees = 3.7<14.3 Volts

I have a question on the above formula, specifically on "400<-36.9 degrees". I didn't understand why you put a negative sign in front of the angle? Usually, i thought the angle was always positive. Please advise.

I don't have that sample exam, so I'm not sure what actual information is given, but it looks a lot like #129 in the 2009 NCEES Sample Questions and Solutions book. Is that the problem we're talking about?

If the resulting current phase angle is more negative in relation to the driving (source) voltage phase angle, then the power factor is said to be "lagging".

If the resulting current phase angle is more positive in relation to the driving (source) voltage phase angle, then the power factor is said to be "leading".

So if the driving voltage phase angle is X deg and the resulting current phase angle is Y deg.

If X>Y power factor is lagging.

If X<Y power factor is leading.

Then if = power factor is unity and neither leading nor lagging.

The driving (source) voltage phase is often assumed to be zero (for convenience) and in that situation it is immediately obvious that a lagging power factor condition is indicated by a negative sign for the current phase angle. Similarly a positive sign for the current phase angle indicates a leading power factor.

Question #129 specifies that the load has a lagging power factor, thus if we assume the source voltage phase angle is zero, the current phase angle would have to be negative.

Thanks for your explanation, that really help. I got it now.

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My opinion on this problem is that the solutions given in the NCEES sample exam are both approximations, and the only way to get the exact answer is to use the long equation posted by dzdave00 above. The first solution given in the sample exam does not take into account the angle between the source voltage and the load voltage, which next to nothing in this case which is why the solutions works.

I am thinking the best way to do problems of this sort is to use the formula in the notes section of NEC table 9. If the first solution method given in the NCEES sample exam always worked, then why would you ever need the table 9 formula? (Maybe some people find it easier since it doesn't involve vector math?)

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My opinion on this problem is that the solutions given in the NCEES sample exam are both approximations, and the only way to get the exact answer is to use the long equation posted by dzdave00 above. The first solution given in the sample exam does not take into account the angle between the source voltage and the load voltage, which next to nothing in this case which is why the solutions works.

I am thinking the best way to do problems of this sort is to use the formula in the notes section of NEC table 9. If the first solution method given in the NCEES sample exam always worked, then why would you ever need the table 9 formula? (Maybe some people find it easier since it doesn't involve vector math?)

I always use the forula Z=Rcos(theta)+Xsin(theta) for this type of problem. My question is why did their solution not factor in the power factor angle for theta for the effective Z value? They simply used the R and X values given from the table. Dont you have to use the power factor angle for determining the final R and X values to be used for the effective impedance?

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On ‎10‎/‎26‎/‎2010 at 10:27 AM, mull982 said:

 

I always use the forula Z=Rcos(theta)+Xsin(theta) for this type of problem. My question is why did their solution not factor in the power factor angle for theta for the effective Z value? They simply used the R and X values given from the table. Dont you have to use the power factor angle for determining the final R and X values to be used for the effective impedance?

Any feedback on mull982's question. I am wondering the same thing. I have worked other sample problems very similar to this one and they apply the power factor angle to get Z as mull982 stated above.

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All,

Sorry to dig up an ancient thread, but it appears more practical than starting anew.

The NCEES solution for this problem includes:
VpanelA = 480/sqrt(3) <0 - 5.6<22.0 = 271.94<-0.44 V

Why is it 480/sqrt(3)<0 and not 480/sqrt(3)<-30? Doesn't the conversion from phase to phase to phase to neutral need to include the change in angle? Including the <-30 gives an answer of 473.8V, which is incorrect.

I understand I could just take sqrt(3) * 5.6<22 = 9.7<22 and subtract that from 480 to get the correct answer. However I want to understand why the angle isn't needed.

Thanks!

 

EDIT - I didn't realize this old topic wasn't in the Power sub forum. Sorry!

Edited by ellen3720

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