Complex imaginary #2, problem 12

[Jabert]

Problem is related to voltage drop:

In the 3 phase system, a bus serves a 230A lagging power factor of .75

(3) 350 wag copper conductors are held in a PVC conduit.

With a 500' feeder, calculate voltage at the load (600V source)

Table 9 provides feeder impedance for unity power factor and for .85 power factor. The solution uses unity power factor values...

When I worked the problem, I calculated the effective impedance for .75 power factor and got 592.7V which is incorrect.

Not converting to effective impedance values yields 593.7V which is marked as the correct answer

Can anyone explain?

 

[EEpowerOK]

Problem says uncoated copper wire not Eff Z. Eff Z has a pf requirement that is not listed in the uncoated conductor.

 

[iahim]

Here is how I would solve this problem:

From Table 9:

XL=0.040 ohm/1000 ft

R= 0.038 ohm/1000 ft

For 500 ft:

XL=0.02 ohm

R=0.019 ohm

Z= 0.019+j0.02 ohm

pf=0.75 lag, so angle is (-cos-1(0.75))= -41.41

I=230<(-41.41)

So line voltage drop, IZ=6.345<5.1

(Up to here is the same as the CI solution)

At source: VLL=600V so VLN=346.41V

At load: VLN=(346.41<0)-(6.345<5.1)=340.1<(-0.1)

So VLL at load is 589V.

I think the CI solution is incorrect because they subtract the voltage drop from the line to line voltage. If you draw the equivalent circuit you can see that is incorrect.

Am I wrong?

 

[Ship Wreck PE]

The problem shows the voltage as 600 volts phase. And they are asking for phase voltage at the end, so 600 - 6.34 = 593.68. So the solution is correct.

 

[iahim]

In my version of the exam it is not specified that the voltage is phase voltage. If 600V, is phase voltage, the CI solution is correct.

 

[Ship Wreck PE]

What version do you have?? I got mine from CI this year and it says 2011 at the bottom of page and 2012 in front of book.

In the drawing of the system, just beneath the voltage source it says 600 V / Phase.

 

[iahim]

I bought mine used, this summer. There is no date on the front cover, but the copyright at the bottom of the page is 2011 as well.

Under the voltage source it says 600V. You probably have a newer version. The solution in my version is also incorrect (595.6V).

Also, all problems refer to NEC 2008, not 2011.

 

[iahim]

Additionally, in my version they ask for bus voltage, not phase voltage. They have probably realized the mistake and changed the problem to make the solution correct.

 

[Redskinsdb21]

In my version of the exam it is not specified that the voltage is phase voltage. If 600V, is phase voltage, the CI solution is correct.

Most places will tell you assume phase voltage unless told otherwise...or thats what I remember anyway

 

[iahim]

It's the other way around: if not specified, assume line voltage. For example, if you have a 480V, 3-phase system, the line voltage is 480V and phase voltage is 277V.

 

[EEpowerOK]

My book says bus phase voltage (V) is most nearly. Then displays "600 V/phase" in the diagram

 

[Redskinsdb21]

It's the other way around: if not specified, assume line voltage. For example, if you have a 480V, 3-phase system, the line voltage is 480V and phase voltage is 277V.

line voltage = phase voltage..depends on if its WYE or DELTA actually....you assume line voltage not line to neutral voltage

 

[iahim]

It's the other way around: if not specified, assume line voltage. For example, if you have a 480V, 3-phase system, the line voltage is 480V and phase voltage is 277V.

line voltage = phase voltage..depends on if its WYE or DELTA actually....you assume line voltage not line to neutral voltage

Now you got me confused...

For a wye system:

Line voltage = voltage line to line

Phase voltage = voltage line to neutral

For a delta system:

Line voltage = phase voltage = voltage line to line

If not specified, you assume wye system and line to line voltage.

 

[EEpowerOK]

Adding to this discussion, in Spin-up 3-70 the problem states;

A 3 phase 480 V Wye generator has a load of 24KVA @.9 lagging PF. What is the current in Line A, if the load is between line A and line B?

In a Wye the line current and phase current are the same so, I= 24KVA/(sq rt 3 * 480 V)

book has I=24KVA/480 V

Load is between line A and line B so it is 3 phase load.

 

[iahim]

If the problem states that the load is connected between phases A and B, the load is single phase. It doesn't matter how the source is configured. The only thing you need to know to solve the problem is that the voltage between lines A and B is 480V. So the solution in the book is correct.

 

[EEpowerOK]

Connect the load between A and B, you get the following. The book said the formula was S=VI, that is true if the load was connected between phase to neutral (second drawing). Is that right or am I missing something?

A

\ /----------------------

\ / | |

\ / | |

\ / V line Load S=Sq Rt 3 * V-line * I-line

| | |

| | |

|----------------------------|

B

A

\ /----------------------

\ / | |

\ / V phase Load S=V-single phase * I-single phase

\ / _________|_______|

| N

|

|

B

 

[Ship Wreck PE]

You can have a single phase load on two phase legs or one phase leg and the neutral and you do not need the sq root 3 because the load does not have three phase conductors.

 

[Redskinsdb21]

You can have a single phase load on two phase legs or one phase leg and the neutral and you do not need the sq root 3 because the load does not have three phase conductors.

This is an important concept and there are a couple similar problems in the NCEES practice exam

 

[EEpowerOK]

Thanks for clarifying.

3 phase delta or wye are both = sq rt 3 * V(line) * I (line) ?

 

[Ship Wreck PE]

Have you watched the free complex imaginary videos on their website??