Spin Up #4-62

[DetroitEE]

Is it just me or does the spin-up book incorrectly show how to sum three phase currents?

This problem (4-62) is the second time I have noticed this, and I want to know if the way I'm thinking about it is incorrect, or if the book is incorrect.

To find the solution to this problem, you have to find the current in the neutral. Neutral current is In = Ia + Ib + Ic.

You have to find current Ia, which is a single phase motor between A and B with a power factor of .75 lagging, 9.6kVA.

Current Ib is given at .25 < 10 degrees.

Current Ic must be calcuated. The load is a single phase motor with a power factor of .8 lagging, 30kVA.

The solution shows ((9,600/240) < -36.89 deg) + (.25 < -10 deg) + ((30000/240) < -41.41 deg)

The magnitudes are correct, but I do not understand why the angle of the Ib and Ic currents are not being adjusted based on the fact that voltages B and C are 120 and 240 degrees out of phase with A. The angle determined by the power factor is the angle between the current and the voltage, not just the absolute angle of the current, right?

Let me pose another situation to better explain my point: Same wye system, with (3) 10kVA, 240V single phase motors (connected AB, BC, CA), all having a power factor of .8. The neutral current should be 0, since the system is balanced right? If we were to solve this problem using the spin-up method, you would end up with this: ((10000/240) < -36.9 deg) + ((10000/240) < -36.9 deg) + ((10000/240) < -36.9 deg), which makes no sense at all, we know the neutral current is 0.

If anyone could shed some light on this it would be appreciated, thanks!

 

[spinup]

Is it just me or does the spin-up book incorrectly show how to sum three phase currents?

This problem (4-62) is the second time I have noticed this, and I want to know if the way I'm thinking about it is incorrect, or if the book is incorrect.

To find the solution to this problem, you have to find the current in the neutral. Neutral current is In = Ia + Ib + Ic.

You have to find current Ia, which is a single phase motor between A and B with a power factor of .75 lagging, 9.6kVA.

Current Ib is given at .25 < 10 degrees.

Current Ic must be calcuated. The load is a single phase motor with a power factor of .8 lagging, 30kVA.

The solution shows ((9,600/240) < -36.89 deg) + (.25 < -10 deg) + ((30000/240) < -41.41 deg)

The magnitudes are correct, but I do not understand why the angle of the Ib and Ic currents are not being adjusted based on the fact that voltages B and C are 120 and 240 degrees out of phase with A. The angle determined by the power factor is the angle between the current and the voltage, not just the absolute angle of the current, right?

Let me pose another situation to better explain my point: Same wye system, with (3) 10kVA, 240V single phase motors (connected AB, BC, CA), all having a power factor of .8. The neutral current should be 0, since the system is balanced right? If we were to solve this problem using the spin-up method, you would end up with this: ((10000/240) < -36.9 deg) + ((10000/240) < -36.9 deg) + ((10000/240) < -36.9 deg), which makes no sense at all, we know the neutral current is 0.

If anyone could shed some light on this it would be appreciated, thanks!

DetroitEE, thanks for the feedback. I will forward your question on.

Joan

 

[spinup]

DetroitEE you are correct, there appears to be an issue with 4-62. There are two other problems (3-17,3-18) that have related issues. There will be an updated errata with the solutions placed on www.spinupexams.com in the next few days.

The correct answers to these problems are:

4-62 = E

3-17 = E

3-18 = E

Thanks DetroitEE for bringing this to our attention.

Sorry for the inconvenience.

Joan

 

[lizzy]

DetroitEE you are correct, there appears to be an issue with 4-62. There are two other problems (3-17,3-18) that have related issues. There will be an updated errata with the solutions placed on www.spinupexams.com in the next few days.

The correct answers to these problems are:

4-62 = E

3-17 = E

3-18 = E

Thanks DetroitEE for bringing this to our attention.

Sorry for the inconvenience.

Joan

Have the errors in problems 4-62, 3-17 and 3-18 been corrected? The errata on your website does not list a revised solution. Can you please post the correct solution?

 

[spinup]

DetroitEE you are correct, there appears to be an issue with 4-62. There are two other problems (3-17,3-18) that have related issues. There will be an updated errata with the solutions placed on www.spinupexams.com in the next few days.

The correct answers to these problems are:

4-62 = E

3-17 = E

3-18 = E

Thanks DetroitEE for bringing this to our attention.

Sorry for the inconvenience.

Joan

Have the errors in problems 4-62, 3-17 and 3-18 been corrected? The errata on your website does not list a revised solution. Can you please post the correct solution?

Lizzy, it appears you have an old version (1st Edition, 1st Printing). The Errata on the website (spinupexams.com) for the 1st edition contains the corrections.

 

[lizzy]

DetroitEE you are correct, there appears to be an issue with 4-62. There are two other problems (3-17,3-18) that have related issues. There will be an updated errata with the solutions placed on www.spinupexams.com in the next few days.

The correct answers to these problems are:

4-62 = E

3-17 = E

3-18 = E

Thanks DetroitEE for bringing this to our attention.

Sorry for the inconvenience.

Joan

Have the errors in problems 4-62, 3-17 and 3-18 been corrected? The errata on your website does not list a revised solution. Can you please post the correct solution?

Lizzy, it appears you have an old version (1st Edition, 1st Printing). The Errata on the website (spinupexams.com) for the 1st edition contains the corrections.

Actually I have the second edition and my concerns are regarding the answers in the second edition. The equation is correct, but the results seem to be wrong.