NCEES problem 512

[BebeshKing PE]

Why they did not consider the power factor? Rcos(theta) +jXrsin(theta)?

is this because the formula above is for the effective impedance? And the problem is asking for the actual impedance?

that being said, is this mean that the chapter 9 table 9 are the actual impedance values of the wires?

thanks

 

[Orchid PE]

Yes. Table 9 contains the actual impedance values of the conductors.

Note 2 says:

Effective Z is defined as R cos(θ) + X sin(θ), where θ is the power factor angle of the circuit. Multiplying current by
effective impedance gives a good approximation for line-to-neutral voltage drop. Effective impedance values shown in
this table are valid only at 0.85 power factor.

Power factor of the circuit includes PF of voltage and current (load). In the problem, you were only given the PF of the load, so we cannot calculate the circuit PF. But this is not needed, since the effective Z is only used for an approximation of voltage drop. Actual voltage drop can still be calculated using Current*(actual R+jX)=Vdrop. It is mainly used when you are given the PF of the circuit and a voltage.

 

[DLD PE]

Pretty sure the NCEES mentions the 0.90 power factor of the  motor FLC as a red herring, don't you think?

 

[BebeshKing PE]

Yes. Table 9 contains the actual impedance values of the conductors.

Note 2 says:

Effective Z is defined as R cos(θ) + X sin(θ), where θ is the power factor angle of the circuit. Multiplying current by
effective impedance gives a good approximation for line-to-neutral voltage drop. Effective impedance values shown in
this table are valid only at 0.85 power factor.

Power factor of the circuit includes PF of voltage and current (load). In the problem, you were only given the PF of the load, so we cannot calculate the circuit PF. But this is not needed, since the effective Z is only used for an approximation of voltage drop. Actual voltage drop can still be calculated using Current*(actual R+jX)=Vdrop. It is mainly used when you are given the PF of the circuit and a voltage.

Thanks for the explanation. quick question though, is the circuit power factor angle is the angle between the "source voltage" and the current? 

 

[Sparky Bill PE]

Pretty sure the NCEES mentions the 0.90 power factor of the  motor FLC as a red herring, don't you think?

Wouldn't you need to know the PF is .9 AND NOT .85 so you know exactly which impedance's to use on the NEC table 9 chart? 

 

[Orchid PE]

Wouldn't you need to know the PF is .9 AND NOT .85 so you know exactly which impedance's to use on the NEC table 9 chart? 

Not for this particular question. Only conductor size, conductor material, and conduit material are needed to find R and X. PF is needed for effective Z, but effective Z is not needed for this question.

 

[Orchid PE]

Thanks for the explanation. quick question though, is the circuit power factor angle is the angle between the "source voltage" and the current? 

Correct.

 

[Sparky Bill PE]

Not for this particular question. Only conductor size, conductor material, and conduit material are needed to find R and X. PF is needed for effective Z, but effective Z is not needed for this question.

I'm still lost. If they question had a 0.85 PF you would still solve it simply looking over at the R and X values? 

 

[Orchid PE]

I'm still lost. If they question had a 0.85 PF you would still solve it simply looking over at the R and X values? 

Correct. Even if PF was 0.25, we look at the R and X values. PF comes into play with voltage drop.

 

[DLD PE]

The power factor in this problem is a red herring.

 

[SaltySteve PE]

This is going to sound dumb but can someone explain the logic behind throwing that 2 in the denominator. I get that there are 2 sets of wires but I'm not getting the connection of doubling the 1000 number.

Impedance of one wire = x ohm

Impedance of 2 wire = x ohm /2

Impedance of 3 wire = x ohm /3?

 

[DLD PE]

2 sets of conductors.

 

[Orchid PE]

Impedance is given per 1000ft. So Z is per 1000ft.

Total Z of the line is Z * 500 / 1000. But, since there are two (parallel) lines, it's like resistors in parallel. So with two lines, Z * Z / (Z + Z) = Z / 2.

So, (Z / 2) * 500 / 1000 == Z * 500 / (1000 * 2).

 

[Orchid PE]

Impedance of 3 wire = x ohm /3?

Impedance of 3 conductors is like 3 resistors in parallel. So yes.

 

[SaltySteve PE]

Impedance is given per 1000ft. So Z is per 1000ft.

Total Z of the line is Z * 500 / 1000. But, since there are two (parallel) lines, it's like resistors in parallel. So with two lines, Z * Z / (Z + Z) = Z / 2.

So, (Z / 2) * 500 / 1000 == Z * 500 / (1000 * 2).

That makes sense. Thanks.