NCEES problem 512

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BebeshKing PE

BebeshKing PE

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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

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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.

 
Last edited by a moderator:
Chattaneer PE said:
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.
Click to expand...
Thanks for the explanation. quick question though, is the circuit power factor angle is the angle between the "source voltage" and the current? 

 
MEtoEE said:
Pretty sure the NCEES mentions the 0.90 power factor of the  motor FLC as a red herring, don't you think?
Click to expand...
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? 

 
SparkyBill said:
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? 
Click to expand...
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.

 
Chattaneer PE said:
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.
Click to expand...
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? 

 
SparkyBill said:
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? 
Click to expand...
Correct. Even if PF was 0.25, we look at the R and X values. PF comes into play with voltage drop.

 
Last edited by a moderator:
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?

 
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).

 
Chattaneer PE said:
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).
Click to expand...
That makes sense. Thanks.

 

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