Code question from PE Exam Review Book

[jcook980]

Here is a very basic Code question from a well-known, highly regarded Electrical PE exam review book. I throughly and completely disagree with the published answer. Before I reveal their answer in a future post, you may wish to work through the problem yourself:

A 200 hp, 460V, three-phase, continuous duty, squirrel cage induction motor is designed to operate at a 0.85 power factor. The supplying branch circuit is 30.5m long and utilizes THHN copper conductors in a steel conduit. The voltage drop is limited to 3% in the branch circuit, leaving 2% for the feeder circuit. NEC Table 430-150 and Table 9 list the applicable data. What size cable should be used to meet the voltage drop and current requirements of the load?

What is your answer?

[ December 08, 2003, 12:58 PM: Message edited by: jcook980 ]

 

[charlie b]

Re: Code question from PE Exam Review Book

Is there a typo in your question? Table 430.150 does not address motors rated for a 260 volt system. Did you mean 460V?

Also, are you sure this came from a PE exam book? I?ve never known the PE exam to get into NEC issues. Just curious.

 

[tom baker]

Re: Code question from PE Exam Review Book

Interesing question.

THHN copper conductors .... What size cable?
Do they want cable in conduit or is this another typo?

 

[jcook980]

Re: Code question from PE Exam Review Book

Sorry, typo, 460V. The rest is quoted accurately. Assume they did not really mean "cable" as we mean it. THHN conductors in steel conduit.

 

[jim dungar]

Re: Code question from PE Exam Review Book

I have always liked the voltage drop table from the Shawmut Electircal Handbook. These tables take into account the power factor of the circuit as well as the magnetic properties of the conduit.

http://www.ferrazshawmut.com/resources/electricalhandbook.html

 

[bennie]

Re: Code question from PE Exam Review Book

I get 250 Kcmil, what is the test answer?

 

[electricman2]

Re: Code question from PE Exam Review Book

How about 350kcmil? I assumed it was continuous and applied the 125%.

 

[bennie]

Re: Code question from PE Exam Review Book

I think you are correct electricman2.

 

[ron]

Re: Code question from PE Exam Review Book

are you sure this came from a PE exam book? I?ve never known the PE exam to get into NEC issues.

There is a section of the PE exam (second of two) that are specific questions regarding application of the NEC.

 

[steve66]

Re: Code question from PE Exam Review Book

Charlie: I took my Electrical PE exam only about 6 months ago. It may have changed since you took yours. Everyone taking the electrical PE exam takes the same morning test that covers basics common to all fields. The examinee then selects from one of 3 specialities for the afternoon exam. This could be either (I am quoting the categories from memory here):
1. Electrical power and distribution (which does include questions on the NEC).
2. Communications
3. Comuputers and programming.

Incidently, there is also now an industrial controls PE exam.

Steve

 

[charlie b]

Re: Code question from PE Exam Review Book

Steve66: I took my PE exam in 1987. At that time, we could select any four questions for the morning exam, and select any three for the afternoon (everyone had to do the one mandatory ?engineering economics? question). There were no code questions in the exam book. But there was no restriction that you stay within your own discipline. So two of the seven questions that I selected were in the mechanical area, and one was in fluid dynamics.
- - - - - - - - - - - - - - - - - -
Here?s my approach to the question.
Full load amps (from Table 430.150) is 240. Multiply by 125% and you get 300. Table 310.16 tells me that a 350 MCM will handle the 300 amps. Table 9 (?Effective Z at .85 pf for uncoated copper wires in steel conduit) gives me a resistance of 0.197 ohms per kilometer.

Voltage drop for three phase is found from:
VD = 1.732 x (R x L x I) / 1000
VD = 1.732 x (0.197 x 30.5 x 240) / 1000
(Note that you use the actual current of 240 amps, not the ?biggie sized? current of 300 amps ? not the 125% of 240 amps)
VD = 2.50 volts
VD = 0.5% of a 460 volt system

Since the voltage drop on a 350 MCM is within acceptable limits, I conclude that a 350 MCM is an acceptable size conductor for this application. So I agree with electricman2 and Bennie.

 

[jcook980]

Re: Code question from PE Exam Review Book

I, too, feel that 350kcmil is the appropriate size.

However...

The short answer: 4/0

The long answer:

The key to determining the proper size conductors is calculating the maximum allowable effective impedance. The effective impedance is calculated from

Z0.85 = VDln/(circuit length/1000 ft)*(load in A) = VDln/(L/1000)* Iload

NEC Art. 210-19(a), fine print note (FPN) 4, recommends that conductors be sized to allow for a 3% voltage drop, with no more than 5% in the branch and feeder circuits combined. (NEC Art. 215-2(d), FPN 2, recommends the same for a feeder circuit.) The branch-circuit voltage drop allowed was given as 3%. The voltage drop, line-to-line, is

VDll = 0.03 * 460 = 13.8V

The voltage drop, line-to-neutral is

VDll = SQRT(3*VDln)
VDln = VDll/SQRT(3) = 13.8V/1.732 = 7.9674V [II]

The circuit length is 30.5m. Converting US units, the circuit length is

L = 30.5m * 1ft/.3048m = 100.06 ft [III]

The circuit load on the conductors must be a minimum of 125% of the continuous load per NEC Art. 210.19(a). This requirement is echoed in NEC Art. 430-22(a) for a motor, which states that the supplying conductors must be rated for 125% of the motor's full-load rating. NEC Art. 430-6(a)(1) requires that the ampere ratings be determined from a series of tables unless both the horsepower rating and the ampere rating are given, in which case the nameplate ampere rating is used. The applicable table for three-phase alternating current motors is NEC Table 430-150. Using Table 430-150, the 200hp, 460V motor has a full-load ampere rating of 240A. Thus, the conductors must be sized for

Iload = 1.25 * Irated = 1.25 * 240 = 300 [IV]

Substituting Eqs. II, III and IV into Eq. I gives

Z0.85 = 7.9674/((100.06/1000) * 300) [V]
= 0.222ohm/1000ft

The information calculated in Eq. V is now used to determin the conductor size using NEC Ch. 9, Table 9. The table is given in ohms-to-neutral per 1000ft, hence the use of the line-to-neutral voltages and the per-1000ft reference. Also, the reference power factor is 0.85. Should this not be the case, NEC Ch. 9, Table 9 provides the method for calculating the effective impedance. The effective impedance must be 0.266ohm/1000ft or less. For Z0.85 for uncoated copper wires in steel conduit, a cable of AWG 3 or larger is required.

The current capability is found from

P = 1.732 * I * V * PF

I = P / (1.732 * V * PF)
I = (200hp * 745.7W/HP)/(1.732 * 460V * .85)
I = 220A

The cable size required is either AWG 3/0 or 4/0, depending on the type of cable used, per NEC Table 310-16.

The answer is 4/0.

Six-Minute Solutions for Electrical and Computer PE Exam Problems John A. Camara, PE

Comments?

[ December 08, 2003, 12:56 PM: Message edited by: jcook980 ]

 

[steve66]

Re: Code question from PE Exam Review Book

Although the author even mentions the wire must be sized for 300A, tt looks like he got wrapped up in the voltage drop calculation and forgot about ampacity. I used books by this author for my exam review. I found them to be generally good, but I did find a lot of errors similar to this one.

Professional Publications (who prints this book) is very good about providing corrections to their books. I checked their website, and they do have a correction to this problem, but it seems like they still have the wrong answer. I think they would appreciate it if you let them know.

PPI

 

[charlie b]

Re: Code question from PE Exam Review Book

The approach used in ?The Book? is similar to my approach, in that it first determines a current value (upon which to base the conductor size), and then cross-checks the selected conductor size for the acceptability of voltage drop. My cross-check was a calculation of actual voltage drop. ?The Book? calculated a minimum size of cable to limit the voltage drop to a given value. Both cross-check methods are equally acceptable. My way is simpler and more straight forward.

However, ?The Book? made three errors.

(1) It calculated current on the basis of the horsepower rating. The NEC does not allow this approach. Article 430.22(A) requires that the conductors be sized on the basis of 125% of the current that is determined from 430.6(A)(1). That article, in turn, says that you must use the tables, and not use the nameplate rating.

(2) ?The Book? failed to account for the motor efficiency, when it calculated amps from horsepower. If you divide by a 90% efficiency factor (a reasonable value), ?The Book? answer of 220 amps becomes 244 amps, a value that is close to the NEC value of 240 amps.

(3) ?The Book? failed to account for the 125% requirement of NEC 430.22(A). I find it interesting that ?The Book? mentioned, and used, the 125% (i.e., to get the 300 amp value for current). But it used that value only in the cross-check for voltage drop, and then ignored the 125% in its final answer.

 

[bob]

Re: Code question from PE Exam Review Book

What size cable should be used to meet the voltage drop and current requirements of the load?

You actually have 2 problems to solve. First lets get the correct size for the 200hp motor. As was stated in the solution the 200 hp motor has a FLA
of 240 amps. Since this is continuous load we need to use the 1.25 factor. 240a x 1.25 = 300 a.
Form table 310.16 this load requires 350 thhn at 75C.
To caculate the VD, we need the impedance of the cable at the given .85 power factor. The given solution

Z0.85 = 7.9674/((100.06/1000) * 300) [V]
= 0.222ohm/1000ft

is incorrect. You would not use 300 amps but the actual FLA of 240 amps. No need to include the 1.25 Factor in this caculation. WE have already determined that the minimum size for load is #350 cu. The information obtained for the following caculation is incorrect also.

The current capability is found from

P = 1.732 * I * V * PF

I = P / (1.732 * V * PF)
I = (200hp * 745.7W/HP)/(1.732 * 460V * .85)
I = 220A

The cable size required is either AWG 3/0 or 4/0, depending on the type of cable used, per NEC Table 310-16.

The answer is 4/0.

Unless Mr. Camera has discoverd a 100% motor the equasion should be assiming 80 % efficiency
P = 1.732 * I * V * PF

I = P / (1.732 * V * PF)
I = (200hp * 745.7W/HP)/(1.732 * 460V * .85 * .80)
I = 234 amps.

 

[jcook980]

Re: Code question from PE Exam Review Book

I see I wasn't the only one to spot the bad assumptions!

Like steve66 suggested, I checked the on-line errata from the publisher and I was surprised to see that there was a correction made to this problem, except the correction was incorrect. I, too, believe the author got so worked up about the voltage drop calculation that he forgot to keep the conductor ampacity high enough for 125% of the full load current.

charlie b described exactly the way I would have solved the problem. I also spotted the assumption that the motor was 100% efficient. But, how about the assumption that the power factor was going to be .85? The question clearly indicated that it was the motor design PF, not the actual system PF. How many motors are precisely loaded to their design load? Anything less than full-load and the power factor gets worse.

I guess what really bothers me is that a young electrical engineer would be dangerously led astray about how to size motor conductors after reading this. Another electrical engineer (not licensed, thank goodness) told me recently that motor conductors need to be sized to six times the full-load current so they wouldn't burn up when the motor was started!

Frankly, this sort of stuff scares me...