ECN Forum Forums Code Related Discussion NEC & other Code issues Calculation rounding

Are there any formal rules as to rounding when doing calculations?

In a practice workbook, the calculations are rounded to the tenths. When I do the calculations precisely (to the thousandths) my fractional answer is below .5, but the book's fractional calculation is .5 exactly.

This changes the answer I get versus what the book says is the answer.

I don't see anything in the NEC that calculations must be rouded to the tenths before applying 220-3(b).

I think the amount of rounding up or down depends on the magnitude of the answer.
For example if an answer is 120.6 volts,it would be ok to round up to 121.If the answer is 0.555 amps,I would round it up to 0.56.I guess a lot depends on the required accuracy.
Chris

This is what Annex D in the 2002 NEC says:

The exact question is this--

Calculate the minimum size Type THW copper conductors required to serve the following motor loads from a 480/277, 3-phase, 4-wire service: Two 40-horsepower, 460 volt, 3-phase synchronous type, 90 percent P.F. and one 10-horsepower squirrel-cage, 460 volt, 3-phase motor at 100 percent P.F.

A. No. 3
B. No. 2
C. No. 1
D. No. 1/0

Steve, I'm a little confused as to where 220.3(B) comes in to play with motor feeders.

The answer to the question would be "B" #2

Table 430.150 shows 41 amps for your 40 hp
and 14 amps for the 10 hp.

41 amps with a 1.1 mutiplier = 45.1

largest motor x 125%
45.1 x 1.25 = 56.38

56.38 + 45.1 + 14 = 115.48

115 amps, .48 could be dropped, but, if you know the test is going to force you to round to the tenth, you would have to raise the .48 to .5 and take you to 116 amps meaning #1 THW.

Now, in reality if you converted 116 amps at 460 v to watts, you could then divide by your true voltage of 480 x 1.732 and get down to 111 amps.

Sorry about the rambling.


Roger



[This message has been edited by Roger (edited 02-21-2003).]

A motor with a 100% PF?
My friends would like to buy one...S

Er, synchronous {versus induction} motors can run at 1.0 or leading PF, but require a DC excitation source and a wound rotor.

Roger,
You used the 575 volt amp for the 40 hp. The 480 volt current is 52. Also what is the 1.1 multiplier for?
I come up with 1.25(52) + 52 + 14 = 131 amps. This will require a #1 THW copper feeder.
Don

Geeze, Bjarney.
The last synchronous motor I saw was 30 yrs. ago.(and then was 30 yrs old)
You must admit they're in not common use.
Giant Compressor.

Don, wouldn't we use the Sychronous-Type Unity Power Factor coloums and then apply the multiplier of the foot note?

Roger