Art. 430.24 calculation question

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Benaround, I see (hear) what you are saying. I didn't balance the load. Now I think I have it.


Code:
Leg                A           B         C
---------------------------------------------
M1 @ 125%         57.75     57.75     57.75
M2                46.20     46.20     46.20
M3                46.20     46.20     46.20
M4                18.70     18.70
M5                          18.70     18.70
M6                18.70               18.70
M7                16.00
M8                          16.00
M9                                    16.00
--------------------------------------------
Total Amps       203.55     203.55    203.55

204 on the 75C column is 4/0.

Thanks for the lesson.

Roy
 
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benaround said:
VA is the only way engineer's will do it.

I tried doing the VA calculations, but I came up with 177A. Could you please explain or show the proper way to convert to VA and then back to A?

Thanks.
 
46.2 + 46.2 + 57.7 = 150.1
18.7 x sqrt(3) = 32.4
16

150.1 + 32.4 + 16 = 198.5

Originally I used 48.3A for the 15hp motor (and got 205.4A)

After that I used 46.2A from the 208V column, however I think this is incorrect.

Motors on 480V systems are rated at 460V
Motors on 240V systems are rated at 230V
Motors on 208V systems are rated at 200V

We should use the 200V column in Table 430.250 for our calculations, not the 208V column.
That being said, what voltage system is the 208V column for?
This applies to the single-phase table too.

Using the 200V columns:
48.3 + 48.3 + 60.4 = 157
19.6 x sqrt(3) = 33.9
16

157 + 33.9 + 16 = 206.9A
 
wasasparky said:
(2) 18.7A 208V 1ph loads on A-B an B-C

What is the current on phase B?
Hint: It is not 37.4

The hint is not helping me at all. I have no idea what you're trying to get at, but I do want to learn. I tried the 2ph-4w forumula to get VA and then use the 3ph formula to go back to A, but I came up with 43A, which I can't imagine is right. Then I used the 1ph formula to get VA and then use the 3ph formula to get back to A, and it gives me 21.6A, which I can't imagine is right either.

Attempt 1
E x I x 2 = VA
208 x 18.7 = 7779.2

VA / E x 1.73 = I
7779.2 / 360 = 21.6 per B phase = 43.2



Attempt 2
E x I = VA
208 x 18.7 = 3889.6

VA / E x 1.73 = I
3889.6 / 360 = 10.8 per B phase = 21.6


I don't understand the theory, and I don't know which formulas to use. Throw me a bone here. :)
 
wasasparky said:
46.2 + 46.2 + 57.7 = 150.1
18.7 x sqrt(3) = 32.4
16

150.1 + 32.4 + 16 = 198.5

Originally I used 48.3A for the 15hp motor (and got 205.4A)

After that I used 46.2A from the 208V column, however I think this is incorrect.

Motors on 480V systems are rated at 460V
Motors on 240V systems are rated at 230V
Motors on 208V systems are rated at 200V

We should use the 200V column in Table 430.250 for our calculations, not the 208V column.
That being said, what voltage system is the 208V column for?
This applies to the single-phase table too.

Using the 200V columns:
48.3 + 48.3 + 60.4 = 157
19.6 x sqrt(3) = 33.9
16

157 + 33.9 + 16 = 206.9A

OK, I see what you're doing, but I don't know why you're doing it, or the theory behind it. Why, where, etc.? Thanks.
 
Let's consider two 208V 1phase loads circuited A-B, B-C.

The two loads are out of phase by 120 degrees (their max values don't peak at the same time)

This means you cannot just add their currents together on B phase (so something less than two times their individual currents)

The number happens to be the square root of three (approx 1.732)

Don't have time to show all the math, but there are a lot of resources out there...
 
wasasparky said:
We should use the 200V column in Table 430.250 for our calculations, not the 208V column.


If a motor is rated at 208 volts and the column lists ampacities for 208 volts why wouldn't you use the 208 volt ampacity listed in the table?
 
infinity said:
If a motor is rated at 208 volts and the column lists ampacities for 208 volts why wouldn't you use the 208 volt ampacity listed in the table?

I just started a new thread...
 
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