Help a newbie out on this 3Ph to Single phase questions - please!

[downtownfool]

Ok, here are the facts:

I have 3PH 480 coming in from the power company (kVA is irrelevant at this point).

I want to install the following:

Three (3) single phase 480 to 240/120 25kVA transformers
One (1) single phase 480 to 240/120 10kVA transformer
One (1) 15HP 480V 3PH motor

According to Ugly's Book which references NEC, a single phase 25kVA has a FLA of 52.1, a single phase 10kVA transformer has a FLA of 20.8, and a 3PH 15HP motor has an FLA of 24.15 (FLA's obtained from Ugly's Book).


They will be wired as follows:

1phase 1phase 1phase 1phase 3phase
25kVA 25kVA 10kVA 25kVA 15HP
Phase - A X X X X

Phase - B X X X

Phase - C X X X X


Does this mean EACH phase will see the following amperage?

Phase - A - 52.1 + 20.8 + 52.1 + 24.15 = 149.15A

Phase - B - 52.1 + 52.1 + 24.15 = 149.15A = 128.35A

Phase - C - 52.1 + 52.1 + 20.8 +24.15 = 149.15A

For a given 480V to 240/120 single phase transformer...will BOTH primary 480V legs ( say A and B for the first 25kVA listed above) see the same current (52.1 amps assuming the transformer is 100% loaded)?

OR

will BOTH primary 480V legs see 26.05A (half of the 52.1 amps assuming the transformer is 100% loaded)?

 

[augie47]

(bump)

 

[cadpoint]

Weres a road map for this? You could get a panel put in six throws then transform as required. This is better if all things are close, well maybe. :roll:

I frankly also thought of a fused disconnects and a throuf, then disconnents.

Ok yours numbers just presented wrong as I read them.

What gets me is your accounting for a summary total for/of transforming on a straight in service. IE you added
backing in the 480 motor as transformed.(I 've read it wrong)

This is the loading of your service! There is five items at three points your math doesn't look right.

 

[jim dungar]

For all intents your numbers look correct:

TX1 connected AB
TX2 connected BC
TX3 connected CA
tx4 connected AC
M1 connected ABC

A = 52.1+0+52.1+20.8+24.15
B = 52.1+52.1+0+0+24.15
C = 0+ 52.1+52.1+20.8+24.15

 

[Smart $]

...
For a given 480V to 240/120 single phase transformer...will BOTH primary 480V legs ( say A and B for the first 25kVA listed above) see the same current (52.1 amps assuming the transformer is 100% loaded)?

OR

will BOTH primary 480V legs see 26.05A (half of the 52.1 amps assuming the transformer is 100% loaded)?

Each transformer primary wire will see the same current as its counterpart.

The currents at the service will not sum up arithmetically as you have shown.. because the transformers' currents are out of phase. To get an approximation, simply add up the kVA values and divide by 480 then divide by sqrt(3). However, the result would be slightly low on two lines because of the imbalance the 10kVA TX presents. An approximation of the higher current would be to pretend there are two more 10kVA TX's balancing out the system.

 

[Smart $]

For all intents your numbers look correct:

TX1 connected AB
TX2 connected BC
TX3 connected CA
tx4 connected AC
M1 connected ABC

A = 52.1+0+52.1+20.8+24.15
B = 52.1+52.1+0+0+24.15
C = 0+ 52.1+52.1+20.8+24.15

Yeah... but they don't add up arithmetically.

 

[david luchini]

Yeah... but they don't add up arithmetically.

I agree...adding the single phase loads and then multiplying those by 0.866 (cos(30)) before adding the three phase load would provide a pretty close approximation:

A = (52.1+0+52.1+20.8)*0.866+24.15 = 132.4
B =(52.1+52.1+0+0)*0.866+24.15 = 114.4
C =(0+ 52.1+52.1+20.8)*0.866+24.15 = 132.4

 

[jim dungar]

Yeah... but they don't add up arithmetically.

If you want to be the expert, show the math instead of just snipping.

For the purpose of the OP there is nothing wrong with my explanation. They asked for an "about this amount" answer.

An easy explanation (no trig required) would be: the (3) 25kVA single phase transformers will present as (1) 75kVA three phase unit at 90A per leg. Plus the motor at 24.15 gives 114.15 per leg. Now add the (1) 10kVA unit to A and C wand you get about 135A per leg.

 

[Smart $]

If you want to be the expert, show the math instead of just snipping.

Getting a bit testy, are we... :huh:

 

[kwired]

downtownfool, lets start with just saying you have a 10 amp load connected from A to B and another 10 amp load connected from C to B. Current measured on A would be 10 amps, current measured on C would be 10 amps, but the current measured on B is not double the value of A or C. Because of phase angles it is sq. root of 3 (1.73) times A or C. So in that example B would be carrying 17.3 amps. Now add another 10 amp load between A and C and measured current on all three phases will be 17.3.

So to go back to your installation, balance the current across all three phases a much as possible (which you did) and add up all the 'balanced' load between two phases and multiply by 1.73 and that will give you the balanced current on all three phases. Then add the unbalanced load to the two phases it is connected to to get a net for each phase.

 

[mivey]

For a given 480V to 240/120 single phase transformer...will BOTH primary 480V legs ( say A and B for the first 25kVA listed above) see the same current (52.1 amps assuming the transformer is 100% loaded)?

OR

will BOTH primary 480V legs see 26.05A (half of the 52.1 amps assuming the transformer is 100% loaded)?

Each transformer primary wire will see the same current as its counterpart.

But at the transformation ratio. The 480 volt primary will see 1/2 of the current we see at the 240 volt secondary. As Smart noted, kVA is an easy way to keep up with primary vs secondary especially if we want the max current as he described.

 

[mivey]

Yeah... but they don't add up arithmetically.

If you want to be the expert, show the math instead of just snipping.

He did describe how to calculate it like an expert. Read his post #6.

For the purpose of the OP there is nothing wrong with my explanation. They asked for an "about this amount" answer.

Other than it was wrong. At best you could say he identified which currents were involved with which transformer. But the OP was asking about the numbers and the numbers were not correct. You did follow up with the correct answer so no harm done.

An easy explanation (no trig required) would be: the (3) 25kVA single phase transformers will present as (1) 75kVA three phase unit at 90A per leg. Plus the motor at 24.15 gives 114.15 per leg. Now add the (1) 10kVA unit to A and C wand you get about 135A per leg.

And that is answer the OP needed.

Getting a bit testy, are we... :huh:

Seemed like it to me also. Just one of those days I suppose.

 

[jim dungar]

He did describe how to calculate it like an expert. Read his post #6.

Actually it is an all but a useless post. No instruction is offered.

But the OP was asking about the numbers and the numbers were not correct.

Go back and read the original post.

The OP was asking about a method based on made up numbers. He was not looking for specific values, he was questioning how the currents might divide up. Pretty much he was looking for a sanity check. His method certainly is usable for estimating purposes.

 

[mivey]

Actually it is an all but a useless post. No instruction is offered.

Oops. I meant Smart's post #5 where he posted:

To get an approximation, simply add up the kVA values and divide by 480 then divide by sqrt(3). However, the result would be slightly low on two lines because of the imbalance the 10kVA TX presents. An approximation of the higher current would be to pretend there are two more 10kVA TX's balancing out the system.

However, he meant use the kVA sum with kV (0.480) or the VA with volts (480). An example calc would have made it clearer.

The OP was asking about a method based on made up numbers. He was not looking for specific values, he was questioning how the currents might divide up. Pretty much he was looking for a sanity check. His method certainly is usable for estimating purposes.

I will disagree. They may be good for identifying which currents are involved, but not good for estimating.

I'll stick with your second answer.