Three Phase Delta Load

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fifty60

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USA
The calculation sequence (formula?) I use for 3 phase delta loads is to take the total KW and divide by line voltage and then divide that by the square root of three to get the line current.

I then have to divide by the square root of three to get the phase current.

I am confused by the first division by the square root of three. I'll look at it another way, starting at each load. Say the overall KW in my delta connected load is 30KW. The KW at each load will be 30/3= 10KW. I then take 10KW divide that by the line voltage (480V) and I get 20.83A. This would be the current through each load of the 3 phase delta. I multiply that by the square root of 3 and I get a line current of 36.08A.

Now is where I get confused. If I multiply it one more time by the square root of 3, I get 62.5A. This is the same amperage I get if I divide the entire load by 30000W by 480V = 62.5.

In summary, I understand the Line and Phase calculation but I do not understand why I have to divide 62.5A by the square root of three to get the line voltage, and then divide by the square root of three again to get the phase current. What is the initial 62.5A divided by the square root of three?
 

david luchini

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Now is where I get confused. If I multiply it one more time by the square root of 3, I get 62.5A. This is the same amperage I get if I divide the entire load by 30000W by 480V = 62.5.

What is the initial 62.5A divided by the square root of three?

62.5A is not relevant to the circuit you described. You took a 10kW, 480V single phase load and got a load current of 20.83. You then multiplied that by 3, and got an answer of 62.5. IF you had three 10kw single phase loads connected to the same 480V single phase source the load current would be 62.5, but you have a 480v three phase source and a connecting the loads in a balanced delta.
 

fifty60

Senior Member
Location
USA
Thanks. I understand that it is not relevent, but I am trying to understand why the first division by the square root of three gives me the correct line current.
 

GoldDigger

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Placerville, CA, USA
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Look at the first division by sqrt3 from a different angle and it might make more sense:
You are not really dividing by sqrt3. You are dividing by 3 because your load is spread over 3 phases and then multiplying by sqrt3 to get the line current because of the way the two phase currents add as vectors on each line. (Multiply by (sqrt3)/2 and then double it.)
It is just a coincidence that dividing by 3 and then multiplying by sqrt3 gives the same result as dividing by sqrt3. :)
That analysis works for me anyway....

Tapatalk!
 

iceworm

Curmudgeon still using printed IEEE Color Books
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North of the 65 parallel
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EE (Field - as little design as possible)
Thanks. I understand that it is not relevent, but I am trying to understand why the first division by the square root of three gives me the correct line current.
Adding to david's and GD's responses:
I think you got most all of it. It is just a little more algebra. Look at the power equations from the other direction. The load is just a box. You can't tell if it is delta or wye. You can see the 3 phase voltage, you can see the 3 phase line currents.

Now, consider just the basic balanced 3 phase power equation - doesn't matter if the load is Wye or Delta.

P = E x I x Sqrt(3) or I = P/E/sqrt(3)
where I is what you called the "line current" and
E is the line to line voltage​

This is true for any balance 3 phase load. For a 30kw, 480V load, this will give you the line current of 36.1A - as you calculated. From here, you showed how to get the delta leg current

So where did the current values lining up concerning the x3 calcs and and the x sqrt(3) calcs. If i tell you, you wil do a :slaphead: cause I think you got it already

Well, for one, you took the total power (all three delta legs), divided by 3 to get the power in each leg, divided by 480 to get the leg current, then multiplied the leg current by sqrt(3) to get the line current.

If you divide by 3, then multiply by sqrt(3), that is the same as just dividing by sqrt(3).

So, to get the line current, take the total power, divide by 480, divide by sqrt(3).

hope this helped

ice
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140518-1318 EDT

fifty60:

Another way to view the problem.

First, assume the load is resistive. You did this, but did not say so.

Second, assume the load is a balanced wye of 30 kW. Thus, each leg is 10 kW. With the wye assumption, then it is easy to see that the line to wye mid-point voltage is line-to-line voltage divided by 1.732 or in this case 277 V. The wye line to wye mid-point voltage is the voltage across the 10 kW resistive load.

Third, it is obvious that for this resistive load the line current is in phase with the line to wye midpoint voltage. Thus, the simple power equation P = V * I applies. And Iline = 10,000/277 = 36.1 A.

.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Umm, I think you are simply seeing a mathematical identity.

(X / 3) * sqrt(3) = X / sqrt(3). Sometimes stated as 1/sqrt(3) = sqrt(3)/3

For a balanced system, the power in each load is total power divided by 3.

For a balanced delta system, the current in each line is the single phase current * sqrt(3). Total power / 3 / voltage * sqrt(3) gives line current.

Now re-arrange the math: total power / voltage / 3 * sqrt(3) = total power / voltage / sqrt (3).

-Jon

(which is pretty much what the others said....)
 
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