3-phase load with different configuration

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hardworkingstiff

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Assume a 208Y/120 supply, 3 10-ohm resistors.

Scenario 1: The resistors are connected in a Delta configuration. What is the amperage per wire going back to the breaker?

Scenario 2: The resistors are connected in a Wye configuration (with no neutral pulled to the common point of the resistors). What is the amperage per wire going back to the breaker?

Thank you to all that answer.

I believe the answer for the Wye connection is 12-amps, but I'm unsure of how to calculate the Delta connection.
 
hardworkingstiff said:
Thanks money, that's what I came up with but was too unsure to post it because it's the same resistors connected in a different way with one heck of a current difference.
Click to expand...

It's the same ohm resistor, but its a different load when connected in the different configuration.

In the delta, you have 12,979VA (12,979VA/208V/1.732 = 36A)

In the wye, you have 4,320VA (4,320VA/208V/1.732 = 12A)
 
I know this is the long way around, but is this a viable way to look at the Delta connection.

1st: assuming the formulas below are valid,
A delta configuration has three terminals, A, B and C and three resistors, x, y and z. x goes between B and C, y between C and A, z between A and B.
The effective resistance between A and B is Z = 1/(1/z + 1/(x + y)).
The effective resistance between B and C is X = 1/(1/x + 1/(y + z)).
The effective resistance between C and A is Y = 1/(1/y + 1/(z + x)).

The resistance between any 2 points (AB, BC, or AC) can be calculated as:

1/(1/10+1/(10+10))
1/(1/10 + 1/20)
1/(.1+.05)
1/.15
6.67 Ohms

E/R=I, 208/6.67 = 31.18 amps. This is for just the relationship for 2 points (let's say A,B). Then if we look at A,C we do the same calculation.

So, 31.18*2 = 62.36. Since this is 3-phase, 62.36/1.732 = 36 amps on the conductor at point A, and since the resistors are all the same value, it will also be the same for points B and C.

Is the reasoning sound, or does it just happen to come up with the correct answer?
 
hardworkingstiff said:
Is the reasoning sound, or does it just happen to come up with the correct answer?
Click to expand...

I think it just happens to come up with the correct answer.

hardworkingstiff said:
A delta configuration has three terminals, A, B and C and three resistors, x, y and z. x goes between B and C, y between C and A, z between A and B.
The effective resistance between A and B is Z = 1/(1/z + 1/(x + y)).
The effective resistance between B and C is X = 1/(1/x + 1/(y + z)).
The effective resistance between C and A is Y = 1/(1/y + 1/(z + x)).
Click to expand...

If you connect a resistor x between B and C, then the resistance between B and C is simply x.
hardworkingstiff said:
The resistance between any 2 points (AB, BC, or AC) can be calculated as:

1/(1/10+1/(10+10))
1/(1/10 + 1/20)
1/(.1+.05)
1/.15
6.67 Ohms

E/R=I, 208/6.67 = 31.18 amps. This is for just the relationship for 2 points (let's say A,B). Then if we look at A,C we do the same calculation.
Click to expand...

Using 10 ohm resistors, the resistance between A-B, B-C & A-C will all be 10 ohm. You would therefore see 208/10 = 20.8Amps "between" any of the two points. The current on any leg will be the sum of the currents flowing in the load resistors. For instance, the current on A will be the sum of the current from B-A and the current from C-A, but don't forget that these currents have different angles (120 deg apart.)

With the 120deg difference, the current on any leg will be 20.8 * 1.732 = 36A.
 
hardworkingstiff said:
What about the path from B to A then to C, isn't that a parallel path to B to C?
Click to expand...
Yes.

Rather than me rethinking my answer, test your theory using unequal resistor values and convert to total VA. Say 8, 10, and 12. Then compare with simple totaling of VA between lines...

208??8=5408VA
208??10=4326.4VA
208??12=3605.3VA
Total VA = 13339.7VA
 
hardworkingstiff said:
Even if there are resistors connected between AB and BC? What about the path from B to A then to C, isn't that a parallel path to B to C?
Click to expand...

What about the "parallel path"? If you were going to figure out the parallel path resistance from B-C by including the path from B to A then to C, would you also have to apply the parallel path voltage (rather than applying 208V) by including the voltage from B to A then to C?

Its over complicating things. Start with three 10 ohm resistors each individually connected between A-B, B-C and C-A. For a single phase, 208V load you will see 20.8Amps flowing in the conductor from A, through the resistor and in the conductor to B. You will see the same with the load from B-C and with the load from C-A. If you put an ammeter on both A conductors, however, you would see 36A.

Now connect interconnect each A, B and C point at the load to form a delta, and remove the extra conductor from A, B and C. You will still see 20.8A flowing in each resistor, and you will still see 36A in each supply conductor.
 
david luchini said:
I think it just happens to come up with the correct answer.
Click to expand...

One of the handy things for calculating the wye current and power is to look at an equivalent circuit:

Since the resistors are all the same value, and the phase voltages are assumed to be equal, you can calculate that the voltage at the neutral point of the wye network is zero even though there is no physical neutral connection to that point.
Since the common point voltage is zero, the current in the three resistors will not be changed by connecting the common point to the real neutral. And once you have done that, you have simply three resistors, each connected from 120 volts to zero volts. Phase angle now becomes unimportant.

You could also work out the currents in the three resistors by considering each resistor as being in series with the parallel network composed of the other two resistors, each connected to a different phase, but why bother. :)
 
Reading some more on the electronic highway, I'm of the opinion that if I really want to understand this, I'm going to have to do some extensive reading on material that I don't have a good enough foundation to truly understand.

I want to thank Smart$ and David for taking the time to try to help me.
 
hardworkingstiff said:
I know this is the long way around, but is this a viable way to look at the Delta connection.

1st: assuming the formulas below are valid,
A delta configuration has three terminals, A, B and C and three resistors, x, y and z. x goes between B and C, y between C and A, z between A and B.
The effective resistance between A and B is Z = 1/(1/z + 1/(x + y)).
The effective resistance between B and C is X = 1/(1/x + 1/(y + z)).
The effective resistance between C and A is Y = 1/(1/y + 1/(z + x)).

The resistance between any 2 points (AB, BC, or AC) can be calculated as:

1/(1/10+1/(10+10))
1/(1/10 + 1/20)
1/(.1+.05)
1/.15
6.67 Ohms

E/R=I, 208/6.67 = 31.18 amps. This is for just the relationship for 2 points (let's say A,B). Then if we look at A,C we do the same calculation.

So, 31.18*2 = 62.36. Since this is 3-phase, 62.36/1.732 = 36 amps on the conductor at point A, and since the resistors are all the same value, it will also be the same for points B and C.

Is the reasoning sound, or does it just happen to come up with the correct answer?
Click to expand...

I think it's convoluted and not sound.
In delta each resistor has 208V across it so the current in it is 20.8A. The current in the line is sqrt(3)*20.8 or 36A.
In star each resistor has 120V across it so the current is 12A.

It's no more complicated than that. The 6.67ohms is of no relevance.
 
hardworkingstiff said:
Thanks money, that's what I came up with but was too unsure to post it because it's the same resistors connected in a different way with one heck of a current difference.
Click to expand...

When they're connected in delta, each resistor gets line-to-line voltage. In wye, they get ph-to-ph/(root of 3).

Older motor starters sometimes use contacts to connect motors to line in wye configuration, then rearrange to delta to reduce starting current.
 
GoldDigger said:
One of the handy things for calculating the wye current and power is to look at an equivalent circuit:
Click to expand...

That's great...but he's trying to calculate a delta circuit, not a wye circuit.

If he wanted to use an equivalent circuit, he could use a d-y transform to change his three 10 ohm delta connected resistors to three 3.33 ohm wye-connected resistors. Then he'd get his line current by: 120V/3.33ohm=36Amps.
 
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