three phase calculation

[Larry Bohn]

This question always leads to a "discussion" in class about how to interpret and solve this problem.
"If you are given the line-to-line voltage to be applied to a three phase resistance heater, and given the measured resistance between any two of the three terminals on that heater, the amperage on each line can be calculated using ohms law without using 1.732 as a factor."
If the information above is all that is given, this seems correct. If the resistance is measured after any connections are made internally, either Delta or Wye, it shouldn't matter, and we're just going to calculate current flow.
What do you think?

 

[Dennis Alwon]

I am out of my league on this but I will try. I agree that the 1.73 is not needed for the calculation of the amperage however I do believe it is needed for the calculation of the wattage. I don't understand why.

Get ready cause all the ee will be jumping in soon. lol

 

[wwhitney]

Interesting question, and it prompted me to work out one case. Since the answer scales with resistance R and voltage V in the obvious way, I'll just pick convenient values to work it out:

Say we have (3) 1 ohm resistors in a delta configuration. If we measure the resistance between any two terminals, it's 2/3 ohm, as we have a 1 ohm resistor in parallel with a 2 ohm resistor.

If we apply 1 volt L-L 3 phase to this delta configuration, then each resistor sees a current of 1 amp. In the line conductors, these load currents add in the usual way for 3 phase, and so the line currents are going to be sqrt(3) amps.

So no, we don't have "L-L voltage" = "Line Current" * "Measured Resistance Between 2 Terminals." That would be 1 = sqrt(3) * 2/3, which is not true.

The discrepancy for wye-connected resistors works out to be the same factor, which I checked with a similar computation. I expect there must be some principle one could apply to conclude the computation has to work out that way.

Cheers, Wayne

 

[ggunn]

Interesting question, and it prompted me to work out one case. Since the answer scales with resistance R and voltage V in the obvious way, I'll just pick convenient values to work it out:

Say we have (3) 1 ohm resistors in a delta configuration. If we measure the resistance between any two terminals, it's 2/3 ohm, as we have a 1 ohm resistor in parallel with a 2 ohm resistor.

If we apply 1 volt L-L 3 phase to this delta configuration, then each resistor sees a current of 1 amp. In the line conductors, these load currents add in the usual way for 3 phase, and so the line currents are going to be sqrt(3) amps.

So no, we don't have "L-L voltage" = "Line Current" * "Measured Resistance Between 2 Terminals." That would be 1 = sqrt(3) * 2/3, which is not true.

The discrepancy for wye-connected resistors works out to be the same factor, which I checked with a similar computation. I expect there must be some principle one could apply to conclude the computation has to work out that way.

Cheers, Wayne

With wye connected resistors you just use the line to neutral voltage and the resistances V=IR. That's how I do Vd calculations; the sqrt(3) terms cancel.

 

[Larry Bohn]

I see your logic and have gone down that path myself and couldn't make anything work. That being said, there isn't a way to solve this problem given a three phase voltage and a resistance between two wires? It seems logical to me, given a three phase heater, I could measure the resistance between any two wires, apply a three phase voltage, and be able to calculate the wattage of the heater using some type of multiplier. I've gone at this problem three different ways, gotten three different answers, talked myself into all of them and then talked myself out of all of them.
Given the three phase voltage drop formula, could it be applied to this circuit, calculating amperage and using the three phase voltage as the voltage drop? Now wer're back to single phase which doesn't seem to work

 

[wwhitney]

With wye connected resistors you just use the line to neutral voltage and the resistances V=IR. That's how I do Vd calculations; the sqrt(3) terms cancel.

In the context of the OP, they don't cancel, because (a) the OP specifies using the L-L voltage, which is a sqrt(3) times the L-N voltage and (b) the OP specifies measuring the resistance between two terminals, which in a wye configuration of resistors of resistance R gives you a measurement of 2 * R. Together these give you the ratio of 2/sqrt(3), which is the same ratio as in my initial post.

Cheers, Wayne

 

[wwhitney]

I see your logic and have gone down that path myself and couldn't make anything work. That being said, there isn't a way to solve this problem given a three phase voltage and a resistance between two wires?

There's a difference between (a) a fixed resistor R supplied L-L and (b) a configuration of resistors with 3 terminals, and a measured resistance of R' between any two terminals.

It seems logical to me, given a three phase heater, I could measure the resistance between any two wires, apply a three phase voltage, and be able to calculate the wattage of the heater using some type of multiplier.

You can, my first post demonstrates the multiplier is 2/sqrt(3). So if you combine that with P = I * V_LL * sqrt(3) for 3 phase power and I = V/R, you get P = 2 * V²_LL / R', where R' is your measured resistance between two terminals.

I've gone at this problem three different ways, gotten three different answers

If each of the 3 ways are valid, and you do it carefully, you should get the same answer 3 times. Like I did with a wye vs a delta arrangement of resistors.

Cheers, Wayne

 

[Larry Bohn]

I'll study a little more on the math this weekend. I'm still a little puzzled that measuring the resistance from two lines works when the resistances change when the third line is connected, both in a delta or a wye. Not saying you're incorrect, I just need to get a few things straight in my brain. My brother-in-law was an electrical engineer, (now retired), I'll see how much he remembers and maybe he can help me.
Thanks for the informationl
Larry

 

[wwhitney]

I'll study a little more on the math this weekend. I'm still a little puzzled that measuring the resistance from two lines works when the resistances change when the third line is connected, both in a delta or a wye.

The base assumption in my posts yesterday is that you have 3 resistors connected together in a delta (or as a separate class of examples, a wye) configuration, and that you know all the resistors are the same. That's a one parameter family of configurations (the parameter is that individual resistance), so taking a single measurement of the system is enough to determine which configuration you have.

If instead you just hand me a box with 3 terminals and tell me that some unknown configuration of resistors is inside, then I can't possibly determine the configuration or 3 phase circuit behavior just by measuring the resistance once between 2 terminals. I'd obviously need to measure all 3 of the pairwise resistances, and then consider how that would translate into circuit behavior.

Cheers, Wayne