dumb question - 200 amp, 3 phase, 208 volt

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mull982 said:
I dont follow this one can someone explain?
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I'll try. But it wasn't my statement, so I might not be interpreting the comment correctly.

First, losses are equal to the resistance times the square of the current. They are even, sometimes, called "I (squared) R Losses."

Suppose you start with balanced currents of 2, 2, and 2. Each phase will have a loss that is associated with 2 (squared). So for a simple comparison, you can describe it as 4 plus 4 plus 4 (ignoring the resistance value, which we can assume is the same in all phases anyway, so it won't change the answer). That gives you a total loss that is related to the total of 12.

Now suppose you unbalance the currents, so that you get 1, 2, and 3. If you square each number, and add them, you get 1 plus 4 plus 9. That total is 14, and is higher than the original 12.

The point, I believe, is that by making one phase current higher, its losses are much bigger, because you are squaring a bigger number. At the same time, you are making one phase current smaller. But when you square the smaller number it does not have as big an impact on the overall answer.

Does that help?
 
600 amps at 120?

600 amps at 120?

Okay, so let me see if I have this right. If his service is 208 volts 200 amps will he get 600 amps at 120 volts?

208 x 200 x 1.73 = 71968

71968 / 120 = 599.733

If the phrase "per phase" is an issue, would the phrase "has the capacity" be better? For example, ?In a 208 volt 200 amp service each phase has the capacity of 200 amps.? I say that because I read that as saying at 120 volts there are 200 amps present. The total capacity is 600 amps at 120 volts, but it is not 600 amps at 208 volts.
 
richard.bessey said:
Wow, so much response from a simple question, but I understand it, mostly. The 3 phases and how the relation from phase to phase varies in total output.

Anyway, one final question.

All of our loads are single phase, 120 volts. None of them are 3 phase. I checked the amps of each phase, and they are close to the same:
Phase 1 - 30.5 AMPS
Phase 2 - 46 AMPS
Phase 3 - 35.5 AMPS

I remember someone told me once this is bad and not efficient, and causes our electrical bill to go up, but I believe that is due to the inefficiency of a 3 phase motor due to un-equal load. In our case with all single phase loads, that makes no difference. Can anyone confirm or deny that?

I don't think we need to move our loads around so they are all the same, but if it would save on our electric bill, we will do it.
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From your figures, it would appear you are using an average of 19% of capacity of your system.

I seriously doubt that these readings would affect the operation or efficiency of a 3-phase motor at all, unless its operation would overload one of more of your phases.

The thing about electric bills and such may be related to your demand billing more than anything else. Having an evenly balanced load can minimize excess billing demand kWh and this could probably be what your friends were referring to about making your bills to go up.

IF you could move some of those loads from the "B" phase to the "A" phase, this might save a little, depending on what you are being charged for billing demand.

A close examination of some recent electric bills might offer some insight as to whether it would be worthwhile or not.

Others might say that there is not enough difference between phase loads to warrant any action at all.
 
charlie b said:

I'll try. But it wasn't my statement, so I might not be interpreting the comment correctly.

First, losses are equal to the resistance times the square of the current. They are even, sometimes, called "I (squared) R Losses."

Suppose you start with balanced currents of 2, 2, and 2. Each phase will have a loss that is associated with 2 (squared). So for a simple comparison, you can describe it as 4 plus 4 plus 4 (ignoring the resistance value, which we can assume is the same in all phases anyway, so it won't change the answer). That gives you a total loss that is related to the total of 12.

Now suppose you unbalance the currents, so that you get 1, 2, and 3. If you square each number, and add them, you get 1 plus 4 plus 9. That total is 14, and is higher than the original 12.

The point, I believe, is that by making one phase current higher, its losses are much bigger, because you are squaring a bigger number. At the same time, you are making one phase current smaller. But when you square the smaller number it does not have as big an impact on the overall answer.

Does that help?
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That's a good example of what I meant.
 
richard.bessey said:
All of our loads are single phase, 120 volts. None of them are 3 phase. I checked the amps of each phase, and they are close to the same:
Phase 1 - 30.5 AMPS
Phase 2 - 46 AMPS
Phase 3 - 35.5 AMPS

I remember someone told me once this is bad and not efficient, and causes our electrical bill to go up, but I believe that is due to the inefficiency of a 3 phase motor due to un-equal load. In our case with all single phase loads, that makes no difference. Can anyone confirm or deny that?
Click to expand...
That really aint all that bad, and may even be as close as you can get. Depending on what the loads were that got added up to get each.... Anyway your bill would not change in anyway that you would notice - you would just feel better about it.... ;)
 
Besoeker said:
mull982 said:
I have heard that unbalanced currents can also effect power factor is this true? If so how?
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It's not true.
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Ahh... but it is.

Take for example a single resistive-only 1? 208V load on a 208Y/120 system. The current of said load will be 30? out-of-phase with the applied voltage (i.e. Line to Neutral, that of the windings), resulting in a power factor of .87. Balance that with two other identical loads and the power factor is then 1.00.
 
Smart $ said:
Ahh... but it is.

Take for example a single resistive-only 1? 208V load on a 208Y/120 system. The current of said load will be 30? out-of-phase with the applied voltage (i.e. Line to Neutral, that of the windings), resulting in a power factor of .87. Balance that with two other identical loads and the power factor is then 1.00.
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If the resistive load is fed with 208V, then the applied voltage is line to line, not line to neutral.
 
Besoeker said:
If the resistive load is fed with 208V, then the applied voltage is line to line, not line to neutral.
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Line to Line voltage on a wye system is derived from two single-phase voltage sources 120? out-of-phase and with a common connection.

If you measure power factor using Line to Line voltage as the reference, then single phase resistive currents will have a power factor...???
 
Smart $ said:
...

If you measure power factor using Line to Line voltage as the reference, then single phase resistive currents will have a power factor...???
Click to expand...

That line should read, "If you measure power factor using Line to Line voltage as the reference, then Line-to-Neutral resistive loads will have the .87 power factor...???"
 
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charlie b said:
I'll try. But it wasn't my statement, so I might not be interpreting the comment correctly.

First, losses are equal to the resistance times the square of the current. They are even, sometimes, called "I (squared) R Losses."

Suppose you start with balanced currents of 2, 2, and 2. Each phase will have a loss that is associated with 2 (squared). So for a simple comparison, you can describe it as 4 plus 4 plus 4 (ignoring the resistance value, which we can assume is the same in all phases anyway, so it won't change the answer). That gives you a total loss that is related to the total of 12.

Now suppose you unbalance the currents, so that you get 1, 2, and 3. If you square each number, and add them, you get 1 plus 4 plus 9. That total is 14, and is higher than the original 12.

The point, I believe, is that by making one phase current higher, its losses are much bigger, because you are squaring a bigger number. At the same time, you are making one phase current smaller. But when you square the smaller number it does not have as big an impact on the overall answer.

Does that help?
Click to expand...

OK, my ignorance is showing but here goes (based on my understanding of your example):

If you have currents of 2,2,2 on the 3 phases, then each phase sees a resistance of 60 (E/I=R). I*I*R = 240. 240 + 240 + 240 = 720.

If you have currents of 1,2,3 on the 3 phases, then the R values would be 120,60,40. Using I*I*R I get loss values of 120 + 240 + 360 = 720.

This is in conflict with your statement. What did I do wrong?
 
Smart $ said:
Ahh... but it is.

Take for example a single resistive-only 1? 208V load on a 208Y/120 system. The current of said load will be 30? out-of-phase with the applied voltage (i.e. Line to Neutral, that of the windings), resulting in a power factor of .87. Balance that with two other identical loads and the power factor is then 1.00.
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You're right on a single phase basis, but the overall power factor will be 100%. You will have leading power factor in one phase and lagging in the other phase.
 
hardworkingstiff said:
OK, my ignorance is showing but here goes (based on my understanding of your example):

If you have currents of 2,2,2 on the 3 phases, then each phase sees a resistance of 60 (E/I=R). I*I*R = 240. 240 + 240 + 240 = 720.

If you have currents of 1,2,3 on the 3 phases, then the R values would be 120,60,40. Using I*I*R I get loss values of 120 + 240 + 360 = 720.

This is in conflict with your statement. What did I do wrong?
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You're using the load resistance. Load watts will be the same, balanced or not, as you have shown. What will be different is the losses. Assume conductor and transformer resistance in each phase is 1 ohm total. In the balanced case, losses will be:

LB = 2*2*1 + 2*2*1 + 2*2*1 = 12 W

In the unbalanced case:

LU = 1*1*1 + 2*2*1 + 3*3*1 = 14 W
 
hardworkingstiff said:
This is in conflict with your statement. What did I do wrong?
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Your math is OK. But you are talking about the resistance of the load. Or more precisely, your values of resistance correspond to the total resistance seen by the source. That will include the resistance of the load (the lion's share of the total resistance) plus the resistance of the wires (a small percentage). When I speak of the wires, I mean the wires that are internal to the transformer and the wires between the transformer and the load.

If you change the loads from a 2, 2, 2 configuration to a 1, 2, 3, configuration, you have not changed the resitances of the wires. The I2R losses within the wires will change from (2x2xR) plus (2x2xR) plus (2x2xR), for a total of 12R, to (1x1xR) plus (2x2xR) plus (3x3xR), for a total of 14R. Thus, the losses in the wires are higher, when the load is not balanced.
 
I have to learn to type faster, or to read the thread once more before I post, so I don't wind up posting something identical to someone who typed it faster. :D
 
081201-115 EST

Consider this:

Your load is resistive and the goal is the have the same total power dissipated in the load independent of line balance. Going from the balanced load of 2, 2, 2 to 1, 2, ? we get 2.645 for the third current. Thus, for the same total load power the total line losses are independent of the amount of unbalance.

I have no objection to "current per phase" but I think "current per line" is a better description, and I also have no problem with "infinite resistance".

From dictionary.com
in⋅fi⋅nite   /ˈɪnfənɪt/ Show Spelled Pronunciation [in-fuh-nit] Show IPA Pronunciation

?adjective 1. immeasurably great: an infinite capacity for forgiveness.
2. indefinitely or exceedingly great: infinite sums of money.
3. unlimited or unmeasurable in extent of space, duration of time, etc.: the infinite nature of outer space.
4. unbounded or unlimited; boundless; endless: God's infinite mercy.
5. Mathematics. a. not finite.
b. (of a set) having elements that can be put into one-to-one correspondence with a subset that is not the given set.
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.
 
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