208V 1phase loads

[al hildenbrand]

Well I am still clueless, wanna elaborate?

The "red monkey" (don't think of one) is the idea that the 33.85 A of supply current to bus A can go to the neutral. It doesn't. The current is only going phase to phase (the heaters are 208 Volt).

The current in the supply to bus A is 60? off from the current in the A - B heaters and is also 60? off from the current in the A - C heaters. All three currents have a different direction, and the supply current to A is the vector sum of the pair of phase to phase currents.

 

[david luchini]

The "red monkey" (don't think of one) is the idea that the 33.85 A of supply current to bus A can go to the neutral. It doesn't. The current is only going phase to phase (the heaters are 208 Volt).

The current in the supply to bus A is 60? off from the current in the A - B heaters and is also 60? off from the current in the A - C heaters. All three currents have a different direction, and the supply current to A is the vector sum of the pair of phase to phase currents.

This statement is incorrect, and may be why you are arriving at a current of 33.85A on each bus, instead of 50A.

The current in the bus A would be 30? from the current in the A-B heaters and 30? from the current in the A-C heaters.

 

[jumper]

The loads are not 120v they are 208v SINGLE PHASE

Yeah, I know, which means you cannot use 1.732 in calc.unless all three phases are accounted for.

What numbers did you use?

 

[jumper]

The "red monkey" (don't think of one) is the idea that the 33.85 A of supply current to bus A can go to the neutral. It doesn't. The current is only going phase to phase (the heaters are 208 Volt).

The current in the supply to bus A is 60? off from the current in the A - B heaters and is also 60? off from the current in the A - C heaters. All three currents have a different direction, and the supply current to A is the vector sum of the pair of phase to phase currents.

Working on the math, still do not get it. Did I mention I am slow.

 

[jumper]

This statement is incorrect, and may be why you are arriving at a current of 33.85A on each bus, instead of 50A.

The current in the bus A would be 30? from the current in the A-B heaters and 30? from the current in the A-C heaters.

Oy vey! The numbers I just started punching are wrong?

 

[Dennis Alwon]

Yeah, I know, which means you cannot use 1.732 in calc.unless all three phases are accounted for.

What numbers did you use?

Well I got it to the 28.8 but never went further. I was trying to figure the 1.73 and I couldn't make it work. Then I said if there were 18000 watts / 3 phases as in 3 phases loads- you would get 50 amps-- so how could you get more amps per phase on 3 phase then you would on single phase. I guess I didn't get it.

I just saw David jumped in with an argument-- let's see where this goes. I thought this was simple. :grin:

 

[al hildenbrand]

The current in the bus A would be 30? from the current in the A-B heaters and 30? from the current in the A-C heaters.

Is the A-B load current only 60? from the A-C load current?

Think about it. 208 V 3 ? supply with a balanced load in a delta configuration. That is:

Load AB = Load BC = Load AC = 6 kVA resistive. The current in each load is 120? from the next current.

 

[jumper]

Well I got it to the 28.8 but never went further. I was trying to figure the 1.73 and I couldn't make it work. Then I said if there were 18000 watts / 3 phases as in 3 phases loads- you would get 50 amps-- so how could you get more amps per phase on 3 phase then you would on single phase. I guess I didn't get it.

I just saw David jumped in with an argument-- let's see where this goes. I thought this was simple. :grin:

I am ready to throw calculator into trash can.

 

[david luchini]

Working on the math, still do not get it. Did I mention I am slow.

Here is the math - hope it helps.

There are six 3kW, 208V heaters connected balanced between the phases, so there is a 6kW load between A-B, B-C and C-A. The load current is 6000W/208V=28.85A. The three load currents are 120? out of phase with each other, so lets say Iab=28.85<0, Ibc=28.85<-120, Ica=28.85<-240.

The current flowing on Bus A will be Ia=Iab-Ica (likewise, Ib=Ibc-Iab & Ic=Ica-Ibc.)

Converting: Iab=28.85<0 = 28.85 + j0
.................Ibc=28.85<-120 = -14.43-j24.98
.................Ica=28.85<-240 = -14.43+j24.98

So calculating for current on Bus A: Ia=Iab-Ica =>
.................Ia= (28.85 + j0) - (-14.43+j24.98)
.................Ia= 43.28 - j24.98

Converting back: Ia=50<-30

The current on Bus A (Ia) has a magnitude of 50 (and is 30? separated from Iab)

 

[david luchini]

Well I got it to the 28.8 but never went further. I was trying to figure the 1.73 and I couldn't make it work. Then I said if there were 18000 watts / 3 phases as in 3 phases loads- you would get 50 amps-- so how could you get more amps per phase on 3 phase then you would on single phase. I guess I didn't get it.

I just saw David jumped in with an argument-- let's see where this goes. I thought this was simple. :grin:

It is simple. 18000W/208V/1.732 = 50A.

 

[al hildenbrand]

It is simple. 18000W/208V/1.732 = 50A.

Since we are sizing the Main Breaker, we only need the current on any one leg.

6000 of those 18000W are not connected on leg A.

 

[Dennis Alwon]

It is simple. 18000W/208V/1.732 = 50A.

Obviously that was my opinion originally but when an EE jumps in and says otherwise I have to wonder. Now we have 2 EE's arguing it out. That makes it not simple.

 

[david luchini]

Is the A-B load current only 60? from the A-C load current?

Think about it. 208 V 3 ? supply with a balanced load in a delta configuration. That is:

Load AB = Load BC = Load AC = 6 kVA resistive. The current in each load is 120? from the next current.

Yes, the A-B load current is only 60? from the A-C load current. But the A-B load current is 120? from the C-A load current. This is an important point. See my post #69: Iab=28.85<0, Ibc=28.85<-120, Ica=28.85<-240. These three load current are all 120? apart.

But Iac=-Ica, so Iac=28.85<-60.

 

[david luchini]

Since we are sizing the Main Breaker, we only need the current on any one leg.

6000 of those 18000W are not connected on leg A.

I don't understand your point. the 50A is the current on ANY of the 3 legs.

It doesn't matter if 6000 of the 18000W are not connected on leg A. The current on each of the legs for the 18000W balanced load is 50A.

If we removed the load from B-C so that Iab=28.85<0, Ibc=0, and Ica=28.85<-240, then the loads on the busses would be:

Ia=50<-30
Ib=28.85<180
Ic=28.85<-240

Even removing the 6000W that are not connected on leg A, the maximum leg is still 50.

 

[david luchini]

Obviously that was my opinion originally but when an EE jumps in and says otherwise I have to wonder. Now we have 2 EE's arguing it out. That makes it not simple.

Those darn gEEks. Nothing is ever simple with them. :grin:

Btw Dennis, haven't been on in a while. Congrats on the promotion!

 

[al hildenbrand]

Yes, the A-B load current is only 60? from the A-C load current. But the A-B load current is 120? from the C-A load current. This is an important point. See my post #69: Iab=28.85<0, Ibc=28.85<-120, Ica=28.85<-240. These three load current are all 120? apart.

But Iac=-Ica, so Iac=28.85<-60.

This is where vector conventions, and my not being a PE gets in the way. It's been too many years since my class work.

When I write "A-C load current" and you write "C-A load current" I feel by the seat of my pants that we are describing the same current. However, you are telling me that the load current in the 6 kVA resistive load connected between phase A and phase C has two different angular relationships to the load current between A and B depending upon whether it is labeled "A-C" or "C-A".

 

[Dennis Alwon]

This is where vector conventions, and my not being a PE gets in the way. It's been too many years since my class work.

Don't feel bad I have no idea what either of you are saying. :grin:

Btw Dennis, haven't been on in a while. Congrats on the promotion!

Thanks David

 

[al hildenbrand]

Ok. It's an isoceles triangle, not the equilateral that I was fixated on. That's where the 30? comes in.

 

[david luchini]

This is where vector conventions, and my not being a PE gets in the way. It's been too many years since my class work.

When I write "A-C load current" and you write "C-A load current" I feel by the seat of my pants that we are describing the same current. However, you are telling me that the load current in the 6 kVA resistive load connected between phase A and phase C has two different angular relationships to the load current between A and B depending upon whether it is labeled "A-C" or "C-A".

I would have "understood" your writing A-C load current as being the same as what I call C-A, but somehow you came up with the wrong angle separation in post #61, so I thought is was important to make the distinction.

If you think of it, if three phase voltages are 120? apart such that Vab=208<-60, Vbc=208<-180 and Vca=208<-300, then the voltages Vac=208<-120, Vcb=208<0 and Vba=<120. In other words Vab is 180? from Vba.

 

[al hildenbrand]

Sheesh. It really is the Fool Moon.