Three Phase Equipment - Different amps on each phase

[chris009]

Hi,

I was trying to figure out what type of three phase breaker I need to size for an electric griddle. It's a 208V, 3 phase, 6.7KW piece of equipment. Normally, I'd size it for 30A, but there was additional information. X=27.95A, Y=16.11A, Z=16.11A. I wasn't sure if 30A would cut it if one of the phases is already running at 28A.

Can anyone clarify what is going on with this? Or is this normal for most pieces of equipment and just not put on the specification sheets?

Thanks in advance.

 

[Phil Corso]

Chris...

If the Appliance Nameplate indicates only one current value then the data presented indicates severe unbalance. And if the heating elements are wye connected, then the unbalance could lead to damage!

Can you provude additional information or detail?

Regards, Phil Corso

 

[chris009]

Thanks for replying. I see that I'm a moderated user, so hopefully I can post a link. Spec Sheet. If you look under three phase for 208V that's what it lists. If this is really unbalanced, maybe getting the single phase version of this equipment would be better.

 

[jim dungar]

Just because it is an unbalanced load does not mean it will be a problem.
If you order it as single phase and then install it on 3-phase, you will still have an unbalance.

If you have 3-phase order the appropriate breaker (sized for the largest current) and just rearrange other single phase circuits if the unbalance causes problems.

 

[Phil Corso]

Chris...

In theory, the Yaki maxum 'X' line-current is based on a 'Cold' element resistance or a higher than normal supply voltage! As the element heats up, its temperature increases, and "nominal" current will decrease to about 22A presuming supply remains at 208V!

Since it's temperature controlled, current will vary, so that the nominal RMS value, over time, will probably be somewhat lower.

Also, since the heating elements are Delta-connected, my caveat about the Wye-connection is unwarranted!

I would go with Jim's suggestion!

Regards, Phil

 

[chris009]

Okay, This is good information and makes a lot of sense. I take it that I need to size the breaker for the highest amperage phase, so I would need 40A.

In theory, the Yaki maxum 'X' line-current is based on a 'Cold' element resistance or a higher than normal supply voltage! As the element heats up, its temperature increases, and "nominal" current will decrease to about 22A presuming supply remains at 208V!

Since it's temperature controlled, current will vary, so that the nominal RMS value, over time, will probably be somewhat lower.

Also, since the heating elements are Delta-connected, my caveat about the Wye-connection is unwarranted!

How can you tell it is delta connected? Also do you know where I can read up about some of the math. In particular how you came up with 22A.

Thanks again for everyone taking the time to read and respond.

 

[Besoeker]

How can you tell it is delta connected? Also do you know where I can read up about some of the math. In particular how you came up with 22A.

Thanks again for everyone taking the time to read and respond.

I's guess that there might be two heating elements. One between X and Y and the other between X and Z with the controls connected from Y to Z. That would neatly explain Y and Z having equal currents and the sum of the two being slightly higher than the current in X.
In short, I don't see the unbalance as an issue.

I'm not sure that I agree with Phil Corso regarding the variation in resistance of the heating element. We make power electronic equipment and routinely use an array of heating elements as a test load. The elements are individually rated at 240V, 1kW. Cold, they measure about 60ohms. At operating temperature, they still have to be about 60ohms to get the 1kW dissipation.

 

[kwired]

I's guess that there might be two heating elements. One between X and Y and the other between X and Z with the controls connected from Y to Z. That would neatly explain Y and Z having equal currents and the sum of the two being slightly higher than the current in X.
In short, I don't see the unbalance as an issue.

I think this is the most likely scenario. If you multiply the lower current value by 1.73 you get exactly what the higher value current is and should be if that is what the situation is.

If there were wye or full delta connected elements of equal ratings then there is something wrong with one of the elements.

This is a common way of producing an appliance that is suitable for single or three phase, Just make both elements parallel to one another to use on single phase.

 

[Phil Corso]

Chris...

Why Delta was chosen:
Besoeker has explained why the configuration is a Delta.

Regarding my conjecture about resistance change:
The spec shows that if Supply Volts and kW/per phase were used, then resistance is V?/P or 12.91Ω.
Back calculating Line-Currents, then resistance is 16.11Ω, hence my explanation!
Regards, Phil Corso

 

[ggunn]

Hi,

I was trying to figure out what type of three phase breaker I need to size for an electric griddle. It's a 208V, 3 phase, 6.7KW piece of equipment. Normally, I'd size it for 30A, but there was additional information. X=27.95A, Y=16.11A, Z=16.11A. I wasn't sure if 30A would cut it if one of the phases is already running at 28A.

Can anyone clarify what is going on with this? Or is this normal for most pieces of equipment and just not put on the specification sheets?

Thanks in advance.

Is it maybe for Wye connection to a high leg supply - Y and Z to neutral 120V and X to neutral 208V? Assuming a resistive load and the same resistance on all three legs, the numbers work out about right, i.e., 208/120 * 16.11A = 27.92A

 

[Besoeker]

Chris...

Why Delta was chosen:
Besoeker has explained why the configuration is a Delta.

I think it's like this:

It isn't really delta.
For single phase operation it's easy to common terminals L2 and L3.

Regarding my conjecture about resistance change:
The spec shows that if Supply Volts and kW/per phase were used, then resistance is V²/P or 12.91Ω, hence my explanation!

I don't know where you get that from.
Two voltages are offered. The specification shows the power rating is 3.35kW for each heating element whether the voltage is 208V or 240V. Clearly this means the resistance of the element provided differs depending on specified voltage.

For 208V it would be 208^2/3350 giving 12.9 ohms.
For 240V it would be 240^2/3350 giving 17.2 ohms.

This difference has nothing whatsoever to do with temperature change.

 

[Phil Corso]

Besoker... the vendor spec lists 6 models. I used the 3-Ph, 208V, 3.35kW one, as Chris desribed.

The fact that phase L2-L3 has no heating-element doesn't make it any less of a Delta-Connected circuit!

Regards, Phil

 

[Besoeker]

Besoker... the vendor spec lists 6 models.

But only four electrical specifications to be considered:

240V single phase
208V single phase

240V three phase
208V three phase

I used the 3-Ph, 208V, 3.35kW one, as Chris desribed.

It doesn't actually matter whether you connect it as single phase or three phase. At 208V 3.35kW, the resistance in each element is 15.9Ω.

But I don't know how you got to this:

Back calculating Line-Currents, then resistance is 16.11Ω, hence my explanation!

The fact that phase L2-L3 has no heating-element doesn't make it any less of a Delta-Connected circuit!

I suppose you could call it that if you wanted to. From my perspective, it's simply two line to line connected loads. Or, for the UK market, two 240V line to neutral loads.

 

[Phil Corso]

Besoeker...

1) # of Choices!
You are correct!

2) Resistance Calc
R=V^2/P = 12.9 Ohms. Same as your 21-Jun-12, 01:03 pm calc!

3) Back-Calc of Line-Currents
Using vector analysis, given supply-voltage, supply-sequence, and line-current magnitudes!

Phil

 

[Besoeker]

Besoeker...

1) # of Choices!
You are correct!

TY.

2) Resistance Calc
R=V^2/P = 12.9 Ohms. Same as your 21-Jun-12, 01:03 pm calc!

Quite right. Don't know where I got the 15.9. Brain phart obviously.
BTW, my post couldn't have been made at 21-Jun-12, 01:03 pm.
At that point I was driving along the A52 in Derbyshire, UK. I think you mean post #11 which was made at 18:03 (6:03pm) yesterday.
Depending on where you are (Mrs B is from GA) a five hour difference would make that 13:03 (1:03pm) local time there if you are in that time zone.
Posts show local time for me.

) Back-Calc of Line-Currents
Using vector analysis, given supply-voltage, supply-sequence, and line-current magnitudes!

C'mon!
Pull the other one!
We're looking at two resistive loads.

 

[Phil Corso]

Besoeker...

Having lived a while in Hampton Court, quite distant from your neck-of-the-woods, l grew to love Brutish humor! :lol:

So, what solution wold you use? :?

Phil

 

[Besoeker]

Besoeker...

Having lived a while in Hampton Court, quite distant from your neck-of-the-woods, l grew to love Brutish humor! :lol:

Actually, not so distant. About 30 miles from where we live.

So, what solution wold you use? :?

Phil

To solve what?

 

[Phil Corso]

Besoeker... reur 06:05pm (edt):

Given: Source Vab, Vbc, Vca;Seq ABC;Line-current magnetudes |Ia|, |Ib|, |Ic|.

Find: Delta-connected resistances, Zab = Zbc, Zca = infinite

Regards, Phil

PS; One son is a Cockney!

 

[Besoeker]

Besoeker... reur 06:05pm (edt):

Given: Source Vab, Vbc, Vca;Seq ABC;Line-current magnetudes |Ia|, |Ib|, |Ic|.

Find: Delta-connected resistances, Zab = Zbc, Zca = infinite

Regards, Phil

What difference does the sequence make to the magnitude of the currents in the resistive loads?
If you reversed it, would that change anything?

PS; One son is a Cockney!

He was born within the sound of Bow Bells?

I wasn't. But then I'm not English, you see.

 

[Phil Corso]

Besoeker...

Ur 1st response: Excellent. 100%

Phil