Load summary

[weressl]

In all likelihood the omission of power factor (PF) in a "POWER" calculation is going to more than offset most voltage differences.

The premise was that we are dealing with unity power factor.

Otherwise the pf in each phase should also be used.

Offset is perhaps not the most accurate way to describe the resultant error and it could go both ways. Unbalanced voltages have absolutely nothing to do with power factor.

 

[Smart $]

The premise was that we are dealing with unity power factor.

I must have missed that statement

Drawing conclusions on the premise that all line-to-line voltages are equal is just as plausible as one assuming untiy power factor :grin:

Unbalanced voltages have absolutely nothing to do with power factor.

Now don't be inferring I put some hidden message in what I wrote

 

[gar]

090115-2138 EST

Consider this problem:

1. Three equal resistors each with a value of R connected in a Y. Three identical voltages in a Y source with a value V from line to neutral. The load neutral point and the Y neutral are not connected together, but by the design of the question are at the same potential. So connected together or not does not matter. Power total = 3 * V^2 / R . The magnitude of each line current is Iline = V / R .

2. Modify the load to two resistors in an open delta such that the total power is the same. What is the magnitude of each line current?

3. Is (2) a better or worse load than (1)?

.

 

[Smart $]

What is the magnitude of each line current?

Vₐ = Vᵦ/√3, or
Vᵦ = √3Vₐ _voltage relationship
3Vₐ?/Rₐ = 2Vᵦ?/Rᵦ _total power relationship

Therefore, solving for Rᵦ

3(Vᵦ/√3)?/Rₐ = 2Vᵦ?/Rᵦ
3(Vᵦ?/3)/Rₐ = 2Vᵦ?/Rᵦ
1/Rₐ = 2/Rᵦ
Rᵦ = 2Rₐ

Hence,

IlineA = VᵦRᵦ = 2√3VₐRₐ
IlineB = √3VᵦRᵦ = 6VₐRₐ
IlineC = VᵦRᵦ = 2√3VₐRₐ

Is (2) a better or worse load than (1)?

Lacking the parameters to make that determination. If higher line current is worse, then (2) is worse than (1).

Is there a point to this?

 

[kencoel]

It seems that after reading all the input on my question, for general reference I should take a reading on each phase (leg) and determine that the highest phase (in this case the A phase that is 78.84 amps) would be the total load for this panel. ?? I s this correct? Thanks.

 

[Cold Fusion]

It seems that after reading all the input on my question, for general reference I should take a reading on each phase (leg) and determine that the highest phase (in this case the A phase that is 78.84 amps) would be the total load for this panel. ?? I s this correct? Thanks.

In my mind - an absolute "could be":roll:

Load is usually in VA, not amps.

But it really depends on the context. What are you going to do with the number? Are you looking at how much load you can add to a panel? Are you looking at how close a panel or the xfm feeding it is to overload? The panel is three phase with likely all single phase loads - so you are not asking what 3ph loads you can add.

If I asked a tech to get readings on a panel, I would expect the get all three - not an average, not the highest.

The way you are phrasing your question, there is no decent answer - except the one you already got: Don't add the three together.

cf

 

[gar]

090116-0946 EST

Smart $:

Yes there is a point.

I was interested to see what would happen to the line currents if I gradually reduced the resistance of one of the elements of the Y load to 0 while maintaining the same total power output. I chose the extreme of zero resistance for one resistor as the limiting condition. Obviously with the neutral point of the Y load unconnected to anything else as was my statement of the question.

Unfortunately I can not follow your analysis because square boxes show at many points in your equations.

It does appear that you calculate each resistance in the open delta load as 2R. I agree with that.

However, Iline A = 1.732*V/2*R for the open delta vs V/R for the Y connection.

Thus, the line A current for open delta is 1.732/2 = 0.866 that of line A current for the Y connection. And line C is the same value by symmetry. This would imply that the distribution line losses are less in line A and C with the open delta load than with the Y load.

Line B current is the sum of line A and line C. A and C are displaced by 120 deg. What is the magnitude of line B relative to line A or C? I have my answer, but what do others get?

.

 

[jim dungar]

It seems that after reading all the input on my question, for general reference I should take a reading on each phase (leg) and determine that the highest phase (in this case the A phase that is 78.84 amps) would be the total load for this panel. ?? I s this correct? Thanks.

Ken,

You can pretty much ignore the thread jackers. Their discussion is how to add unbalanced currents at the neutral point.

If your highest measured phase conductor reading is 78.84A, you could say that that is the maximum current drawn by the load. You could use this value to make judgments about the sizing of conductors and protective devices.

While not technically correct (because loads should really be expressed in kVA or kW), most people would understand the phrase "the load for my panel is 78.84A".

What you probably shouldn't do is use the word total.

 

[weressl]

Ken,

You can pretty much ignore the thread jackers. Their discussion is how to add unbalanced currents at the neutral point.

If your highest measured phase conductor reading is 78.84A, you could say that that is the maximum current drawn by the load. You could use this value to make judgments about the sizing of conductors and protective devices.

While not technically correct (because loads should really be expressed in kVA or kW), most people would understand the phrase "the load for my panel is 78.84A".

What you probably shouldn't do is use the word total.

Jim, you are correct. I would suggest that talking about ampere loading of a panel is appropriate, since the panels are rated in amperes that refers to the main bus/lug/CB size.

 

[Smart $]

Unfortunately I can not follow your analysis because square boxes show at many points in your equations.

If you used a unicode font, with those glyphs, in your browser those boxes would be "alpha" and "beta" subscripts.

Probably just as well that you can't follow, as my Iline's are incorrect. I see you caught that :grin:

However, Iline A = 1.732*V/2*R for the open delta vs V/R for the Y connection.

I have my answer, but what do others get?

IlineB2 = 1.732 ? A2 or C2 [or 0.866 ? (A2 + C2)].
IlineB2 = 1.5 ? A1, B1, or C1

 

[gar]

090116-1353 EST

Smart $:

I get a different value for Iline B for the open delta than you get.

Two ways to view this.

1. Line B current is the sum of two sine waves of magnitude 0.866*V/R that are 120 deg apart. We know that the sum of sine waves of the same frequency produce a sine wave of that frequency. Draw the two sine waves 120 deg apart. These intersect at 150 deg relative to the zero crossing of one. What is the value of the sine of 150? It is 1/2. By symmetry the resultant of the sum of the two sine waves is at this 150 deg point. Thus, the peak of the new sine wave is equal to the peak of either of the two that added to make the new one. Thus, Iline B = Iline A = Iline C. These line currents are less than the line currents when the load is a Y connection of the same total power.

2 The other way is to add two vectors 120 deg apart. Same result.

My conclusion is the open delta load has less supply line loss, but intuitively this does not seem correct. Have I made a mistake somewhere?

.

 

[Smart $]

Have I made a mistake somewhere?

Yep!

The currents are ?60? with respect to each other where they sum.

 

[kencoel]

What is an explaination of a Continuous Load and a Non-Continuous Load?

 

[Smart $]

What is an explaination of a Continuous Load and a Non-Continuous Load?

It can be found in Article 100 Definitions...

A continuous load is one where the maximum current is expected to continue for 3 or more hours.

 

[Pierre C Belarge]

(...continued from a previous post.) I am trying to determine how to figure the full load amps of an existing commercial kitchen panel. I have read each leg of each phase during peak hours and added in some new load/amps for some kitchen equipment. I need to submit this information for a permit. The A phase reads 78.84 amps, the B phase reads 61.19 amps and the C phase reads 50.14 all included with the new equipment turned on. How do you determine what the Total load amps for this panel would be? Thanks.

This portion of the post sticks out like a sore thumb.
In the permit process that we use, the person pulling the permit is required to list the loads and their nameplate ratings.

Go back to the site and start documenting the nameplate ratings of the equipment installed, including the items that may be plugged into receptacles. This would include the lighting load if the panel supplies lighting.
One way to find the existing total load on the panel, take the amperage you have metered and multipy by the voltage of the system...this will give you the Kva. Take all 3 legs (100 x nominal voltage) and you will have the total Kva the panel can handle.
This method may not be completely accurate, depending on the consideration of whether or not all loads were on during the metered testing.