Relative size and Arc Flash Hazards of 50 Hz versus 60 Hz transformers at same KVA

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Besoeker said:
But, for equal volt drops, the IZ products have to be equal whatever combination of I and Z you use.
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True, which tells you that the Z part of the term would have to be frequency dependent for the same transformer or different for different designs of the equivalent transformer, or the currents would have to be different.

It is not terribly useful to make a "mutatis mutandi" statement when in fact everything is changing. I am not sure Sahib comprehends that. :)
 
GoldDigger said:
True, which tells you that the Z part of the term would have to be frequency dependent for the same transformer or different for different designs of the equivalent transformer, or the currents would have to be different.

It is not terribly useful to make a "mutatis mutandi" statement when in fact everything is changing. I am not sure Sahib comprehends that. :)
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Well, the thread has meandered around from claims that a 50Hz transformer being smaller for the same rating as 60 Hz to the arc flash hazard being greater, non-like for like comparisons to voltage drop being equal to the nominal voltage...............so I don't think mutatis mutandi has been in evidence.
 
Besoeker said:
But, for equal volt drops, the IZ products have to be equal whatever combination of I and Z you use.
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Not necessarily, because IZ product does not denote arc voltage drop here but internal voltage drop of transformer and that of connecting wires if any.
GoldDigger said:
It is not terribly useful to make a "mutatis mutandi" statement when in fact everything is changing. I am not sure Sahib comprehends that. :)
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I think we may assume the arc to be stable for some time for the sake of discussion here.
 
Sahib said:
Not necessarily, because IZ product does not denote arc voltage drop here but internal voltage drop of transformer and that of connecting wires if any.

I think we may assume the arc to be stable for some time for the sake of discussion here.
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That is not the type of change I was referring to. Never mind....
 
Below are given formulas for calculating arc flash boundary and incident arc flash energy. For calculating arc flash boundary, bolted fault MVA is used as a factor in the formula for it and for calculating incident arc flash energy, bolted short circuit current factor is used. In our case the 11500/400V, 50hz transformer has a higher bolted short circuit current factor and so a higher incident arc flash energy than when used as 13800/480V, 60hz transformer. But a 13800/480V, 60hz transformer has a higher bolted fault MVA and so a higher arc flash boundary than when used as 11500/400V, 50hz transformer. So it appears to be a contradiction.

The Electrical Arc Flash Hazard Boundary is the distance from the arc source at which the Arc Thermal Energy becomes less than or equal to 1.2 calories/Cm2. The boundaries are established for the safety of the worker who must perform tasks near known arc flash hazards.


Formula for Calculating Electrical Arc Flash Hazard Boundaries

For calculating the Electrical Arc Flash Hazard Boundary, the formula below can be used. Or use a software package as referenced in this section to calculate the Electrical Arc Flash Hazard Boundary. Or use Table 1 in this section to establish the Electrical Arc Flash Hazard Boundary.
The Formula

Dc = [2.65 x MVABF x t}1/2 This formula is used when Short Circuit Current (Isc) is known.
Variables:
? Dc represents the distance (in feet) of a person from an arc source for curable second-degree burns
? t equals the time (in seconds) of arc exposure.
? MVABF is the bolted fault MVA (megavolt-amperes) at the point involved.
? MVA equals the MVA rating of transformer (for transformers with MVA ratings below 0.75 MVA, multiply the transformer rating by 1.25). The MVA value is used only in simple, one-transformer calculations.
Performing an Electrical Arc Flash Hazard Boundary Calculation

Given the following information, calculate the Arc Flash Hazard Boundary.
? 50 MVA transformer (base MVA of 50)
? Impedance (Z) of 6.75% on a 13,800 volt bus.
? Fault clearance time (t) of 6 cycles (0.1 seconds for 60Hz).
Table 1.

Step 1- Calculate the bolted fault MVABFMVABF = MVABASE/Z50/.0675 = 741
Step 2- Calculate the flash hazard boundary (Dc)Dc = [2.65 x MVABF x t]1/2[2.65 x 741F x 0.1]1/2 = 14
Calculate the Short Circuit CurrentIsc = (MVABF / (?3 x bus voltage)) x 1,000,000(741/(?3 x 13,800)) x 1,000,000 = 31,000


Variables:
? Dc represents the distance (in feet) of a person from an arc source for curable second-degree burns
? t equals the time (in seconds) of arc exposure.
? MVABF is the bolted fault MVA (megavolt-amperes) at the point involved.
? Z = Transformer Impedance
? Isc = Short Circuit Current
? ? is the square root

Formula for Calculating Arc Thermal Energy for 600 volts or less in Open Air

EMA = 5271 x DA-1.9593 x tA x [0.0016F2 - 0.0076F + 0.8938]
where:
EMA = maximum open Arc Thermal Energy in calories/cm2
DA= distance in inches from the arc source to the worker's torso. Valid for distance of 18" and greater.
TA = arc duration in seconds
F = bolted fault short circuit current in kA (valid for 16 to 50 kA).
Formula for Calculating Arc Thermal Energy for 600 volts or less in a Box

EMB = 1038.7 x DB-1.4738 x tA x [0-0093F2 - 0.3453F + 5.9675]
where:
EMB = maximum 20-inch cubic box Arc Thermal Energy in calories/cm2
DB = distance in inches from the arc source to the worker's torso. Valid for distance of 18" and greater.
TA = arc duration in seconds
F = bolted fault short circuit current in kA (valid for 16 to 50 kA).
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The only valid way to resolve the contradiction is to assume that 11500/400V, 50hz transformer has less bolted short circuit current despite having lower internal impedance so that the formulas given in the previous post does not give conflicting results for 11500/400V, 50hz and 13800/480V 60hz transformers.
So the 13800/480V 60hz transformer, when operated as 11500/400V, 50hz transformer has not only lower internal impedance but also have lower three phase bolted short circuit current and so lower arc flash hazard at its terminals.
 
Sahib said:
Below are given formulas for calculating arc flash boundary and incident arc flash energy. For calculating arc flash boundary, bolted fault MVA is used as a factor in the formula for it and for calculating incident arc flash energy, bolted short circuit current factor is used. In our case the 11500/400V, 50hz transformer has a higher bolted short circuit current factor and so a higher incident arc flash energy than when used as 13800/480V, 60hz transformer. But a 13800/480V, 60hz transformer has a higher bolted fault MVA and so a higher arc flash boundary than when used as 11500/400V, 50hz transformer. So it appears to be a contradiction.
Click to expand...

Sahib, please cite a source for this information/formulas or you are in violation of copyright laws/forum rules.
 
Sahib said:
The only valid way to resolve the contradiction is to assume that 11500/400V, 50hz transformer has less bolted short circuit current despite having lower internal impedance so that the formulas given in the previous post does not give conflicting results for 11500/400V, 50hz and 13800/480V 60hz transformers.
So the 13800/480V 60hz transformer, when operated as 11500/400V, 50hz transformer has not only lower internal impedance but also have lower three phase bolted short circuit current and so lower arc flash hazard at its terminals.
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Did you try to model the transformers? I believe I suggested you do some modeling a while back to provide insight.
 
mivey said:
Did you try to model the transformers? I believe I suggested you do some modeling a while back to provide insight.
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I think you have equivalent circuit diagram of transformer in mind. But I believe that modeling so would not deny that the 13800/480V, 60hz transformer, when used as 11500/400V, 50hz transformer, has lower incident arc flash energy at its terminals.
 
Sahib said:
I think you have equivalent circuit diagram of transformer in mind. But I believe that modeling so would not deny that the 13800/480V, 60hz transformer, when used as 11500/400V, 50hz transformer, has lower incident arc flash energy at its terminals.
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Then model it and see what you get so you will have some facts to draw on.
 
mivey said:
Such as the relative size of the impedance difference, arc flash, and safe distance. These will reveal the relative value of bringing up arc flash in the first place.
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The formulas given in post#126 may be used to find out the relative size of arc flash, and safe distance.
 
mivey said:
Then model the transformer under both conditions and make your calculations.
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Sorry, mivey. It does not motivate me enough: I am interested in understanding underlying principles and not doing numerical calculation. You are welcome to do it and I am ready to understand it.:)
 
Sahib said:
Sorry, mivey. It does not motivate me enough: I am interested in understanding underlying principles and not doing numerical calculation.
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Suit yourself.

Understanding quantities is key. For example, we all know your relative mass increases with velocity: that is the principle. Without the quantity, you may not realize it is not significant until you get really fast.

How fast? Moving from zero to 95 million mph gives a relative mass increase of about 1%. So now you think the principle is millions of mph faster is relatively a small impact.

But not so fast, pardon the pun. When your velocity makes your relativistic mass 1 million times your rest mass, an increase of 1 foot per hour in velocity increases your relativistic mass 52%.

Sahib said:
You are welcome to do it and I am ready to understand it.:)
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I already have but that teaches you little. I have a suspicion you learn better when the information comes from within.
 
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