Can anyone lay out a method to calculate the equivalent %z of two single phase transformers wired as autotransformers in open delta when the %Z impedance as an isolation transformer is given on the nameplate. I need to calculate fault current on the load side of the transformers.
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How to calculate %Z of autotransformers
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With a three phase bolted fault, test with a balanced three phase supply. Be sure that the transformer is up to temperature before the test. You will get low test results on a cold (room temperature) transformer.
Bill
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"Why not the best?"
Jimmy Carter
Bill
--------------------
"Why not the best?"
Jimmy Carter
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No, this is real life. I've involved Square D as its their transformers, they've been working on it for a week but can't find anyone who can answer the question. I've found two white papers on the subject, but they arrive at different answers. So I am here asking for help.
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% imp voltage is related to the combined impedance of both the primary and secondary windings.
For a really stiff source you can ignore the auto transformer contribution and use the source impedance limited by the impedance of the secondary winding. But we don't know the impedance of the secondary winding, just the impedance of the primary/secondary combination.
Next, the auto transformer will limit the current on A and C phases but will not limit the current on B phase.
As a practical solution I suggest using the available short circuit current of the source transformer. The answer will be on the high side, but if that is is a value that you can live with, you need go no further.
Be aware that the line to ground fault on a solidly grounded system will be the source available short circuit current on B phase and will be limited by the auto transformer on A and C phase.
Bill
--------------------
"Why not the best?"
Jimmy Carter
For a really stiff source you can ignore the auto transformer contribution and use the source impedance limited by the impedance of the secondary winding. But we don't know the impedance of the secondary winding, just the impedance of the primary/secondary combination.
Next, the auto transformer will limit the current on A and C phases but will not limit the current on B phase.
As a practical solution I suggest using the available short circuit current of the source transformer. The answer will be on the high side, but if that is is a value that you can live with, you need go no further.
Be aware that the line to ground fault on a solidly grounded system will be the source available short circuit current on B phase and will be limited by the auto transformer on A and C phase.
Bill
--------------------
"Why not the best?"
Jimmy Carter
I believe you have the transformers with you. Why dont you just measure impedance? Apply a small voltage on primary (say 2 %) and measure the shorted line currents on secondary.Extrapolate voltage for full load secondary current.Express this extrapolated voltage as % of rated primary voltage and that is the line % impedance.
Generally the rule for auto-transformers is % impedance on line MVA= self impedance x co-rtaio where co-ratio = (HV-LV/HV) voltages.With open delta, Iam not sure.Can you post the white papers or at least the titles?
Generally the rule for auto-transformers is % impedance on line MVA= self impedance x co-rtaio where co-ratio = (HV-LV/HV) voltages.With open delta, Iam not sure.Can you post the white papers or at least the titles?
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Here is the link to one of them.
id=Sf1ppwGI6uYC&pg=PA122&lpg=PA122&dq=impedance+autotransformer&source=bl&ots=NsdFOtv4Bb&sig=n2bB2r1QNsOP5wuAG3MvOZq4BZE&hl=en&sa=X&ei=V29FVdXBKcWZNuGigPAO&ved=0CDAQ6AEwAg#v=onepage&q=impedance%20autotransformer&f=false
Here is another:
Both of these is only looking at one transformer in single phase.
My problem is that I need to perform an Arc Flash evaluation on a panelboard supplied by the open Delta autotransformers in 3ph. You can merely assume the 'highest' fault current value, say from the source lineside of the xfrms, as high fault current often cause the breaker to open in its instantaneous region and produce low arc heat values, whereas, low falut currents move the breaker into its short time region, which holds the breaker closed longer, and produces high arc heat values. I would evaluate each condition and take the higher heat value. Fault current without the autotransformers present is easy, as its all symetrical and straight forward, its the fault current with the transformers in open delta that present the issue. If I had a good %z value I would calculate the equivalent capacity of the autotransformer bank, and then solve for short current with the %Z. That value wouldn't be totally right as there are only two transformers, not three, so the B phase would have less impedance and hence higher fault current. But in this case it would be okay as I just need to find the highest arc heat value, and if I have an ultimate high and an ultimate low, choose the highest resultant heat value - I don't care whats in between. However, finding the equivalent %z is question that seems to stump everyone.
id=Sf1ppwGI6uYC&pg=PA122&lpg=PA122&dq=impedance+autotransformer&source=bl&ots=NsdFOtv4Bb&sig=n2bB2r1QNsOP5wuAG3MvOZq4BZE&hl=en&sa=X&ei=V29FVdXBKcWZNuGigPAO&ved=0CDAQ6AEwAg#v=onepage&q=impedance%20autotransformer&f=false
Here is another:
Both of these is only looking at one transformer in single phase.
My problem is that I need to perform an Arc Flash evaluation on a panelboard supplied by the open Delta autotransformers in 3ph. You can merely assume the 'highest' fault current value, say from the source lineside of the xfrms, as high fault current often cause the breaker to open in its instantaneous region and produce low arc heat values, whereas, low falut currents move the breaker into its short time region, which holds the breaker closed longer, and produces high arc heat values. I would evaluate each condition and take the higher heat value. Fault current without the autotransformers present is easy, as its all symetrical and straight forward, its the fault current with the transformers in open delta that present the issue. If I had a good %z value I would calculate the equivalent capacity of the autotransformer bank, and then solve for short current with the %Z. That value wouldn't be totally right as there are only two transformers, not three, so the B phase would have less impedance and hence higher fault current. But in this case it would be okay as I just need to find the highest arc heat value, and if I have an ultimate high and an ultimate low, choose the highest resultant heat value - I don't care whats in between. However, finding the equivalent %z is question that seems to stump everyone.
I think the concept of %Z is not applicable here. It is useful for single phase and for single phase representations of balanced source three phase systems. As you point out, the phases do not contribute equally in the case of a three phase fault, so the single phase representation does not work. Other complications would include open deltas with one transformer secondary center tapped and grounded and closed deltas with a single larger "lighter" transformer. I don't know how to calculate these by hand. Perhaps there is some software that will do it.
Waross, %impedance is same from primary side or secondary side or step down or step up.
With open delta auto connection, the rating kVA to line output ratio = 1.15 (E1-E2)/ E1 =0.23, in this case. ie out put in this case will be 2x50x0.23 = 86.6 kVA. Based on this, I 'gusstimate' the line impedance as 1.1 % on 86.6 KVA.
Geometrical neutrals of the HV and LV circuits will not coincide. A difference in voltage exists between them at fundamental frequency,equal to 0.57(E1-E2) = 68.4 V ,in this case. Hence it is possible to ground the neutral of only one circuit,but not both. If both neutrals are grounded, a low impedance path for the neutral displacement voltage results,leading to a single phase short circuit current to flow through the entire length of the system,returning through the ground connections,(ref Transformer Engineering (book) - L F Blume.)
With open delta auto connection, the rating kVA to line output ratio = 1.15 (E1-E2)/ E1 =0.23, in this case. ie out put in this case will be 2x50x0.23 = 86.6 kVA. Based on this, I 'gusstimate' the line impedance as 1.1 % on 86.6 KVA.
Geometrical neutrals of the HV and LV circuits will not coincide. A difference in voltage exists between them at fundamental frequency,equal to 0.57(E1-E2) = 68.4 V ,in this case. Hence it is possible to ground the neutral of only one circuit,but not both. If both neutrals are grounded, a low impedance path for the neutral displacement voltage results,leading to a single phase short circuit current to flow through the entire length of the system,returning through the ground connections,(ref Transformer Engineering (book) - L F Blume.)
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To make it more challenging, the AB current and the CA current will be limited by one auto transformer winding and the BC current will be limited by both auto transformer windings.
The currents may be at a phase angle determined by the inequality of the currents.
Bill
--------------------
"Why not the best?"
Jimmy Carter
The currents may be at a phase angle determined by the inequality of the currents.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Let’s try to separate the primary[600V] and secondary[480 v] impedance.
At first let’s neglect the resistance and say X=Z
N1/N2=E1/E2=~V1/V2=a
N1- total number of turns N2-secondary[common part] winding number of turns. The primary windings turns=N1-N2
X=X1+((N1-N2)/N2)^2*X2=X1+(a-1)^2*X2 [Steinmetz].
Following general relation L[leakage inductivity]=miuo*Llmt *(d/3+s/2)/Lc*N^2 where:
Llmt=length of one turn ;d=width of the coil;s=distance to the magnetic core;Lc coil length;N=number of turns and presuming these data are the same for both coils except Lc [let’s say Lc1=Kl*(N1-N2) and Lc2=Kl*N2] then:X1/X2=(N1-N2)/N2=a-1; X2=X1/(a-1)
X=X1+(a-1)^2/(a-1)*X1=X1(1+a-1)=a*X1
X=5%*600^2/50000=0.36 a=600/480=1.25
X1=0.36/1.25=0.288 ohm ; X2=0.288/(1.25-1)=1.152 ohm
Actually Z1||Z2 then X||=X1*X2/(X1+X2); Ish=V2/X1/X2*(X1+X2) on 480 V.
Ish480=480/0.288/1.152*(0.288+1.152)= 2083 A
From 600 V winding the current will be: V2/X1=480/0.288=1666.7 A
If we would use resistance also [for 1000 w copper losses] the short-circuit current will be 10% less[approx.].
At first let’s neglect the resistance and say X=Z
N1/N2=E1/E2=~V1/V2=a
N1- total number of turns N2-secondary[common part] winding number of turns. The primary windings turns=N1-N2
X=X1+((N1-N2)/N2)^2*X2=X1+(a-1)^2*X2 [Steinmetz].
Following general relation L[leakage inductivity]=miuo*Llmt *(d/3+s/2)/Lc*N^2 where:
Llmt=length of one turn ;d=width of the coil;s=distance to the magnetic core;Lc coil length;N=number of turns and presuming these data are the same for both coils except Lc [let’s say Lc1=Kl*(N1-N2) and Lc2=Kl*N2] then:X1/X2=(N1-N2)/N2=a-1; X2=X1/(a-1)
X=X1+(a-1)^2/(a-1)*X1=X1(1+a-1)=a*X1
X=5%*600^2/50000=0.36 a=600/480=1.25
X1=0.36/1.25=0.288 ohm ; X2=0.288/(1.25-1)=1.152 ohm
Actually Z1||Z2 then X||=X1*X2/(X1+X2); Ish=V2/X1/X2*(X1+X2) on 480 V.
Ish480=480/0.288/1.152*(0.288+1.152)= 2083 A
From 600 V winding the current will be: V2/X1=480/0.288=1666.7 A
If we would use resistance also [for 1000 w copper losses] the short-circuit current will be 10% less[approx.].
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