Section 110.9 & 110.10
- Thread starter gman2003
- Start date
- Status
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
Re: Section 110.9 & 110.10
This is the reference that I use
IEEE 141-1993 RED BOOK Recommended Practice for Electric Power Distribution for Industrial Plants
This is the reference that I use
IEEE 141-1993 RED BOOK Recommended Practice for Electric Power Distribution for Industrial Plants
Re: Section 110.9 & 110.10
Just questioning;
On plans submitted to the State, Do engineers always make calculations for the installation when they mark the prints with Kair values for the OCPD , panelboards , MCC's etc.?
Recently looked at the Engineer stamped plans and installation , up to this time period, noted the plans called for panelboards with 22Kair OCPD's, but the most that I could calculate for the Available Short Circuit Current (ASCC) at the closest panelboard to the SDS (208Y/120V) was 4176 ASCC.
Of course I am not privy to future plans, but this is a new installation, is energized, but not ready for occupancy.
The Service (480Y/277V) portion of the system at 65Kair was appropriate, I thought, for the Poco supply.
It seems as 'over-kill' of using all 22Kair OCPD's in the 208Y/120Vpanelboards when 10Kair would have been adequate, or maybe series ratings could have been applied, since these are supplied off of SDS transformer with over 4.0 %Z (Percent Impedance).
Of course, I am not an engineer either.
gwz2
Just questioning;
On plans submitted to the State, Do engineers always make calculations for the installation when they mark the prints with Kair values for the OCPD , panelboards , MCC's etc.?
Recently looked at the Engineer stamped plans and installation , up to this time period, noted the plans called for panelboards with 22Kair OCPD's, but the most that I could calculate for the Available Short Circuit Current (ASCC) at the closest panelboard to the SDS (208Y/120V) was 4176 ASCC.
Of course I am not privy to future plans, but this is a new installation, is energized, but not ready for occupancy.
The Service (480Y/277V) portion of the system at 65Kair was appropriate, I thought, for the Poco supply.
It seems as 'over-kill' of using all 22Kair OCPD's in the 208Y/120Vpanelboards when 10Kair would have been adequate, or maybe series ratings could have been applied, since these are supplied off of SDS transformer with over 4.0 %Z (Percent Impedance).
Of course, I am not an engineer either.
gwz2
charlie
Senior Member
- Location
- Indianapolis
Re: Section 110.9 & 110.10
Glenn, we post our standard fault currents on the Internet and tell the ECs, architects, and engineers that they should use the next higher available fault current in case we have to replace the transformer. If they do the same thing with their SDS, they may have the same results since they don't have a clue what the impedance will be on a future transformer. Also, the available fault current will go up when you are transforming to a lower voltage.
When I am doing short circuit calculations, I use the Bussmann SPD for the information to do the calculations.
Glenn, we post our standard fault currents on the Internet and tell the ECs, architects, and engineers that they should use the next higher available fault current in case we have to replace the transformer. If they do the same thing with their SDS, they may have the same results since they don't have a clue what the impedance will be on a future transformer. Also, the available fault current will go up when you are transforming to a lower voltage.
When I am doing short circuit calculations, I use the Bussmann SPD for the information to do the calculations.
- Location
- Illinois
- Occupation
- retired electrician
Re: Section 110.9 & 110.10
Just remember that it causes no problems, other than a cost increase, to use a higher available fault current for this purpose, but that the actual available fault current should always be used when making the arc/blast/flash calculations. This selection of PPE is partly based on the clearing time for the OCPD in the circuit. The use of a fault currents that are artificially high may result in a shorter clearing time and the selection of PPE that is not adequate for the actual fault conditions.
Don
Just remember that it causes no problems, other than a cost increase, to use a higher available fault current for this purpose, but that the actual available fault current should always be used when making the arc/blast/flash calculations. This selection of PPE is partly based on the clearing time for the OCPD in the circuit. The use of a fault currents that are artificially high may result in a shorter clearing time and the selection of PPE that is not adequate for the actual fault conditions.
Don
Re: Section 110.9 & 110.10
Charlie,
The SDS 's I see are not Poco and do not have infinite primary but have a definite ASCC at the primary connection.
Example;
75 KVA , 480V Pri, 208V Sec, 4.6 %Z, 25000A ASCC at Pri,.
By re-writing the Bussmann Point-To-Point method for a second transformer in system, the ASCC at the Sec. terminals would be and using continous hand calculator steps ( ie where * = Multiply, + = add, M+ = memory in, Mr = memory recall, / = divide, 1.732 = 3? multiplier and = = equal: ( the final = is the ASCC at the Sec. terminals.;
Pri ASCC / 100,000 * Pri V * 1.732 * %Z / KVA = + 1 = M+ 1 / Mr = * Pri ASCC / Sec V * Pri V = _____________ Sec ASCC.
( remember to clear calculator before starting the following sequence of steps.)
25000A / 100,000 * 480V * 1.732 * 4.6 / 75 = + 1 = M+ 1 / Mr = * 25000A / 208V * 480V = _____________ Sec ASCC. ( which is 4196.5557 Amperes .
This is a lot LESS than the Pri. ASCC.
gwz2
edit; had some spelling errors.
Remember these are calculator continous steps.
[ November 29, 2003, 11:37 AM: Message edited by: gwz2 ]
Charlie,
The SDS 's I see are not Poco and do not have infinite primary but have a definite ASCC at the primary connection.
Example;
75 KVA , 480V Pri, 208V Sec, 4.6 %Z, 25000A ASCC at Pri,.
By re-writing the Bussmann Point-To-Point method for a second transformer in system, the ASCC at the Sec. terminals would be and using continous hand calculator steps ( ie where * = Multiply, + = add, M+ = memory in, Mr = memory recall, / = divide, 1.732 = 3? multiplier and = = equal: ( the final = is the ASCC at the Sec. terminals.;
Pri ASCC / 100,000 * Pri V * 1.732 * %Z / KVA = + 1 = M+ 1 / Mr = * Pri ASCC / Sec V * Pri V = _____________ Sec ASCC.
( remember to clear calculator before starting the following sequence of steps.)
25000A / 100,000 * 480V * 1.732 * 4.6 / 75 = + 1 = M+ 1 / Mr = * 25000A / 208V * 480V = _____________ Sec ASCC. ( which is 4196.5557 Amperes .
This is a lot LESS than the Pri. ASCC.
gwz2
edit; had some spelling errors.
Remember these are calculator continous steps.
[ November 29, 2003, 11:37 AM: Message edited by: gwz2 ]
charlie
Senior Member
- Location
- Indianapolis
Re: Section 110.9 & 110.10
Glenn, I finally had some time to do this but I have not checked your steps. I recommend you use the step-by-step method and go step by step to get your answer.
This is what I got:
Step 1
f=(25000)(480)(1.732)(4.6)/(100000)(75)
f=12.74789
I multiply all the terms together before dividing them
Step 2
M=1/1+12.74789
M=.999734
Step 3
Isc=(V pri/V sec)(M)(Isc pri)
Isc=(480/208)(.999734)(25000)
Isc=4196.446 amperes or 5 kA
I trust this helps and that I have not erred in my calculations (I just changed my answeragain, I was wrong the first three times).
[ December 04, 2003, 12:40 PM: Message edited by: charlie ]
Glenn, I finally had some time to do this but I have not checked your steps. I recommend you use the step-by-step method and go step by step to get your answer.
This is what I got:
Step 1
f=(25000)(480)(1.732)(4.6)/(100000)(75)
f=12.74789
I multiply all the terms together before dividing them
Step 2
M=1/1+12.74789
M=.999734
Step 3
Isc=(V pri/V sec)(M)(Isc pri)
Isc=(480/208)(.999734)(25000)
Isc=4196.446 amperes or 5 kA
I trust this helps and that I have not erred in my calculations (I just changed my answeragain, I was wrong the first three times).
[ December 04, 2003, 12:40 PM: Message edited by: charlie ]
Re: Section 110.9 & 110.10
Charlie,
The Point to Point method ( for System B , on page 30 of the Bussmann 2001 copywrite SPD booklet ) does not reduce the Percent Impedance of 1.2 to 0.012 in their example.
ie., in my example of 4.6 Percent Impedance is 4.6 , not .046 .
Which one is correct ?
Glenn
Charlie,
The Point to Point method ( for System B , on page 30 of the Bussmann 2001 copywrite SPD booklet ) does not reduce the Percent Impedance of 1.2 to 0.012 in their example.
ie., in my example of 4.6 Percent Impedance is 4.6 , not .046 .
Which one is correct ?
Glenn
pierre
Senior Member
- Location
- Westchester County, New York
Re: Section 110.9 & 110.10
Glen
Using 4.6 for 4.6% is not the correct method of using percentages.
100 (100%)has two zeros, so for4.6% you move the decimal point two places to the left. = 0.046.
10% = 0.10
20% = 0.20
5% = 0.05
75% = 0.75
125% = 1.25
200% = 2.00
etc... etc...
Pierre
Glen
Using 4.6 for 4.6% is not the correct method of using percentages.
100 (100%)has two zeros, so for4.6% you move the decimal point two places to the left. = 0.046.
10% = 0.10
20% = 0.20
5% = 0.05
75% = 0.75
125% = 1.25
200% = 2.00
etc... etc...
Pierre
Re: Section 110.9 & 110.10
At present, I am still sticking to the 4.6 Percent Impedance of the 75 KVA, 480V at 25000 ASCC to 208V 3? 2nd transformer in the system as per pages 44 and 45 ( Fault X3 in System B on page 45 ) of the Bussmann SPD 2002 booklet reorder item No. 3002 on the back outside cover.
The 100,000 instead of 1,000 in finding " f " compensates for the ( 0.046 % Z ) to 4.6%Z.
The 1000 of the 100,000 is or the KVA compensation and the 100, of the 100,000 is for the %Z .
The Formula I used is the same as the Bussmann formula on page 45.
It is just re-arranged so that continuous in-put to the calculator until the last equal sign.
No copying down numbers and reusing those numbers in a subsequent formula.
Try it a couple of times.
I also use another sequence for not over-flowing the calculator with large numbers.
Can also do the other Point-To-Point formulas with continuous calculator inputs.
gwz2
[ December 05, 2003, 05:07 AM: Message edited by: gwz2 ]
At present, I am still sticking to the 4.6 Percent Impedance of the 75 KVA, 480V at 25000 ASCC to 208V 3? 2nd transformer in the system as per pages 44 and 45 ( Fault X3 in System B on page 45 ) of the Bussmann SPD 2002 booklet reorder item No. 3002 on the back outside cover.
The 100,000 instead of 1,000 in finding " f " compensates for the ( 0.046 % Z ) to 4.6%Z.
The 1000 of the 100,000 is or the KVA compensation and the 100, of the 100,000 is for the %Z .
The Formula I used is the same as the Bussmann formula on page 45.
It is just re-arranged so that continuous in-put to the calculator until the last equal sign.
No copying down numbers and reusing those numbers in a subsequent formula.
Try it a couple of times.
I also use another sequence for not over-flowing the calculator with large numbers.
Can also do the other Point-To-Point formulas with continuous calculator inputs.
gwz2
[ December 05, 2003, 05:07 AM: Message edited by: gwz2 ]
Re: Section 110.9 & 110.10
Let's do the Bussmann example for a second transformer in a system as on page 44 and 45 of their 2003 SPD booklet, Fault X3, but using the 'continious step ' formula where:
Pascc = ASCC at Primary is 30,059A.
Kva = 225 KVA.
Pv = Primary voltage = 480V.
3? = 1.732 multiplier.
Sv = Secondary voltage = 208V.
Sascc = ASCC at the Secondary terminals.
%Z = Percent Impedance = 1.2 (Charlies would be 0.012 ).
M+ = Memory in.
Mr = Memory Recall.
/ = divide.
* = multiply.
+ = add.
Now using the formula;
Pascc / 100,000 * Pv * 1.732 * %Z / Kva = + 1 = M+ 1 / Mr = * Pascc / Sv * Pv = _______________ Sascc.
30,059A / 100,000 * 480V * 1.732 * 1.2 / 225 = + 1 = M+ 1 / Mr = * 30,059A / 208V * 480V = 29,735.572 Sascc.
This is slightly 'off' of the SPD booklet value of 29,731A because of not losing or adding by rounding-up or rounding-down in-between steps.
Note this is LESS than the Primary ASCC of 30,059A.
Now using the %Z value of 0.012 ;
30,059A / 100,000 * 480V * 1.732 * 0.012 / 225 = + 1 = M+ 1 / Mr = * 30,059A / 208V * 480V = 68454.58 Sascc,
which is more than twice the Primary ASCC and is NOT what the SPD example shows of 29,731A.
With the Continious Step formula, there is not rounding-up or rounding-down except for the final answer.
gwz2
Let's do the Bussmann example for a second transformer in a system as on page 44 and 45 of their 2003 SPD booklet, Fault X3, but using the 'continious step ' formula where:
Pascc = ASCC at Primary is 30,059A.
Kva = 225 KVA.
Pv = Primary voltage = 480V.
3? = 1.732 multiplier.
Sv = Secondary voltage = 208V.
Sascc = ASCC at the Secondary terminals.
%Z = Percent Impedance = 1.2 (Charlies would be 0.012 ).
M+ = Memory in.
Mr = Memory Recall.
/ = divide.
* = multiply.
+ = add.
Now using the formula;
Pascc / 100,000 * Pv * 1.732 * %Z / Kva = + 1 = M+ 1 / Mr = * Pascc / Sv * Pv = _______________ Sascc.
30,059A / 100,000 * 480V * 1.732 * 1.2 / 225 = + 1 = M+ 1 / Mr = * 30,059A / 208V * 480V = 29,735.572 Sascc.
This is slightly 'off' of the SPD booklet value of 29,731A because of not losing or adding by rounding-up or rounding-down in-between steps.
Note this is LESS than the Primary ASCC of 30,059A.
Now using the %Z value of 0.012 ;
30,059A / 100,000 * 480V * 1.732 * 0.012 / 225 = + 1 = M+ 1 / Mr = * 30,059A / 208V * 480V = 68454.58 Sascc,
which is more than twice the Primary ASCC and is NOT what the SPD example shows of 29,731A.
With the Continious Step formula, there is not rounding-up or rounding-down except for the final answer.
gwz2
charlie
Senior Member
- Location
- Indianapolis
Re: Section 110.9 & 110.10
I sent an e-mail to a person at Bussmann stating "I think your example is incorrect for fault X3 on Page 45 (6 of 10), bottom R/H of the SPD." The answer I got back said, "The formula is correct as written. It already accounts for the decimal change."
Therefore, I have edited my calculations again.
I sent an e-mail to a person at Bussmann stating "I think your example is incorrect for fault X3 on Page 45 (6 of 10), bottom R/H of the SPD." The answer I got back said, "The formula is correct as written. It already accounts for the decimal change."
Therefore, I have edited my calculations again.
charlie
Senior Member
- Location
- Indianapolis
Re: Section 110.9 & 110.10
Glenn, I am not sure what I had a problem with but it is fixed now and matches your calculations. I suspect it is the problem of having lower available fault current on the secondary than is at the primary of the transformer. That would be because of the high impedance of the transformer.
Glenn, I am not sure what I had a problem with but it is fixed now and matches your calculations. I suspect it is the problem of having lower available fault current on the secondary than is at the primary of the transformer. That would be because of the high impedance of the transformer.
Re: Section 110.9 & 110.10
Just for checking the " Secondary ASCC always being higher than the Primary ASCC, I used a value of 0.5 %Z in the calculation (which I have no idea such a low value %Z is realistic ) in the November 29, 2003, 11:33AM post and got a Secondary ASCC value of 24183.558, which is less than the primary value of 25000.
I also did edit the December 01, 2003, 09:08PM post.
gwz2
Just for checking the " Secondary ASCC always being higher than the Primary ASCC, I used a value of 0.5 %Z in the calculation (which I have no idea such a low value %Z is realistic ) in the November 29, 2003, 11:33AM post and got a Secondary ASCC value of 24183.558, which is less than the primary value of 25000.
I also did edit the December 01, 2003, 09:08PM post.
gwz2
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