Voltage Drop - Continuous loads?

[gilly]

So I have a question on how to figure vd on a continuous load. My understanding of the code is that I would do my voltage drop calc and compare that back to the size of wire needed for My continuous load and verify my wire is larger. If that is not explained very well, here is my math.

480v 3phase load - 30kva XFMR - so I end up with 36.1 amps
I am 3105' distance.

so when I start my voltage drop calc I used 37amps just to make sure I am covered. When all the math is finished I figured a 4/0 is needed to feed this XFMR. I just had an engineer tell me that I am missing the 125% on the wire size. Where is this coming from? Is this just something that this particular firm would do? Has anyone else run into this? I am a little frustrated since origionally the IE had a wire size of 500mcm - so being a dumb electrition I sent our an RFI and proposed a solution of 4/0. The wording of the reply did not sit well with me, but the IE did state that they would lower the size to 350's, but I can not get that math to work out. Any input would be great...

 

[GoldDigger]

So I have a question on how to figure vd on a continuous load. My understanding of the code is that I would do my voltage drop calc and compare that back to the size of wire needed for My continuous load and verify my wire is larger. If that is not explained very well, here is my math.

480v 3phase load - 30kva XFMR - so I end up with 36.1 amps
I am 3105' distance.

so when I start my voltage drop calc I used 37amps just to make sure I am covered. When all the math is finished I figured a 4/0 is needed to feed this XFMR. I just had an engineer tell me that I am missing the 125% on the wire size. Where is this coming from? Is this just something that this particular firm would do? Has anyone else run into this? I am a little frustrated since origionally the IE had a wire size of 500mcm - so being a dumb electrition I sent our an RFI and proposed a solution of 4/0. The wording of the reply did not sit well with me, but the IE did state that they would lower the size to 350's, but I can not get that math to work out. Any input would be great...

Keep the ampacity determination (minimum wire size for conductor heating) separate in you mind from the Voltage Drop (VD) calculation.
Meeting either one alone does not guarantee meeting both.

Your starting point should be the required ampacity determination, taking into account all of the various derating factors that apply. Conductor length will NOT be one of them.
The size conductor this calculation gives you is the minimum size you can use (with a few very limited exceptions for short segments that are part of a longer run.)
Then figure out what voltage drop the engineer will accept as maximum for his particular application, since the NEC does not set a requirement for this.
Next figure out what the voltage drop is for 3105 feet of the code minimum size wire. Depending on the details of the wiring, you may have to double the 3105 foot distance to determine the total length of current carrying wire in the circuit. You can do the resistance per foot calculation or you can use an online VD calculator like the one at the Southwire website.
Almost certainly the voltage drop will be too high, so now you enter the target voltage drop (in volts or percent) and see what wire size it gives you.

If the engineer actually ran the rough calculations himself, you will probably find that the calculator gives 500 if you used his inputs. But maybe the engineer went back and rethought his VD requirements and relaxed things to 350.

When I do the math for 3% VD on a 3100 foot 3-phase circuit in conduit at 37A, I get 250mcm, not 4/0.
But the transformer inrush current will be several times that 37 amps and if the engineer has a large motor load on the secondary of the transformer he may well need to minimize the starting voltage drop, with a current of 4 or more times the continuous current limit. The transformer will probably handle that short term overload just fine, but the 12% or more VD, plus transformer VD, may be hard on the motor if it has to start under load!

If you want to maintain a 3% VD with a current of 120A, the calculator says two 500s per phase.
If the engineer is expecting large unbalanced line to ground loads, the voltage drop would double because the drop in the neutral would have to be counted too.

You cannot make this determination yourself without knowing what the engineer's underlying requirements are.

 

[don_resqcapt19]

The on line calculator that I use gives a voltage drop of 2.2% for a 3 phase, 480 volt, 37 amp load at 3105' using 4/0 copper.

 

[Smart $]

Also keep in mind that continuous loads are "exaggerated" to 125% of actual current. If you must supply a load at 125% for continuous loading, it is only the 100% portion of the current that counts towards the voltage drop.

 

[GoldDigger]

The on line calculator that I use gives a voltage drop of 2.2% for a 3 phase, 480 volt, 37 amp load at 3105' using 4/0 copper.

The Southwire calculator gives a 2.679% voltage drop with exactly the same numbers, using 250, and assuming a PF of .9.
Possibly the difference is the assumed PF, or possibly the assumed temperature (although going to 90C in your calculator only increases the VD to 2.6% with 4/0.
If you increase the allowed VD to 3.1%, Southwire then switches to 4/0 and calculates a 3.06% VD.

If you turn the Southwire calculator around and specify 4/0, it tells you that the maximum distance for 3% VD is 3050 feet.

Anybody feel like analyzing the differences in calculation methods between the two calculators?
I would be very cynical to suggest that the Southwire calculator is biased to sell more copper.

 

[Julius Right]

I'm very sorry Don ,but I don't agree with your "on line calculator"
1)rca it is not the ac resistance but "rca = thermal resistance in thermal ohm feet".Neher [and Mc Grath] notation is not "Rca" but "Rac" for ac resistance of a cable [see "The Calculation of Temperature Rise and Load Capability of Cable System" equation no.14 - for instance].rac depends on cable insulation and jacket and surrounding medium thermal resistance [air , Earth],number conductors and other.:happysad:
2)In order to calculate the voltage drop one needs to know cable reactance and p.f. also.

 

[ramsy]

You cannot make this determination yourself without knowing what the engineer's underlying requirements are.

A 2nd opinion from another engineering firm would help, since engineers don't concern themselves with contractor complaints.

Another engineer might use a better xfmr..

A 3?, 30Kva, 1.7%z, 480Y/4160, with #6 Medium Voltage cable, is near 2.5% voltage drop, @ 37A, 3105-feet away.

The medium-voltage hardware might even be cheaper than 3000 feet of #500mcm cable @ 480 volts.

 

[Julius Right]

According to NEC Table 450.3(B) Maximum Rating or Setting of Overcurrent Protection for Transformers 600 Volts and Less (as a Percentage of Transformer-Rated Current) for "current of 9 A or more" primary protection the maximum setting of overcurrent protection of transformer has to be not less than 125% of rated.
Let's say 90 dgr.C rated conductor temperature for 4/0 copper rac=0.07 and X=0.09 ohm/1000 ft and for 350 mcm rac=0.046 and X=0.083 [Kerite Catalogue]. Then
for 3105 ft. cable length and pf=1 VD[4/0]=3.54% but if pf=0.8 VD[4/0]=5.56%.
For 350 mcm pf=1 VD=2.32% and if pf=0.8 VD=4.38%.

 

[gilly]

So if I am understanding everyone correctly I use the amps needed at 100% not 125%. So let's use a motor as an example this time.. 15hp motor still 3phase 480v at 3105' away. Looking at table 430.250 this gives me 21amps. What I would typically do is use 21amps as the voltage to put into the calculator. So if I would put that into my program I get a 2/0 - 2.38%. The southwire app is the same size wire with the % a little different. The calculator that I use has the option to calculate with or without impedance figured, so that could be some of the variation. Do this seem correct?

My next question is what calculation does everyone use when you bid jobs and all you are given are motor horse powers? So if we were to get a print that has a 480v 3phase 100hp motor 1500' away from mcc room. What size of wire would everyone use? What power factor would you use? Do you figure the calculation with impedance or without? Back in the day I was taught to use..... VD=1.732*K*I*D/CM... From what everyone is telling me is that you have to figure more in than what this calculation figures..

Thoughts???

 

[Smart $]

So if I am understanding everyone correctly I use the amps needed at 100% not 125%. So let's use a motor as an example this time.. 15hp motor still 3phase 480v at 3105' away. Looking at table 430.250 this gives me 21amps. What I would typically do is use 21amps as the voltage to put into the calculator. So if I would put that into my program I get a 2/0 - 2.38%. The southwire app is the same size wire with the % a little different. The calculator that I use has the option to calculate with or without impedance figured, so that could be some of the variation. Do this seem correct?

My next question is what calculation does everyone use when you bid jobs and all you are given are motor horse powers? So if we were to get a print that has a 480v 3phase 100hp motor 1500' away from mcc room. What size of wire would everyone use? What power factor would you use? Do you figure the calculation with impedance or without? Back in the day I was taught to use..... VD=1.732*K*I*D/CM... From what everyone is telling me is that you have to figure more in than what this calculation figures..

Thoughts???

When figuring size of wire for voltage drop, you should always use the actual current (amperes) value.

From there, it's as they say, the devil is in the details. Your original post starts out with voltage drop when fed from a specific, but the general output rating of a transformer and no specific load. We can only surmise that you want to supply a load at the rated output.

With a motor, there are other factors involved, most notably the inrush or starting current. You can figure your wire size for voltage drop at FLC, but you also have to account for starting current and voltage drop of both the wire and the supply transformer. I'll defer the detail discussion to a more willing participant...

The voltage drop formula you cite is quite basic but the most widely accepted and used. However, the value of K can easily change depending on the conditions of installation and other details, but that fact seems to be overlooked in general discussions and use. It typically will provide a conservative result. Including other parameters in the calculation is basically an attempt to justify using a smaller conductor.

Your reference to impedance actually amounts to including consideration for power factor in the calculation.

 

[kwired]

So if I am understanding everyone correctly I use the amps needed at 100% not 125%. So let's use a motor as an example this time.. 15hp motor still 3phase 480v at 3105' away. Looking at table 430.250 this gives me 21amps. What I would typically do is use 21amps as the voltage to put into the calculator. So if I would put that into my program I get a 2/0 - 2.38%. The southwire app is the same size wire with the % a little different. The calculator that I use has the option to calculate with or without impedance figured, so that could be some of the variation. Do this seem correct?

My next question is what calculation does everyone use when you bid jobs and all you are given are motor horse powers? So if we were to get a print that has a 480v 3phase 100hp motor 1500' away from mcc room. What size of wire would everyone use? What power factor would you use? Do you figure the calculation with impedance or without? Back in the day I was taught to use..... VD=1.732*K*I*D/CM... From what everyone is telling me is that you have to figure more in than what this calculation figures..

Thoughts???

NEC does not prescribe an acceptable voltage drop. It does suggest 3% and 5% values in an informational note, or "fine print note" in earlier than 2011 editions.

That said you may have specifications, or local amendments to the code that do give you an acceptable level. What is sometimes missed is what are the conditions for achieving that level? Is it acceptable for more short term voltage drop during motor starting? Must a motor have max allowed voltage drop applied when it is running at full load, or at "normal load"? If you have a motor only loaded to 85% under normal conditions allowable voltage drop on the circuit will not be the same as if it were drawing 100% of rated load. Power factor, temperature, phase balance, other loads on same service / feeder will make some difference also.

 

[SMTech]

Yup

Yup

Keep the ampacity determination (minimum wire size for conductor heating) separate in you mind from the Voltage Drop (VD) calculation.
Meeting either one alone does not guarantee meeting both.

Your starting point should be the required ampacity determination, taking into account all of the various derating factors that apply. Conductor length will NOT be one of them.
The size conductor this calculation gives you is the minimum size you can use (with a few very limited exceptions for short segments that are part of a longer run.)
Then figure out what voltage drop the engineer will accept as maximum for his particular application, since the NEC does not set a requirement for this.
Next figure out what the voltage drop is for 3105 feet of the code minimum size wire. Depending on the details of the wiring, you may have to double the 3105 foot distance to determine the total length of current carrying wire in the circuit. You can do the resistance per foot calculation or you can use an online VD calculator like the one at the Southwire website.
Almost certainly the voltage drop will be too high, so now you enter the target voltage drop (in volts or percent) and see what wire size it gives you.

If the engineer actually ran the rough calculations himself, you will probably find that the calculator gives 500 if you used his inputs. But maybe the engineer went back and rethought his VD requirements and relaxed things to 350.

When I do the math for 3% VD on a 3100 foot 3-phase circuit in conduit at 37A, I get 250mcm, not 4/0.
But the transformer inrush current will be several times that 37 amps and if the engineer has a large motor load on the secondary of the transformer he may well need to minimize the starting voltage drop, with a current of 4 or more times the continuous current limit. The transformer will probably handle that short term overload just fine, but the 12% or more VD, plus transformer VD, may be hard on the motor if it has to start under load!

If you want to maintain a 3% VD with a current of 120A, the calculator says two 500s per phase.
If the engineer is expecting large unbalanced line to ground loads, the voltage drop would double because the drop in the neutral would have to be counted too.

You cannot make this determination yourself without knowing what the engineer's underlying requirements are.

Gotta go with this, my only concern would be load, Inrush :?

 

[Julius Right]

At first I agree with smart, of course.
From IEEE Std 141-1993 ch.3.11 Calculation of voltage drops 3.11.1 General mathematical formulas:
The approximate formula for the voltage drop is:
V = IRcosfi? + IX sinfi?
where:
V is the voltage drop in circuit, line to neutral
I is the current flowing in conductor
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
?fi is the angle whose cosine is the load power factor
cosfi? is the load power factor, in decimals
sinfi??is the load reactive factor, in decimals

The voltage drop V obtained from this formula is the voltage drop in one conductor, one way,commonly called the line-to-neutral voltage drop. The reason for using the line-to-neutral voltage is to permit the line-to-line voltage to be computed by multiplying by the following constants:
Single-phase 2 ; Three-phase 1.732
For exact calculations, the following formula may be used:
actual voltage drop= es+ IRcosfi?+ IX sinfi?+SQRT( es^2-(IXcosfi?-IRsinfi?)^2)
where es=sending end or bus voltage

 

[kwired]

So I have a question on how to figure vd on a continuous load. My understanding of the code is that I would do my voltage drop calc and compare that back to the size of wire needed for My continuous load and verify my wire is larger. If that is not explained very well, here is my math.

480v 3phase load - 30kva XFMR - so I end up with 36.1 amps
I am 3105' distance.

so when I start my voltage drop calc I used 37amps just to make sure I am covered. When all the math is finished I figured a 4/0 is needed to feed this XFMR. I just had an engineer tell me that I am missing the 125% on the wire size. Where is this coming from? Is this just something that this particular firm would do? Has anyone else run into this? I am a little frustrated since origionally the IE had a wire size of 500mcm - so being a dumb electrition I sent our an RFI and proposed a solution of 4/0. The wording of the reply did not sit well with me, but the IE did state that they would lower the size to 350's, but I can not get that math to work out. Any input would be great...

I think I see what you are asking here - how to treat the continuous load. Voltage drop doesn't care if the load is continuous or not, it is what it is regardless of duration. Continuous load changes minimum ampacity selection of your conductor - primarily concerning temperature rating of terminations. So if you have a continuous load of 36.1 amps you must have a conductor that is rated at least 36.1 x 1.25 =45(rounded) amps @ 60 or 75 deg C (whichever applies to the terminations). Any conductor larger than that is only required to be larger because of ampacity adjustments or for voltage drop purposes.

But when calculating voltage drop you only need to base it on actual load, which in your case you say the load is a transformer. What kind of load does the transformer supply? If it isn't even fully loaded you may base it on less than 36.1 amps.

 

[gilly]

Thanks for all the input... This is a great forum...