Drop Voltage 240V 60A

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KrisT

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Hi Guys , I have never ran through drop voltage calculations before and now I have a job which requires some calculations.
A customer needs a car charger at distance of 150 feet away from main panel. Usually I use #6awg copper wire and 60A breaker for much shorter runs since the load is 48A. But with a distance of 150 feet my guess is I have to use 1 inch pipe and #4awg or #3 awg wire. I am asking for an advice ,because there is so much information online which I do not trust. Some of the online info says #6 wire is fine up to 188 feet for 240V ,some says you need #3 ,I am ttying to do it right!
Thanks!
 
KrisT said:
Hi Guys , I have never ran through drop voltage calculations before and now I have a job which requires some calculations.
A customer needs a car charger at distance of 150 feet away from main panel. Usually I use #6awg copper wire and 60A breaker for much shorter runs since the load is 48A. But with a distance of 150 feet my guess is I have to use 1 inch pipe and #4awg or #3 awg wire. I am asking for an advice ,because there is so much information online which I do not trust. Some of the online info says #6 wire is fine up to 188 feet for 240V ,some says you need #3 ,I am ttying to do it right!
Thanks!
Click to expand...
You can use a couple different formulas. VD = 2KIL/CM. Use 12.9 for “K” for copper. I = amps L= one way length VD = 2*12.9*48*150 = 187,500/26,240 = 7.07 volts. 7.07/240 = 2.95%. Chapter 9 table 8 gives you the circular mils. You can also find the resistance for #6 under noncoated copper per thousand feet in the same table! The other formula is VD = 2LRI/1,000. L = one way length R = resistance for 1,000 feet of #6 non coated copper from chapter 9 table 8.
VD = 2*150*.491*48/1,000 = 7.07. 7.07/240 = 2.95%
This is for single phase!
 
Sorry ,I should have to tell you it is single phase residential service ,so I use 2 pole breaker for 240V. I am a little bit confused when have to put parameters in calculator. Especially should I do it for 120V or 240V ,since each hot wire carry 120V and then become 240V at the load. Not sure how calculator is set up for if 240V means for what I have or 240V for Hot to Neutral in 3 phase services. Sorry if I make it complicated ,just need to make sure I am doing it right!
 
HEYDOG said:
You can use a couple different formulas. VD = 2KIL/CM. Use 12.9 for “K” for copper. I = amps L= one way length VD = 2*12.9*48*150 = 187,500/26,240 = 7.07 volts. 7.07/240 = 2.95%. Chapter 9 table 8 gives you the circular mils. You can also find the resistance for #6 under noncoated copper per thousand feet in the same table! The other formula is VD = 2LRI/1,000. L = one way length R = resistance for 1,000 feet of #6 non coated copper from chapter 9 table 8.
VD = 2*150*.491*48/1,000 = 7.07. 7.07/240 = 2.95%
This is
Click to expand...
Thanks for explanation ,it is super helpful! since VD is less than 3% is safe to use #6 copper wire.
 
221114-0009 EST

For a charger to do its job, charge a battery, I might expect the input voltage might be able to vary over a wide range. If this is the case, then the current rating of the wire would be the limiting factor. Or possibly how much power was wasted in the supply wire. So voltage drop might not be a factor at all.

.
 
electrofelon said:
I assume the commercial chargers are three phase no?
Click to expand...
The cars accept DC or single phase AC.

The home and standard commercial charger are not really chargers. The actual charger is built into the car. The standard basic device that is plugged in or hardwired and has a cord is not really a charger. It just extends the branch cirucit to the car and has signal wires that tell the cars charger the maximum rate it can charge at.
 
KrisT said:
Sorry ,I should have to tell you it is single phase residential service ,so I use 2 pole breaker for 240V. I am a little bit confused when have to put parameters in calculator. Especially should I do it for 120V or 240V ,since each hot wire carry 120V and then become 240V at the load.
Click to expand...

Simple answer: for 240V loads you use 240V and for 120V loads you use 120V. The voltage to neutral doesn't matter, just the voltage 'seen' by the load.

It sounds like you don't have a solid grasp of what causes voltage drop.

The issue is that the wires themselves are resistors, somewhat blocking the flow of current. Standard 'Ohm's law' applies: voltage = current * resistance

Next the basic rule for voltage is that the total voltage in any complete loop is zero. You have a source (the utility transformer) that creates voltage, and you have a load that consumes voltage, and you have the connecting wires also consuming voltage. (To understand this point it is easier to pretend that you are using DC power, because totaling up the voltages in a loop means considering the sign of the voltage. )

Your wire has a resistance per foot value, the longer it is, the more ohms it has, and the more of the available voltage consumed by the wires.

Jon
 
The first thing when it comes to selecting increased conductor size is determining how much voltage drop is deemed acceptable then design around that.
 
HEYDOG said:
You can use a couple different formulas. VD = 2KIL/CM. Use 12.9 for “K” for copper. I = amps L= one way length VD = 2*12.9*48*150 = 187,500/26,240 = 7.07 volts. 7.07/240 = 2.95%. Chapter 9 table 8 gives you the circular mils. You can also find the resistance for #6 under noncoated copper per thousand feet in the same table! The other formula is VD = 2LRI/1,000. L = one way length R = resistance for 1,000 feet of #6 non coated copper from chapter 9 table 8.
VD = 2*150*.491*48/1,000 = 7.07. 7.07/240 = 2.95%
This is for single phase!
Click to expand...
If the load is balanced you can just use the one way distance and calculate Vd = DIr where r = ohms/1000' and compare it to the line to neutral voltage to get %Vd whether it is single phase or three phase.

No current on the neutral = no contribution to Vd. In three phase the sqrt(3) factors cancel as does the factor of 2 in single phase for the ratios of line to line vs line to neutral voltages.
 
Also is 60 amps actual current or the overcurrent device setting? If actual draw is say 45 amps you will have a 60 amp OCPD but your voltage drop is based on the 45 amps being drawn.
 
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