24vdc wire sizing

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lilsparky

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Hello everybody, glad to be back.

I have a 50 amp load that will be comprised of control voltages (digital inputs and digit outputs) and 50 amps may be a bit conservative it could be more like 30 amps. The problem is that the 24vdc source will be about 70 ft away. This blows up the cable sizing or does it? The instrumentation guy says there shouldn't be a problem with a #10 AWG. I'm not comfortable with that, either I size it at #2 or we provide a 24vdc converter at the load and bring the 110v the 70 ft. He doesn't want to do that because there are spacing issues at the load.

Am I off base?
 
lilsparky said:
Hello everybody, glad to be back.

I have a 50 amp load that will be comprised of control voltages (digital inputs and digit outputs) and 50 amps may be a bit conservative it could be more like 30 amps. The problem is that the 24vdc source will be about 70 ft away. This blows up the cable sizing or does it? The instrumentation guy says there shouldn't be a problem with a #10 AWG. I'm not comfortable with that, either I size it at #2 or we provide a 24vdc converter at the load and bring the 110v the 70 ft. He doesn't want to do that because there are spacing issues at the load.

Am I off base?
Click to expand...

NEC tables and instructions tell you how to calculate voltage drop due to cable length. #2 sounds a bit overkill.
And you are correct,it is wiser to have the DC supply near load; DC power supplies are getting smaller and smaller these days anyways.
 
120514-1050 EDT

#10 copper is about 1 ohm per 1000 ft. Thus, your wire resistance is about 0.14 ohms (2*70 = 140 ft). From 0 load current to a 50 A load is a change of 50 A, and this will produce about a 7 V change at the load end. Is that tolerable?

.
 
most Dc power supplies of that size have a sensing line just for this problem.

you run a 3rd and maybe a 4th wire back from the load and the power supply uses that line as the feedback rather than the output at the terminals.

Most of them can run upto about 28VDC at the terminals. If it really is a 30A load, #10 should be OK if you run the sensing line. personally, I would run #8.
 
As others sugest, it would probably be better to place the DC power supply nearer the load.
7 volts drop as calculated above is almost certainly too much.
In the absence of more detailed information, control circuits should be OK with a voltage of from 90% up to 110% of nominal. But even if you start with 26.4 volts (110% of 24) you will still have much less than 21.6 volts (90% of 24) at the load. And that is presuming that the power supply can be set to 110% of nominal.

I would aim for a minimum of 92% of nominal voltage at the input terminals of the 24 volt load, thereby allowing 2% for drop within the equipment.

Finally are you certain that the control power requirement is so much ? well over 1KVA seems an awful lot for controls ?
 
Ok I'm following the NEC 215 for DC volt drop calculations and chapter 9 table 8 for ohms/1000ft: 70 ft lenght and 50 amps of load

#10 wire: 2x70x1.21x50/1000=8.47 VD=8.47/24 x100=35%

#8 wire: 2x70x.764x50/1000=5.35 VD=5.35/24x100=22%

#6 wire: 2x70x.491x50/1000=3.44 VD=3.44/24x100=14%


I don't see this getting any better. Am I doing this right? Am I missing something?

GAR, can you explain to me how you get a 7v drop at the load end with a #10 wire?
 
Ok I checked with the instrumentation guy and 30amps is his total load. He did a more accurate account of loading on his coils, tranducers and switches. He's going to see if he can live with a 13% drop in voltage (#8 AWG).

I gave him 2 options, one (the one I liked all along) put the dc supply at the load or two boost the supply up to account for the voltage drop.

Thanks again everyone!! as always you have been very helpful!!
 
lilsparky said:
Ok I'm following the NEC 215 for DC volt drop calculations and chapter 9 table 8 for ohms/1000ft: 70 ft lenght and 50 amps of load

#10 wire: 2x70x1.21x50/1000=8.47 VD=8.47/24 x100=35%

#8 wire: 2x70x.764x50/1000=5.35 VD=5.35/24x100=22%

#6 wire: 2x70x.491x50/1000=3.44 VD=3.44/24x100=14%


I don't see this getting any better. Am I doing this right? Am I missing something?

GAR, can you explain to me how you get a 7v drop at the load end with a #10 wire?
Click to expand...

If 0.14 ohms is the correct number, 50 amps* 0.14 ohms = 7 volts. maybe he used the wrong number for resistance.
 
120514-1241 EDT

From my copper wire table the resistance of #10 wire at 20 deg C (room temperature) is about 1 ohm per 1000 ft of wire length. At full load current the resistance will be a little higher. NEC uses a higher temperature than 20 deg C for voltage drop calculations. This makes good sense.

If you have a distance from the source to the load of 70 ft, then there is 140 ft of wire over which the voltage drops. Thus, your loop circuit resistance will be 1 * 140 / 1000 = 0.14 ohms. The voltage drop across an 0.14 ohm resistor with 50 A flowing thru the wire is Vdrop = I * R = 50 * 0.14 = 7 V.

There is 3.5 V drop on each 70 ft length of of #10, but both the + and - wires have a 3.5 V drop. Thus, the total drop is 7.0 V.

.
 
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