I'm trying to install a 3000 Amp, 6 V DC Rectifier with some existing bus bars to some existing electoplating vats. The Rectifiers have bus bar connetions but there is about five feet between the existing bus bars and the rectifier. What size conductors (multiple) could be used to make the connection and what table is that located in in the NEC book?
DC Rectifier
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Re: DC Rectifier
Sluker, how many rectifiers are involved, and what are there sizes in amps? I suspect there are several rectifiers operating in parallel. If that is the case you use the size of the rectifier to determine the size cable for each rectifier. You might be able to use free air for the conductor between rectifier and bus bar Table 310.17.
For example if you have 6 rectifiers rated at 500 amps each you would use a 500 MCM to connect rectifier to the bus.
Sluker, how many rectifiers are involved, and what are there sizes in amps? I suspect there are several rectifiers operating in parallel. If that is the case you use the size of the rectifier to determine the size cable for each rectifier. You might be able to use free air for the conductor between rectifier and bus bar Table 310.17.
For example if you have 6 rectifiers rated at 500 amps each you would use a 500 MCM to connect rectifier to the bus.
Re: DC Rectifier
Sluker, here is my reply to your e-mail.
I design DC plants for telephone offices at 48 VDC @ 10,000 amps. We determine the cable size by voltage drop at DC and not current. In your case 6 volts is extremely low so voltage drop will be critical with the high current. The formula is:
CM = (22.2 x I x L)/VD
Where
CM = Circular Mills of Cable
I = Load current in Amps
L = Length one-way distance in feet
VD = Voltage Drop
After you perform the calculation you will need to compare it to the appropriate table in 310 to make sure you do not exceed the requirements. This is not normally a problem unless the length is very short; say less than 10 to 20 feet.
By looking at Table 310.17 (free-air) the minimum size using four cables is 700 MCM. My guess for voltage drop would be 1/8 volt or less.
So lets run a sample.
I = 3000
VD = 1/8 = .125
Length one way = 10 feet
CM = (22.2 x 3000 x 10)/.125 = 5328 KCM
Take 5328 KCM divide by 4 = 1332 KCM. Closest match is 4-1250 KCM cables. This exceeds the minimum requirement of even Table 310.16 @ 75 degree table.
Hope this helps.
Sluker, here is my reply to your e-mail.
I design DC plants for telephone offices at 48 VDC @ 10,000 amps. We determine the cable size by voltage drop at DC and not current. In your case 6 volts is extremely low so voltage drop will be critical with the high current. The formula is:
CM = (22.2 x I x L)/VD
Where
CM = Circular Mills of Cable
I = Load current in Amps
L = Length one-way distance in feet
VD = Voltage Drop
After you perform the calculation you will need to compare it to the appropriate table in 310 to make sure you do not exceed the requirements. This is not normally a problem unless the length is very short; say less than 10 to 20 feet.
By looking at Table 310.17 (free-air) the minimum size using four cables is 700 MCM. My guess for voltage drop would be 1/8 volt or less.
So lets run a sample.
I = 3000
VD = 1/8 = .125
Length one way = 10 feet
CM = (22.2 x 3000 x 10)/.125 = 5328 KCM
Take 5328 KCM divide by 4 = 1332 KCM. Closest match is 4-1250 KCM cables. This exceeds the minimum requirement of even Table 310.16 @ 75 degree table.
Hope this helps.
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