Voltage Drop Formula Question

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vergeront

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Need help solving this problem
A 3-hp pump is installed 220 feet from the 2220-volt main service. What is the voltage drop when using a #12 AWG solid wire.

I need with the formulas

I found the Table 430.248 and Table 8 in chapter 9 but still need help with the formula
 
vergeront said:
Need help solving this problem
A 3-hp pump is installed 220 feet from the 2220-volt main service. What is the voltage drop when using a #12 AWG solid wire.

I need with the formulas

I found the Table 430.248 and Table 8 in chapter 9 but still need help with the formula
Click to expand...

They make a 3-hp motor that operates on 2220-volt service? Did you mean 220-volt?
 
The NEC® recommends a maximum 3% voltage drop for either the
branch circuit or the feeder.
Voltage drop can be reduced by limiting the length of the conductors.
VD = Volts (voltage drop of the circuit)
R = 12.9 Ohms/Copper or 21.2 Ohms/Aluminum (resistance constants for a conductor that is
1 circular mil in diameter and 1 foot long at an operating temperature of 75O C.)
I = Amps (load at 100 percent)
L = Feet (length of circuit from load to power supply)
CM = Circular-Mils (conductor wire size)
2 = Single-Phase Constant
1.732 = Three-Phase Constant

VD = _(_1.732 x R x I x L_)_/ CM
 
vergeront said:
Need help solving this problem
A 3-hp pump is installed 220 feet from the 2220-volt main service. What is the voltage drop when using a #12 AWG solid wire.

I need with the formulas

I found the Table 430.248 and Table 8 in chapter 9 but still need help with the formula
Click to expand...

VD (1-Phase) = 2KID/CM

3% of 230 = 2*12.9*17*220/CM

6.9 = 96492 / CM

CM = 96492/6.9 = 13984

Per table 8, #10 is too small so a #8 would be required to keep the VD to 3% or less.


To your question,

VD = 2*12.9*17*220/6530 = 14.78

14.78/230 is 6.43%

14.78/240 is 6.16%
 
There is a very important issue in using the above voltage drop formulas. You need to figure out what the load current is, and that can be tricky with motors.

You say '3-hp pump', and based on the voltage and tables you can easily estimate the full load current of the motor. From this current you can calculate a value of voltage drop.

However the actual current will depend on the load connected to the motor, and could be less or more than the rated full load current.

More critically: the current drawn by a motor at starting is much greater than the full load current, and thus there will be considerably more voltage drop at starting. The voltage drop will reduce the starting torque, which for some pumps can mean the motor doesn't start.

-Jon
 
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