Im trying to plan a large scale solar project . The racks that the solar modules mount to are trackers , they follow the sun. The motors are 12v dc , run off a battery ,with a charger. I have groups of 27 motors per batt location. the total starting current is 15amps per motor . 4a normal load, the battery must be rated for at least 405a . Should I use a large fused disconect at the battery , then use a terminal block with fingersafe fuse holders for each motor . The motors all have quick connect cords [disconect]Im just not familiar with using a battery for such an application and the code does not give any specific info on 12v systems for motors. The size and cost of a 600 a disconect are huge abd i would like to avoid this . Do i need to protect the wiring off of the battery? any advice?
12v dc motors
- Thread starter joshelect
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joshelect said:
The NEC does not give specific info about 12 VDC systems as the normal rules apply.
Motor protection, conductor protection remain the same 480 AC or 12 DC.
opinion t only: I guess my first question would be do all 27 motors start at once. If not, then you are dealing with a running load of less than 12 amps for all motors. IF it is necessary for all motors to run (start) simultaneously, then I would break it up into smaller circuits.
gar
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080622-1133 EST
joshelect:
More information is needed.
What are the shortest and longest cable lengths?
Is it correct that running current is 4/10 A which means the motor load mechanical power is less than 4.8 W or 0.006 HP? This seems very low.
Or was the running current 4 A which is 48 W and 0.06 HP?
What is the ON and OFF time for a motor?
I know how you got 405 A. But what do you mean the battery must be rated for 405 A. Batteries are generally rated for X ampere-hours, energy storage, and at a specific discharge rate. In addition there may be additional ratings for self-dischage time, and high current discharge time.
I have no idea how much average power you will require to move these 27 arrays, but consider some wild estimates. The RPM at each array is maybe 1/2 revolution per 16 hours or RPM = (1/2)*(1/(16*60)) = 1/(2*960) = 0.0005 RPM.
Next assume 10 #-ft of torque, then on a continuous rotation basis HP = 10*0.0005/5252 = 10^-6. Next assume 1% efficiency, then input per array is 10^-4 HP. Next assume 100 arrays and average HP = 10^-2 for all the arrays. That is an average of 7 W for all the arrays.
My guesses may be way off and I might have made a mistake. However, a 400 A supply at 12 V makes no sense. There needs to be much more consideration to the total system design.
If the motor duty cycle is very short, like 1% or less, then never run more than one motor at a time. The total amount of energy is so small that a full size car battery would be adequate.
My wild estimates do sort of confirm your values for motor power.
I have to leave for a while. Please comment.
.
joshelect:
More information is needed.
What are the shortest and longest cable lengths?
Is it correct that running current is 4/10 A which means the motor load mechanical power is less than 4.8 W or 0.006 HP? This seems very low.
Or was the running current 4 A which is 48 W and 0.06 HP?
What is the ON and OFF time for a motor?
I know how you got 405 A. But what do you mean the battery must be rated for 405 A. Batteries are generally rated for X ampere-hours, energy storage, and at a specific discharge rate. In addition there may be additional ratings for self-dischage time, and high current discharge time.
I have no idea how much average power you will require to move these 27 arrays, but consider some wild estimates. The RPM at each array is maybe 1/2 revolution per 16 hours or RPM = (1/2)*(1/(16*60)) = 1/(2*960) = 0.0005 RPM.
Next assume 10 #-ft of torque, then on a continuous rotation basis HP = 10*0.0005/5252 = 10^-6. Next assume 1% efficiency, then input per array is 10^-4 HP. Next assume 100 arrays and average HP = 10^-2 for all the arrays. That is an average of 7 W for all the arrays.
My guesses may be way off and I might have made a mistake. However, a 400 A supply at 12 V makes no sense. There needs to be much more consideration to the total system design.
If the motor duty cycle is very short, like 1% or less, then never run more than one motor at a time. The total amount of energy is so small that a full size car battery would be adequate.
My wild estimates do sort of confirm your values for motor power.
I have to leave for a while. Please comment.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080622-1242 EST
Continuing.
If your 0.4 A running current is correct, then the next question is what is the minimum current that you can apply to the motor and have the motor overcome the breakaway torque of the system. Suppose that this is 0.8 A. Then I might put a series resistor of 12/2 = 6 ohms in series with the battery output. You might need to look at the locked rotor resistance of the motor. It should be less than 12/15 = 0.8 ohms.
This 6 ohm resistor would limit maximum current to 2 A at 12 V, or with the charger running maybe 14/6 or 2.3 A. Following this resistor you have a sequencer that only connects to one motor at a time. 500 ft to the furthest motor and #12 wire would add 1.6 ohms. So make the resistor 5 ohms @ 50 W and this should be a standard item from Ohmite. If you are having to do a fairly accurate calculation, then include the motor resistance. But here I think ballpark figures are sufficiently good.
Use an AGC 5 A fuse at the battery. 10 A if you want.
.
Continuing.
If your 0.4 A running current is correct, then the next question is what is the minimum current that you can apply to the motor and have the motor overcome the breakaway torque of the system. Suppose that this is 0.8 A. Then I might put a series resistor of 12/2 = 6 ohms in series with the battery output. You might need to look at the locked rotor resistance of the motor. It should be less than 12/15 = 0.8 ohms.
This 6 ohm resistor would limit maximum current to 2 A at 12 V, or with the charger running maybe 14/6 or 2.3 A. Following this resistor you have a sequencer that only connects to one motor at a time. 500 ft to the furthest motor and #12 wire would add 1.6 ohms. So make the resistor 5 ohms @ 50 W and this should be a standard item from Ohmite. If you are having to do a fairly accurate calculation, then include the motor resistance. But here I think ballpark figures are sufficiently good.
Use an AGC 5 A fuse at the battery. 10 A if you want.
.
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