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lukas86

I was asked by my supervisor what was going on here, and was looking for some additional help from anyone who may have thoughts.

I can't really draw a picture here, so I'll describe it as best as possible. There is 120 Volt power supply leading out to a machine/load. Basically, one power supply, one resistor. Theoretically, the voltage on the ground end of this load should be zero; however, it is measured as 6.7 volts. So this is causing problems. The current measured was to be 16 amps just after the load, so on the ground end of the load. The length of the cable going from the load to ground is 600 feet.

I am to look at, and comment on why there is still voltage on the line when it is directly connected to ground, or what can be done to get the 6.7 volts down to 1-1.5 volts which the instruments on the ground end normally function at without malfunctioning. The wire is copper, and "1 ott, (the size of my thumb)" so anywhere from 10mm-20mm in diameter. Any thoughts would be appreciated.

Thanks.
 
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I was asked by my supervisor what was going on here, and was looking for some additional help from anyone who may have thoughts.

I can't really draw a picture here, so I'll describe it as best as possible. There is 120 Volt power supply leading out to a machine/load. Basically, one power supply, one resistor. Theoretically, the voltage on the ground end of this load should be zero; however, it is measured as 6.7 volts. So this is causing problems. The current measured was to be 16 amps just after the load, so on the ground end of the load. The length of the cable going from the load to ground is 600 feet.

I am to look at, and comment on why there is still voltage on the line when it is directly connected to ground, or what can be done to get the 6.7 volts down to 1-1.5 volts which the instruments on the ground end normally function at without malfunctioning. The wire is copper, and "1 ott, (the size of my thumb)" so anywhere from 10mm-20mm in diameter. Any thoughts would be appreciated.

Thanks.

hi ,
im trying toi get my head around this problem you have ,how are the instruments connected ,and what do they do ,and what sort of machine does it supply ,let me know and then i may have an answer for you
 
The machine that requires the power I believe a saw timmer line, or a stacker. Both very large, the timmer line basically cuts edges off the logs that are coming in. The stacker brings the logs from a lower conveyor up onto a higher level conveyor. The meter that was reading the voltage and current readings were just after the machine I believe, just after the connection of the machine to the ground. One electrician did the measurement readings so other than that I am not too sure.

I am not too sure what instruments are on the ground end. He just said "it needs to be close to zero, and not above 1.5V because things malfunction if it is above that."
 
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What supply are we talking about Is it AC or Dc ?
If its Ac what are the earthing arrangements
What is the supply voltage ?
If the machine has a connection to ground What are you comparing it to that enables this voltage to be obseved
I'm confused
 
The machine that requires the power I believe a saw timmer line, or a stacker. Both very large, the timmer line basically cuts edges off the logs that are coming in. The stacker brings the logs from a lower conveyor up onto a higher level conveyor. The meter that was reading the voltage and current readings were just after the machine I believe, just after the connection of the machine to the ground. One electrician did the measurement readings so other than that I am not too sure.

I am not too sure what instruments are on the ground end. He just said "it needs to be close to zero, and not above 1.5V because things malfunction if it is above that."

hi
i am going to attempt an answer now .i presume the motors on the machine is 3 phase,now ,are you monitoring the neutral current ?, ie if it rises then it shows an inballance between the phases ,which would indicate ,for example one phase drawing more current than the rest, is the motor controlled by an inverter?
because this would be a good way of providing a monitoring circuit.

i would test the resistance of all the motor windings,with the windings open ,ie all links removed, use a megger on the ohms scale ,not a multi meter, then do a load test with a clip on ammeter on each phase and study the data,you might get a clue as to whats causing it ,i think

also if its a saw ,make sure its sharp! ,a blunt blade will obviously affect your current readings and be very carefull if the instuments use current transformers ,dont go disconnecting them to substitute with your meter, c.ts are letal,THAT MEANS THAT YOU CAN GET KILLED ,be very careful,please ;)

and let me know how it goes , all the best:)
 
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This is how the circuit was drawn for me by my supervisor,

Code:
 |-----------------------------------------------------|
 |                                                   A |
 |                                                     |
Vs                                                     L
 |                                                     |
 |                                                   B |---J---|
 |                                                     |       |
 |-----------------------------------------------------|       |
 |                                                             | 
 |-------------------------------------------------------------|
 |
GND

Vs is the source power, 120 volt supply. L stands for the load which requires this power. Now apparently between A and B there is 113.3V, leaving 6.7V on the ground side. So J stands for the jumper where another ground cable is temporarily in place to lower that 6.7V. So on the diagram he showed me, there is 6.7V across J, the jumper.


The setup for the load itself looks like this... which when my supervisor gave to me confused me for a few minutes because I had to understand it as well.

Code:
               |----------------------|
               | Optimizer End Column |
|--------------|                      |
|              |                      |
|   A|--N      |                      |
|    |         |                      |
|----O---------|                      |
               |                      |
               |                      |
               |                      |
        |------O                      |
        |      |----------------------|
      B |
        |
        |             16.7A @ 60.0 Hz               To PDC
<-------O----------------------------------------------->

"N" is where the neutral wire is, and the lead coming off from that "A" connects to the machine itself (bolted). "N" also is where the ground and power (from the first diagram) are located. "A" in this diagram is the jumper from the first diagram, and has 5.7A across it, and "B" has 4.2A across it which is the machine ground. This machine ground connects to the main mill ground which has 16.7A @ 60Hz and goes to the PDC.

Now, something is wrong with this because at the moment, the 'ground' should not have current but does, and the jumper is not supposed to be there, but is required otherwise the voltage is too high. Why is this?
 
Ya, my thoughts right there... this is a little extreme for me being here 2 weeks... a student none-the-less haha. I keep getting told different things/aspects of this every day. Like "oh ya, what I told you yesterday is a bit wrong." Then I get confused on what EXACTLY I am supposed to be trying to figure out.
 

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