appliance PROBLEM voltage drop 220-230ft run HELP!

[gar]

090418-1525 EST

mivey:

Why is experiment #3 so different? Because it was something convenient for me to show the large voltage drop that occurs at a typical motor under startup conditions and a long #12 wire. With the current blade on the machine startup is quite quick, probably less than 1 second. With a maximum blade the startup time is several seconds and maybe 70 amps.

I am going to guess that the voltage at the freezer may drop to a value in the range of 70 to 90 V during startup. If this is the case and it lasts for a number of seconds, then it might also trip the breaker at the end of startup.

Some voltage measurements from the freezer installation will be important information.

I can not tell you what the voltage is when under full load on the saw. However, on thick wood you can easily stall the motor if you want. This has nothing like the capability of some guys that connected a chain saw to a V-8 engine and cut thru about a 2' diameter tree in few seconds.
See: http://www.youtube.com/watch?v=_owXd_MsdXI

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[mivey]

090418-1525 EST

mivey:

Why is experiment #3 so different? Because it was something convenient for me to show the large voltage drop that occurs at a typical motor under startup conditions and a long #12 wire. With the current blade on the machine startup is quite quick, probably less than 1 second. With a maximum blade the startup time is several seconds and maybe 70 amps.

I am going to guess that the voltage at the freezer may drop to a value in the range of 70 to 90 V during startup. If this is the case and it lasts for a number of seconds, then it might also trip the breaker at the end of startup.

Some voltage measurements from the freezer installation will be important information.

I can not tell you what the voltage is when under full load on the saw. However, on thick wood you can easily stall the motor if you want. This has nothing like the capability of some guys that connected a chain saw to a V-8 engine and cut thru about a 2' diameter tree in few seconds.
See: http://www.youtube.com/watch?v=_owXd_MsdXI

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I completely missed the note about the saw being different. I'm guessing 116-117 volts when sawing normally. I guess the stall gets you back to 92 volts.

I would think that continually staring the motor/compressor at such a low voltage would dramatically reduce the life. I don't have my book handy, but it probably is on the order of the square of the voltage. I'm actually surprised that it was running for very long and I would not put my stuff in that freezer until the supply was fixed.

 

[gar]

090418-2018 EST

mivey:

I will take my handy 1500 W heater and measure the resistance of that #12 circuit.

We really should run a 240 circuit out to that location for the saw and convert it to 240. But we only limitedly use the saw and therefore anything like that gets put off. Also there is the inconvenience of changing the motor wiring.

My guess is the loop resistance is about 1.5*0.25 = 0.375 ohms, and that puts the loop voltage drop for full load at about 17*0.375 = 6.38 V. No problem for the DeWalt for cutting, however, it is a big problem with a large blade and startup.

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[Rick Christopherson]

Hey Rick.

What inputs did you use? I used the IEEE exact VD method ...

I prefer to use Ohm's Law directly instead of some shortcut. Even though there is not complete agreement by all sources for the resistivity of #10 copper, it is pretty universal as being just below 1 ohm per 1000 feet.

Regardless of the exact numbers, the voltage drop in the line is actually trivial, so I would be looking for another problem, or at least a combination of voltage problems beyond just this single run.

I don't know if I am reading this correctly, but if the OP is seeing 90 volts at the motor after the motor is up to speed, then I strongly believe that the start cap is not dropping out after run-speed is reached.

 

[gar]

090418-2120 EST

Rick:

I believe the OP referred to a calculated 90 V at the motor using an LRA (locked rotor amps -- I assume is his meaning) of an unspecified value.

I do not know that 90 V under locked rotor conditions should be considered bad.

It would be helpful if there were some actual voltage measurements made.

I agree with you on 1 ohm at 1000 ft for #10 copper. This is the 20 deg C value given in "Reference Data for Radio Engineers" by ITT. I believe Belden will have the same value for solid copper. Slightly higher for stranded.

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[gar]

090419-1208 EST

Experiments on my#12 circuit to the DeWalt saw.

The load current from the 1500 W heater is close to 12 A @ 120 V and I will use 12.0 for this in all the calculations.

A long extension cord, actually two, is used as a voltage test lead extension. Since there is negligible current flow thru the extension cord its resistance is of no importance.

The voltmeter is a Fluke 87 III.


Direct voltage drop measurements from main panel to outlet:

With no load current the voltage reading of the open loop consisting of the neutral wire and the test lead is about 4 MV. This may be inductive coupling due to the inherent loop created by the neutral wire and the test lead wire which creates a one turn coil with a center window of about 15 ft by 100 ft. This is obviously a possible error voltage in the measurements, but for our experiments can be ignored. 4MV in 2500 MV (2.5 V) is insignificant.

At 12 A load the neutral drop was 2.44 V, and the hot drop 2.47 V. The drop across the breaker is 0.07 V.

The calculated4ed resistances are:
Neutral 2.44/12 = 0.203 ohms,
Hot wire 2.47/12 = 0.206 ohms,
Breaker 0.07/12 = 0.006 ohms.


A less accurate method, but easier to perform, using changes in voltage from no load to a known load:

I will call the circuit to be measured phase A, and the opposite phase is B.

First, find an outlet on phase B that has no changing load on it. This effectively provides test leads into the main panel and directly to the neutral bus and the phase B hot bus because there is no current flowing in this circuit other than the voltmeter current.

With my 12 A load on phase A this produced a 0.1 V rise in the phase B reading from 122.5 to 122.6 V. This has a large error because my resolution is only 0.1 V. From this change I can estimate the resistance of the neutral wire from the main panel to the pole transformer center tap at 0.1/12 = 0.08 ohms.

Next I connect my long test lead (extension cord) to a phase A circuit that is not supplied by the circuit breaker to the DeWalt outlet, and on which there is no changing load. I am encountering very little supply line voltage variation and thus I am using one meter that I switch between the points to be measured.

Meter plugged into the DeWalt outlet. No load the voltage is 122.8 and with the 12 A load drops to 117.6 or 5.2 V drop. This drop includes all the impedances back to the pole transformer.

From the extension cord that effectively connects me to the main panel buses the change in voltage is 0.3 V. If I were connected to the output of the DeWalt breaker, then the change is about 0.4 V. Thus, the voltage drop due to the resistance of the #12 wires is about 5.2 - 0.4 = 4.8 V. The calculated loop resistance is 4.8/12 = 0.4 ohms. This correlates well with the direct measurement of 0.203 + 0.206 = 0.409 ohms.

If we assumed that all the transformer impedance was resistive, then we can estimate its impedance as follows:
0.0083 + 0.0083 = 0.0166 ohms for the service drop loop resistance. This is assuming that both the neutral and hot service wires are the same resistance. The voltage drop, with 12 A load, between the phase A hot and neutral buses in the main panel is 0.3 V. 0,2 V of this is in wire resistance and therefore 0.1 is from the transformer. Thus, the transformer internal impedance for 1/2 of the secondary is about 0.083 ohms.

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[mivey]

The calculated loop resistance is 4.8/12 = 0.4 ohms. This correlates well with the direct measurement of 0.203 + 0.206 = 0.409 ohms.

And is closer to the 75 deg C listing in Tables 8 & 9 of the NEC instead of the 0.1588x2 you would get from the 20 deg C value given in "Reference Data for Radio Engineers" by ITT (or 0.1580x2 if you adjusted Table 8 to 20 deg C)

 

[gar]

090418-1552 EST

mivey:

If I use 1.588 ohms per 1000 ft from The ITT book, and 0.409 ohms loop resistance, then I calculate a loop length of 258 feet or a length of 129 ft.

When I traced the approximate conduit length today it appears to be on the order of 120 ft to 130 ft. The building is 100 ft front to rear. The outlet being tested is about 15 ft from one wall and about 12 ft to the ceiling. That gives me 100 ft front to rear. Down the wall to the main is about 8 ft, and there is a jog of 20 ft or so midway across the building. So a reasonably good correlation with the resistance calculation of 129 ft based on 20 deg C. There was very little heating of the wire during the test.

My calculation of a loop for inductive coupling of stray magnetic fields and the use of a very approximate value of 100 ft in that description for one side of a rectangular loop was based on two extension cords totaling 100 ft.

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[mivey]

090418-1552 EST

mivey:

If I use 1.588 ohms per 1000 ft from The ITT book, and 0.409 ohms loop resistance, then I calculate a loop length of 258 feet or a length of 129 ft.

When I traced the approximate conduit length today it appears to be on the order of 120 ft to 130 ft. The building is 100 ft front to rear. The outlet being tested is about 15 ft from one wall and about 12 ft to the ceiling. That gives me 100 ft front to rear. Down the wall to the main is about 8 ft, and there is a jog of 20 ft or so midway across the building. So a reasonably good correlation with the resistance calculation of 129 ft based on 20 deg C. There was very little heating of the wire during the test.

My calculation of a loop for inductive coupling of stray magnetic fields and the use of a very approximate value of 100 ft in that description for one side of a rectangular loop was based on two extension cords totaling 100 ft.

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Makes better sense now.