Request for data

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090427-0818 EST

I would like to get experimental data from anyone willing to run the experiment.

The experiment is as follows:
1. Residential 50 to 200 A service with 240/120 transformer. In other words has a neutral (a center tapped transformer secondary).
2. Test load about 12 A at 120 V (a typical 1500 W heater).
3. A digital meter capable of resolving 0.1 V on a range that includes 120 V capability.
4. Classify one phase as A and the opposite one as B.
5. Pick any outlet for phase A.
6. Connect the meter to this phase A outlet.
7. Measure and record the voltage with and without the test load plugged into this same outlet. Implication is that this is a duplex so that we are not measuring the load caused voltage drop across the load 's plug. This does measure all voltage drops starting at the transformer and including the rear connections to the duplex outlet.
8. Find a different phase A outlet that is not on the same breaker as the outlet in (7). Measure the voltage at this second phase A outlet with and without the test load connected to its outlet of (7). What this test does is essentially measure the phase A voltage at the main panel with and without the load connected to phase A. In other words all the drops from the transformer to the main panel.
9. Last find a phase B circuit and connect the voltmeter to it. Then do the same as in (8).
10. If possible record the pole transformer size, wire size, kind (Al or Cu), and length from transformer to meter.

What do I expect:

In experiment 9 the reading should not change or it should increase with the load connected. At my home I get between 0 and 0.1 V increase. At two other homes I got 3.5 V, and 1.4 V. These seem high to me. The transformer to main panel wire length is about 70 ft in both cases.

In experiments 7 and 8 I expect the voltage to drop, and experiment 8 should drop less than experiment 7. In my home and experiment 7 and an outlet very close to the main panel I get a 1.1 V drop. Exp 8 produces 0.6 to 0.7 V drop.

The voltage drop in experiment 9 provides a measure of the resistance of the neutral path. So for for my home it is less than 0.1/12 = 0.008 ohms. For my home transformer impedance, wire resistance, meter, and fuse resistances total about 0.7/12 = 0.058 ohms and subtracting the neutral path produces 0.050 for the hot path.

In both of the other homes the total wire length from the transformer to the main panel was about 70 ft. Based on this length I believe the neutral drops are too high. So is there something wrong with the design of my experiment or are there some high resistance connections on the neutrals of these two other locations?

I believe my experimental method is a useful tool for troubleshooting supply problems. Data from anyone else and thus different locations can be useful in determining the effectiveness of this approach.

.
 
#7: phase A = 121.4V, 118.2V at 11.5A
#8: phase A = 121.3V, 120.5V with 11.8A on #7
#9: phase B = 123.4V, 123.9V with 13.1A on #7

75 kVA pad, 190 ft to lot stub point (? wire size/type), another 225 to meter (? wire size/type), another 6 ft to (2) 200A panels (each fed with 4/0 AL).
 
090427-1126 EST

mivey:

Thank you. Looks interesting.

At the present I am ignoring transformer primary impedance and assuming that it has no effect on the #9 results. Basically #9 provides a means to measure the neutral resistance from the transformer center tap to neutral bus in the main panel.

From your measurements I calculate the neutral resistance as 0.5/13.1 = 0.038 ohms.

The calculated resistance of the hot side from the transformer to phase A hot bus and including neutral is 0.8/11.8 = 0.068 ohms. So the hot alone is 0.03 ohms. I might expect it to be slightly higher than the neutral alone.

From the main panel phase A hot bus to phase A outlet is about 4.2/11.5 = 0.365 - 0.03 or about 0.335 ohms. Some of this is breaker drop.

Based on the size of the transformer and the length of the service lines I suspect that the transformer impedance is a small part of the above calculated values.

420 ft of 0000 copper is 0.049*0.42 = 0.0206 and calculated for Al is 0.206/0.62 = 0.033 ohms. Close to the above calculations from your voltage change measurements.

Thanks again.

.
 
Here are the readings from my SureTest 61-165:
Receptacle A1 (load-toaster oven):
VD = 3.2% @ 12 A
VD = 4.0% @ 15 A
Zhot = 0.21 ohms
Zneutral = 0.11 ohms
Zground = N/A (GFI trips)
ASCC1 = VHN / (Hot ohms + Neu ohms) = 390 A
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = N/A (GFI trips)

Receptacle A2 (microwave):
VD = 2.9% @ 12 A
VD = 3.9% @ 15 A
Zhot = 0.17 ohms
Zneutral = 0.07 ohms
Zground = 0.06 ohms
ASCC1 = VHN / (Hot ohms + Neu ohms) = 510 A
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = 598 A

Receptacle B (kitchen counter reference):
VD = 3.4% @ 12 A
VD = 4.5% @ 15 A
Zhot = 0.19 ohms
Zneutral = 0.09 ohms
Zground = N/A (GFI trips)
ASCC1 = VHN / (Hot ohms + Neu ohms) = 443 A
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = N/A (GFI trips)
 
Here are the readings from my SureTest 61-165:...
I may have written the Vdrops down wrong so I re-read them and added some panel readings from the load side of the main breaker:

Here are the readings from my SureTest 61-165:
Receptacle A1 (load-toaster oven):
VD = 3.1% @ 12 A
VD = 3.9% @ 15 A
VD = 5.2% @ 20 A
Zhot = 0.21 ohms
Zneutral = 0.11 ohms
Zground = N/A (GFI trips)
ASCC1 = VHN / (Hot ohms + Neu ohms) = 390 A
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = N/A (GFI trips)

Receptacle A2 (microwave):
VD = 2.4% @ 12 A
VD = 3.0% @ 15 A
VD = 4.0% @ 20 A
Zhot = 0.17 ohms
Zneutral = 0.07 ohms
Zground = 0.06 ohms
ASCC1 = VHN / (Hot ohms + Neu ohms) = 510 A
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = 598 A

Receptacle B (kitchen counter reference):
VD = 2.7% @ 12 A
VD = 3.4% @ 15 A
VD = 4.6% @ 20 A
Zhot = 0.19 ohms
Zneutral = 0.09 ohms
Zground = N/A (GFI trips)
ASCC1 = VHN / (Hot ohms + Neu ohms) = 443 A
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = N/A (GFI trips)

Panel phase A:
VD = 1.2% @ 12 A
VD = 1.5% @ 15 A
VD = 2.0% @ 20 A
Zhot = 0.11 ohms
Zneutral = 0.01 ohms
Zground = 0.00 ohms
ASCC1 = VHN / (Hot ohms + Neu ohms) = 1.0 kA
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = 1.1 kA

Panel phase B:
VD = 1.0% @ 12 A
VD = 1.3% @ 15 A
VD = 1.7% @ 20 A
Zhot = 0.09 ohms
Zneutral = 0.01 ohms
Zground = 0.00 ohms
ASCC1 = VHN / (Hot ohms + Neu ohms) = 1.2 kA
ASCC2 = VHN / (Hot ohms + 1/(1/Neu ohms + 1/ Grd ohms) = 1.3 kA
 
...From the main panel phase A hot bus to phase A outlet is about 4.2/11.5 = 0.365 - 0.03 or about 0.335 ohms. Some of this is breaker drop...
Should be: 3.2/11.5 => 0.273-0.03 => 0.249 ohms
 
you might find this interesting

you might find this interesting

gar,

I took some more volt & amp readings at the panel with added load on phase B:

Reference:
phase A: 123.0 V & 2.33 A
phase B: 124.0 V & 0.52 A
neutral: 2.02 A

With added load:
phase A: 123.5 & 2.33 A
phase B: 123.0 V & 11.82 A
neutral: 9.5 A
 
gar:

Here is my whole house mapped with a Ideal Suretest 61-165 sometime ago.

Ideal_1.jpg


Ideal_2.jpg
 
gar,

I took some more volt & amp readings at the panel with added load on phase B:

Reference:
phase A: 123.0 V & 2.33 A
phase B: 124.0 V & 0.52 A
neutral: 2.02 A

With added load:
phase A: 123.5 & 2.33 A
phase B: 123.0 V & 11.82 A
neutral: 9.5 A
The neutral current to start with is 2.33 - 0.52 ~ 2.

Note here that the change in neutral current is not the change in load current. If I had added the load to phase A, that would be the case (2.33 + 11.3 = 13.63).

Since I added the load to B, which originally had the smaller current, the unbalanced neutral current is now 11.82 - 2.33 = 9.5.

When the highest current shifts from one phase to the other, we can't just use delta V / delta I at the outlet. We have to go back to the common point of the shared neutral to find the source neutral delta current.

Not sure what to do with that info yet.
 
retraction

retraction

When the highest current shifts from one phase to the other, we can't just use delta V / delta I at the outlet. We have to go back to the common point of the shared neutral to find the source neutral delta current.
I had to draw out a circuit to get it clear in my mind and it works even if the maximum current changes phases. So delta V from one phase divided by delta I in the other does give you the primary neutral impedance.

The values I used:
ZpriHot = 0.11
ZpriNeu = 0.08
Ia_pri = 2.3
Ib_pri = 0.5
ZsecA = 0.07
ZsecB = 0.05

Beginning Values:
In = 1.800
Vn = 0.144
Va_panel = 119.747
Vb_panel = (119.945)
Van_panel = 119.603
Vbn_panel = (120.089)
IA = 0.000
IB = 0.000
VA = 119.603
VB = 120.089

Add load on B:
In = 1.800
Vn = (0.736)
Va_panel = 119.747
Vb_panel = (118.735)
Van_panel = 120.483
Vbn_panel = (117.999)
IA = 0.000
IB = 11.000
VA = 120.483
VB = 116.899
Delta VA = 0.880
Delta VB = (3.190)
Delta IA = 0.000
Delta IB = 11.000
Delta VB / Delta IA = #DIV/0!
Delta VA / Delta IB = 0.080

Add load on A:
In = 1.800
Vn = 1.024
Va_panel = 118.537
Vb_panel = (119.945)
Van_panel = 117.513
Vbn_panel = (120.969)
IA = 11.000
IB = 0.000
VA = 115.973
VB = 120.969
Delta VA = (3.630)
Delta VB = 0.880
Delta IA = 11.000
Delta IB = 0.000
Delta VB / Delta IA = 0.080
Delta VA / Delta IB = #DIV/0!

add: The Zsec impedance is the same for the hot & neutral
 
Last edited:
090427-2038 EST

Bill:

I will get back to you later. Too much to digest at the moment.


mivey:

I am ignoring the SureTest for the moment.

Your post #7 correction of my mistake is good. I wish there was an easy way to make corrections in old posts.
"Should be: 3.2/11.5 => 0.273-0.03 => 0.249 ohms". This is maybe about 70 ft from the main panel to the outlet if the wire is #12 copper.

In your post #8 you provide the values:

"Reference:
phase A: 123.0 V & 2.33 A
phase B: 124.0 V & 0.52 A
neutral: 2.02 A"

If all the loads were resistive, then neutral should be 2.33 - 0.52 = 1.81 . Without other information on the phase angles of the load currents we have to guess on what to use in the next part.

I received your PM and yes the majority of the 2.02 neutral current is in phase with phase A voltage, and therefore out of phase with the phase B.

When you add load to phase B:

"With added load:
phase A: 123.5 & 2.33 A ............ no change in load on this phase
phase B: 123.0 V & 11.82 A
neutral: 9.5 A"

Now 11.82 - 2.33 = 9.49 and that is a good correlation with the present neutral reading.

The change in neutral current due to the added load on phase B is 11.82 - 0.52 = 11.3 A, and the change in voltage seen at phase A is an increase of 0.5 V and thus the neutral drop changed by 0.5 V. Thus, calculated neutral resistance is 0.5/11.3 = 0.044 ohms. Not far from the previous calculation of 0.038 using the change in load on phase A. Considering the voltage resolution is 0.1 V this is very good correlation.

An assumption I make is that the neutral wire is the same material and size for the neutral and the hot wires. When I do this I can then assume that the neutral resistance is the same as the hot wire (ungrounded) resistance. Thus, my means of measuring neutral resistance is a means of measuring the ungrounded wire's resistance. If the wire sizes are not the same, then one can estimate on the basis of their known size ratio. Then one has to estimate meter and main breaker resistances. Knowing these one may be able go backward and estimate the transformer impedance if it is a significant factor.


I do not know anything about the SureTest so that data will require some study on my part and questions to find out how it works.


It is late and I have not reread this for errors.

I will try to study some of the other data tomorrow.

.




Reference:
phase A: 123.0 V & 2.33 A
phase B: 124.0 V & 0.52 A
neutral: 2.02 A
current below
 
Gar,

Phase (leg) A (open cir. ) 120.9Vac
Phase (leg) A ( 12 amp load) 119.5Vac
delta V = 1.4 V
Z= 0.12 ohms

Phase (leg) B (open cir. - no load on phase A) 120.7Vac
Phase (leg) B ( open cir- Phase A loaded to 12 amps) 121.5Vac
delta V = 0.8 V
Z (neutral) = .062 ohms

Approx. 100 ft to transformer pad ( no KVA marking on transformer)
4/0 cable to service -AL
Note: Readings taken from two duplex recepticals mounted directly under load center. - No downstream loads -
 
090428-0956 EST

I am Honda power at the moment. On a DTE map our area is indicated as having 5000 to 10,000 locations without power. I do not believe that information. In our area I think it is only our local substation. When load is constant the Honda voltage is very constant but has an instantaneous fluctuation of 0.2 to 0.3 V based on the "TED" voltage monitor in second graphing mode. Actually at the generator the voltage is very good under varying load.


ELA:

I estimate your neutral resistance as:
0.049/0.62 = 0.079 ohms per 1000 ft for Al
or
0.079*0.1 = 0.008 ohms for 100 ft.

Is your neutral 0000?

Following is the theory of my technique for this measurement of neutral impedance.

A power transformer is designed to have low leakage flux relative to all coils. Thus, as a first order approximation the output voltage of one half of a center tapped secondary should be exactly equal to the other half except for a 180 deg phase shift.

If we assume all voltage drop in a secondary is due to the coil resistance, then the equivalent circuit is two equal voltage sources of the same magnitude but opposite phase. In series with each of the voltage sources is the coil internal resistance.

Since I do not load one half of the secondary its hot output voltage is a translation of the voltage at the center tap and therefore a means to get a voltage measurement relative to the center tap. When I load the opposite side a current flows in the neutral producing a voltage drop across the neutral. Measuring the change of voltage from the main panel neutral to the unloaded hot line gives me the change in voltage across the neutral from the main panel to the transformer center tap.

Using this change in voltage and knowing the current change thru the neutral allows a calculation of the neutral resistance.

In your case you have measured 0.8 V change for a 12 A change, or 0.067 ohms. Thus, the question becomes why the difference relative to my calculated value of 0.008 ohms for the neutral wire?

.
 
Gar,
Not sure my neutral is '0000' (4/0) . Been too long since I opened the load center. I am pretty sure the hot legs are 4/0.

I think this method is a good rough approximation of the total impedance but I would not be too concerned with absolute accuracy there.

I am guessing that terminations are accounting for the additional impedance.
 
090429-0652 EST

ELA:

I ran a quick test on a Signal Transformer A41-175. This is a 175 VA unit. With no external lead resistance the voltage of the unloaded side dropped about 1/3 of the drop on the loaded side. This results from the common primary impedance.

If the transformer was wound 1 to 1 from primary to a single secondary, then I would expect primary and secondary resistances to be equal. Next split the secondary into two equal coils. This means each secondary coil has twice the resistance of the primary.

This results in a voltage divider of 1 to 2 when only one secondary is loaded. This correlates with my reading change of 1/3.

In my measurement method the for neutral resistance from transformer center tap to the main panel the error from this reduction of source voltage due to the common primary impedance would produce a lower predicted neutral impedance than the actual impedance. Therefore, this effect should not be the origin of the difference in your measured neutral resistance, large, compared to my estimate of the neutral resistance based on wire size and length, small.

Connection points may be a factor or something I am overlooking in my analysis.


mivey and bill:

I am thinking about the SureTest measurements.


Yesterday was power outage problems and then my wife tripped on somebody's trailer tongue and we had to make a hospital trip.

.
 
090429-0652 EST

mivey and bill:

Yesterday was power outage problems and then my wife tripped on somebody's trailer tongue and we had to make a hospital trip.

.
gar:

Had a friend rushing to retreive his crashed model plane do that too! Broke his nose.
 
090429-2102 EST

I have printed several of the Ideal references for the 61-165.

This apparently has three input leads and is designed to plug into a standard outlet. One input lead is neutral, a second one is EGC, and the third is the hot lead. In many cases one would use a test cord with a 3 prong plug to plug into a 3 prong socket. In fewer cases one would use a test cord with alligator clips.

There is no major power source in this tester. Thus, the hot lead has to be one connection for any high current test. Thus, to measure EGC impedance the test current is supplied from the hot lead.

The test current 12, 15, or 20 A is probably something between 1/2 to several 60 Hz cycles. This would be to minimize power dissipation in whatever current limiting resistor is used. They might even use a very short time pulse. It can not be too long or it would trip the breaker or fuse supplying the circuit under test. Breaker tripping is in addition to the problem of power dissipation within the tester.

When the 3 prong plug connects the tester to the outlet it is including the outlet contact resistance in whatever measurement is made. When the clip leads are used then the clip contact resistance is in the measurement.

There are three possible sources of a voltage measurement --- L-N, L-G, and N-G.

From the combination of the current injection points and the voltage measurements all of the output data has to be derived.

Is my analysis of this instrument correct to this point?

More later.

.
 
gar:

The Suretest doen't display L-G voltage. One issue/problem that it has is that the G impedance test trips a GFCI. I was told that they were trying to get around that but my upgraded unit still does it, so the issue is still there.

I think that I tested the duration and it was around 8ms or about a 1/2 cycle.
 
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