Resistivity of Romex (NM) cable

The voltage drop values it gives are not correct. It assumes that you have adjusted the supply voltage to force the current to stay the same. With the supply voltage constant, increased resistance in the wiring causes the current to drop. That calculator does not account for that.

The extreme case I posted shows it more obviously in that the voltage drop exceeds the supply voltage. But even in non-extreme cases, it is still assuming a constant current.

If you already had a situation where you can measure the existing current then you can plug that current in and see what the voltage drop is. But that is a situation where you are perhaps operating with the wrong voltage at the load, and thus the wrong current. If you know what voltage drop is with the wrong current, it's not telling you want to do to get the right current.

|> The program you described is probably overkill (does it |> model the cooling level of airflow). | | Actually, it is quite simple to use. It does not calculate the actual | conductor temperature. It allows you to select the temperature of the | wire (60, 75, or 90 C) and then calculates the voltage drop at that | temperature.

So how will you know what temperature the wire will rise to? It's asking you to actually set up the situation, measure the current and temperature, then it will tell you how much voltage drop you have? If I did set it up and could measure the current and temperature, don't you think I could just as easily measure the voltage at the load and do a simple subtraction?

The usefulness of a calculator is for predicting a situation where you know certain starting parameters like what the load is, what the supply voltage is, and what kind of wire you have (or are considering using). Temperature knowledge would at this point be the ambient temperature. The numbers you gave look more like common insulation ratings rather than expected operating temperatures. A more useful calculator would model the temperature rise over time and how that affects the circuit. But to do that you also have to consider the thermodynamics of the situation, and that is the hard part as you now have to consider so many factors about how the wiring is enclosed. It might be better to just take load characteristics, supply voltage, wire size, and minimum ambient temperature, and plot a graph over temperature rises showing running current, load voltage, voltage drop, and wattage loss.

| In many cases this gives a conservative solution, and further precision is | not | necessary.

But the calculator on that web page isn't giving the correct answer to begin with, because it isn't working in the real situation most people have.

Or plot a graph across the range of lowest case ambient to highest case of insulation breakdown. Or at least list a table for every 5C step.

I think you folks are trying too hard to beat up a guy over a program you didn't pay for. These things are simply to select a wire size and not a test question from a pedantic university bound professor who never saw the real world. If you start with a realistic wire size this will tell you if you are close and which way to go if your choice is wrong. That is all most people want. "Acceptable" voltage drop is a relative thing anyway. It is broadly defined as

Temperature

That is exactly what they are. You are doing a worst-case calculation at the temperature limit of the wire. If it runs cooler, you have a margin of safety.

Bear in mind that the loads that you use as input for most of these calculations are estimates to begin with (i.e. va/sq ft lighting & receptacles, diversity factors, etc.) Actual measurements almost never agree with them. All of the extra complexity and precision that you are proposing for the calculator won't provide results any more accurate than these estimates.

Ben Miller

On 23 Feb 2004 17:41:15 GMT Greg wrote: | I think you folks are trying too hard to beat up a guy over a program you | didn't pay for. These things are simply to select a wire size and not a test | question from a pedantic university bound professor who never saw the real | world.

The program is being pushed on people for a purpose it does not appear to actually be designed for. I simply stated that it is not giving answers that are correct for what people want, by showing cases where it is very far off (and thus not using the correct formula). I was challenged on that and so I will defend my position, unless and until someone shows me where I am wrong (it has happened before, but not yet in this case).

I've done voltage drop calculations before using ohms law. I get different answers than what the online calculator gives.

| If you start with a realistic wire size this will tell you if you are close and | which way to go if your choice is wrong. That is all most people want. | "Acceptable" voltage drop is a relative thing anyway. It is broadly defined as | 3-5% but it really depends on the nature of the load and the supply voltage you | started with. In my neighborhood I see around 125VAC at the lugs in my service | so I can lose several volts without really hurting anything. If my service was | closer to 115 it would be a lot more critical that I don't lose much in my | feeders and branch circuits.

I'll also leave it for people to decide for themselves what is acceptable. They can also measure the exact supply voltage they have before starting any of this. What they can't measure is what the current will be within the planned wiring and expected load, yet the calculator needs that current to be right, even though it asks for the current of the load (which one would expect to read off the nameplate of the device).

| Bear in mind that the loads that you use as input for most of these | calculations are estimates to begin with (i.e. va/sq ft lighting & | receptacles, diversity factors, etc.) Actual measurements almost never agree | with them. All of the extra complexity and precision that you are proposing | for the calculator won't provide results any more accurate than these | estimates.

But you're still saying that a wrong formula is acceptable just because there are other factors that make answer fuzzy. I simply do not agree with that practice (and will refuse to ever hire any engineer who would work that way). Yes, there are times when fuzzy answers have to do, or estimates have to do. But everything those answers are based on needs to be absolutely correct math. If you estimate the load, you should at least get an answer that does not further add more error when a formula can be used that maintains error without adding to it. If your estimate is plus or minus 10%, why add another 10% of fuzziness with the wrong formula?

| Bear in mind that the loads that you use as input for most of these | calculations are estimates to begin with (i.e. va/sq ft lighting & | receptacles, diversity factors, etc.) Actual measurements almost never agree | with them. All of the extra complexity and precision that you are proposing | for the calculator won't provide results any more accurate than these | estimates.

I've decided that my voltage drop calculator program will take in the following information:

1. Supply voltage/configuration.
2. Distance by wire to load
3. Load design nominal voltage
4. Load amperage or wattage

It will then produce a table of figures for each of the common wire sizes in both copper and aluminum. Each row will show wire size, resistance at nominal temperature (suggestions for that?), actual current of the circuit, voltage at load, wattage at load, voltage drop in wiring, and wattage loss in wiring (overall and per meter). Each row will then be a hyperlink to click to get another table listing the same figures over a temperature range of 0C to 100C in 5C steps. A hyperlink for each table will produce a graph of that table in a GIF image. Another hyperlink will produce the table in a spreadsheet loadable form (tab delimited).

At present I will be assuming a power factor of 1. I'll upgrade that later to allow for other power factors. My concern is that power factors less than 1 are usually motors which have large starting currents, and I have not decided how best to have users describe that load. Maybe it would be to specify idle, loaded, and starting current, and worst case power factor. Often loads are a mix. I'm not going to worry about any fuzziness in load estimation based on things like square footage; the user can supply their load information summarized based on their best estimate or actual knowledge.

This would assume the supply voltage is itself unaffected by this new load. A better program would consider even the present supply from its source and the wiring feeding it to show the voltage drop in the present system caused by the added load. Getting this complete will have to consider a lot more issues like multiple wiring gauges, transformer characteristics, etc.

Big snip.

How will your program calculate the actual current?

Let's suppose I have a 300 watt computer power supply rated at 115 volts,

How much current will these components draw, singly, at 120 volts? What will be the total? And what will these values be at 115 volts? At 125 volts?

Hint: None of these loads will obey Ohm's law. Current will NOT be proportional to voltage.

?????????

|> I've decided that my voltage drop calculator program will take in the |> following information: |>

|> 1. Supply voltage/configuration. |> 2. Distance by wire to load |> 3. Load design nominal voltage |> 4. Load amperage or wattage |>

|> It will then produce a table of figures for each of the common wire sizes |> in both copper and aluminum. Each row will show wire size, resistance at |> nominal temperature (suggestions for that?), actual current of the circuit | | Big snip. | | | How will your program calculate the actual current? | | Let's suppose I have a 300 watt computer power supply rated at 115 volts, | 500 watts of incandescent lamps rated at 130 volts, and a 1/3 hp | refrigeration compressor rated at 117 volts. Typical home office circuit, | including a fridge. | | How much current will these components draw, singly, at 120 volts? What will | be the total? And what will these values be at 115 volts? At 125 volts? | | Hint: None of these loads will obey Ohm's law. Current will NOT be | proportional to voltage.

To start with I'm only going to base in on resistive loads, the go from there to upgrade it. Issues will include how to characterize the load ... that is, how a user will describe what they have. That's not an easy thing to figure out.

...sounds like a rather trivial spreadsheet to me. Perhaps I don't understand the issue?

But here you'll run into problems similar to thos of the voltage drop calculator at

On Wed, 25 Feb 2004 12:10:42 -0800 Paul Hovnanian P.E. wrote: | snipped-for-privacy@ipal.net wrote: |> |> On Mon, 23 Feb 2004 19:19:07 GMT Ben Miller wrote: |> |> | Bear in mind that the loads that you use as input for most of these |> | calculations are estimates to begin with (i.e. va/sq ft lighting & |> | receptacles, diversity factors, etc.) Actual measurements almost never agree |> | with them. All of the extra complexity and precision that you are proposing |> | for the calculator won't provide results any more accurate than these |> | estimates. |> |> I've decided that my voltage drop calculator program will take in the |> following information: |> |> 1. Supply voltage/configuration. |> 2. Distance by wire to load |> 3. Load design nominal voltage |> 4. Load amperage or wattage | | But here you'll run into problems similar to thos of the voltage drop | calculator at

Or maybe multiple inputs for how much of each load type in a mix.

Any idea how best to characterize motors? PC power supplies will be fun, too (especially on 3 phase). Imagine that triplens overloading the neutral issue affecting the calculations.