Voltage Drop Question

Location
MD
Formula:

Single Phase--- (2xKxIxD)/VD

K=Copper 12.90
I= amperage

Question: What does the "K" and the "I" stand for???? Sorry for the stupid question. The math is easy, and it would help me remember the order of the formula if I knew what the "K" and the "I" was short for.
Thanks in advance,
Lisa
 

david

Senior Member
Location
Pennsylvania
Formula:

Single Phase--- (2xKxIxD)/VD

K=Copper 12.90
I= amperage

Question: What does the "K" and the "I" stand for???? Sorry for the stupid question. The math is easy, and it would help me remember the order of the formula if I knew what the "K" and the "I" was short for.
Thanks in advance,
Lisa


https://iaeimagazine.org/magazine/2017/04/05/voltage-drop-formulas/

The k-factor is a multiplier representing the DC resistance for a given size conductor 1,000 feet long and operating at 75°C.
 

Carultch

Senior Member
Location
Massachusetts
Formula:

Single Phase--- (2xKxIxD)/VD

K=Copper 12.90
I= amperage

Question: What does the "K" and the "I" stand for???? Sorry for the stupid question. The math is easy, and it would help me remember the order of the formula if I knew what the "K" and the "I" was short for.
Thanks in advance,
Lisa
What K really is, is the electrical resistivity of the metal in question, with the units re-worked for how we use them in our application (kcmil for area and feet for length, instead of unified on meters). K has units of Ohm-kcmil/ft, while resistivity (rho) has units of Ohm-meters.

K and C are commonly used for generic constants, because constant starts with a K auf Deutsch.
 

kwired

Electron manager
Location
NE Nebraska
https://iaeimagazine.org/magazine/2017/04/05/voltage-drop-formulas/

The k-factor is a multiplier representing the DC resistance for a given size conductor 1,000 feet long and operating at 75°C.
Yes, reality is operating temp won't be 75C but I imagine it is common practice to go with that anyway as that is typical termination temp rating and an assumed worst case scenario. Most VD calculations are done to figure voltage drop with a worst case targeted low volts allowed. If you want precision VD for some reason you would need to know actual temperature and adjust K accordingly.

If you ran 90C conductor in an actual 76 - 90C ambient, it should result in even more VD than what you calculated at 75C.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
I don't remember formulas very well, so I just do the calculation from first principles, i.e., Ohm's Law.

Vd = IR, where I is current and R is equal to the resistance per 1000 feet divided by 1000 (r) times twice the distance (D).
Vd = 2DIr
%Vd = (Vd/V)(100%)
 

Gary11734

Senior Member
Location
Florida
I don't remember formulas very well, so I just do the calculation from first principles, i.e., Ohm's Law.

Vd = IR, where I is current and R is equal to the resistance per 1000 feet divided by 1000 (r) times twice the distance (D).
Vd = 2DIr
%Vd = (Vd/V)(100%)
When I got in the trade, we used 10.4 for copper, and 17.0 for aluminum for k.

Anybody else used this? Was it based on a different temperature?

I use to think it was the copper not as pure as before but now...
 

kwired

Electron manager
Location
NE Nebraska
When I got in the trade, we used 10.4 for copper, and 17.0 for aluminum for k.

Anybody else used this? Was it based on a different temperature?

I use to think it was the copper not as pure as before but now...
Probably mostly due to different temperature. Change in conductor alloy could factor in, though I think that chance of change over the years is better with aluminum than it is with copper.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
When I got in the trade, we used 10.4 for copper, and 17.0 for aluminum for k.

Anybody else used this? Was it based on a different temperature?

I use to think it was the copper not as pure as before but now...
Why wouldn't you just calculate Vd from Ohm's Law and Table 9 data? It only takes a couple of minutes max, the calculation is simple, you don't have to memorize k values, and when it's right you know it.
 

Gary11734

Senior Member
Location
Florida
Why wouldn't you just calculate Vd from Ohm's Law and Table 9 data? It only takes a couple of minutes max, the calculation is simple, you don't have to memorize k values, and when it's right you know it.

Your right.
I can't remember if a 1968 codebook had a different k for the wire in Chapter nine for 1000 MCM or not.
I did google those numbers, 10.4 and 17.0 K and found some references that have been used but can't find why...
I just assumed for years that when I saw the change it was a purity change of Alum and copper being established.
 

kwired

Electron manager
Location
NE Nebraska
Why wouldn't you just calculate Vd from Ohm's Law and Table 9 data? It only takes a couple of minutes max, the calculation is simple, you don't have to memorize k values, and when it's right you know it.
Table 9 or table 8?

I tried a hypothetical situation out both ways and got same result, rounded to nearest tenth anyway.

One thing one must remember is the table 8 resistance values are for 1000 feet, so if you are figuring drop for 300 feet like I did in my hypothetical situation, you need to multiply table 8 resistance by .300 before solving your ohm's law based equation.
 

Gary11734

Senior Member
Location
Florida
Table 9 or table 8?

I tried a hypothetical situation out both ways and got same result, rounded to nearest tenth anyway.

One thing one must remember is the table 8 resistance values are for 1000 feet, so if you are figuring drop for 300 feet like I did in my hypothetical situation, you need to multiply table 8 resistance by .300 before solving your ohm's law based equation.
All of my apprenticeship classes we used 10.4 and 17.0 for all the exams. It was the k for testing all problems. I wonder where they got that from?

When I googled it, I found some 10.4 and 17.0 references. So, It cant be some bastard number. They had a reason for using these numbers. Hmmmm. I'm sure all my teachers are dead by now or I would ask them! I may never know...
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Table 9 or table 8?

I tried a hypothetical situation out both ways and got same result, rounded to nearest tenth anyway.

One thing one must remember is the table 8 resistance values are for 1000 feet, so if you are figuring drop for 300 feet like I did in my hypothetical situation, you need to multiply table 8 resistance by .300 before solving your ohm's law based equation.
Not really; the resistance per 1000' (r) has a divide by 1000 built into it.

If D = 150', I =10A, and r = 0.31 Ohms/1000' for Cu AWG#4
Vd = IR = 2DIr = (2)(150')(10A)(0.31 Ohms/1000') = 0.930V
If V = 240V
%Vd = (Vd/V)(100%) = (0.93V/240V)(100%) = 0.388%

Easy peasy.
 

Carultch

Senior Member
Location
Massachusetts
Table 9 or table 8?

I tried a hypothetical situation out both ways and got same result, rounded to nearest tenth anyway.

One thing one must remember is the table 8 resistance values are for 1000 feet, so if you are figuring drop for 300 feet like I did in my hypothetical situation, you need to multiply table 8 resistance by .300 before solving your ohm's law based equation.
Table 9 includes factors that are specific to AC.
Table 8 is DC resistance only.

The DC resistance is the value that is resistance by strict definition, the one that Ohm's law in its pure form considers. The "AC resistance" accounts for the fact that inductance effects play a role in voltage drop, when current is time-varying. For sizes #1 and smaller, you generally can neglect this difference. For sizes #1/0 and larger, it becomes significant.

The reason one might want to use the K-factor instead of the tables, is so you can keep track of the general trend in how resistance depends on wire size. Rather than solving for Ohms/kft and doing a reverse lookup, you solve for kcmil as a direct algebra problem.
 

kwired

Electron manager
Location
NE Nebraska
'K' comes from the resistance of a wire 1 foot long and 1 circular mil cross section. It changes with temperature.

-Jon
I don't fully understand table 9, in particular the heading with "ohms to neutral per 1000 feet".

With a three phase three wire circuit I could care less about any neutral, so what exactly is the values in the table?
 
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