units of K constant in volt. drop calcs ???

[ToolHound]

Re: voltage drop K constants, units of...

K = 12.9 copper

K = 21.2 aluminum


For you who are enthusiasts for keeping up with the 'units' in your electrical calculations, what are the units of the 'k factor', as
listed above and as used in voltage drop calculations?


My guess is that the units are:

(ohms cmils)/feet

In other words, for copper

K = 12.9 (ohms cmils)/feet


in other words, 'ohms cmils' in numerator, 'feet' in denominator.



What say you ? What do you say are the units ?


--Thanks.

 

[Hv&Lv]

Here is a previous thread that may help.

http://forums.mikeholt.com/showthread.php?t=143399

 

[GoldDigger]

Re: voltage drop K constants, units of...

K = 12.9 copper

K = 21.2 aluminum


For you who are enthusiasts for keeping up with the 'units' in your electrical calculations, what are the units of the 'k factor', as
listed above and as used in voltage drop calculations?


My guess is that the units are:

(ohms cmils)/feet

In other words, for copper

K = 12.9 (ohms cmils)/feet


in other words, 'ohms cmils' in numerator, 'feet' in denominator.



What say you ? What do you say are the units ?


--Thanks.

The constant K depends on the resistivity of the conductor material. The standard units for resistivity (ohm-meters (not a joke!) or ohm-centimeters) assume that you are measuring area and length in the same units. Since we find it easier to measure conductor area in cmils and distance in feet, the K used here has a mixture of units and is therefore a constant multiple of the material resistivity.

 

[ToolHound]

The constant K depends on the resistivity of the conductor material. The standard units for resistivity (ohm-meters (not a joke!) or ohm-centimeters) assume that you are measuring area and length in the same units. Since we find it easier to measure conductor area in cmils and distance in feet, the K used here has a mixture of units and is therefore a constant multiple of the material resistivity.

GoldDigger. Thanks. You are ahead of me. I gotta go look up Resistivity. As far as units of measurement for the K constant, here's some fiddling around with the units in the overall Voltage Drop equation, and the units seem to cancel out to make sense as show below. I'm got my fingers crossed. I may be way off. Not sure. ??????????

 

[ToolHound]

Here is a previous thread that may help.

http://forums.mikeholt.com/showthread.php?t=143399

Hv&Lv. I looked at that other thread. Good information. It's amazing what is stored in all the threads on this site. Thanks. --ToolHound

 

[GoldDigger]

GoldDigger. Thanks. You are ahead of me. I gotta go look up Resistivity. As far as units of measurement for the K constant, here's some fiddling around with the units in the overall Voltage Drop equation, and the units seem to cancel out to make sense as show below. I'm got my fingers crossed. I may be way off. Not sure. ??????????

That is absolutely right. K has the dimensions of ohms (resistance) times circmils (area) divided by feet (length).

If the same system of units were used for all of the measurements, the area and length would partially cancel out and you get (ohms) times (the length unit). And that is exactly how the resistivity of a material is dimensioned. You can verify that from the tables.

If you are ever interested in a real brain teaser, you should look at the units for the resistivity of a thin layer on a surface. It comes out to "ohms per square". When you then ask "per square what" the answer is that it does not matter.
That comes up when you are trying to design a resistor with a deposited film and you want to figure out what the path width and length need to be.

 

[George Stolz]

I say abolish K and just use R. Quicker and easier.

Vd =2RID or 1.73RID

You don't have to think about K if you use Table 8.

 

[Smart $]

...just use R. Quicker and easier.

Vd =2RID or 1.73RID

...

I agree!

 

[ToolHound]

I say abolish K and just use R. Quicker and easier.

Vd =2RID or 1.73RID

You don't have to think about K if you use Table 8.

Seorge Stolz. Thanks. I like quicker and easier, that's always better.

 

[ToolHound]

Example from NECH 215.2

Example from NECH 215.2

Same as you are saying, here's exapmle from 215.2 :

 

[ToolHound]

That is absolutely right. K has the dimensions of ohms (resistance) times circmils (area) divided by feet (length).

If the same system of units were used for all of the measurements, the area and length would partially cancel out and you get (ohms) times (the length unit). And that is exactly how the resistivity of a material is dimensioned. You can verify that from the tables.

If you are ever interested in a real brain teaser, you should look at the units for the resistivity of a thin layer on a surface. It comes out to "ohms per square". When you then ask "per square what" the answer is that it does not matter.
That comes up when you are trying to design a resistor with a deposited film and you want to figure out what the path width and length need to be.

GoldDigger. Thanks. I am surprised sometimes how an equation can start out as big as the Empire State Building but end up as small as a phone booth. The 'Units for the resistivity of a thin layer on a surface' sounds interesting. I probably will need a Big Mac and Milk Shake before I start in on that one.

--ToolHound

 

[ToolHound]

I agree!

Thanks.