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Default Find the correct wire size for a load or the load for a selected wire size

I have seen a few of your posters over at alt.engineering.electrical
asking questions about sizing conductors. I have written two very
useful 2005 NEC compliant web page calculators that do what you are
looking for and they are out of the beta stage and in the alpha stage.
In other words, known bugs are fixed. It has taken me about 10 years
on and off and about 2000 hours to write these programs. I am a
licensed electrician in two states and a licensed electrical
administrator in the state of Alaska and have a degree in Mathematics.
The programs are free to use.
The programs are at:
http://www.electrician2.com/calculat...cpd_ver_1.html
http://www.electrician2.com/calculat..._1_reverse.htm

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Default Find the correct wire size for a load or the load for a selected wire size


wrote in message
oups.com...
I have seen a few of your posters over at alt.engineering.electrical
asking questions about sizing conductors. I have written two very
useful 2005 NEC compliant web page calculators that do what you are
looking for and they are out of the beta stage and in the alpha stage.
In other words, known bugs are fixed. It has taken me about 10 years
on and off and about 2000 hours to write these programs. I am a
licensed electrician in two states and a licensed electrical
administrator in the state of Alaska and have a degree in Mathematics.
The programs are free to use.
The programs are at:
http://www.electrician2.com/calculat...cpd_ver_1.html
http://www.electrician2.com/calculat..._1_reverse.htm

At first glance it looks like a good job. Thanks


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Default Find the correct wire size for a load or the load for a selected wire size

What is your source for the cable impedance R+jX values? Do you use a
default power factor or is it user defined?



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Default Find the correct wire size for a load or the load for a selected wire size


# Fred # wrote:
What is your source for the cable impedance R+jX values? Do you use a
default power factor or is it user defined?


The voltage drop calculator uses the equation:
vd=2*K*L*I/CMA
Where K = 12.9 for copper and 21.2 for Aluminum
K is the resistance in ohms of one foot of a one mil conductor at 75
degrees C.
L is one way distance from supply to load in feet
I is current in amperes
CMA is circular mil area of conductor
vd is voltage drop in volts
This is a a very conservative calculation since the load is assumed to
be the same as the OCPD value and the conductor is assumed to be at 75
degrees C.
This is the standard method of calculating voltage drop used by
electricians for the last 50 years and appears in journeyman
electrician tests for several states.
Using the impedance method and knowing the precise load can be more
accurate, but voltage drop is a best guess for the most part anyway.
At the $300 million Healy Power Plant Job in Alaska the engineers
simply specified 460 volt motors to be used on a 480 volt system and
that solved all their voltage drop problems.

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Default Find the correct wire size for a load or the load for a selected wire size


"Jerry" wrote in message
oups.com...

# Fred # wrote:
What is your source for the cable impedance R+jX values? Do you use a
default power factor or is it user defined?


The voltage drop calculator uses the equation:
vd=2*K*L*I/CMA
Where K = 12.9 for copper and 21.2 for Aluminum
K is the resistance in ohms of one foot of a one mil conductor at 75
degrees C.
L is one way distance from supply to load in feet
I is current in amperes
CMA is circular mil area of conductor
vd is voltage drop in volts
This is a a very conservative calculation since the load is assumed to
be the same as the OCPD value and the conductor is assumed to be at 75
degrees C.
This is the standard method of calculating voltage drop used by
electricians for the last 50 years and appears in journeyman
electrician tests for several states.
Using the impedance method and knowing the precise load can be more
accurate, but voltage drop is a best guess for the most part anyway.


The more exact formula incorporating impedance and power factor is no more
complicated than the above equation but the system power factor maybe a
little tricky to get - measurement, calculation or assumption. When
conductors get larger, the Z (impedance) will be significantly larger than
the R (dc resistance) so you expect a significant different between the
equations that incorporate only R values and equation with Z values. The
different in voltage drop is even more so between a 60Hz system and a 400Hz
system like an aircraft generator.

At the $300 million Healy Power Plant Job in Alaska the engineers
simply specified 460 volt motors to be used on a 480 volt system and
that solved all their voltage drop problems.


You mean 460V synchronous motors?




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Default Find the correct wire size for a load or the load for a selected wire size


# Fred # wrote:
"Jerry" wrote in message
oups.com...

# Fred # wrote:
What is your source for the cable impedance R+jX values? Do you use a
default power factor or is it user defined?


The voltage drop calculator uses the equation:
vd=2*K*L*I/CMA
Where K = 12.9 for copper and 21.2 for Aluminum
K is the resistance in ohms of one foot of a one mil conductor at 75
degrees C.
L is one way distance from supply to load in feet
I is current in amperes
CMA is circular mil area of conductor
vd is voltage drop in volts
This is a a very conservative calculation since the load is assumed to
be the same as the OCPD value and the conductor is assumed to be at 75
degrees C.
This is the standard method of calculating voltage drop used by
electricians for the last 50 years and appears in journeyman
electrician tests for several states.
Using the impedance method and knowing the precise load can be more
accurate, but voltage drop is a best guess for the most part anyway.


The more exact formula incorporating impedance and power factor is no more
complicated than the above equation but the system power factor maybe a
little tricky to get - measurement, calculation or assumption. When
conductors get larger, the Z (impedance) will be significantly larger than
the R (dc resistance) so you expect a significant different between the
equations that incorporate only R values and equation with Z values. The
different in voltage drop is even more so between a 60Hz system and a 400Hz
system like an aircraft generator.

At the $300 million Healy Power Plant Job in Alaska the engineers
simply specified 460 volt motors to be used on a 480 volt system and
that solved all their voltage drop problems.


You mean 460V synchronous motors?

No, they were regular squirrel cage induction 3-phase motors. The
onsite engineer called me at the beginning of the job to tell me they
had a major voltage drop problem. She later called and said they
either initially ordered the 460 volt motors or switched to correct the
voltage drop problem.
Electricians do not do engineering for good reasons, but in many
situations small contractors do a lot of design bid build and use
electrician math. This is especially true in some of the remote parts
of Alaska such as Nome, Kotzebue, and Barrow. I live in Alaska.

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