# Thread: IEEE std Voltage Drop

1. Member
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## IEEE std Voltage Drop

I have been trying to calculate the voltage drop using the IEEE std.141
Vd = es + IRcos(theta) + IX sin(theta) + sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)

Inputs:
Number of conductors: Single
Conductor size: 1/0
Power factor: 80%
System voltage: 480
Length of Run (feet) 460
Conduit: EMT

I am coming up with 1.62% Voltage Drop

Am I correct?

Using the simplified formula V = I Rcos(theta) + I Xsin(theta) I get 1.95% Voltage Drop
Don't know which one is correct.

Thanks a lot

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Originally Posted by schicco
I have been trying to calculate the voltage drop using the IEEE std.141
Vd = es + IRcos(theta) + IX sin(theta) + sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)

Inputs:
Number of conductors: Single
Conductor size: 1/0
Power factor: 80%
System voltage: 480
Length of Run (feet) 460
Conduit: EMT

I am coming up with 1.62% Voltage Drop

Am I correct?

Using the simplified formula V = I Rcos(theta) + I Xsin(theta) I get 1.95% Voltage Drop
Don't know which one is correct.

Thanks a lot
At first the actual voltage drop formula is:
Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
and the result is 1.9247% and from the approximate formula is 1.9233%.
The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

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Julius
Thanks a lot, I see where I was going wrong. Now I am getting your numbers.
Again thanks

Steve

Originally Posted by Julius Right
At first the actual voltage drop formula is:
Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
and the result is 1.9247% and from the approximate formula is 1.9233%.
The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

4. Member
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Location
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Posts
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Julius
Just want to check again
What numbers did you use for the cos(theta) and sin(theta)
Thanks a lot

Steve

Originally Posted by Julius Right
At first the actual voltage drop formula is:
Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
and the result is 1.9247% and from the approximate formula is 1.9233%.
The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

5. Senior Member
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Dec 2010
Posts
1,381
Originally Posted by schicco
Julius
Just want to check again
What numbers did you use for the cos(theta) and sin(theta)
Thanks a lot

Steve
Theta is .8. Cos theta is 0.99990252400930421155123810793869. Sin theta is 0.01396218033914527162191138769678 (roughly )

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Thanks a lot
Originally Posted by dkidd
Theta is .8. Cos theta is 0.99990252400930421155123810793869. Sin theta is 0.01396218033914527162191138769678 (roughly )

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Since I am feeding a fire pump motor; I should then multiply by 1.73 to get the Line to Line?
Then it would mean I have gone way over 3% voltage drop, right?
The 2%(feeder)+3%(branch)=5%vd applies?
Not sure if Fire Pump falls under branch or feeder (definition of branch is wiring that comes after the final protection device)

Thanks

Originally Posted by Julius Right
At first the actual voltage drop formula is:
Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
and the result is 1.9247% and from the approximate formula is 1.9233%.
The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

8. Senior Member
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Posts
1,381
Originally Posted by schicco
Since I am feeding a fire pump motor; I should then multiply by 1.73 to get the Line to Line?
Yes
Originally Posted by schicco
Then it would mean I have gone way over 3% voltage drop, right?
The 2%(feeder)+3%(branch)=5%vd applies?
Not sure if Fire Pump falls under branch or feeder (definition of branch is wiring that comes after the final protection device)

Thanks
Codes limit the running voltage drop to 5%, and starting to 15%.

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What confuses me about the code definition of branch ckt ("the conductors between the branch circuit final overcurrent device protecting the circuit and the outlets") is;
Why does it only mention outlets?
If I am going to hard wire a motor or heating, it cannot be branch?

From the same panel you can have an OCPD feeding outlets and another OCPD feeding a motor. Are they all not branch ckts?

Thanks

Andy

Originally Posted by dkidd
Yes

Codes limit the running voltage drop to 5%, and starting to 15%.
Last edited by aprice44; 05-16-17 at 11:56 AM.

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Originally Posted by aprice44
What confuses me about the code definition of branch ckt ("the conductors between the branch circuit final overcurrent device protecting the circuit and the outlets") is;
Why does it only mention outlets?
If I am going to hard wire a motor or heating, it cannot be branch?

From the same panel you can have an OCPD feeding outlets and another OCPD feeding a motor. Are they all not branch ckts?

Thanks

Andy
Definition from Article 100

Outlet. A point on the wiring system at which current is
taken to supply utilization equipment.

Don't confuse outlet with receptacle.

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