IEEE std Voltage Drop

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schicco

Member
Location
New York
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%
Load Current(Amps) 91
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
 
Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
schicco said:
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%
Load Current(Amps) 91
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
Click to expand...
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.
 
S

schicco

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

Steve

Julius Right said:
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.
Click to expand...
 
S

schicco

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

Steve


Julius Right said:
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.
Click to expand...
 
S

schicco

Member
Location
New York
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

Julius Right said:
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.
Click to expand...
 
dkidd

dkidd

Senior Member
Location
here
Occupation
PE
schicco said:
Since I am feeding a fire pump motor; I should then multiply by 1.73 to get the Line to Line?
Click to expand...
Yes
schicco said:
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
Click to expand...

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

aprice44

Member
Location
New Jersey
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

dkidd said:
Yes


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

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dkidd

dkidd

Senior Member
Location
here
Occupation
PE
aprice44 said:
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
Click to expand...

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.
 
Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
schicco said:
Julius
Just want to check again
What numbers did you use for the cos(theta) and sin(theta)
Thanks a lot

Steve
Click to expand...
The angle theta has to be acos(0.8).I did now the same calculation on the other computer and the VD is close to yours indeed : 1.94943 accurate and 1.948536 approximate. Since I was 100 miles away I don't know what happened. I am sorry.
 
Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
The angle theta has to be acos(0.8).I did now the same calculation on the other computer and the VD is close to yours indeed : 1.94943 accurate and 1.948536 approximate. Since I was 100 miles away I don't know what happened. I am sorry.:ashamed1:
NEC 2014 art. 695.7 Voltage Drop. A) Starting 15% start. B) Running not more than 5% voltage drop from rated motor voltage at 115% load.
 
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