I need help calculating the voltage drop across

500mcm cable @ 100 ft length.
4/O cable @ 60 ft length.

V = 6300
I = 1400 In-rush

I am trying to explain to my people that DOL of a pump that large is a bad idea and to guarantee a Voltage drop of no more tha 20% is also a bad idea.

Couls you guys help me out with the proper method for figuring the Voltage drop.

 

You really don't want this forum to do Owm's law calculations for you?

Meir

 

I assume you mean that you have a three phase run of 500mcm for 100ft, then 4/0 for another 60ft. It looks like you're talking about 2-3MW size motor, so I can see why you would want to soft start it, but I don't see voltage drop being a problem with those size conductors if my stated assumptions are correct.

--
Steve Myres, PE
Automation Solutions
(480) 813-1145

 

Responding to C. Hardracker's query... I believe you have the cart before the (excuse the pun) horse(power).

Unless the load or driven machine is negatively affected by DOL starting, then reduced-voltage starting for medium-voltage motors should be dictated by the short-circuit capacity of the supply. (If you need additional info, contact me, via the List!)

That said, I propose providing a "rule-of-thumb"' approach in the following form:

% V-drop (per ft) = Kc x A / kV, where:

Kc = constant related to cable size and load power factor.

A = Amperes that the cable is to carry.

kV = System kiloVolts.

BTW, replying to your direct query... the 4/0 cable voltage-drop's about 10%, while the 500 kcmil cable's is about 8%.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Further to my earlier reply to C. Hardacker's query:

Oops... the most feared word in an operating room... I erred. Yes, engineers occasionally make mistakes (but very, very few will admit it)!

The earlier % voltage drop values are wrong. The voltage drop values for the 500 kcmil and AWG 4/0 cases are 0.13% and 0.10%respectively.

If anyone wants details, please contact me... via the List!

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Responding to C. Hardacker's Jan 14, 2005 query:

Regarding your position on DOL (Direct-On-Line starting), I believe you have the cart before the (excuse the pun) horsepower. Unless the load or driven machine is negatively affected by DOL starting, then reduced-voltage
starting for medium- or high-voltage motors should be dictated by the short-circuit capacity of the supply. A project I was involved with had DOL starting of a 22,000 Hp induction motor. The supply voltage was 11kV. If you need additional info, contact me via the List!

Returning to your original query about Voltage drop. If desired, I can provide you and the List, a "rule-of-thumb" approach in the following form:

% V-drop (per ft) = (Kc x A x L) / kV, where:

Kc = constant related to cable size and load power factor.

A = Amperes that the cable is to carry.

L = Circuit length, feet.

kV = System voltage, kV.

For your V-drop query... the 500 kcmil cable voltage-drop is about 0.13%, while the voltage drop for the Awg 4/0 cable is about 0.10%. Calculation parameters are available.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])