Calculating % VD( Motor Starting Voltage Drop)

[Ron11]

Hello,

I am trying to calculate the % voltage drop and figure out the best approach to the scenario below

I have a SCE Overhead line 16.5KV feeding a transformer which steps the voltage down to 480V which then supplies a control panel about a 1000 feet away. From the panel, we have a firewater pump(150P) connected about 30 feet away. The FLA is 180 and LRA is 1085. Starting power factor is 30 percent and also we assuming normal impedance for the transformer at 5.75%.

I am having a hard time finding the voltage drop given the conditions. Also, I need to know if the 300KVA transformer is viable for the application taking into account the percent of nameplate min is 85%. Please can someone help me figure this out. Currently, I have found applications showing calculations without taking into account the distance. However, this is not accurate considering the 1000'. Please share your comments. Appreciate any help provided

 

[magoo2]

What have you tried and what are your results?

This is really a T&D type question, but regardless, what equations are you using?

And what results did you get?

What impedance did you use for the 1000' distance of conductor?

 

[Ron11]

magoo2,

I am using a program created by someone else currently. I got it from the web. I am actually looking for specific equations to do the hand calculation

For Example,

the 300KVA transformer I tried at normal impedance
is giving me
Volts @ 150 HP motor Design 480V, Calculated 409.3V
Percent of nameplate: Calculated is 85.28% Min is 85%
Note that this is without taking into account the 1000'. That is where I have a problem as I am not too sure about how to incorporate that value to get a final answer

Thanks for your help

 

[magoo2]

This sure sounds like homework to me. These forums are intended for practicing engineers. If some have legitimate questions, we encourage them to post what they have done and then, and only then, will we provide assistance.

In any event, go to the forum on Electric Power and T&D. Search for voltage drop or motor starting and go from there. There have been a number of examples relating to voltage drop. Good luck.

 

[Ron11]

magoo2,

Definitely not a homework problem but an actual scenario I need to justify backing use of the 300KVA transformer. SCE believes the 150KVA should work but I feel this may not be the case considering that distance. In any case, thanks for your help.

 

[7anoter4]

First of all you have to know the permissible voltage drop for your motor.
The motor starting current [LRA] is direct proportional with the supply voltage but the motor torque is direct proportional with the square of supply voltage. The motor torque has to stay above the load torque always in the starting process.
A thumb rule says the transformer power has to be at least 3 times the motor power for DOL starting. If your motor is 150 hp[110 kw] the transformer has to be at least 330 kVA. Then 500 kVA will be better.
Let's say the short-circuit power at 16.5 kV could be 200 MVA then the impedance [saw from 480 V transformer side] will be 0.48^2/200=0.001152 ohm.
Let's say the transformer will be full loaded at 0.8 pf and the remote panel which supplies the motor will receive all the transformer full current. So you start with NEC to find the minimum cable cross section let's say 750 MCM copper [60 oC UF] NEC art 340 and 315]. The short cable of 30 ft is dedicated only to your motor and 3*2/0 copper will be enough.
Now let's check the total voltage drop. The 1000 ft cable will be 3 single core cables at 8" apart the dc-resistance at 20oC will be 0.0141 ohm/1000 ft. and the reactance 0.0536 ohm/1000 ft. You may use the Okonite links as:

http://www.okonite.com/engineering/index.html

The no-load transformer voltage is somewhere of 1.05*.48 [kV]
You can reduce the LRA current proportional with the remaining voltage at motor terminals-let's say if the drop will be 20% the reduction factor will be 0.48*0.8/0.48=0.8[80%].
It seems 1 cable per phase is not enough so you have to take 2 cables per phase.
As a unbalance could occur if the single-core cable are parallel laid an order of A,B,C,C,B,A is good.

 

[dpc]

Ron,

Please don't post the same question in multiple forums. I posted a response in another forum, so you might check there.