Design for 525 V Monorail 2

[sandrasagra]

Hi Guys, Its been a long time since I've done some calculations and I need someone to just verify if what I did was correct.

I am trying to deduce the minimum distance between two trains (3 phase electric motors) on conductor rail (if one exists ?).

Power Supply (3phase, 525V, -10% +15% tolerance, total power consumption 250kva)
Conductor Rail (3phase, 525V rated, 400A rated)
Motor (28*8 kw nominal capacity, 500v rated, 41a rated)

I know the impedance of the conductor as 0.00369 ohms/m.

The maximum allowable voltage drop for the motor is 475 V (5% of rated). The voltage drop formula is V=LIZ*1.732 where L is the length and varies depending on the power source location.

I plan on feeding the conductor with one power source in the middle. So L = 0.5totallength.

Q1. How would one calculate the minimum distances between the trains ? Stupid question, but is there a need to do this ?

To make things more complicated I am using a 600/387 Transformer to supply the 525 V, since its within the tolerances.

Q2. Is this something okay to do ?

I appreciate any help on this... I've been looking at the Canadian Electric Code book but haven't found much help.

 

[wareagle]

The max VD occurs at the end of the rail(max distance between the trains). I don't understand the idea of the
Minimum Distance. Your formula is correct.
You said "The maximum allowable voltage drop for the motor is 475 V (5% of rated)." If the voltage supply is 500 volts
5% drop = 25 volts. 500 volts - 25 volts = 475 volts. Is this correct?

VD = 41 amps x Z x 1.73 = 25 volts if you are to maintain the minimum voltage level.
0.50 Max Z rail distance = 25 volts/(41 amps 1.73)
1/2 Max Z = 0.353 ohms
If Rail Z = 0.000369 ohms/m, then 1/2L = 0.353/0.000369
1/2L = approx 1000 M. I assume M is meters.
This does not include the VD in the cables from the transformer.

 

[sandrasagra]

Thanks wareagle, please excuse my grammer.

- Your correct in the voltage drop being onluy 25 volts and that's why the minimum voltage tolerance is 475 volts.
- M is in meters
- However z = 0.00369 therefore 1/2L = 100m.

I understand this calculation means that one power feed can be connected in the middle of a conductor rail approx. 200m in length (L).

QUESTION: Since my supply voltage is 540 V (600v*efficiency of my transformer[0.95]) then could my conductors then be able to withstand a voltage drop of 65V (540-475) and then redo my calculations to have a longer conductor length ?

QUESTION: Would there be any adverse affects when two trains are on the same conductor rail ?
Is there a minimum distance if so ?
How would one calculate this ?

 

[wareagle]

QUESTION: Since my supply voltage is 540 V (600v*efficiency of my transformer[0.95]) then could my conductors then be able to withstand a voltage drop of 65V (540-475) and then redo my calculations to have a longer conductor length ?

You stated that the distance is only 100 meters. My calculations show that you can go 1000 meters using the load of one train. You will not have a problem with VD. You need to be concerned with high voltage at no load or lightly loaded. Your motor is rated at 500 volts and rail at 525 volts.
"Conductor Rail (3phase, 525V rated, 400A rated)
Motor (28*8 kw nominal capacity, 500v rated, 41a rated)"

QUESTION: Would there be any adverse affects when two trains are on the same conductor rail ?
Is there a minimum distance if so ?

When two trains are in use the load would be 2x41 amps=82.
VD on your conductors would be calculated using 82 amps.
My understanding is that you will only have one train on each side of the supply conductors.
I do not understand the minimum distance question. Explain in detail.

 

[waross]

Are you checking for compliance with a code or standard? If so, use the parameters given in the standard. For example, for compliance with the CEC, use the impedances given in table D3 and extrapolate to find the voltage and percentage voltage drop that you are limited to.
If you want to know the actual voltage at the trains, it gets a little more complex.
Total voltage drop is a combination of in-phase voltage drop and reactive voltage drop.
Step 1; Find the current and power factor of the load and resolve into both in-phase current and reactive current.
step 2; Resolve the impedances of all sections of the conductors into in-phase current and reactive current.
Sep 3;Resolve the transformer impedance into resistance and reactance.
Step 4; Calculate the resistive voltage drops of all resistive components.
Step 5; Sum the resistive voltage drops.
Step 6; Calculate the reactive voltage drops of all reactive components.
Step 7; Sum the reactive voltage drops.
Step 8; Combine the resistive and reactive voltage drops with the applied voltage to determine the actual voltage drop at the load.
This assumes an infinite supply at the transformer. If there is an significant voltage drop in the transformer feeders the feeder voltage drop must be calculated in a similar manner.
These rigorous methods require information concerning the X/R ratios of circuit components that are often either difficult to obtain or not readily available. The simpler estimations based on the voltage drop tables are generally adequate for most work. I think that it is well to be aware of the methods involved in a rigorous solution even if you never actually solve a rigorous example.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sandrasagra]

Compliance with standard CEC.

I tried to use Table D3 however its only for use in voltage drops on nominal 120V. My conductor rails are rated 525V.

Is there a technique to apply the table to a voltage of 525V ?

 

[sandrasagra]

Wareagle, thank you for your help. However I don't think I am explaining the entire picture to you,

I have a 6 trains, each is comprised with 4 drive units (each with 2 motors/drive unit). Each train is 44m long.

Each motor is rated at 500V, current 41A, nominal capacity of 28kw.

What I need to do is design a single line diagram of the entire power distribution for the conductor rails. The voltage drop that you calculated earlier used a factor of Z=0.000369 instead of 0.00369. Therefore using the Z=0.00369 you would get 100m.

Because each train is 44m long and one conductor rail is 200m long there is a chance that two trains might be on the same rail. However looking at the fact that each train is drawing 328A (41A*8motors) and the conductor rails are rated for 400A. There is no way two trains can be on the same track.... ie. breaker will open.

Thoughts ?