voltage drop calculation

[rrzzan]

Experts,

I'm new to my job.

How would you find the total voltage drop from a main swich yard(at 480V) to a load at 120V? The details are follows.

Conductor at 480V is 100ft(impecence= 0.000777),
then a transformer(480/120V, 3 kVA, 3.5% impedence)and
load(500 Watt) at a distance of 50 ft(impedence=0.0019)at 120V. I am assuming the transformer does not has taps.

Looking for your help.
Thank You for reading.

 

[rrzzan]

This is not a home work problem.

I was thinking of transferring the voltage drop on primary side of the transformer and the voltage drop on the 480V conductor to secondary side by the transformation ratio (120/480) And then to calculate the voltage drop on the 120V conductor, and add up total voltage drop.

Is this right?

I would appreciate any help.
Thanks

 

[dpc]

Current flow through an impedance causes a voltage drop. How accurate an answer do you require? There are several approaches.

Why don't you give us at least an attempt...

 

[waross]

Without knowing either the regulation of the transformer or the X:R ratio of the transformer you don't have a hope of finding the real actual voltage drop. Possibly your instructor does not understand that the transformer impedance is only valid for short circuit current calculations, not voltage drop under normal load conditions.

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

 

[rrzzan]

Here is my approach.

Current at Secondary side

Isec = 500 watts/120V = 4.17A (I've skipped the power factor)

Iprimary= (120/480)*4.17 =1.04A

Voltage drop on conductor at 480V
=I*Z*L
= 1.04*0.000777*100 (length is 100ft)
= 0.081V
For Transformer (3kVA, 3.5%Z)
At primary, Zbase = 480^2/3000 = 76.8 Ohm
Therfore, Z primary = 0.0035*76.8 = 0.27ohm
Voltage Drop=I*Z = 1.04*0.27 = 0.218V
Total Voltage Drop(VD) on Primary Side
= VD on conductor+VD on Transformer
= 0.081+0.281 = 0.362V
Now the VD refered to secondary (120V) side
= 0.362 * (120/480) = 0.091V
For VD on conductor at 120V
VD= 2*I*Z*L
= 2* 4.17*0.0019*50
= 0.79V
Total voltage drop on the conductor at 480V, Transformer and the conductor at 120V is
=0.091+0.79 = 0.881V

%VD = (0.881/120)*100 = 0.7% which meets limits set be NEA

This is what I have done.
I would appreciate anybody who would correct me, if wrong.I have been delaying on this matter so much and it is so embarrassing to ask such basic questions to my supervisor as well.
Thanks so much

 

[waross]

Simple.
Find your reactances.
Find your reactive current.
Calculate the reactive voltage drop.
Find your resistances.
Find your real current.
Calculate your in phase voltage drop.
Use Pythagoras' theorem to calculate;
Impedance.
Actual current.
Actual voltage drop.

(I've skipped the power factor)

Not good

For Transformer (3kVA, 3.5%Z)

X:R ratio??

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

 

[waross]

Hi rrzzan, what part of the world are you in?

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

 

[rrzzan]

I'm from SE Asia, but work here in US.

 

[waross]

The source has impedance, the cable has impedance, the transformer has impedance, the load has impedance.
Unfortunately these impedances are seldom at the same phase angle or power factor.
For a rigorous solution, you must add the impedances vectorily. A simple solution may show a greater voltage drop than actually exists. However, this is invariably a conservative solution.
Possibly the worst source of error will be the X:R ratio of the transformer which may show a calculated voltage drop of several times the actual voltage drop if impedace is used instead of regulation. A lagging power factor may lessen the error a small amount.

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

 

[jghrist]

This is what I have done.
I would appreciate anybody who would correct me, if wrong.I have been delaying on this matter so much and it is so embarrassing to ask such basic questions to my supervisor as well.

If you use the VD=I·Z approach, you have to use vector math. Ignoring the power factor of the load and the impedance angle of the elements will not get you close to the right answer.

You can use the following formula to get a close approximation of the voltage drop:

VD = I·(R·cosØ + X·sinØ) where cosØ = power factor

I, R, and X are referred to the same side of the transformer.

 

[waross]

rrzzan;
Can you look your transformer up in the catalog and find a regulation value for it? We are missing R and X for the transformer. Other than that, use jghrist's method.
jghrist;
Thanks for the help.


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

 

[Sideswiper]

"I have been delaying on this matter so much and it is so embarrassing to ask such basic questions to my supervisor as well."

I ask basic questions to my supervisor all the time. He knows I'm an idiot until I get some more basic real world experience under my belt.