×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

reducing drop voltage

reducing drop voltage

reducing drop voltage

(OP)
how  capacitor to reduce dropping voltage line
i mean can you using to reducing dropping voltage by conection capacitor and how caculate power off this capacitor

RE: reducing drop voltage

I think capacitors if connected in series or parallel will reduce voltage drop since to Xl ,-jXc is added.
There may be equations for finding this if you know model of circuit
you want to find out power eg:Line Model

RE: reducing drop voltage

(OP)
thank you dear
i need caculate dropping voltage acrosing power lines or cables or ground cable also i need improve or reducing this dropping by connection capacitor across load
for example i have motor 30kw ,400v, p.f=0.86 cable 4x35mm^2
length cable 200m to inbox sorce   connect to sorce size cable 4x50mm^2 length=500m
before working voltage in box sorce 395v but when working this motor voltage at box sorce 360v and current 60A
i need connection capacitor at box sorce to rising voltage to 380v
how calculate rating capacitor

RE: reducing drop voltage

formulae here as;
Calculation and selection of required capacitor rating
Qc = P * {tan [acos (pf1)] - tan [ acos (pf2)]}
Qc = required capacitor output (kVAr)
pf1 = actual power factor
pf2 = target power factor
P = real power (kW)

If you improve p.f-->load current will reduce-->so voltage drop

If not solved then use high csa cables

hope this helps

RE: reducing drop voltage

The capacitor could improve the p.f. from 0.84 to 1.In this case you may increase the voltage only by 8 V.
In order to get 380 V at source box you need to use 3*120+70 sqr.mm instead of 3*50+25 cable .No capacitor may help you.
 

RE: reducing drop voltage

The voltage rise caused by the capacitors will be:

% volt rise = (kvar·XL)/(10·kV²)

Where:  XL = the inductive reactance of the line (ohms)
        kvar = three-phase capacitor size
        kV = line-to-line voltage

This voltage rise is above the voltage as reduced by load voltage drop.  That is, if there is 5% voltage drop with no capacitors, and 6% voltage rise caused by the capacitors, then the net voltage rise will be 1%.

 

RE: reducing drop voltage

(OP)
hi sir
if using for example Q=30kvar=p
Qc = P * {tan [acos (pf1)] - tan [ acos (pf2)]}
then cos (pf1)] =  cos (pf2) but become leading
vd=1.73xIxL(Rxcosq+X x sinq) if pf Lag
vd=1.73xIxL(Rxcosq-X x sinq) if pf lead
then vload=vsource-vd=vsource-vd=1.73xIxL(Rxcosq-X x sinq)
if R=0
vload=vsource+1.73xIxLxXxsinq

now rising voltage this speakig ok or no
Mr.jghrist  said

% volt rise = (kvar·XL)/(10·kV²)=30x1/10x0.4^2,assume xL=1


% volt rise=18.75%
above vd%= 395-360/395=8.8%
using capacitor Qc=30 XL=1  then vd%=8.8%x18.75%=1.65%
vload=(1-1.65%) x 395=388v

is ok or no please  

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources