reducing drop voltage 1

[hasanhadi]

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

 

[abrahamJP]

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

 

[hasanhadi]

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

 

[abrahamJP]

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

 

[7anoter4]

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.

 

[jghrist]

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%.

 

[hasanhadi]

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