What the capacitor Current formula?, For 1 KVAr how amps
its required? for using the KVAr for boosting the PF Design
Calculation.( Eg: for 30 KVAr required how many KW?)

Answers were Sorted based on User's Feedback

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

As per my knowledge,
On my site bellow capacitor bank I am using.
I have 350 Kvar capacitor bank in this bank 25 kvar
capacitors are placed.

Capacitor specification

Volt = 440
Kvar = 25
Three Phase

KVA
I = ----------
Volt *1.732

25000
I = ----------
440 *1.732

I = 32.80 Amps

Eg 30 Kvar

KVA
I = ----------
Volt *1.732

30000
I = ----------
440 *1.732

I = 39.36 Amps

 Is This Answer Correct ? 402 Yes 162 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

Kvar= 1.73*V*I*Sin\$ (\$ IS INSTESD OF PHASE ANGLE PHI)

so I= kVAR/1.73*V*SIN\$

IN A PURELY CAPACITIVE CIRCUIT \$ IS 90 DEGREE
SO TERM SIN 90 = 1
HENCE WE CAN CALCULATE CAPACITOR CURRENT BY kVAR/1.73*V

 Is This Answer Correct ? 131 Yes 15 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

You explained the answer in KVA? where do get the kva from
this variables, the formula for capacitor in KVAr, KVAr
into KW, KVA & Amps

 Is This Answer Correct ? 89 Yes 26 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

Hi every one how are you
if u need to know three phase capacitor rating than

I= kvar x 1000 / V x 1.73

Example:

i have 25 Kvar capacitor and line to line voltage is 440. i want to know current rating of capacitor

I = KVAR x 1000 / V x 1.73
I = 25 x 1000 / 440 x 1.73
I = 25000 / 762.08
I = 32.80 Amp

So my Ans is (32.80 Amp)

 Is This Answer Correct ? 46 Yes 2 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

Its 1.3A/KVAR

 Is This Answer Correct ? 58 Yes 15 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

VOLT = 400
KVAr = 30
PHASE = 3

I = KVAr
-----
400 x 1.732
=43amps.
Kw = I x 0.624
= 26.8KW.

 Is This Answer Correct ? 57 Yes 37 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

Calculate Size of Capacitor Bank / Annual Saving & Payback Period

Calculate Size of Capacitor Bank Annual Saving in Bills and Payback Period for Capacitor Bank.
Electrical Load of (1) 2 No’s of 18.5KW,415V motor ,90% efficiency,0.82 Power Factor ,(2) 2 No’s of 7.5KW,415V motor ,90% efficiency,0.82 Power Factor,(3) 10KW ,415V Lighting Load. The Targeted Power Factor for System is 0.98.
Electrical Load is connected 24 Hours, Electricity Charge is 100Rs/KVA and 10Rs/KW.
Calculate size of Discharge Resistor for discharging of capacitor Bank. Discharge rate of Capacitor is 50v in less than 1 minute.
Also Calculate reduction in KVAR rating of Capacitor if Capacitor Bank is operated at frequency of 40Hz instead of 50Hz and If Operating Voltage 400V instead of 415V.
Capacitor is connected in star Connection, Capacitor voltage 415V, Capacitor Cost is 60Rs/Kvar. Annual Deprecation Cost of Capacitor is 12%.
Calculation:

For Connection (1):
Total Load KW for Connection(1) =Kw / Efficiency=(18.5×2) / 90%=41.1KW
Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 41.1 /0.82=50.1 KVA
Total Load KVA (new) for Connection(1)= KW /New Power Factor= 41.1 /0.98= 41.9KVA
Total Load KVAR= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
Total Load KVAR1=41.1x([(√1-(0.82)2) / 0.82]- [(√1-(0.98)2) / 0.98])
OR
tanǾ1=Arcos(0.82)=0.69
tanǾ2=Arcos(0.98)=0.20
Total Load KVAR1= KWX (tanǾ1- tanǾ2) =41.1(0.69-0.20)=20.35KVAR
For Connection (2):
Total Load KW for Connection(2) =Kw / Efficiency=(7.5×2) / 90%=16.66KW
Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 16.66 /0.83=20.08 KVA
Total Load KVA (new) for Connection(1)= KW /New Power Factor= 16.66 /0.98= 17.01KVA
Total Load KVAR2= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
Total Load KVAR2=20.35x([(√1-(0.83)2) / 0.83]- [(√1-(0.98)2) / 0.98])
For Connection (3):
Total Load KW for Connection(3) =Kw =10KW
Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 10/0.85=11.76 KVA
Total Load KVA (new) for Connection(1)= KW /New Power Factor= 10 /0.98= 10.20KVA
Total Load KVAR3= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
Total Load KVAR3=20.35x([(√1-(0.85)2) / 0.85]- [(√1-(0.98)2) / 0.98])
Total KVAR=KVAR1+ KVAR2+KVAR3
Total KVAR=20.35+7.82+4.17
Total KVAR=32 Kvar
Size of Capacitor Bank:

Site of Capacitor Bank=32 Kvar.
Leading KVAR supplied by each Phase= Kvar/No of Phase
Leading KVAR supplied by each Phase =32/3=10.8Kvar/Phase
Capacitor Charging Current (Ic)= (Kvar/Phase x1000)/Volt
Capacitor Charging Current (Ic)= (10.8×1000)/(415/√3)
Capacitor Charging Current (Ic)=44.9Amp
Capacitance of Capacitor = Capacitor Charging Current (Ic)/ Xc
Xc=2 x 3.14 x f x v=2×3.14x50x(415/√3)=75362
Capacitance of Capacitor=44.9/75362= 5.96µF
Required 3 No’s of 10.8 Kvar Capacitors and
Total Size of Capacitor Bank is 32Kvar

Protection of Capacitor Bank

Size of HRC Fuse for Capacitor Bank Protection:
Size of the fuse =165% to 200% of Capacitor Charging current.
Size of the fuse=2×44.9Amp
Size of the fuse=90Amp
Size of Circuit Breaker for Capacitor Protection:
Size of the Circuit Breaker =135% to 150% of Capacitor Charging current.
Size of the Circuit Breaker=1.5×44.9Amp
Size of the Circuit Breaker=67Amp
Thermal relay setting between 1.3 and 1.5of Capacitor Charging current.
Thermal relay setting of C.B=1.5×44.9 Amp
Thermal relay setting of C.B=67 Amp
Magnetic relay setting between 5 and 10 of Capacitor Charging current.
Magnetic relay setting of C.B=10×44.9Amp
Magnetic relay setting of C.B=449Amp
Sizing of cables for capacitor Connection:
Capacitors can withstand a permanent over current of 30% +tolerance of 10% on capacitor Current.
Cables size for Capacitor Connection= 1.3 x1.1 x nominal capacitor Current
Cables size for Capacitor Connection = 1.43 x nominal capacitor Current
Cables size for Capacitor Connection=1.43×44.9Amp
Cables size for Capacitor Connection=64 Amp
Maximum size of discharge Resistor for Capacitor:
Capacitors will be discharge by discharging resistors.
After the capacitor is disconnected from the source of supply, discharge resistors are required for discharging each unit within 3 min to 75 V or less from initial nominal peak voltage (according IEC-standard 60831).
Discharge resistors have to be connected directly to the capacitors. There shall be no switch, fuse cut-out or any other isolating device between the capacitor unit and the discharge resistors.
Max. Discharge resistance Value (Star Connection) = Ct / Cn x Log (Un x√2/ Dv).
Max. Discharge resistance Value (Delta Connection)= Ct / 1/3xCn x Log (Un x√2/ Dv)
Where Ct =Capacitor Discharge Time (sec)
Un = Line Voltage
Dv=Capacitor Discharge voltage.
Maximum Discharge resistance =60 / ((5.96/1000000)x log ( 415x√2 /50)
Maximum Discharge resistance=4087 KΩ
Effect of Decreasing Voltage & Frequency on Rating of Capacitor:
The kvar of capacitor will not be same if voltage applied to the capacitor and frequency changes
Reduced in Kvar size of Capacitor when operating 50 Hz unit at 40 Hz
Actual KVAR = Rated KVAR x(Operating Frequency / Rated Frequency)
Actual KVAR = Rated KVAR x(40/50)
Actual KVAR = 80% of Rated KVAR
Hence 32 Kvar Capacitor works as 80%x32Kvar= 26.6Kvar
Reduced in Kvar size of Capacitor when operating 415V unit at 400V
Actual KVAR = Rated KVAR x(Operating voltage / Rated voltage)^2
Actual KVAR = Rated KVAR x(400/415)^2
Actual KVAR=93% of Rated KVAR
Hence 32 Kvar Capacitor works as 93%x32Kvar= 23.0Kvar
Annual Saving and Pay Back Period

Before Power Factor Correction:
Total electrical load KVA (old)= KVA1+KVA2+KVA3
KVA Demand Charge=KVA X Charge
KVA Demand Charge=82x60Rs
KVA Demand Charge=8198 Rs
Annual Unit Consumption=KWx Daily usesx365
Annual Unit Consumption=62x24x365 =543120 Kwh
Annual charges =543120×10=5431200 Rs
Total Annual Cost= 8198+5431200
Total Annual Cost before Power Factor Correction= 5439398 Rs
After Power Factor Correction:
Total electrical load KVA (new)= KVA1+KVA2+KVA3
KVA Demand Charge=KVA X Charge
KVA Demand Charge=69x60Rs =6916 Rs————-(1)
Annual Unit Consumption=KWx Daily usesx365
Annual Unit Consumption=62x24x365 =543120 Kwh
Annual charges =543120×10=5431200 Rs—————–(2)
Capital Cost of capacitor= Kvar x Capacitor cost/Kvar = 82 x 60= 4919 Rs—(3)
Annual Interest and Deprecation Cost =4919 x 12%=590 Rs—–(4)
Total Annual Cost= 6916+5431200+4919+590
Total Annual Cost After Power Factor Correction =5438706 Rs
Pay Back Period:

Total Annual Cost before Power Factor Correction= 5439398 Rs
Total Annual Cost After Power Factor Correction =5438706 Rs
Annual Saving= 5439398-5438706 Rs
Annual Saving= 692 Rs
Payback Period= Capital Cost of Capacitor / Annual Saving
Payback Period= 4912 / 692
Payback Period = 7.1 Years

 Is This Answer Correct ? 18 Yes 0 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

For correcting PF KVAr is introduced, thus formula

pf=cos A is significant

there you should have a desired PF. If for example you need
a power factor of .90 to maintain for a 30kVAR the watts
that would come will be:

A=arccos .90
A= 25.84

WATTs= KVAR/tan 25.84

= 62500 watts approx.

remember that for PF correnction it is the KVAR that is in
need of computation and not the watts so the question would
rather be, how much capacitor kVAR you need to correct the
power factor.

 Is This Answer Correct ? 40 Yes 24 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

The question requires more details.
First the "K" represents kilo or 1000.
The real power (kW) can be thought of as the x-axis and the
reactive power (kVar) from inductance or capacitance can be
thought of as the y-axis. The hypotenuse is the
Apparent power (KVA). The relationship of Real to apparent
power gives the Power Factor, where a power factor of 1.0
simply means that you only have real power, no reactive
power. The leading and lagging VARs from capacitance and
inductance, respectively can be added numerically since
they are on the same axis.
To ask, how muck kVars are required for a given kW doesn’t
make a lot of sense because you need to include the desired
power factor.

 Is This Answer Correct ? 24 Yes 11 No

What the capacitor Current formula?, For 1 KVAr how amps its required? for using the KVAr for boos..

** Capacitor Current formula is

I= kVAr/kV/1.732 ( for 3 ph capacitor )

**for 1 kVar capacitor... suppose voltage is 440 volts than,

I=1/0.440/1.732
I=1.31 Amps.

For all your information,generally 1 kVAr capacitor takes
0.5 watts...henceforth..

30 kvar capacitor required kw=30X0.5=15 watts and is 0.015kw

thanks!!!!!!!

 Is This Answer Correct ? 23 Yes 14 No

More Electrical Engineering Interview Questions

what size steel armoured cable is required to support 1500kva capacity. would 70mm squared be enough or would I need 90 or 120mm squared 3c plus earth

What is LMU and LMDU. in PLCC communication of 132 kv transmission line. is it possible to use only one LMU with out LMDU ?

why dose not transformer not rotate ?

what is the clearances between phease to phease & phease to earth in 400 kv transmission line?

what is the use of tertiary winding?

IF we have three 30KW motor, these motors are connected in DOL, star-delta and VFD respectively, what will be the fuse rating for the above three?

As an electrical engineer what are the responsibilities which i can hold in TCS?

What is it mean Transformer 100/133MVA Power Transformer, significance of 100/133 MVA and 230/115KV please

6. A DC AMMETER HAS R=0.1 OHM & CURRENT RANGE IS 0-100 A IF THE RANGE IS TO BE EXTENDED TO 0-500 A THEN METER REQUIRES SHUNT RESISTANCE OF ..........

what is difference between neutral& grounding?

How to calculate flow of water delivered by pump when pressure & size of pipe line is given

what is the significant for using two types of voltage(line voltage & phase voltage)

Categories
• Civil Engineering (4645)
• Mechanical Engineering (3704)
• Electrical Engineering (15785)
• Electronics Communications (2638)
• Chemical Engineering (762)
• Aeronautical Engineering (79)
• Bio Engineering (30)
• Metallurgy (95)
• Industrial Engineering (199)
• Instrumentation (2882)
• Automobile Engineering (171)
• Mechatronics Engineering (41)
• Marine Engineering (62)
• Power Plant Engineering (170)
• Engineering AllOther (1336)