How To Calculate The Size OfCapacitor Bank

How To Calculate The Size Of Capacitor Bank ??? Example: 1: A 3 Phase, 5 kW Induction Motor has a P.F (Power factor) of 0.75 lagging. What size of Capacitor in kVAR is required to improve the P.F (Power Factor) to 0.90? Motor input = P = 5 kW Original P.F = Cosθ 1 = 0.75 Final P.F = Cosθ 2 = 0.90 θ 1 = Cos -1 = (0.75) = 41°.41; Tan θ 1 = Tan (41°.41) = 0.8819 θ 2 = Cos -1 = (0.90) = 25°.84; Tan θ 2 = Tan (25°.50) = 0.4843 Required Capacitor kVAR to improve P.F from 0.75 to 0.90 Required Capacitor kVAR = P (Tan θ 1 – Tan θ 2) = 5kW (0.8819 – 0.4843) = 1.99 kVAR And Rating of Capacitors connected in each Phase 1.99/3 = 0.663 kVAR ******************************************************* **************** Example: 2: An Alternator is supplying a load of 650 kW at a P.F (Power factor) of 0.65. What size of Capacitor in kVAR is required to raise the P.F (Power Factor) to unity (1)? And how many more kW can the alternator supply for the same kVA loading when P.F improved. Supplying kW = 650 kW Original P.F = Cosθ 1 = 0.65 Final P.F = Cosθ 2 = 1 θ 1 = Cos -1 = (0.65) = 49°.45; Tan θ 1 = Tan (41°.24) = 1.169 θ 2 = Cos -1 = (1) = 0°; Tan θ 2 = Tan (0°) = 0 Required Capacitor kVAR to improve P.F from 0.75 to 0.90 Required Capacitor kVAR = P (Tan θ 1 – Tan θ 2) = 650kW (1.169– 0) = 759.85 kVAR ******************************************************* **************** How to Calculate the Required Capacitor bank value in both kVAR andFarads? (How to Convert Farads into kVAR and Vice Versa) Example: 3 A Single phase 400V, 50Hz, motor takes a supply current of 50A at a P.F (Power factor) of 0.6. The motor power factor has to be improved to 0.9 by connecting a capacitor in parallel with it. Calculate the required capacity of Capacitor in both kVAR and Farads. Motor Input = P = V x I x Cosθ = 400V x 50A x 0.6 = 12kW Actual P.F = Cosθ 1 = 0..6 Required P.F = Cosθ 2 = 0.90 θ 1 = Cos -1 = (0.60) = 53°.13; Tan θ 1 = Tan (53°.13) = 1.3333 θ 2 = Cos -1 = (0.90) = 25°.84; Tan θ 2 = Tan (25°.50) = 0.4843 Required Capacitor kVAR to improve P.F from 0.60 to 0.90 Required Capacitor kVAR = P (Tan θ 1 – Tan θ 2) = 5kW (1.3333– 0.4843) = 10.188 kVAR ******************************************************* ********************** To find the required capacity of Capacitance in Faradsto improve P.F from 0.6 to 0.9 (Two Methods) Solution #1 (Using a Simple Formula) We have already calculated the required Capacity of Capacitor in kVAR, so we can easily convert it into Farads by using this simple formula Required Capacity of Capacitor in Farads/Microfarads C = kVAR / (2 π f V 2) in microfarad Putting the Values in the above formula = (10.188kVAR) / (2 x π x 50 x 400 2) = 2.0268 x 10 -4 = 202.7 x 10 -6 = 202.7μF Solution # 2 (Simple Calculation Method) kVAR = 10.188 … (i) We know that; I C = V/ X C Whereas X C = 1 / 2 π F C I C = V / (1 / 2 π F C) IC = V 2 F C = (400) x 2π x (50) x C I C = 125663.7 x C And, kVAR = (V x I C) / 1000 … [kVAR =( V x I)/ 1000 ] = 400 x 125663.7 x C I C = 50265.48 x C … (ii) Equating Equation (i) & (ii), we get, 50265.48 x C = 10.188C C = 10.188 / 50265.48 C = 2.0268 x 10 -4 C = 202.7 x 10-6 C = 202.7μF ******************************************************* ********************* Example 4 : What value of Capacitance must be connected in parallel with a load drawing 1kW at 70% lagging power factor from a 208V, 60H S i d i h ll f 91% 60Hz Source in order to raise the overall power factor to 91%. Solution: You can use either Table method or Simple Calculation method to find the required value of Capacitance in Farads or kVAR to improve Power factor from 0.71 to 0.97. So I used table method in this case. P = 1000W Actual Power factor = Cosθ 1 = 0.71 Desired Power factor = Cosθ 2 = 0.97 From Table, Multiplier to improve PF from 0.71 to 0.97 is 0.783 Required Capacitor kVAR to improve P.F from 0.71 to 0.97 Required Capacitor kVAR = kW x Table Multiplier of 0.71 and 0.97 = 1kW x 0.783 =783 VAR (required Capacitance Value in kVAR) Current in the Capacitor = IC = QC / V = 783 / 208 = 3.76A And XC= V / I C = 208 / 3.76 = 55.25Ω C = 1/ (2 π f X C) C = 1 (2 π x 60 x 55.25) C = 48 μF (required Capacitance Value in Farads) Good to Know: Important formulas which is used for Power factor improvement calculation as well as used in the above calculation Power in Watts kW = kVA x Cosθ kW = kVA x Cosθ kW = HP x 0.746 or (HP x 0.746) / Efficiency … (HP = Motor Power) kW = √ ( kVA 2– kVAR 2) kW = P = VI Cosθ … (Single Phase) kW = P =√3x V x I Cosθ … (Three Phase) Apparent Power in VA kVA= √(kW 2+ kVAR 2) kVA = kW/ Cosθ Reactive Power in VA kVAR= √(kVA2– kW2) kVAR = C x (2 π f V 2) Power factor (from 0.1 to 1) Power Factor = Cosθ = P / V I … (Single Phase) Power Factor = Cosθ = P / (√3x V x I) … (Three Phase) Power Factor = Cosθ = kW / kVA … (Both Single Phase & Three Phase) Power Factor = Cosθ = R/Z … (Resistance / Impedance) X C = 1/ (2 π f C) … (X C = Capacitive reactance) I C = V/ X C … (I = V / R) Required Capacity of Capacitor in Farads/Microfarads C = kVAR / (2 π f V 2) in microfarad Required Capacity of Capacitor in kVAR kVAR = C x (2 π f V 2) ******************************************************* ********************** Example 5 : § 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]) § Total Load KVAR1=20.35 KVAR § OR § tanǾ1=Arcos(0.82)=0.69 § tanǾ2=Arcos(0.98)=0.20 § Total Load KVAR1= KWX (tanǾ1- tanǾ2) =41.1(0.690.20)=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]) § Total Load KVAR2=7.82 KVAR 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 Load KVAR1=4.17 KVAR § 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 Amps ******************************************************* ********************** 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) § Cn=Capacitance Farad. § Un = Line Voltage § Dv=Capacitor Discharge voltage. § Maximum Discharge resistance =60 / ((5 96/1000000)x log ( 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 § Total electrical load= 50.1+20.08+11.76 § Total electrical load=82 KVA § Total electrical Load KW=kW1+KW2+KW3 § Total electrical Load KW=37+15+10 § Total electrical Load KW =62kw § Load Current=KVA/V=80×1000/(415/1.732) § Load Current=114.1 Amp § 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 § Total electrical load= 41.95+17.01+10.20 § Total electrical load=69 KVA § Total electrical Load KW=kW1+KW2+KW3 § Total electrical Load KW=37+15+10 § Total electrical Load KW =62kw § Load Current=KVA/V=69×1000/(415/1.732) § Load Current=96.2 Amp § 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 **************************************************** ******************** You may also see...  Why is electrical power transmitted at low frequencies and not at high frequencies..???  Why do we hear a hissing noise while passing a high voltage line...???  Introduction to Conductors and their types...  Formulas for electrical engineers...  What are the Causes of Poor Power Factor ???  What is True, Reactive, and Apparent power ... 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