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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.
Solution.: (1) To find the required capacity of Capacitance in kVAR (i.e, Convert Capacitor Farads into kVAR) to improve P.F from 0.6 to 0.9 (Two Methods) Solution #1 (By Simple Table Method) Motor Input = P = V x I x Cosθ = 400V x 50A x 0.6 = 12kW From Table, Multiplier to improve PF from 0.60 to 0.90 is 0.849 Required Capacitor kVAR to improve P.F from 0.60 to 0.90 Required Capacitor kVAR = kW x Table Multiplier of 0.60 and 0.90 = 12kW x 0.849 = 10.188 kVAR Solution # 2 (Classical Calculation Method) Motor Input = P = V x I x Cosθ = 400V x 50A x 0.6 = 12kW ActualP.F = Cosθ1 = 0..6 Required P.F = Cosθ2 = 0.90 θ1 = Cos-1 = (0.60) = 53°.13; Tan θ1 = Tan (53°.13) = 1.3333 θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843 Required Capacitor kVAR to improve P.F from 0.60 to 0.90 Required Capacitor kVAR = P (Tan θ1 – Tan θ2) = 12kW (1.3333– 0.4843) = 10.188 kVAR (2) To find the required capacity of Capacitance in Farads (i.e, Convert Capacitor Farads into kVAR) to 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 V2) in microfarad Putting the Values in the above formula = (10.188kVAR) / (2 x π x 50 x 4002) = 2.0268 x 10-4 = 202.7 x 10-6 = 202.7μF |