Explanation Given data: The voltage of the three-phase inductive load is 440 V . The real power ( P 1 ) is 51 kW . And the apparent power is S = 60 kVA . The operating frequency ( f ) is 60 Hz . The power factor to be achieved to 0.95 (lagging). Formula used: Consider that the real power of the newly connecting load is equal to the present load. The real power of the present and newly connected load is, P 1 = P 2 = 51 kW . Here, P 1 is the real power in present load. P 2 is the real power in newly connected load at 0.95 factor.. Write the expression to find the real power ( P 1 ) . P 1 = S 1 cos θ 1 (1) Here S 1 is the apparent power of the present load, and cos θ 1 is the power factor of the inductive load. Rewrite the equation (1) to find the power factor. cos θ 1 = P 1 S 1 (2) Write the expression to find the reactive power ( Q 1 ) . Q 1 = S 1 sin θ 1 (3) Write the expression to find the real power ( P 2 ) . P 2 = S 2 cos θ 2 (4) Here S 2 is the apparent power newly connected, and cos θ 2 is the power factor of the inductive load. Write the expression to find the new reactive power ( Q 2 ) . Q 2 = S 2 sin θ 2 (5) Write the expression to find the reactive power of the shunt capacitor. Q C = Q 1 − Q 2 (6) Here Q 1 is the old reactive power, and Q 2 is the new reactive power at 0.95 power factor. Write the expression to find the reactive power for each load is, Q C 1 = Q C 3 (7) Here, Q C is the reactive power of the shunt capacitor. Write the expression to find the capacitance. C = Q C 1 ω V rms 2 (8) Here Q C 1 is the reactive power of the shunt capacitor. ω is the angular frequency, and V rms is the rms voltage. Write the expression to find the angular frequency. ω = 2 π f (9) Here, f is the operating frequency. Calculation: Substitute 51 kW for P 1 , and 60 kVA for S 1 in equation (2) to find the power factor. cos θ 1 = 51 kW 60 kVA = 0.85 { ∵ 1 W=1 VA } Rewrite the above equation to find the angle ( θ 1 ) . θ 1 = cos − 1 ( 0.85 ) θ 1 = 31.79 ° Substitute 31

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ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 12, Problem 78CP

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Find the value of capacitor that should be connected in parallel.

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Chapter 12, Problem 77CP

Chapter 12, Problem 79CP

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Find the value of capacitor that should be connected in parallel.

Solution Summary: The author explains how to find the value of a capacitor that should be connected in parallel. The real power of the newly connecting load is equal to the present load.

Find the value of capacitor that should be connected in parallel.

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Chapter 12 Solutions