Mathcad - 02-12

Mathcad - 02-12 - then solve for the line current Recall...

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Since the circuit is capacitive, the reactive power is negative Q 1.694 - kVAR = Q Im S 3ph ( 29 = reactive power is the imag part of S 3ph : P 2.824kW = P Re S 3ph ( 29 = real power is the real part of S 3ph : S 3ph 3.293 kVA = the apparent power is the magnitude of S 3ph : S 3ph 3 S ph = S ph 0.941 0.565j - kVA = S ph V ph I = We can use the complex power formula on a per-phase basis arg I ( ) 0.964deg = I 7.921A = I 7.92 0.133j + A = I V ph Z = arg V ph ( 29 30 - deg = V ph 138.564V = V ph V AB 3 e j - 30 deg = kVAR kW kVA kW define Z 15 9j - = Z R j X C - = Since this is a wye-load with a series impedance, we need to find the total impedance and
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Unformatted text preview: then solve for the line current. Recall that the phase voltage lags the line voltage by 30 deg Solution: V AB 240 V ⋅ = X C 9 Ω ⋅ = R 15 Ω ⋅ = P2-12. Three of the series-connected impedances of P2-5 are connected in a three-phase, wye load. A balanced, three-phase set of voltages is applied to the load, and V AB = 240 p o V. Calculate the total, three-phase real, reactive, and apparent power....
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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