EE 255 - Chapter 9i

# EE 255 - Chapter 9i - Thus the line voltage lags the...

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285 Thus the line voltage lags the corresponding phase voltage by 30 ° . The phasor diagram for this connection is shown below. V an V bn V cn V ab V bc A-5. Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure P2- 3. S OLUTION Note that because this load is Δ -connected, the line and phase voltages are identical. 120 0 V 120 120 V 208 30 V ab an bn - - = - = ∠ ° ° = ° V V V 120 120 V 120 240 V 208 90 V bc bn cn - - - = - = ∠ - ° ° = ° V V V 120 240 V 120 0 V 208 150 V ca cn an - = - = ∠ - ° ∠ ° = ° V V V

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286 208 30 V 20.8 10 A 10 20 ab ab Z φ ° = = = ° ° Ω V I 208 90 V 20.8 110 A 10 20 bc bc Z ∠ - ° = = = ∠ - ° ° Ω V I 208 150 V 20.8 130 A 10 20 ca ca Z ° = = = ° ° Ω V I 20.8 10 A 20.8 130 A 36 20 A a ab ca - - = - = ∠ ° ° = ° I I I 20.8 110 A 20.8 10 A 36 140 A b bc ab - - = - = ∠ - ° ∠ ° = ° I I I 20.8 130 A 20.8 -110 A 36 100 A c ca bc - = - = ° ° = ° I I I A-6. Figure PA-4 shows a small 480-V distribution system. Assume that the lines in the system have zero impedance. (a) If the switch shown is open, find the real, reactive, and apparent powers in the system. Find the total current supplied to the distribution system by the utility. (b) Repeat part (a) with the switch closed. What happened to the total current supplied? Why? S OLUTION (a) With the switch open, the power supplied to each load is ( ) kW 86 . 9 5 30 cos 10 V 80 4 3 cos 3 2 2 1 = ° Ω = = θ Z V P ( ) 2 2 1 480 V 3 sin 3 sin 30 34.56 kvar 10 V Q Z = = ° = Ω ( ) kW 04 . 46 36.87 cos 4 V 277 3 cos 3 2 2 2 = ° Ω = = Z V P ( ) 2 2 2 277 V 3 sin 3 sin 36.87 34.53 kvar 4 V Q Z = = ° = Ω kW 105.9 kW 46.04 kW 86 . 59 2 1 TOT = + = + = P P P TOT 1 2 34.56 kvar 34.53 kvar 69.09 kvar Q Q Q = + = + = The apparent power supplied by the utility is 2 2 TOT TOT TOT 126.4 kVA S P Q = + = The power factor supplied by the utility is
287 -1 1 TOT TOT 69.09 kvar PF cos tan cos tan 0.838 lagging 105.9 kW Q P - = = = The current supplied by the utility is ( ) ( ) TOT 105.9 kW 152 A 3 PF 3 480 V 0.838 L T P I V = = = (b) With the switch closed, 3 P is added to the circuit. The real and reactive power of 3 P is ( ) ( ) kW 0 90 cos 5 V 277 3 cos 3 2 2 3 = ° Ω = = - Z V P θ φ ( ) ( ) 2 2 3 277 V 3 sin 3 sin 90 46.06 kvar 5 V P - Z = = ° = - Ω TOT 1 2 3 59.86 kW 46.04 kW 0 kW 105.9 kW P P P P = + + = + + = TOT 1 2 3 34.56 kvar 34.53 kvar 46.06 kvar 23.03 kvar Q Q Q Q = + + = + - = The apparent power supplied by the utility is 2 2 TOT TOT TOT 108.4 kVA S P Q = + = The power factor supplied by the utility is -1 1 TOT TOT 23.03 kVAR PF cos tan cos tan 0.977 lagging 105.9 kW Q P - = = = The current supplied by the utility is ( ) ( ) TOT 105.9 kW 130.4 A 3 PF 3 480 V 0.977 L T P I V = = = (c) The total current supplied by the power system drops when the switch is closed because the capacitor bank is supplying some of the reactive power being consumed by loads 1 and 2.

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EE 255 - Chapter 9i - Thus the line voltage lags the...

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