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EENG 457 Power System Analysis I HOMEWORK 1 1. A single-phase voltage source is applied to an impedance Z 4.0 30 The source current is measured as i(t ) 5cos(t ) A . Determine (a) the instantaneous power. (b) the real and reactive power delivered by the source. (c) the power factor at the terminals of the source. 2. A voltage source v(t ) 100cos(100 t 60 ) V is connected to an impedance Z 5.030 . (a) Determine the expressions for the current and the instantaneous power delivered by the source as functions of time. (b) Find the frequency and the average value of the instantaneous power. 3. The real power delivered by a source to two impedances Z1 4 j5 and Z 2 8 connected in parallel is 4000 W. Determine (a) the real power absorbed by each impedance. (b) the total source current Is. 4. A load consumes 60 kW at a power factor of 0.7 lagging from a 240-V 50-Hz source. The power factor at the source terminals is to be raised to 0.95 lagging by placing a capacitor in parallel with the load. Find the capacitance required in μF. 5. Two single-phase sources are connected by a line of impedance Z L 0.7 j 2.4 as shown in the figure. The source voltages are E1 50016.26 V and E2 5850 V . Find the complex power for each source and determine whether they are supplying or receiving real and reactive power. Also find the real and reactive power loss on the line. ZL=0.7+j2.4 Ω + + I E1 E2 6. A 380-V (line-to-line) three-phase source delivers power to a balanced delta-connected load of Z 10 j5 per phase through a three-phase line as shown below. Find (a) Current in phase ab of the load. (b) Total complex power supplied by the source. (c) Magnitude of the line-to-line voltage at the load terminals. 1+j2Ω a a' V L 1+j2Ω b 1+j2Ω c Z =380 V rms b' Z Z c' SOLUTION 1. (a) The source voltage in phasor form is V Z .I 50 4 30 20 30 V In the time domain v(t ) 20cos(t 30 ) V Instantaneous power p(t ) v(t ).i(t ) 100 cos(t ).cos(t 30 ) W 50 cos(30 ). 1 cos(2t ) +50sin(30 ).sin(2t ) (b) P 50.cos(30 ) 43.3 W, (c) Power factor = cos(30) = 0.866 2. Q 50.sin(30 ) Var Voltage phasor (rms) V (100 / 2)60 V V (100 / 2)60 14.1430 A Z 530 Instantaneous power I i(t ) 14.14 2 cos(t 30 ) p(t ) v(t ).i(t ) 2000cos(t 60 ).cos(t 30 ) W 1000cos(30 ). 1 cos(2t 120 ) 1000sin(30 ).sin(2t 120 ) Frequency of instantaneous power = 2 Average value = 1000cos(30 ) 866 W 3-) (a) Real power absorbed by each impedance V2 4 2 2 P1 Re V 2 2 0.0976 V , Z1 4 5 P1 P2 4000 W 0.2226 V 2 V2 2 1 P2 Re V 0.125 V Z2 8 V 17969 V 2 P1 1753.8 W, P2 2246.2 W 2 (b) Total source current, I s I1 I 2 2 1 V V 1 134.05 29.83 j16.35 A Z1 Z 2 4 j5 8 4-) Reactive power absorbed by the load Q P tan 60 tan(cos1 0.7) 60 1.0202 61.2122 kVar When the power factor at source terminals is raised to 0.95, the source will supply Q P tan 60 tan(cos1 0.95) 19.721 kVar Reactive power to be supplied by the capacitor Qc Q Q 61.2122 19.721 kVar 41.4912 kVar C V 2 41.4912 103 C 2300 F 2 50 2402 5-) I E1 E2 50016.26 5850 42 j56 A 7053.13 A ZL 0.7 j 2.4 Complex power supplied by E1 : S1 E1 I * 50016.26 70 53.13 28 j 21 kVA Complex power absorbed by E2 : S2 E2 I * 5850 70 53.13 24.57 j 32.76 kVA E1 supplies 28 kW real power , and absorbs 21 kVar reactive power E2 absorbs 24.57 kW real power , and supplies 32.76 kVar reactive power Complex power absorbed by the line impedance, 2 Sline Z L I (0.7 j 2.4) 702 3.43 j11.76 kVar Line absorbs 3.43 kW and 11.76 kVar 6-) Convert the to Y: ZY Z / 3 (10 j5) / 3 . Single-phase equivalent circuit: 1+j2Ω a a' Ia ZY n Van 380 / 30 13 1 11 1 j 2 3 (10 j5) 3 j 3 29.5 j 24.97 A 38.65 40.23 A Ia 1 I a . 30 22.31 70.23 A 3 (b) Total complex power supplied by the source I ab S Van I a* Vbn I b* Vcn I c* 19.421 j16.43 kVA (c) 5 Van I a (10 3 j 3 ) 139.96 j34.044 V 144.04 13.67 Vab 3 144.04 249.48 V