Download THREE PHASE CIRCUITS

ELEC3203 Electricity Networks ‐ 2009 Page 1 of 3 THREE PHASE CIRCUITS ­ SOLUTIONS Question 1 (a) By sketching a phasor diagram or by direct calculation, V has a magnitude 173.2 75° V. V √3 greater than V and leads it in phase by 30°. V
lags V by 120, so V
173.2
45° . (b) If V
240√3 0° , V
240
30° so V
240
150° V. Question 2 By sketching a phasor diagram, I has a magnitude 1⁄√3 of the line current and 115.5 30° A. Or, simply use the formula sheet. leads I by 30°, so I
Question 3 The per phase equivalent circuit is as shown. V ⁄Z 100⁄ 3
4
12
16 20 53.1° A I
By symmetry, I
20
66.9° A and I
20 173.1° A. The phasor diagram is shown. For phase ‘a’, V I 100
12
16
1200
1600 VA. For three phases, 6000 , 3600 W and 4800 VAR. tut09s_03ThreePhase.docx ELEC3203 Electricity Networks ‐ 2009 Page 2 of 3 Question 4 11 0° kV. Then (a) Take as reference V
I
V ⁄Z
11000⁄ 12
3
889.3
14.04° A The line current 1540.3 A. I and V determine the √3
complex power absorbed by the relevant impedance, multiplying by 3 gives the total power, and I lags V by 14.04°, so the power factor is cos 14.04° 0.97 lagging. (b) The delta connected load can be replaced by impedances of Z⁄3 4
1 Ω connected in star. The voltage in the equivalent circuit is the line to neutral voltage, so take V
11 ⁄ √3 0° 6.3509 0° kV. Then, I
6350.9⁄ 4
1540.3
14.04° A, etc. Note: If in any doubt, adopt the second approach, i.e., a per phase equivalent circuit using line to neutral voltages etc, in solving this type of problem. Question 5 Note: For a three phase system, quoted values for power are always for three phases unless otherwise stated. (a) The three phase load has a pf cos
0.707 lagging. The three phase active power is cos , so the apparent power is 200⁄0.7071
282.85 kVA and the reactive power sin
200 kVAR. The capacitor bank supplies 50 kVAR (absorbs 50 kVAR) so the net reactive power absorbed 150 kVAR. For the combination of load and capacitors, 250 kVA, and the pf is ⁄
0.8 lagging. Note. Here, we did the calculations on three phase quantities. We could have thought in terms of a per phase equivalent circuit. We would have obtained the same answer for pf (all powers are reduced by a factor 3). (b) The line current can be found in several ways. The simplest is as follows. The per phase apparent power S
250⁄3 83.33 kVA and is equal to the product of the magnitudes of the line to neutral voltage and the line current (look at the per phase equivalent circuit!). Thus, the line current has a magnitude 83.33 kVA/ 440⁄√3 V 328 A. Note. You are asked only for the magnitude of the current, so do not bother with the arg. tut09s_03ThreePhase.docx ELEC3203 Electricity Networks ‐ 2009 Page 3 of 3 Question 6 (a) The equivalent circuit for ‘a’ phase is: The voltage is the line to neutral and the capacitor bank is converted to a star connected equivalent. Three capacitors of impedance 60 Ω connected in delta are equivalent to impedances of 20 Ω connected in star. (b) Taking V as reference, and denoting the total impedance connected across the supply by Z , V
Z
I
V
415⁄√3 0° V 40
20
1 8
15 17.0
40
20
V ⁄Z 14.09 61.93° A 6.633
V
I
252.1
1.51° V ⁄40 6.30 A ⁄20 12.61 A 61.93° Ω 12.436 A and the current in the capacitors connected in Δ is ⁄√3 7.28 A. Note that I in the equivalent circuit is the current in the lines connecting to the capacitor bank; it is not the current flowing in the actual capacitors. (c) The complex power from the source per phase is V I 415√3 14.09
61.93°
so for three phases, 4.768 kW and tut09s_03ThreePhase.docx 1589.2
2979.7 kVA 8.939 kVAR.