Alternating current three-phase circuits

Transcription

1 Unit. C HEE-PHE CCU lternating current three-phase circuits Unit. C HEE-PHE CCU COE: hree-phase systems characteristics Generation of three-phase voltages hree-phase loads -Y and Y- transformation nstantaneous power hree-phase power:, P and Power measurement. aron connection Power factor improvement Electrical measurements Exercises Unit. C HEE-PHE CCU HEE-PHE YEM CHCEC Unit. C HEE-PHE CCU HEE-PHE YEM CHCEC he electricity grid is made up of four main components: GEEO: production of electricity from energy sources such as coal, natural gas, hydropower, wind and solar. MO: the transmission system carries the electric power from power plants over long distances to a distribution system. DBUO: the distribution system brings the power to the customers. COUME: these are the consumers of electric power (industry, service sector and residential uses. 4

2 Unit. C HEE-PHE CCU HEE-PHE YEM CHCEC Unit. C HEE-PHE CCU GEEO OF HEE-PHE OGE nstantaneous electric power has a sinusoidal shape with double frequency than voltage or current. GE-PHE C CCU: instantaneous electric power is negative twice a period (power flows from the load to the generator and positive twice a period, falling to zero. BCED HEE-PHE C CCU: instantaneous electric power is constant. hree-phase power never falls to zero. hree-phase electric motors perform better than single-phase C motors. hree-phase power systems allow two voltage levels (-, -. When electric power is transmitted, three-phase C systems require 5% less Cu/l than single-phase C systems. 5 hree-phase generators contain three sinusoidal voltage sources with voltages of the same frequency but a 0º-phase shift with respect to each other. his is achieved by positioning three coils separated by 0º angles. here is only one rotor. mplitudes of the three phases are also equal. he generator is then balanced. 6 Unit. C HEE-PHE CCU ODUCO Unit. C HEE-PHE CCU ODUCO 0º : neutral point (or B C direct sequence or sequence 7,, : line voltages or line-to-line voltages,, : line-to-neutral voltages line = line-to-neutral v (t = 0 cos( t + 90º v (t = 0 cos( t - 0º v (t = 0 cos( t +0º 8

3 Unit. C HEE-PHE CCU ODUCO Unit. C HEE-PHE CCU HEE-PHE OD CFCO 0º line. line to phase WYE (two voltages Balanced -wires 4-wires Unbalanced -wires 4-wires O= O= 50 Hz Usual system phase = line-to-neutral 0 volt volt line Frequency 50 Hz DE (one voltage Balanced Unbalanced 9 0 Unit. C HEE-PHE CCU BCED WYE-COECED OD he wye or star connection is made by connecting one end of each of the three-phase loads together. he voltage measured across a single load or phase is known as the phase voltage. he voltage measured between the lines is known as the line-to-line voltage or the line voltage. n a wye-connected system, the line voltage is higher than the load phase voltage by a factor of the square root of. n a wye-connected system, the phase current and line current are the same. Unit. C HEE-PHE CCU BCED DE-COECED OD his connection received its name from the fact that a schematic diagram of it resembles the Greek letter delta (. n the delta connection, the line voltage and phase voltage in the load are the same. he line current of a delta connection is higher than the phase current by a factor of the square root of.

4 Unit. C HEE-PHE CCU -Y FOMO Unit. C HEE-PHE CCU BCED HEE/FOU-WE WYE-COECED OD between nodes and : nodes -: nodes -: ( ( nodes -: ( ( (, ( ( ( ( From expressions (, ( and ( the result is: ( + ( - ( ( - ( + ( -( + ( + ( ( ( Y: Y, O= O= ( 0. fase...línia.línia.línia.línia.cos j.línia.línia.sin P º he three currents are balanced. hus the sum of them is always zero. ince the neutral current of a balanced, Y- connected three-phase load is always zero, the neutral conductor may be removed with no change in the results. Balanced loads: Y = / 4 Unit. C HEE-PHE CCU BCED HEE/FOU-WE WYE-COECED OD Example three-phase, system (, 50 Hz, has a three-wire Y-connected load for which = 0 0º Obtain the line currents and the complex power consumption 90º 60º / 0º 0 P O= / 0º 0 / 0º 0 phase 0º 0º 60º 80º (/ 0º (watt l l 6000 l l cos cos 0º 856.4watt l l sin sin 0º 8000r 90º (/ 60º j8000 (r 5 Unit. C HEE-PHE CCU UBCED FOU-WE WYE-COECED OD O= ( 0 6

5 Unit. C HEE-PHE CCU UBCED HEE-WE WYE-COECED OD O O= O O O Y Y 0 O O 0 Y Y Y O 0 Y 7 Unit. C HEE-PHE CCU UBCED HEE-WE WYE-COECED OD O= O O O O O O O O O O 8 Unit. C HEE-PHE CCU UBCED HEE-WE WYE-COECED OD Example three-phase, system (, 50 Hz, has a three-wire unbalanced Y-connected load for which =0 0º, =0 0º and =0 0º Obtain the line currents and the complex power consumption. O Y Y Y O Y Y Y 90º 0º -0º 0º 0º º 0º 0º º.9 O = O = 0 90º º = º O = O = 0-0º º = 64,54-5.0º O = O = 0 0º º = 7,8 -.0º = O / = º /0 0º = 9, º = O / = º /0 0º = 6,45-5.0º = O / = º /0 0º = º tot = O + O + O = W + j8.7 r 4.89º 9 Unit. C HEE-PHE CCU BCED DE-COECED OD fase línia línia línia línia cos j línia línia sin P º 0

6 Unit. C HEE-PHE CCU UBCED DE-COECED OD Unit. C HEE-PHE CCU UBCED HEE-WE -COECED OD Example three-phase, system (, 50 Hz, has an unbalanced - connected load for which =0 0º, =0 0º i =0-0º Obtain the line currents and the complex power consumption 0º 0º 0º 0 0º 0º 0º 0º 0 0º 90º 0º 0 05º º 50º 47.8(W j0 (r Unit. C HEE-PHE CCU POWE MEUME. Four-wire load Balanced wye-connected, four-wire load Unbalanced wye-connected, four-wire load W W W O= W O= W = cos( - W = cos( - W = cos( - P = W + W + W W= cos( - P = W Unit. C HEE-PHE CCU POWE MEUME. O COECO General -wire load. wo-wattmeter method (O connection W Demonstration done for a balanced -wire load W cos( cos(, cos( 0º cos(0º OD 0º PO W W [cos( 0º cos(0º ] cos [W W ] [ cos( 0º cos( 0º ] sin O 4

7 Unit. C HEE-PHE CCU POWE MEUME. BCED OD Balanced load, general (Y/D, /4 wires. wo-wattmeter method (O connection, BCED OD P W W arctg( (W W W W W W Unit. C HEE-PHE CCU POWE MEUME: HE WO-WMEE MEHOD ron cyclic permutations W W Unbalanced wye/delta-connected, three-wire load measurement: cyclic permutations, UBCED OD P = W + W 5 W W cos( line line cos(90º- line line sin O O W 6 Unit. C HEE-PHE CCU EOU HEE-PHE POWE ingle-phase load: cos cosb = 0 5 [cos(+b + cos(-b] p(t = v(t i(t = 0 cos(w t + 0 cos(wt + p(t =/ 0 0 cos( - + / 0 0 cos(wt + + watt Constant Oscillates at twice the mains frequency! Unit. C HEE-PHE CCU EOU POWE: GE-PHE OD v(t hree-phase wye balanced load: p(t = v (t i (t + v (t i (t + v (t i (t = = p cos(wt + p cos(wt + + p cos(wt -0º+ p cos(wt -0º+ + p cos(wt +0º+ p cos(wt +0º+ = p p cos( - + p p cos(wt p p cos( - + p p cos(wt -º+ + + p p cos( - + p p cos(wt +º+ + =/ p p cos( - =/ p p cos = constant! 7 i(t p(t verage power = P 8

8 Unit. C HEE-PHE CCU EOU HEE-PHE POWE Unit. C HEE-PHE CCU POWE OE: HEE-PHE/GE PHE v(t ingle-phase line OD Pload cos Pload Plosses cos i(t hree-phase line Pload cos P losses ( Pload. cos Pload cos upposing same losses l p p p p l p(t ingle-phase line: conductors of length l and section p hree-phase line: conductors of length l and section p = / p p O = p (t + p (t + p (t 9 s a result: weight p-cables = /4weight p-cables 0 Unit. C HEE-PHE CCU Example hree-phase balanced system for which =.6, = 5.77, W = 698. W, W = 000 W, U = 6000 and line =4+j a Obtain the complex power in the loads, as well as the ammeter and the voltmeter U readings. b Obtain the value of C to improve the load s PF to, assuming U = W 000r U U W Balanced load capacitive W Balanced load inductive (P P j( j O.6 90º 45º º 5.º K C C C 90º 6.85º 000r 45º P P W 6000W 5.º 6.87º U sin U cos 000W U cos U sin 48000r 90º 6.85º º U, phase. Uphase (4 j U U, phase 6050 (6000 C.06 F C, / -6000/r - /(.50.C 90º Unit. C HEE-PHE CCU Example hree-phase 50 Hz system for which =, W = W, W = W, =.56. Obtain a the value of. b the reading of. c the value of the inductance W W W ' W W W and W ' W a P b t otal W ' W ' W resulting in : º 60º 60º 60º º t results : º 60º 60º c j5 90º jx X ( 5 X he result is : H X j X

9 Unit. C HEE-PHE CCU Example arley phase-sequence indicator. Calculate the voltage in each element and deduce the practical consequences hree-phase /50 Hz system C = F X C = 8 bulbs = /P = 0 /0 = 590 Y Y Y O Y Y Y 90º -90º -0º 0º 0º 0º 0 /8 0 /590 0 /590 90º 0º 0º / 8 / 590 / º 7.55 O = O = 0 90º º = º 0 O = O = 0-0º º = º 0 0 O = O = 0 0º º 0 0 = ª Conductor is situated where the capacitor is placed, conductor is situated where the brighter bulb is placed and is the remaining conductor. Unit. C HEE-PHE CCU Example 4 and 50 Hz three-phase line feeds two balanced loads through a line which has an internal impedance of =0.5 + j he-connected load has phase impedances whose values are 45+j 0 Ω, whereas the Y-connected load has phase impedances of 5 j 0 Ω Determine: a the reading of the ammeter, b the reading of the voltmeter and c the readings of watt-meters W and W. O W ( W Y / / Y // (0.5 j (6.7 j º / 8.875º a.0 W W 7.5º O 7.65 b ( // ( c POD W W he ron connection results in: OD (W W.0.54 W = 466, W, W = 46,5 W 4 7.5º Unit. C HEE-PHE CCU Example 5 hree-phase -50 Hz line. When switch K is closed, W = 0 W. When K and K are closed, W = 85.6 W and W B = W. Determine: a, b and c. W B a K closed: W = cos( - W 0 = cos(0º+0º =.55 K K K = / = (//.55 = 0 C b K and K closed: P O = W + W = -W B + W = 00 W = / + / = 0 W W c K and K closed: 80º 60º 60 j º 0 j º ( 60 j 4.64 (-0 j 0-80 j his results in = Unit. C HEE-PHE CCU uestion n electrical lineman is connecting three single-phase transformers in a Y(primary- Y(secondary configuration, for a business s power service. Draw the connecting wires necessary between the transformer windings, and those required between the transformer terminals and the lines. ote: fuses have been omitted from this illustration for simplicity. 6

10 Unit. C HEE-PHE CCU uestion dentify the primary-secondary connection configuration of these pole-mounted power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.. Unit. C HEE-PHE CCU uestion dentify the primary-secondary connection configuration of these pole-mounted power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.. H hese transformers are connected in a Yy configuration. 7 hese transformers are connected in a Yd configuration. 8 Unit. C HEE-PHE CCU uestion dentify the primary-secondary connection configuration of these pole-mounted power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.. hese transformers are connected in open-delta configuration. hree single-phase transformers are not normally used because this is more expensive than using one three-phase transformer. However, there is an advantageous method called the open-delta or - connection t functions as follows: a defective single-phase transformer in a Dd three-phase bank can be removed for repair. Partial service can be restored using the open- Delta configuration until a replacement transformer is obtained. hree-phase is still obtained with two transformers, but at 57.7% of the original power. 9 his is a very practical transformer application for emergency conditions. Unit. C HEE-PHE CCU uestion 4 One of the conductors connecting the secondary of a three-phase power distribution transformer to a large office building fails when open. Upon inspection, the source of the failure is obvious: the wire overheated at a point of contact with a terminal block, until it physically separated from the terminal. What is strange is that the overheated wire is the neutral conductor, not any one of the line conductors. Based on this observation, what do you think caused the failure? fter repairing the wire, what would you do to verify the cause of the failure? Here s a hint ( pista : if you were to repair the neutral wire and take current measurements with a digital instrument (using a clamp-on current probe, for safety, you would find that the predominant frequency of the current is 50 Hz, rather than 50 Hz. his scenario is all too common in modern power systems, as non-linear loads such as switching power supplies and electronic power controls have become more prevalent. pecial instruments exist to measure harmonics in power systems, but a simple DMM (digital multimeter may be used as well to make crude assessments.

11 Unit. C HEE-PHE CCU POWE MEUME. O COECO General -wire load. wo-wattmeter method (O connection, OD p(t = v (t i (t + v (t i (t + v (t i (t p(t = v (t i (t + v (t i (t + v (t [-i (t - i (t] p(t = i (t [v (t - v (t] + i (t [v (t - v (t] = v (t i (t + v (t i (t Mean value P = W + W = cos( - + cos( - 4