EXPERIMENT #3 PARALLEL CONNECTION OF TWO LINES

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1 1.2.2 Parallel Connection EXPERIMENT #3 PARALLEL CONNECTION OF TWO LINES This type of arrangement of the operating equipment forms the basis of meshed networks. Ik.rc t00,!11~ rdaliunships ~ flrsi U.C1ivcd for UH; siu1plifie<.i representation (i.e. without shunt elements); then a more exact treatment, in which the shunt elements are considered. is carried out A characteristic property of a parallel circuit is the fact that the same voltage drop occurs across all branches. Thus. the total current in a parallel circuit is always splits so that the ratio of the currents in the branches is inversely proportional to the ratio of the impedances in the branches. The simplified equivalent circuit diagram of two lines in parallel is given in the following figure. Load Key: iongitudinal impedance of line 1, z, = R1 + j ro L1 = R, + jx1 longitudinal impedance of line 2, L = A2 + j ro u '"' R2 + jx2 feeding voltage at the beginning of the lines,!!a load voltage at the end of the lines,.u.e current through line 1,!1 current through line 2,.lz load current, 1E Fig. 9 - Simplified equivalent circuit diagram of two lines in parallel A mixed ohmic-inductive load is assumed at the ends of the two lines in this experiment, in order to simulate realistic conditions with respect to the demand of reactive power. The relationships derived are also principally valid for other load cases. 1

2 The phasor diagram for the circuit shown in Fig. 9 is given in the following figure. Fig. 10 Phasor diagram for the simplified equivalent circuit (for the meanings of Rta and Xlot see below) The tw9 line impedances Za and b form the total impedance Zaoc in accordance with the following equation: Solving the above equation gives '[;.d. = Reoc + j Xtot = =--+-- = + z, Rt + j ro Lt l 1 Rt Rl (Rt + Rl) + Rt X2 + Rl Xt = 2 l {Rt + Rl) + (Xt + X1) +j 2 2 (Rt + R1) + (X1 + Xz) 2

3 The current through line 1 is given by: The current through line 2 is given by: Accordingly, when two identical lines are connected in parallel, the circuit behaves as if it were a single line of half the length (here shunt elements have been ignored):each of the two line branches carries half the load current. Consideration of the shunt elements (operating capacitances) leads to the following equivalent circuit diagram.!, A 1A 1 E -E Load Fig Equivalent circuit diagram of two lines in parallel with shunt elements taken into consideration 3

4 The phasor diagram for the equivalent circuit of Fig. 11 is given in the following figure. Fig Phasor diagram for the equivalent circuit with shunt elements taken into consideration The treatment in the phasoc diagram mentioned above is again qualitative (i.e. not to scale). In contrast to the situation with the series circuit, the relationships here are also quite simple, even when the shunt elements have been taken into consideration. The operating capacitances, which were assumed to be concentrated elements in the 1t equivalent circuit, are added and increase the capacitance of the arrangement accordingly. To draw the phasor diagram the given quantities load current le and voltage lle at the load are again assumed. The addition of IE with the currents!eot and Iem at half the operating capacitance of lines 1 and 2 leads to the fictive current!ae. which is responsible for the voltage drop across the parallel-connected longitudinal impedances Zt and 22. The expressions derived for the simplified representation are t~lso valid fnr thi ~ parallel circuit. In order to obtain the current la which flows into the circuit, the two capacitive currents,laot and!acn are to added to las at the beginning of the lines. 4

5 EXPERIMENT N. 2 PARALLEL CONNECTION OF TWO LINES Objectives: Measurement of the current distribution in the parallel connection of two lines. Investigation of the effect of the operating capacitances on the voltages and currents. Equipments: 2 DL 7901TT 1 DL 1013Tl 1 DL 1017R 1 DL 1017L 1 DL 1080TT 1 DL 2108T02 2 DL 2109T2A5 3 DL 2109T3PV Overhead line model Three-phase power supply Resistive load Inductive load Three-phase transformer Power circuit breaker Moving-iron ammeter (2.5 A) Moving-iron voltmeter ( V) 5

6 EXPERIMENT 2.1: TWO LINES IN PARALLEL H* *' U* ~ co Q; (..) c ~ (..) C'O c.. C'O (..) c CL:.;:: C'O ~ 0) C'O c -... Q) "' c.. Q) 0 -= - - :;:, 0 0 ~._ ~ C'i -c Q). G; c.. >< w 6,.,. f o I I ~ i. ~. ' d ~ dl. I~.

7 Experiment procedure Assemble the circuit according with the foregoing topographic diagram. S~l j..hillla.j. j -Siue o[ UlC U U!X - pha:;~ l.fan:.furmer in ud la cuunectiuu 380 v anj llslllg bridging plugs set the secondary-side to star.un + 5%. Remove all bridging plugs connecting the capacitances to both line models. Connect a three-phase balanced ohmic-inductive load to end terminals of the parallel-connected lines: set the load value to R1 - L1. Adjust the supply voltage in order to obtain the nominal voltage.un = 380 V (phase-to-neutral voltage 220 V) at the beginning of the parallel-connected lines: this value must be kept constant for all the measurements. Beginning from R1 - Lt value change the load in steps for the indicated values. For each step measure the following quantities: current lt at the beginning of line 1, current }2 at the beginning of line 2, current IE and voltage.ue through the load. Enter the measured value into the following table. 1J.A (V) R-L j, (A) b (A) 1e (A).U.e (V) R1- l1 220 R2- L2 220 R3- L3 220 R.- LA 220 Rs - l.s - Temporarily disconnect one of the lines and repeat the above measurements. - lj.a (V) R-L 1, (A) 12 (A) le (A).Ue1 (V) 220 R,- l R2- L R3- L R.. - L Rs- Ls 0 7

8 Plot the load voltage and the current at the beginning lines as a function of the load current in a combined diagram. U!V) I( A) IE (A)

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