Simulation And Analysis Of Current DistributionOn DC Electric Multiple-units With Bus Line

Abstract

DC electric multiple-units are mainly operated on DC 1,500V narrow-gauge railways in Japan. Most of these cars have a high-voltage bus line to reduce damage and wear of pantographs and contact wires by decreasing arcs in between. Among various problems experienced with high-voltage bus lines up to now, we investigated the phenomenon of excessive current. To attain preventive measures against this problem, we have developed a simulation program to analyze the current distribution on DC electric multiple-units. By using this program, we analyze a case of actual installation and make clear the influence of parameters, propose an idea to decrease excessive current that flows in the bus line. 1 Model of calculation and solution 1.1 Model of calculation With the calculation model we made, we calculated the current that flows in the bus line at each moment by sliding the circuit of vehicle. We have developed a simulation program by using this model. This program can calculate the currents in sections with and without insulated overlaps. Figure 1 shows the equivalent circuit for current analysis. The vehicle runs from the section of resistance RB to the that of resistance Rig with the No 3 pantograph at the front . Transactions on the Built Environment vol 34, © 1998 WIT Press, www.witpress.com, ISSN 1743-3509 460 Computers in Railways 1.2 Solution of equivalent circuit We assigned the currents in each mesh of equivalent circuit (Figure 1) and derived the following equations by using Kirchhoff s law. Values of ii~in can be obtained by solving these equations of 17 unknowns. In this program, actual solutions were obtained by using the triangular factorization method. The values of voltage required are obtained by the increase in the voltage from the basis based on the values of current obtained. EL =ii{Ri+Ri2-(Ra+Rb+Rc)+Rpl+RGl+RRL}-i2{Rl2-(Ra+Rb+Rc)} -iiiRpi-inRGi-iieRGi (1) 0 =iz(R2+Rl2)-ii (Rl2-(Ra+Rb+Rc)Hll(Ra+Rb)-il2 Re (2) 0 =l3(R3+Rl3)-il2Rd-i]o(Rl3-Rd ) (3) 0 =J4(R4+Rl4)-ilO Rl4 (4) 0=i5(R5+Ri5)-iioRi5 (5) 0 =l6(R6+Rl6)-iio Rl6 (6) (7) (8) 0=i9(R9+Ri9)-iioRi9 (9) -ER =iio(Ri3Rd+ RM+ RIS+ Ri6+ Rn+ Ris+ Ri9+ Rio+ Ri +Rp3 + Ro3)-il2 RP3-117 RG3+115 Ri (10) 0 =in(Ra+ Rb+ RPI+ RP2+ RsiHi Rpri2(Ra+Rb)-ii2 Rp2-ii6 RBI (1 1) 0 =in(Rc+ Rd+ Rp2+ Rp3+RB2)-in Rp2-i2 Rc-in RB2-i3Rd -iioRps (12) 0 =ii3(RGi+ RMI+ R«)-iiRGi+ii6 Roi+ii6 Re (13) 0 =il4(RG2+ RM2+ Rg)-il6 RG2+il7RG2+il7 Rg (14) 0 =115(RG3+ RM3+ Ri)-ll7 RG3-11Q Rc3 (15) 0 =ii6(Rfii+ RGI+ RG2+ Re+ RfHi Roi+ii3 Re-in RBI -in Rc2+ii3 RGI-IH Rc2 (16) 0 =ii7(RB2+ RG2+ RG3+ Rg + Rh)~il6 RG2'il2 Rfi2 (17) 1 .3 Samples of input data We show the samples of values in the section with an insulated overlap between the substation A and substation B on the line C. Ri =0.137 [Q]R2 =0.028 [0]Rs =0.025[Q]R4 =0.010[Q] Rs =0.012 [Q]R6 =0.2204 [Q]R7=0.006[Q]Rg =0.004[0] Rg -0.004 [Q] Rio =0.083 [Q] Ri2=0.034[Q] Rn=0.028[Q] Ri4=0.020[Q] Ris=0.022[Q] Ri6=0.015[Q] Rn=0.024[Q] Ri8=0.015[Q]Ri9=0.015[Q] RMi=1.875[Q]RM2=2.142[Q] RM3=1.875[Q]Roi =100.0 [Q] RG2=100.0[Q] RG3=100[Q] Rpi=0.001 [Q]Rp2 =0.001 [Q] Rp3=0.001[Q] RBi=0.0132[Q] RB2=0.0203[Q]Re~Ri=0.0[Q] EL=1500[V] ER =1500[V] Transactions on the Built Environment vol 34, © 1998 WIT Press, www.witpress.com, ISSN 1743-3509