. Electric railway journal . ngth of section is L = 4670 ftDistance to station Si is L>i = 5170 ftDistance to station S2 is Z?2 = 4320 ftSize of cable, 1,000,000 circ. current on station 5\ is then, by equation (2)4670 4320 + 1050 -f- 494 amp. 5170 -f 4670 + 4320Load on station 52 is then /, = / — = 1050 — 494 = 556 location of the point P of division of load is readilydetermined. Since the load is uniformly distributed along ABwe get L, = —j- L (3) We then have numerically494 Li = X 4670 = 2185 ft. 1050 that is, at 2185 ft. from the end A of the section each powerstation, d
. Electric railway journal . ngth of section is L = 4670 ftDistance to station Si is L>i = 5170 ftDistance to station S2 is Z?2 = 4320 ftSize of cable, 1,000,000 circ. current on station 5\ is then, by equation (2)4670 4320 + 1050 -f- 494 amp. 5170 -f 4670 + 4320Load on station 52 is then /, = / — = 1050 — 494 = 556 location of the point P of division of load is readilydetermined. Since the load is uniformly distributed along ABwe get L, = —j- L (3) We then have numerically494 Li = X 4670 = 2185 ft. 1050 that is, at 2185 ft. from the end A of the section each powerstation, delivers current. If the load distribution changes,then this point of division shifts, and may be determined foreach distribution. The maximum drop on the section occurs at P, and it is thesame from both stations. We have its value as in equation (1) L = ^57°= volts. (4) 21852 494 A second type of tie section is one in which the main feederbetween stations does not parallel the trolley section, through-. out its length, as shown 111 Fig. 4 The loads on the threeparts of the trolley section arc £,, C%, and C». We then havethe total load I=:Ci + Ci + C» (5)To find the load on we lake moments as before, about S,,which gives CD, + C, (p» I ^ \ + CAIh + U) = /, (D, + U H lh) Multiplying out and factoring we get (C, + C, + C) Pt + C^- +CJL* = U (D, + U + Dt)which by substituting from (5) gives ID, + C, -2 +CL. = U (D, + U + Dt) and hence ID, + C, ^ + CJL,Di + L- + D, (6) The load distributed over the distance x is /, — Cs, and sincethe distribution is uniform, we have h — C,C, (7) The drop from 5*s to the point of division P is then as shownin equation .(1) Bl^(h-Cz)^- + hD,r (8) The load per foot of section = amp. which is the maximum drop on the feeder, but which may beexceeded at the ends A and B of the trolley section. For a particular case we have: Load on section is I =750 amp. Length of section is L = 6740 ft. divided into lengths of3400 ft
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