(25-08-2009, 11:17 AM)mukund Wrote: Solution:
Rail to Rail voltage Vt = pickup current x relay resistance = 0.110 x 20 = 2.2 volts
Voltage at feed end = Vf - Vt = 10 - 2.2 = 7.8 volts
Current at feed end = 7.8 / (20+7) = 288 mA
Current through ballast resistance = current at feed end - current at relay end = 288 - 110 = 178 mA
Ballast resistance = Relay pick up voltage / Ballast current = 2.2/0.178 = 12.35 Ohms
Track circuit length = Ballast resistance in Ohm.Km / Ballast resistance in Ohm = 2/12.35 = 0.162 Km = 162 Mtrs.
Realise not easy to include diagram in an answer here, but suggest you always do so in exam. It may be the lack of that which contributed to your confusion re what you were calculating.
You started off calculating the minimum rail volts for the relay just to pick. I think that you should have added a margin, say 10%, to ensure the track circuit would be reliable, so I'd be calculating on a relay current of 121mA and thus a rail voltage of 2.42V. My assumption would be that the track feed voltage may vary a bit as the voltage on the signalling power distribution changes under conditions of different loadings and fluctuations of incoming supply. Wouldn't want a track circuit to fail to pick because points had recently been operated and thus the charger drawing extra current from the feeder to recharge them!
You didn't state (but obviously are assuming) that the rail resistance is negligible and thus the rail volts at feed end equals that at relay end.
What you state as being "voltage at feed end" seems to be the "voltage dropped across the feed resistor" when track is just picked.
Not sure what you are thinking "current at feed end" is and that this may be where your answer first went adrift; I think that I'd have calculated the "current through the feed resistor" in that state
i.e. (10-2.42)/7 = 1082mA.
Then since we know that 121mA is going via the relay end relay and a further 121mA via the identical feed end relay, this leaves an absolute maximum of 840mA that can go into the ballast before the track is operating too close to bare minimum energisation to be considered reliably energised.
Hence the minimum resistance which can be tolerated is 2.42/0.84 = 2.9 ohms.
Since we are told that the minimum ballast resistance for 1km is 2 ohm, then the greatest length of the track would be
(2/2.9) x 1000 = 690m.
This feels a sensible length for such a track circuit, whereas a maximum length of 162m that you calculated seems very short that it would be impracticable. Yes Im know there are some very short track circuits and in particularly adverse ballast resistance conditions the maximum length of tracks can be severely limited, but it should have been this value that first alerted you to the fact that something might have gone wrong in your calculations.
I'll leave you to follow this through for the second part of the question
PJW