2007 Track circuit calculation, Q3

[Peter]

Right hello again, I am back as promised, with a revised attempt at the 'drop shunt' calculation. Please see what you think, and comments greatly received!!!

OK. One thing that would help a lot is if you could write down the initial conditions from the question so that we know whether or not you have realised that something is included or not. The reason being that the question at the start of this thread refers to a feed end relay being fitted. Your calculations have not included this which, numerically, does not make things wrong, but in terms of answering the question, it does not quite fit the bill. Anyway, on with what you did do.

-PU current plus 10% for reliability=121mA
-DA current = PU current x 68% of PU current
121x0.68=82.28mA
-DA voltage = DA current x relay resistance
0.08228x20=1.645 volts

OK so far.

-Current supplied by feed = feed voltage-DA voltage/relay resistance
(10-1.645)/7=1.193A

Numbers correct but you mean "Feed resistance" not "relay resistance" in your description. Apart from that all OK

-Ballast current = rail voltage/TC length
2.2/0.69=3.188A

Oops.
Several alarm bells here.
1. Dimensions: you are seeking to calculate a current - you won't get amps by dividing a voltage by metres.
2. The value you have used for rail volts is the value that would be present at pick up given the numbers in the original question (110mA through 20ohm relay). You have already said that we are looking at the drop away point and you calculated the rail volts at this point as 1.645v above.
3. Where did you get the 0.69 from?
4. Magnitude of the answer. Think about what you are calculating. From your words, it is the element of the current that leaks through the ballast. How, therefore, can this be bigger than the total current supplied from the feed.

-Current through 'drop shunt' = ballast current-current supplied by feed-DA current
3.188-1.193-0.08228=1.91272A

Right sort of words, but wrong numbers (and remember the point I made above about the feed end relay)

-Value of 'drop shunt'=DA voltage/current through shunt
1.645/1.912=0.86 Ohms

As quoted from the 'SMS, Signalling Maintenance Specifications,' minimum drop shunt value=0.5 Ohms, Preferred drop shunt value 0.8-1.0 Ohm

So this answer given falls into the 'preferred' drop shunt value.

Fingers crossed!!!

Have a look at those bits and try to annotate your steps with a bit more of an explanation to we can understand your thinking a bit more. Don't lose heart, you are starting off well.

Peter