2002 Track circuit cal.

[leochungnet]

I am a bit confused for the calculation of it. Could anyone point out my mistakes?

Can i use the pickup current only as shown in the attachment to calculate the feed resistance? So, it will only include Ballast Resistance, Relay Resistance and Feed Resistance.

From another post, it shows an approach by assuming the drop away % is 65% of Pick-up current. It will include Shunt Resistance, Ballast Resistance, Relay Resistance and Feed Resistance in the calculation. The Feed resistance will be about 3.36ohm.

Which one is correct??

Thanks in advance!!

[Peter]

I am a bit confused for the calculation of it. Could anyone point out my mistakes?

Can i use the pickup current only as shown in the attachment to calculate the feed resistance? So, it will only include Ballast Resistance, Relay Resistance and Feed Resistance.

From another post, it shows an approach by assuming the drop away % is 65% of Pick-up current. It will include Shunt Resistance, Ballast Resistance, Relay Resistance and Feed Resistance in the calculation. The Feed resistance will be about 3.36ohm.

Which one is correct??

Thanks in advance!!

What you have worked out there is the maximum value for the feed resistance at which the relay will pick up. Given all of the other characteristics of the track circuit (including the drop away which, unless it is stated in the question, is generally assumed to be around 65% of the pickup), you would then be able to work out the value of shunt required to drop the track circuit.

Unfortunately, for some reason the 2002 paper is one that I do not have in my collection to see what they actually asked, but you are likely to be asked to set up the track for "reliable operation". I have not done the sums, but that value of Rfeed seem high and I suspect that you would find a high drop shunt value meaning that the track will be prone to drop for small changes in ballast resistance (eg rain making the ballast wet) and hence making it possibly unreliable.

The simple answer to your question is that you may have to consider both the calculations that you speak about in order to prove that the TC will work.

Peter

[leochungnet]

Hi, Peter,

Thanks for your kindly reply.
The question is shown below for your reference.

****************************************************
A d.c. track circuit has the following characteristics:

Length 500.0 metres
Feed transformer-rectifier output: 2.4 volt
Relay coil resistance: 4.0 ohm
Relay pickup current: 100 mA
Ballast resistance: 2.5 ohm km
Feed resistance: adjustable but unknown
Drop shunt: 0.5 ohm

Calculate the value of the feed resistance for these conditions.
Owing to weather conditions, the ballast resistance later rises to 20 ohm km. Calculate the new value of the drop shunt with the same value of feed resistance, and comment on the results.
*****************************************************

As per your reply, I have several questions to ask.
1) As drop shunt resistance is provided in the question, one should use the "drop-away" current approach to find out the feed resistance instead of using "pick-up" current approach?
2) If the drop shunt resistance is not provided, one can use the "pick-up" current to find out the feed resistance?
As you mentioned that the max. feed resistance will be calculated (the relay is just picked up which is not reliable), one can add 10% margin to the pick-up current for reliable operation?
3) If having drop shunt resistance provided, one use the "drop-away" current approach to calculate the feed resistance. Is the calculated resistance is a min. value for the relay to drop away? Should one add a margin on it as well? Or the margin has been already incorporated in multiplying 65% to Current (pick-up) to achieve the drop away current?

My questions may be a bit stupid / redundant....but I hope I could have a better understanding first and use the correct approach to cope with this type of question.
Thanks!

[PJW]

To consider safe operation, need to ensure that the specified drop shunt of 0.5ohm will reduce the current through the relay to less than that needed for the relay to remain up, this being calculated when there is no leakage occuring through the ballast (and thus the ballast resistance is infinite). Hence you want a value which is a bit less than the drop away current.

That calculation permits you to determine the value of the feed resistor; I would be tempted either to calculate this for DS=0.5 and then add about 10% to the resistor value to round up to a integral value so that there is a margin, or alternatively put the margin in another way by calculating the value of the feed resistor for a DS of say 0.7ohm and then making a small rounding to get a suitable value of resistor. [Sometimes the feed resistor is a sliding potentiometer and thus infinitely variable, but more often it is a resistor block with separate resistors of 0.5, 1, 2, 4, 8 etc that can be connected in series to give one of a fixed range of summated values.]

Having decided what the feed resistance would be in that situation, now consider the operation of the track circuit with that setting of feed resistor in the absence of a train but leakage through the poor ballast. For this calculation you want to ensure that the relay will reliably pick even when the ballast is wet, so you'd be looking for a current say 10% more than the pick up value.

Hi, Peter,

Thanks for your kindly reply.
The question is shown below for your reference.

****************************************************
A d.c. track circuit has the following characteristics:

Length 500.0 metres
Feed transformer-rectifier output: 2.4 volt
Relay coil resistance: 4.0 ohm
Relay pickup current: 100 mA
Ballast resistance: 2.5 ohm km
Feed resistance: adjustable but unknown
Drop shunt: 0.5 ohm

Calculate the value of the feed resistance for these conditions.
Owing to weather conditions, the ballast resistance later rises to 20 ohm km. Calculate the new value of the drop shunt with the same value of feed resistance, and comment on the results.
*****************************************************

As per your reply, I have several questions to ask.
1) As drop shunt resistance is provided in the question, one should use the "drop-away" current approach to find out the feed resistance instead of using "pick-up" current approach?
2) If the drop shunt resistance is not provided, one can use the "pick-up" current to find out the feed resistance?
As you mentioned that the max. feed resistance will be calculated (the relay is just picked up which is not reliable), one can add 10% margin to the pick-up current for reliable operation?
3) If having drop shunt resistance provided, one use the "drop-away" current approach to calculate the feed resistance. Is the calculated resistance is a min. value for the relay to drop away? Should one add a margin on it as well? Or the margin has been already incorporated in multiplying 65% to Current (pick-up) to achieve the drop away current?

My questions may be a bit stupid / redundant....but I hope I could have a better understanding first and use the correct approach to cope with this type of question.
Thanks!