2013 TC Calculation

[NJK]

[attachment=1554]Dear Members,

I planned to sit module 5 exam this year. I made my first attempt on TC calculation. Please go through the attached document and give your valuable comments for my further enhancement.

In my attachement, I found the value of drop shunt resistance is 13.63 ohm but the permitted value of drop shunt resistance is 0.5 ohm. I given wrong information reg failure of TC for the higher value of drop shunt resistance.
The higher value of drop shunt resistance may lead to remain the track circuit is energized state after the occupation of train because of the relay voltage not to reduced below to the drop away voltage value.

Thanks & Regards,
NJK

[Peter]

Dear Members,

I planned to sit module 5 exam this year. I made my first attempt on TC calculation. Please go through the attached document and give your valuable comments for my further enhancement.

In my attachement, I found the value of drop shunt resistance is 13.63 ohm but the permitted value of drop shunt resistance is 0.5 ohm. I given wrong information reg failure of TC for the higher value of drop shunt resistance.
The higher value of drop shunt resistance may lead to remain the track circuit is energized state after the occupation of train because of the relay voltage not to reduced below to the drop away voltage value.

Thanks & Regards,
NJK

Hi

Apologies that it has taken me such a long time to look at your answer.

Your calculation, by my reckoning, (and using the models that I put together) your numbers are perfect.

The circuit described by the parameters does give a track circuit that is somewhat ideal and operating far beyond the length that we would normally expect - a length of more like 1000m is more realistic although, the theoretical value of 2250m is exactly what was asked for (ie the maximum).

You have correctly calculated what the drop shunt would be for this length with (as you stated), perfect ballast resistance.

The only flaw in what you have put down is your analysis of the way the track circuit will fail. Remember, if a train axle and the rails were perfect, we would be able to rely on the shunt value being nearly zero and hence design the track circuit accordingly. However, we know that contact resistance and contamination will make the effective shunt value bigger (a larger number of ohms). Thus a HIGH drop shunt value means that you need only to put a shunt across the track that is far from perfect (a high number of ohms) - it is easier to drop the relay. Therefore the 0.5ohms that you quote is indeed the MINIMUM value that we would permit for safety. A track circuit with a value of 13ohms would be "safe" in the sense that it will drop with even the poorest of train shunt, but it will be horribly unreliable as it will be very easy for rain or contamination to achieve a similar value and drop the track, or as will be most likely, prevent it picking after the passage of a train that has correctly dropped it.

So, in summary, you have clearly grasped the electrical modelling and theoretical calculations, but you need to make sure that you understand the application and how a track circuit reacts to changes up or down in these values.

I hope that I have made that clear, but if you have any queries, please feel free to ask.

Peter

[NJK]

Dear Peter,

Thank you very much for your reply.

Regards,
NJK

Dear Members,

I planned to sit module 5 exam this year. I made my first attempt on TC calculation. Please go through the attached document and give your valuable comments for my further enhancement.

In my attachement, I found the value of drop shunt resistance is 13.63 ohm but the permitted value of drop shunt resistance is 0.5 ohm. I given wrong information reg failure of TC for the higher value of drop shunt resistance.
The higher value of drop shunt resistance may lead to remain the track circuit is energized state after the occupation of train because of the relay voltage not to reduced below to the drop away voltage value.

Thanks & Regards,
NJK

Hi

Apologies that it has taken me such a long time to look at your answer.

Your calculation, by my reckoning, (and using the models that I put together) your numbers are perfect.

The circuit described by the parameters does give a track circuit that is somewhat ideal and operating far beyond the length that we would normally expect - a length of more like 1000m is more realistic although, the theoretical value of 2250m is exactly what was asked for (ie the maximum).

You have correctly calculated what the drop shunt would be for this length with (as you stated), perfect ballast resistance.

The only flaw in what you have put down is your analysis of the way the track circuit will fail. Remember, if a train axle and the rails were perfect, we would be able to rely on the shunt value being nearly zero and hence design the track circuit accordingly. However, we know that contact resistance and contamination will make the effective shunt value bigger (a larger number of ohms). Thus a HIGH drop shunt value means that you need only to put a shunt across the track that is far from perfect (a high number of ohms) - it is easier to drop the relay. Therefore the 0.5ohms that you quote is indeed the MINIMUM value that we would permit for safety. A track circuit with a value of 13ohms would be "safe" in the sense that it will drop with even the poorest of train shunt, but it will be horribly unreliable as it will be very easy for rain or contamination to achieve a similar value and drop the track, or as will be most likely, prevent it picking after the passage of a train that has correctly dropped it.

So, in summary, you have clearly grasped the electrical modelling and theoretical calculations, but you need to make sure that you understand the application and how a track circuit reacts to changes up or down in these values.

I hope that I have made that clear, but if you have any queries, please feel free to ask.

Peter