Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
General Track circuit calculations
#11
(01-07-2009, 10:12 PM)Peter Wrote: In 2005 your maths (with one exception, see later) and layout are spot on apart from the fact that you were trying to calculate the drop shunt using the rail voltage value for the pickup.

Thanks for noting these errors, otherwise I might of been scratching my head for hours! I have made an attempt at it - the 2005 TC question, and I didn't use any simultaneous equations ! Just basic ohms law. Have I been too simplistic?

Also I am struggling to understand how the current through the relay changes when there is no train present


Attached Files
.pdf   M5 2005 Q3.pdf (Size: 237.18 KB / Downloads: 133)
Reply
#12
[quote='priyman' pid='970' dateline='1251816124']
Thanks for noting these errors, otherwise I might of been scratching my head for hours! I have made an attempt at it - the 2005 TC question, and I didn't use any simultaneous equations ! Just basic ohms law. Have I been too simplistic?

Also I am struggling to understand how the current through the relay changes when there is no train present
Reply
#13
Hi,

I am going through the calculations and got few doubts. Could you please clarify them.

1. The drop shunt value when the TC length of 1 km is around 2.9 Ohm.
I read that the typical drop shunt value is around 0.5 to 1 ohm where as the value here is a bit different. Is that because of the Track circuit length?
Please clarify.

2.The relay current when no train present as per the calculations is 332 mA.
Yes, this current is fair enough for the relay to pick up.
The calculation seems correct but how to find the exact value as you mentioned that it is not correct.

Please help.

Thanks in advance.

Regards,

JJ.
Reply
#14
(10-07-2010, 10:58 AM)jenni.joseph9 Wrote: Hi,

I am going through the calculations and got few doubts. Could you please clarify them.

1. The drop shunt value when the TC length of 1 km is around 2.9 Ohm.
I read that the typical drop shunt value is around 0.5 to 1 ohm where as the value here is a bit different. Is that because of the Track circuit length?
Please clarify.

2.The relay current when no train present as per the calculations is 332 mA.
Yes, this current is fair enough for the relay to pick up.
The calculation seems correct but how to find the exact value as you mentioned that it is not correct.

Please help.

Thanks in advance.

Regards,

JJ.


Yes, a drop shunt would normally be arranged to be around 1 ohm in average ballast conditions.
If the drop shunt ever becomes as low as 0.5 ohm, there is a chance that the track may not be dropped by a train (wrongside failure).
Conversely if set too high then the track would suffer rightside failure due to deteriorating ballast conditions. This is where the length of the track becomes important- the longer it is, the greater the leakage current which must be expected in such situations.

The actual drop shunt of a track circuit will depend upon the relay characteristics, the feed voltage and feed resistance and the state of the ballast resistance at that time. Generally the "on site variable" is the feed resistance and thus the trick is to set this to make sure that the track will be dropped by the specified minimum drop shunt when ballast resistance is infinite, yet still pick up reliably when the ballast resistance is as bad (low) as one can reasonably expect.

So when you say the drop shunt is 2.9ohm, the important thing is under what ballast conditions.

As Peter pointed out there was an error in the workings that were posted here; you need to be clear what values stay the same and which change between the different scenarios. It is worth re-drawing the new equivalent circuit to avoid making rash assumptions.......


Also look at this attempt
PJW
Reply
#15
Do the IRSE have any books that go through how to carry out track circuit calculations?
Reply
#16
(12-07-2010, 10:16 PM)Archie Wrote: Do the IRSE have any books that go through how to carry out track circuit calculations?

Like most things, not one book that covers the whole range. The questions bring together a whole range of things such as underpinning electrical theory, typical operating parameters, real world limitations, possible non-ideal conditions that lead to some sort of failure and different TC types which is why there is such a range of different variabilities in the questions which, on first reading, appear to be asking essentially the same thing.

Sorry, that is not the answer you were looking for, but the exam is supposed to be challenging!

Peter
Reply
#17
(12-07-2010, 10:16 PM)Archie Wrote: Do the IRSE have any books that go through how to carry out track circuit calculations?

IRSE Green Booklet 9 gives good background on dc track circuits but doesn't go into calculations much. I am sure that you'll find standard electrical / scholl physics texts explaining Ohm's Law and Kirchoff's Law; indeed I am sure that you should find that there is plenty on the internet as well; try this one to start
As far as a worked example, there are several attempts on this website; Aditi's 2002 is pretty close to a text book example.
PJW
Reply
#18
Guys,

I am  attempting 2016 paper Q10 regarding TC calculations and it is driving me round the bend Sad  (it is the first time I have attempted this type of question). I have tried to use the thread and associated spreadsheets to work my way through the a) and b) sections but I am struggling with calculating the drop shunt values for c) and d). Could anyone please point me in the direction of a simple formula I can follow to help me answer it? I also think I may have done something wrong on the first sections as well as I have ended up with 3.9km as an answer for maximum length.

b)

·        V relay PU = Relay resistance (9) x relay pickup current (0.05) = 0.45v
 
·        Assume voltage on the rail to be same as v relay PU = 0.45v
 
·        V Feed – V Rail = 5-0.45 = 4.55v
 
·        I Feed = (V Feed – V Rail 4.55v)/R Feed (6 ohms) = 0.758A
 
·        I R Bal (current lost in ballast) = I Feed (0.758A)-I Relay (0.05)=0.708A
 
·        R Bal (at each meter) = V Rail (0.45)/ IR Bal (0.708) = 0.636 ohms
 
·        Ballast resistance = R Bal x length or Length  = Ballast resistance(2.5)/R Bal (0.636) = 3.93km


Any help would be appreciated!

Thanks in advance

Stu
Reply
#19
(25-08-2017, 09:42 PM)Smaddoc1 Wrote: Guys,

I am  attempting 2016 paper Q10 regarding TC calculations and it is driving me round the bend Sad  (it is the first time I have attempted this type of question). I have tried to use the thread and associated spreadsheets to work my way through the a) and b) sections but I am struggling with calculating the drop shunt values for c) and d). Could anyone please point me in the direction of a simple formula I can follow to help me answer it? I also think I may have done something wrong on the first sections as well as I have ended up with 3.9km as an answer for maximum length.

b)

·        V relay PU = Relay resistance (9) x relay pickup current (0.05) = 0.45v
 
·        Assume voltage on the rail to be same as v relay PU = 0.45v
 
·        V Feed – V Rail = 5-0.45 = 4.55v
 
·        I Feed = (V Feed – V Rail 4.55v)/R Feed (6 ohms) = 0.758A
 
·        I R Bal (current lost in ballast) = I Feed (0.758A)-I Relay (0.05)=0.708A
 
·        R Bal (at each meter) = V Rail (0.45)/ IR Bal (0.708) = 0.636 ohms
 
·        Ballast resistance = R Bal x length or Length  = Ballast resistance(2.5)/R Bal (0.636) = 3.93km


Any help would be appreciated!

Thanks in advance

Stu

Been a bit busy working nights recently, so will add this to the pile if not responded to by others. 
At first sight, calcs don't look unreasonable.

However for the 2017 exam, the chances of a TC calculation being included in the exam is virtually zero given the examiner's comments at last year's exam review, so probably not a priority
PJW
Reply
#20
(28-08-2017, 08:53 AM)PJW Wrote:
(25-08-2017, 09:42 PM)Smaddoc1 Wrote: Guys,

I am  attempting 2016 paper Q10 regarding TC calculations and it is driving me round the bend Sad  (it is the first time I have attempted this type of question). I have tried to use the thread and associated spreadsheets to work my way through the a) and b) sections but I am struggling with calculating the drop shunt values for c) and d). Could anyone please point me in the direction of a simple formula I can follow to help me answer it? I also think I may have done something wrong on the first sections as well as I have ended up with 3.9km as an answer for maximum length.

b)

·        V relay PU = Relay resistance (9) x relay pickup current (0.05) = 0.45v
 
·        Assume voltage on the rail to be same as v relay PU = 0.45v
 
·        V Feed – V Rail = 5-0.45 = 4.55v
 
·        I Feed = (V Feed – V Rail 4.55v)/R Feed (6 ohms) = 0.758A
 
·        I R Bal (current lost in ballast) = I Feed (0.758A)-I Relay (0.05)=0.708A
 
·        R Bal (at each meter) = V Rail (0.45)/ IR Bal (0.708) = 0.636 ohms
 
·        Ballast resistance = R Bal x length or Length  = Ballast resistance(2.5)/R Bal (0.636) = 3.93km


Any help would be appreciated!

Thanks in advance

Stu

Been a bit busy working nights recently, so will add this to the pile if not responded to by others. 
At first sight, calcs don't look unreasonable.

However for the 2017 exam, the chances of a TC calculation being included in the exam is virtually zero given the examiner's comments at last year's exam review, so probably not a priority

Thanks Peter, yes I do remember this being said at the IMechE but wasn't sure how definitive a decision it was. In that case as you say it is not a priority and I will concentrate my efforts elsewhere.

Cheers
Reply


Forum Jump:


Users browsing this thread: 1 Guest(s)