IRSE Exam Forum

Question:
Calculate maximum length of track circuit for reliable operation and calculate minimum drop shunt for that length.
Additional relay connected at feed end, feed voltage 10v, relay resistance 20 ohms, feed resistance 7 ohms, pick up current 110mA, drop current 68% of pick up current, ballast resistance 2 ohm-km.

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.

Drop shunt calculation:
Drop away current = 110 x 0.68 = 74.8 mA = Id
Drop away voltage = 0.0748 x 20 = 1.496 volts
Voltage at feed end = 10 - 1.496 = 8.504 volts
Current through feed resistance = 8.504 / (20+7) = 314 mA = If
Ballast current at this time = Voltage at Relay end / Ballast resistance = Ib = 1.496/12.35 = 121 mA
Current through drop shunt Ish = If -Ib - Id = 314 - 121 - 74.8 = 118.2 mA
Value of drop shunt = Drop away voltage / Current through shunt = 1.496/118.2 = 12.65 Ohms

The value of drop shunt seems to be very high, not sure if I have made any mistakes in calculation.

Please suggest.

(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

Thanks PJW,

I will work out second part tonight and post again for your feedback.

Regards

I think 840mA in your answer is wrong.... How does it come from?
It should be 1082mA-121mA=962mA, but not 840mA, right?
Please comment if I am wrong...

(25-08-2009, 09:22 PM)PJW Wrote: [ -> ]

(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

(07-06-2010, 09:10 AM)greensky52 Wrote: [ -> ]I think 840mA in your answer is wrong.... How does it come from?
It should be 1082mA-121mA=962mA, but not 840mA, right?
Please comment if I am wrong...

There are two relays - the question is about fitting a feed end relay of an identical type. As PJW notes

PJW Wrote: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 total current is 1082mA, this is used up by two relays (at 121mA each) leaving 1082-(2 x 121) = 840mA as the max that can be soaked up by the ballast.

Thanks, Peter~But how do you know the addition relay is parallel with the 1st relay, but not in series with the 1st relay?

(07-06-2010, 04:46 PM)Peter Wrote: [ -> ]

(07-06-2010, 09:10 AM)greensky52 Wrote: [ -> ]I think 840mA in your answer is wrong.... How does it come from?
It should be 1082mA-121mA=962mA, but not 840mA, right?
Please comment if I am wrong...

There are two relays - the question is about fitting a feed end relay of an identical type. As PJW notes

PJW Wrote: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 total current is 1082mA, this is used up by two relays (at 121mA each) leaving 1082-(2 x 121) = 840mA as the max that can be soaked up by the ballast.

(08-06-2010, 06:53 AM)greensky52 Wrote: [ -> ]Thanks, Peter~But how do you know the addition relay is parallel with the 1st relay, but not in series with the 1st relay?

That is where your underpinning knowldge needs to come in. You cannot expect the question to tell you everything. A feed end relay is connected across the rails in the same what that a normal TR is connected across the rails, so assuming the rail impedance is negligable, the circuit model has them in parallel.

Hi I have had a go at the 'drop shunt' calculation so see what you think, please go easy on me as it is only really my first go at something like this, but feed back is greatly received.

Drop away current=PU current x 68% of PU current
121x0.68=82.28mA
Drop away voltage=DA current x relay resistance
0.08228x20=1.645 volts
Voltage at feed end=feed voltage-drop away voltage/feed resistance
10-1.64/7=1.193 volts
Current through feed resistance=voltage at feed end/feed+relay resistance
1.193/27=44mA
Ballast current at this time=voltage at relay end/ballast resistance
1.645/2.9=567mA
Current through 'drop shunt'=567-44-82.28=440.72mA

Value of drop shunt=Drop away voltage/current through shunt
1.645/440.72=3.7 Ohms

Like I have already mentioned this is my first real attempt so go 'easy.' but it does seem like a pretty reasonable answer for a 'drop shunt' in my experience.'

Quote:Hi I have had a go at the 'drop shunt' calculation so see what you think, please go easy on me as it is only really my first go at something like this, but feed back is greatly received.

You make it sound like you are talking to a bunch of ogres! Don't confuse enthusiasm to get people thinking with harsh criticism

Quote:Drop away current=PU current x 68% of PU current
121x0.68=82.28mA

Arithmetic right. You have used the value for pick up taken from the earlier people's calculations which is OK, but remember that this includes a 10% contingency for "reliability". The question gives PU at 110mA. Just be sure why you are using the number that you are.

Quote:Drop away voltage=DA current x relay resistance
0.08228x20=1.645 volts

Correct

Quote:Voltage at feed end=feed voltage-drop away voltage/feed resistance
10-1.64/7=1.193 volts

This is where it goes wobbly.
First, be careful how you write sums. As you have written it, a mathemitician would divide 1.64 by 7 and then take it from 10. You mean (10-1.64)/7.
Second, the sum is not dimensionally correct. (volts - volts)/ohms = amps. You say you are trying to calulate a voltage but your sum would give you a current.
If you amend your words to be "current supplied by feed" and give the answer as 1.193A, then all is well with the world.

Quote:Current through feed resistance=voltage at feed end/feed+relay resistance
1.193/27=44mA
Ballast current at this time=voltage at relay end/ballast resistance
1.645/2.9=567mA
Current through 'drop shunt'=567-44-82.28=440.72mA

Value of drop shunt=Drop away voltage/current through shunt
1.645/440.72=3.7 Ohms

Due to the problems above, this bit does not have any relevance. You have, however, you have made some assumptions in here that are flawed. I always recommend drawing a sketch of what you are trying to work out. If you had this correctly in front of you, you would have seen that some of your statements in this section are contradictory. In the first one you said

Quote:Current through feed resistance=voltage at feed end/feed+relay resistance

Not unreasonable if that is all that was in the circuit, but you later talk about the current through the ballast and the current through the drop shunt. Ask yourself where these resistors were in the circuit when you did the calculation above. Answer - the were there all the time, so need to be taken account of.

Quote:Like I have already mentioned this is my first real attempt so go 'easy.' but it does seem like a pretty reasonable answer for a 'drop shunt' in my experience.'

Gentle enough?

Once you have the values for the components in a given setup, the question is essentially one of static circuit analysis so sketch out your equivalent circuit and have a go on that basis, taking account of the early slip above.

One final thought, you said that the number you came up with was reasonable for a DS in your experience, would you say that is at the upper or lower end of what you would expect and given what the question is asking, would you expect your anwer to be at that end of the spectrum? (Hint, the question is asking us to work at the limit of reliability of the TC).

Peter

Ok many thanks for the detailed response, I will go away and go over what has been said, but do you think my attempt was half decent as a start?

(15-06-2010, 12:43 PM)Peter Wrote: [ -> ]

Quote:Hi I have had a go at the 'drop shunt' calculation so see what you think, please go easy on me as it is only really my first go at something like this, but feed back is greatly received.

You make it sound like you are talking to a bunch of ogres! Don't confuse enthusiasm to get people thinking with harsh criticism

Quote:Drop away current=PU current x 68% of PU current
121x0.68=82.28mA

Arithmetic right. You have used the value for pick up taken from the earlier people's calculations which is OK, but remember that this includes a 10% contingency for "reliability". The question gives PU at 110mA. Just be sure why you are using the number that you are.

Quote:Drop away voltage=DA current x relay resistance
0.08228x20=1.645 volts

Correct

Quote:Voltage at feed end=feed voltage-drop away voltage/feed resistance
10-1.64/7=1.193 volts

This is where it goes wobbly.
First, be careful how you write sums. As you have written it, a mathemitician would divide 1.64 by 7 and then take it from 10. You mean (10-1.64)/7.
Second, the sum is not dimensionally correct. (volts - volts)/ohms = amps. You say you are trying to calulate a voltage but your sum would give you a current.
If you amend your words to be "current supplied by feed" and give the answer as 1.193A, then all is well with the world.

Quote:Current through feed resistance=voltage at feed end/feed+relay resistance
1.193/27=44mA
Ballast current at this time=voltage at relay end/ballast resistance
1.645/2.9=567mA
Current through 'drop shunt'=567-44-82.28=440.72mA

Value of drop shunt=Drop away voltage/current through shunt
1.645/440.72=3.7 Ohms

Due to the problems above, this bit does not have any relevance. You have, however, you have made some assumptions in here that are flawed. I always recommend drawing a sketch of what you are trying to work out. If you had this correctly in front of you, you would have seen that some of your statements in this section are contradictory. In the first one you said

Quote:Current through feed resistance=voltage at feed end/feed+relay resistance

Not unreasonable if that is all that was in the circuit, but you later talk about the current through the ballast and the current through the drop shunt. Ask yourself where these resistors were in the circuit when you did the calculation above. Answer - the were there all the time, so need to be taken account of.

Quote:Like I have already mentioned this is my first real attempt so go 'easy.' but it does seem like a pretty reasonable answer for a 'drop shunt' in my experience.'

Gentle enough?

Once you have the values for the components in a given setup, the question is essentially one of static circuit analysis so sketch out your equivalent circuit and have a go on that basis, taking account of the early slip above.

One final thought, you said that the number you came up with was reasonable for a DS in your experience, would you say that is at the upper or lower end of what you would expect and given what the question is asking, would you expect your anwer to be at that end of the spectrum? (Hint, the question is asking us to work at the limit of reliability of the TC).

Peter