2010 Track Circuit Question

[Peter]

A double rail d.c. track circuit has a fixed feed end resistor connected to the rails and has the following properties:
Relay resistance: 9 ohm
Track relay pick up current: 50 mA
Ballast resistance: 2.5 ohm/km
Feed end resistance: 6 ohm
Feed voltage: 5 V

a) Draw a physical representation of the track circuit and an electrical equivalent circuit;
b) Calculate the maximum length of track circuit for reliable operation (state any assumptions that you make);
c) For this maximum length calculate the drop shunt value.
[20 marks]

If the ballast resistance subsequently changes to 1.5 ohm/km what is the new drop shunt value? [5 marks]

Am I alone in wondering why the ballast resistance give in this years question is again given the wrong units like last year? The older exams (eg 2007) specify it correctly as ohm.km and I expect it would be argued that a good student should know what to expect and know that what was given in this year's paper cannot be right. Increasing the TC length decreases the effective resistance of the ballast, not as implied by the units given that each km presents 2.5 ohm, therefore 2 km would present 2 x 2.5 ohm.

I'm all for challenging students, but on an item where most people tend to struggle on the concept, giving them a duff steer in the specification is a bit off in my view. I wonder whether it was deliberate or a bad error. Throw into the mix that, as far as I can see, when you do the calcs properly, you end up with a surprisingly long TC.

What did any of the candidates find?

[jbrownhill]

You have put my mind at ease!
I made the assumption that ballast resistance was actually 2.5 kmohm and ended up with a 3km track. I did lose some time going back over as I was sure I had to have made a mistake.

[Peter]

You have put my mind at ease!
I made the assumption that ballast resistance was actually 2.5 kmohm and ended up with a 3km track. I did lose some time going back over as I was sure I had to have made a mistake.

Cool.

[adikarina]

I was thinking i did something wrong in my calculation. I got about 3.5km. I also stated that i was concerned about the track circuit length being too long.
That has put me a bit at ease.



Thanks
Aditi

[jenni.joseph9]

Hi all,

Even I ended up with a TC Length of 3Km.

My concern is about the Drop shunt when the Ballast Resistance is around 1.5 ohmkm. The drop shunt current ended up with a negative value. :-(

Regards

[PJW]

Am I alone in wondering why the ballast resistance give in this years question is again given the wrong units like last year? ....

I'm all for challenging students, but on an item where most people tend to struggle on the concept, giving them a duff steer in the specification is a bit off in my view. I wonder whether it was deliberate or a bad error. Throw into the mix that, as far as I can see, when you do the calcs properly, you end up with a surprisingly long TC.

Some months ago I did speak to the examiner responsible for setting the 2009 question and was assured that not only was he "not being given the opportunity of fouling it up again this year", but that the question was being "triple checked". Ummm.

I suppose it shows we are all human, but really I don't think it is good enough- they would rightly criticise a candidate for making an error in their units and in my view this should not have been allowed to get through. However I subscribe to the view "never attriibute to malice that which can adequately be explained by incompetence", so I think it most unlikely to be a deliberate ploy- that I would regard as unfair.

It seems as if there is a broad consensus regarding the length- although a negative shunt value is indicative of having made some wrong assumption or arithmetical error!

[Peter]

Hi all,

Even I ended up with a TC Length of 3Km.

My concern is about the Drop shunt when the Ballast Resistance is around 1.5 ohmkm. The drop shunt current ended up with a negative value. :-(

Regards

Given the values in the question, that is a distinct possibility. That is what happened when I went through the numbers. If you did the calculations with the numbers given not adding in any contingency (as I did at first to work out the worst case to comment on what you would do in reality), you get a TC length that will not pick with the changed value of ballast resistance. I think that if you add 10% margin, you get about 3.4km for the first part and then a track that has a DS of about 9ohms.

Having a TC that does not work when the ballast resistance gets worse is fine if the question is "comment on the effect of the ballast resistance changing to 1.5 ohm km" but as the question was "calculate the DS" I would expect there to be a possible answer.

Provided that you commented on the meaning of a negative number as the answer, I am sure there would be some credit, and if the arithmetic was good, you must get marks for that.

Have a look at the model in this post with the numbers from the exam and you'll find that your 3 and a bit kilometre track is what the numbers deliver.

Above all, don't panic, you cannot change things now!

Peter

[PJW]

Having a TC that does not work when the ballast resistance gets worse is fine if the question is "comment on the effect of the ballast resistance changing to 1.5 ohm km" but as the question was "calculate the DS" I would expect there to be a possible answer.

Provided that you commented on the meaning of a negative number as the answer, I am sure there would be some credit, and if the arithmetic was good, you must get marks for that.

I understand that the examiners are likely to mark the question sympathetically with however the candidate interpreted the ballast resistance figure.

Obviously it should have been quoted in ohm.km but the real issue is that although we all say "ballast resistance" (as that is what it is within the equivalent circuit diagram), the propoperty of the track itself is really "resistiviity".

Hence whether the candidate used their common sense to correct the typo in the question or alternatively interpreted the words exactly literally then I believe that their answer will be assessed on that basis. Perhaps the bigger problem is the last part which is probably somewhat harder to answer if ohm/km was assumed

[PJW]

Having a TC that does not work when the ballast resistance gets worse is fine if the question is "comment on the effect of the ballast resistance changing to 1.5 ohm km" but as the question was "calculate the DS" I would expect there to be a possible answer.

Provided that you commented on the meaning of a negative number as the answer, I am sure there would be some credit, and if the arithmetic was good, you must get marks for that.

I understand that the examiners are likely to mark the question sympathetically with however the candidate interpreted the ballast resistance figure.

Obviously it should have been quoted in ohm.km but the real issue is that although we all say "ballast resistance" (as that is what it is within the equivalent circuit diagram), the propoperty of the track itself is really "resistiviity".

Hence whether the candidate used their common sense to correct the typo in the question or alternatively interpreted the words exactly literally then I believe that their answer will be assessed on that basis.
Whereas an unusually long track is achieved with the one interpretation, an unusually short one is achieved with the other!
Perhaps the bigger problem is the last part re the lower ballast resistance, which is probably somewhat harder to answer if ohm/km was assumed. Should be an interesting Exam Review this year

[greensky52]

Ah, I thought I did a mistake when I got the negative resistance in exam, so I checked for many times and finally did the calculation again, but still got negative.... Because of this, I did not have enough time to answer two long questions..... But I thought although it was a mistake of examiners, it showed that I did not have enough basic knowledge for this. If I was good enough,I should see the error at first glance~
Mod 5 must fail. I will take it next year