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Mod 5 2013 Q4- TC calculations
#1
Hi all,
Find attached my attempt on the track circuit calculation question from 2013. This attempt actually took me an hour- so needs refining for exam conditions. Could anyone offer some tips for streamlining my answers for a similar question otherwise I would need to avoid on the day.

Also if anyone could offer any calculations and/or description of why the relay drop current is 68% of the relay pick current. I understand that it is to do with the electromagnetic properties of the coil, but would be helpful to understand a little more, so I don't just carry on with the old "it is because it is" method of reasoning.

Please review and comment.

Question 4(2013)
Calculate the maximum track circuit length for reliable operation based on the following
parameters:
• Feed voltage 50V
• Feed end resistance 50Ohm
• Relay pick up current 0.4A
• Relay resistance 25Ohm
• Ballast resistivity 75Ohm.km

You should state any assumptions that you make and show your calculations [15 marks]
For the length that you have calculated, calculate the drop shunt value. [5 marks]
A track circuit with the same design parameters as above has had a number of failures of
showing occupied when clear. Briefly discuss the possible causes of the failure. [5 marks]

Thanks

Tommy


Attached Files
.pdf   2013 Q4 attempt.pdf (Size: 1.19 MB / Downloads: 7)
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#2
Your question about the 68% is a good one and I do not have a definitive answer as to why it turns out to be around 68%. As to why it is there at all, that is the simple matter that you have an electrical circuit trying to make mechanical movement via magnetism. To pick up the relay, you have built up enough flux to attract the armature to the coil and having attracted it, it must, by definition, be closer to the magnet. Aside form the time taken for the magnetic flux to die away as you reduce the voltage (which, if you are doing it over a longish period of time you could ignore), for the armature to escape the magnetic influence, you have got to go down to a flux density which is the threshold for the movement at that point in space for it to move back the other way. Even if the spring is perfect and the mechanical linkages are perfect (just like early high school maths) and there was no mechanical hysteresis, the fact that you moved closer to the source of the flux to pick up the relay means that the only way to get to the flux density right for the threshold of movement is to reduce the flux density - in this case by reducing the field by reducing the voltage.

As for your attempt - you spent quite a lot of time writing a lot of assumptions and values. Think about the key items and what best shows off your knowledge. Maybe use a diagram to help you define your terms and add value to your thinking by enabling you to visualise the circuit you are calculating.

In the calculations, you start off really well.

I agree with all of the steps for the part where you conclude about the length of the track. One thing you might improve is to explain why you have added 10% to the PU value. It is not wrong to add 10%, but the examiner will be looking to see that you understand why it is the right sort of value to add (something like the value quoted in the question is a threshold number and hence .......).

In the next part you have used the 68% discussed above but not on the right numbers. You have used it on your ideal for reliable operation instead of on the characteristic value of the relay hence you have a value for drop away current which is 10% too high.

That aside, you have then done most of the calculations correctly until you get to the bottom of page 4. Here, you have used you same value for the current flowing through the ballast as you did for the steady state calculations above. The flaw here being that that value is only valid for the voltage across the rails in the un-shunted state (ie you value of 11v) instead of the value that it would actually be as you have influenced it by introducing the shunt. You might want to re=work the value for the current lost to the ballast and hence work out what you should be taking off your feed current (the total of the shunt, the ballast and the relay).

In your discussion about the causes of failures, you say that the "... voltage and ballast resistance is usually set up to achieve..." whereas, the ballast resistance is a given unless you are going to re-lay the track. One does not tweak this parameter.
You do not say WHY the DS is usually set at 0.5ohm and you make no comment about the magnitude of the value you calculate relative to a normal range.
You then go on to say that if it were set at 0.5ohm, a drop of rain would cause it to fail. The value of the DS on its own would not necessarily mean that and, assuming you have good ballast resistance, the nominal 0.5ohm would probably be expected to give you the required margin for a good spell of rain. The calculated value of >10ohm would me more likely to give you problems.

In your last section, you talk about the addition of TC bonds - this is not the sort of thing that would vary over time in normal operation. If the configuration of the TC has been altered, it would be re-set up.

Let me know if there is anything here that does not make sense.

Peter
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#3
I believe the difference between the PU and DA - with the DA being 68% is like you pointed out due to the electromagnetic properties of the magnetic circuit created in the DC Relay. The IRSE Green booklets on Relays is particularly useful in explaining this.

My understanding it is generally desirable to be able to have some control over the reluctance of a magnetic circuit (similar to controlling the resistance in an electric circuit), there are various reasons why this is required. Not all materials respond the same to magnetisation.

In practice you can't just stick a 'resistor' in a magnetic field - instead air gaps (air has a much lower permeability than soft iron for example) are utilised to reduce the permeability and therefore increase the reluctance of the magnetic circuit under consideration. This has the benefit of keeping the circuit just out of its saturation range and/or in the case of the DC relay control the strength of the MMF applied by the armature to the pole face.

So in a DC relay you have the magnetic circuit mainly made up of soft iron core, yoke, pole face and armature - however the armature has a residual pin (non magnetic material). When the relay is de-energised the air gap is at its biggest and when the relay is energised the armature is attracted to the pole face but is separated with a small air gap due to the risidual pin. The risidual pin creates maintaines an air gap by design of the relay and ensures that the overall reluctance of the circuit is within an acceptable range such that the mmf will collapse rapidly(armature releases and relay drops) soon as as the current is disconnected in the coil.

So there is a substantial difference in air gap between energised and de-energised and hence the magnitude of current to pick up and drop away the relay cannot be the same (more current is required to produce a stronger magnetic flux density to overcome the extra reluctance of the bigger air gap in order to establish a suitable MMF to attrack the aramture when relay is de-energised.)
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#4
Have you read the IRSE green book on T/C's and calcs?
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