15-06-2010, 06:18 PM
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 currentArithmetic 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.
121x0.68=82.28mA
Quote:Drop away voltage=DA current x relay resistanceCorrect
0.08228x20=1.645 volts
Quote:Voltage at feed end=feed voltage-drop away voltage/feed resistanceThis is where it goes wobbly.
10-1.64/7=1.193 volts
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 resistanceDue 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
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
Quote:Current through feed resistance=voltage at feed end/feed+relay resistanceNot 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

