Resistance

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Resistance

Postby Naazima on April 14th, 2014, 6:26 am 

I need your invaluable help.Can you please explain with the help of an analogy that why is the resultant resistance of resistances connected in series equal to the sum of the individual resistances.
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Re: Resistance

Postby BioWizard on April 14th, 2014, 9:12 am 

The simplest analogy I can think of is as follows:

Imagine that you have to drive a car between two points, A and B, and there is only one road that connects them. If there are checkpoints along the way, you will have to stop at each one, and the total delay you will experience along the way is the sum of the individual delays you experience at every individual checkpoint. Since there is only one road, you have no choice but to go through all the checkpoints.

When you connect resistances in series, you're creating just one possible path for electrons to go through. Therefore, they have no choice but to go through all the resistances you put along their path. So if you consider the resistance as a "checkpoint delay" for the electrons rushing through, the overall resistance they experience is the sum all all the resistances along their path.
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Re: Resistance

Postby CanadysPeak on April 14th, 2014, 12:34 pm 

Consider this analogy:

Suppose you are hired to fill a water tank that is on the fifth floor of a building. The water faucet is on the third floor, so you have to walk up two floors (5 – 3).

The boss will come by at noon and at the end of the day and pay you $0.10 for each gallon you carry up to the tank. He doesn’t care about how you do the job, or how fast. He just pays $0.10 per gallon.

You are strong enough to carry 10 gallons at a time. It takes you 5 minutes to walk up each flight of stairs while carrying the water, but there is a circular slide that you can use to get back down in only a second or two.

So, you work hard for the four hours in the morning. Every ten minutes you carry 10 gallons up, slide back down, and repeat. So, you get 60 gallons per hour, or 240 gallons for the morning. At noon, the boss comes by, measures the water and pays you $24.

While you’re eating lunch, the plumbers start work on the second floor, so they turn off the water to the third floor. So, when you go back to work, you find you can only get water down on the first floor. Now you have to carry the water up four floors (5 – 1), so that takes you 20 minutes each trip. You’re as strong as ever, so you still carry 10 gallons at a time.

Every 20 minutes, you carry up 10 gallons, so you get 30 gallons per hour. You work four hours, and get 120 gallons up to the tank. At the end of the day, the boss comes by, checks the water, and pays you $12. He doesn’t know whether you had to carry the water further, or if you got tired and could only carry 5 gallons at a time, or if the steps were partly blocked so it took you ten minutes to walk up each floor. The result is the same. So, double the distance, cut the strength in half, or make it twice as hard to move, and you only get half the water. That’s what equivalent resistance does in electricity – you can have twice the length of resistance wire, half the voltage, or twice the resistivity. The end result is that you get half the water (current) and that’s what equivalent resistance tells you.

Of course, you might have a gullible friend who stops by after lunch. You convince him that carrying water is a lot of fun, and you offer to let him help you if he buys you an ice cream cone. He is as strong as you, so you walk together up the stairs. It still takes 20 minutes to go from the 1st to the 5th floor. You can still carry only 10 gallons at a time, but your friend carries 10 gallons also. So, you get 20 gallons every 20 minutes, or 60 gallons per hour. When the boss comes by 4 hours later, he finds 240 gallons up in the tank, and he pays you $24. Your buddy, who just walked to the corner to get the ice cream, has no idea you were being paid. This is like two resistors in parallel.

The key to thinking about all this is not to think about how much resistance there is. Instead, think about how much water gets moved (electrical people call this conductance).
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Re: Resistance

Postby Naazima on April 16th, 2014, 6:58 am 

Thanks a ton.
I very much got it. Earlier I was thinking the resistances in series as a bigger blockage and a smaller blockage in a water pipe and so the smaller blockage would decide the rate of flow of water.so the resultant resistance as per that analogy should be the largest resistance.
Also in a simple electric circuit when we close the switch the first electron that enters the positive terminal of the battery experiences how much potential drop or where did it at all get its potential from?
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Re: Resistance

Postby ElectraQT on April 17th, 2014, 11:46 am 

Nazima, your post had me thinking for a few minutes. But I suppose you should compare resistance to push-back force rather than flow rate to get a good analogy for this. Good thinking though.
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Re: Resistance

Postby Naazima on April 18th, 2014, 3:02 am 

Thanks ElectraQT. Becoz we define current as the rate of flow of electrons, I was complied to think so.
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Re: Resistance

Postby Naazima on April 19th, 2014, 3:45 am 

No replies yet. Is the question stupid?...... Becoz I'm just a beginner.
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Re: Resistance

Postby CanadysPeak on April 19th, 2014, 6:32 am 

Naazima » Sat Apr 19, 2014 3:45 am wrote:No replies yet. Is the question stupid?...... Becoz I'm just a beginner.


What question are you referring to? I thought your question about the analogy had been answered by several of us. If there's a follow-on question, can you state it again?

What is your field of interest? Physics? Electrical engineering? General?

It is useful to keep in mind that voltage is the potential energy per unit charge gained by separating two charges some distance. If you take two charges, one + and one -, they are attracted to one another. You must use force to separate the charges, so the work done is just Force times Distance (you generally must integrate that product). The work done is matched by an increase in potential energy. For reasons of convenience, we express that potential energy as potential energy per unit charge (voltage). It is as if we had decided to express gravitational potential energy in terms of Joules per kilogram (hydrologists do that all the time).

If we then allow the two separated charges to move in response to the force which attracts them, they will "flow" at a rate determined by the intervening medium (in the same way that a steel ball dropped in air falls at a different rate than a steel ball dropped in a bath of mineral oil).

The take away (for now) is that the only quantities of interest that actually vary, and can be measured, are amount of charge, distance, and time interval. Everything else thus far is just some combination of these quantities. It will turn out that there is not really an independent measure called resistance; it is just a useful concept. We have only charge, distance, time, and some characterization of the flow medium. When we have a specific instance, we can lump all those together into a term we call resistance, but that is very specific, not at all general.

Let me know if I have lost you. If so, I will try again.
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Re: Resistance

Postby Naazima on April 20th, 2014, 1:12 am 

Actually I am a 10th grader (15 yrs old) studying the chapter Electricity.

When I study science I wish I could see everything happen say, electrons in a circuit carrying energy and giving it to the bulb and then coming back to the positive terminal with all their energy unloaded. So I always look for animations. Please suggest a good site where I can get these animations. Your recent post cleared all my doubts and now there are no more questions at present.
Thanks a load.....
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Re: Resistance

Postby Dave_Oblad on April 20th, 2014, 4:09 am 

Hi Naazima,

Actually, your question shows a remarkable level of insight. Your specific question is: "Why is series resistance additive?" (more or less)

I'll show why this is a good question. Imagine a pipe full of water. It is 1 inch in diameter and 10 feet long. I put "P" amount of water pressure at one end and it takes "T" amount of time to fill a bucket at the other end. If I were to ADD another 10 feet of pipe to the first, it would not change any of the parameters. It would still take "T" amount of time to fill that same bucket from the same "P" amount of pressure.

Now Science usually explains this away by saying that adding resistance is effectively reducing the diameter of the pipe. But is it really? No.. it is not (exactly speaking)!

A better model would be that electrons don't flow through the wire like water through a pipe.

Electrons make small quantum jumps from atom to atom. A single electron could take hours to travel down 1 foot of wire. Electrons don't always go the direction you think. They jump any direction that's available, including backwards against the direction of the electrical pressure.

It's a problem of probability. Adding resistance reduces the probability of any single electron working its way from one end of the wire to the other end. Taken as a whole, this is a reduction of current flow due to the addition of "Probability Limiting" aspects.

Electrons tend to travel on the surface of a wire. Greater surface.. more path choices.. greater flow.. more current. Smaller surface.. less path choices.. lesser flow.. lower current.

Adding Resistance in Series reduces the probability of an electron making it from one end to the other end. Adding Resistance in Parallel increases the probability of an electron making from one end to the other end. Seeing this over time defines total current flow in amps.. as a simple Quantum Probability Function.

If you still need a visual model, think many people in a hallway playing Hot Potato. Can't drop.. only pass a Hot Potato (electron). A wider hallway.. more people.. more choices.. more flow down the hallway in general averaging from whoever is tossing new hot potatoes into one end of the hall..lol.

Hope this helps,
Dave :^)
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Re: Resistance

Postby CanadysPeak on April 20th, 2014, 9:39 am 

Naazima » Sun Apr 20, 2014 1:12 am wrote:Actually I am a 10th grader (15 yrs old) studying the chapter Electricity.

When I study science I wish I could see everything happen say, electrons in a circuit carrying energy and giving it to the bulb and then coming back to the positive terminal with all their energy unloaded. So I always look for animations. Please suggest a good site where I can get these animations. Your recent post cleared all my doubts and now there are no more questions at present.
Thanks a load.....


Thanks. I made a mistake, though, when I said we needed charge, distance and time. We really need charge, distance and mass. I was tired and typed the wrong quantity.

Anyway, here's a good animation. You should slow down the current speed just a little tiny bit. Then open and close the various switches. You can actually use a watch and count how many "electrons" (the little dots) go through each path every 10 seconds.

http://www.falstad.com/circuit/e-resistors.html
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Re: Resistance

Postby Naazima on April 22nd, 2014, 2:55 pm 

Thanks Dave and CanadysPeak.Everything is crystal clear to me now.But as I proceed through the chapter more doubts will pop up.So please be there.
Thanks once again
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Re: Resistance

Postby Naazima on April 25th, 2014, 4:20 am 

Why isn't the quantity of charge (in coulombs) present in a cell specified on its label? How can we find it out?
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Re: Resistance

Postby neuro on April 25th, 2014, 6:37 am 

Dave_Oblad » April 20th, 2014, 9:09 am wrote:Imagine a pipe full of water. It is 1 inch in diameter and 10 feet long. I put "P" amount of water pressure at one end and it takes "T" amount of time to fill a bucket at the other end. If I were to ADD another 10 feet of pipe to the first, it would not change any of the parameters. It would still take "T" amount of time to fill that same bucket from the same "P" amount of pressure.

Dave,
I don't want to deny that your atomic/probabilistic model is nicer than the above, but what you have written here is actually incorrect:

Without talking about "resistance" of the pipe, 'cause it seems you don't like it, woould you agree that the pressure drop across the 10-feet pipe is P (assuming atmospheric pressure as 0)?
Would you also agree that if the pressure drop were halved (P/2) the flow would also be halved?
Would you finally agree that the pressure drops across the first 10-feet pipe and across the second 10-feet pipe (in your second condition) would be P/2 each?

If you agree on all this, then the two-pipes-in-series system produces half the flow, twice the time T, and this can be expressed by saying that the resistance of a 10-feet pipe is

R = P.T/(bucket volume) [bar/(L/s)]

and the resistance of two pipes in series is

R2 = P.2.T/(bucket volume) = 2.R
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Re: Resistance

Postby CanadysPeak on April 25th, 2014, 7:01 am 

neuro » Fri Apr 25, 2014 6:37 am wrote:
Dave_Oblad » April 20th, 2014, 9:09 am wrote:Imagine a pipe full of water. It is 1 inch in diameter and 10 feet long. I put "P" amount of water pressure at one end and it takes "T" amount of time to fill a bucket at the other end. If I were to ADD another 10 feet of pipe to the first, it would not change any of the parameters. It would still take "T" amount of time to fill that same bucket from the same "P" amount of pressure.

Dave,
I don't want to deny that your atomic/probabilistic model is nicer than the above, but what you have written here is actually incorrect:

Without talking about "resistance" of the pipe, 'cause it seems you don't like it, woould you agree that the pressure drop across the 10-feet pipe is P (assuming atmospheric pressure as 0)?
Would you also agree that if the pressure drop were halved (P/2) the flow would also be halved?
Would you finally agree that the pressure drops across the first 10-feet pipe and across the second 10-feet pipe (in your second condition) would be P/2 each?

If you agree on all this, then the two-pipes-in-series system produces half the flow, twice the time T, and this can be expressed by saying that the resistance of a 10-feet pipe is

R = P.T/(bucket volume) [bar/(L/s)]

and the resistance of two pipes in series is

R2 = P.2.T/(bucket volume) = 2.R


Neuro,

Forgive me if I am being pedantic. I probably am, but this is an important point.

The water flow analogy works pretty well for electricity (actually, I use the electricity analogy to understand hydrology, but I'm a bit backwards), but you should generally confine it to open channel flow, not closed pipes. Closed pipes just don't work out OK when you actually measure them. As a physician, you no doubt recall that Poiseuille tried to model blood pressure in just that way. It comes close, but not close enough to be useful. The difficulty is in figuring out what "resistance" means in a closed pipe.

Cheers.
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Re: Resistance

Postby CanadysPeak on April 25th, 2014, 7:05 am 

Naazima » Fri Apr 25, 2014 4:20 am wrote:Why isn't the quantity of charge (in coulombs) present in a cell specified on its label? How can we find it out?


It often is. The amp-hour rating can be converted to Coulombs (1 C = 1 As). Some batteries put the amp-hr rating on the outside, some don't.

But, you gotta be careful. The chemical reaction occurs, not as a step function, but more like an exponential curve that is truncated. Integrating the complete It product is more than a little difficult.

You can also get a pretty fair indication by weighing the battery (with a constant that differs for each chemical reaction). I'll leave this one for you to puzzle over for a while.

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Re: Resistance

Postby Naazima on April 25th, 2014, 8:34 am 

No please CanadysPeak.I am already struggling hard with 'electricity'.Please don't put these high level questions or I will faint.Please say the answer.
Please also give an analogy for Ohm's Law.
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Re: Resistance

Postby CanadysPeak on April 25th, 2014, 10:11 am 

Naazima » Fri Apr 25, 2014 8:34 am wrote:No please CanadysPeak.I am already struggling hard with 'electricity'.Please don't put these high level questions or I will faint.Please say the answer.
Please also give an analogy for Ohm's Law.
Thanks


I don't think I can give an analogy for Ohm's Law. Perhaps someone else can.

I know how to give analogies for specific parts of an electric circuit, but Ohm's law is very general, requiring that voltage and current be directly proportional. linear over a wide range, and not dependent on anything else. Almost any analogy I can think of fails to meet at least one of those criteria. Above all else, I do not want to introduce an idea that you might have to unlearn later on. I think it best for a new learner to just learn to accept resistance as the ratio of voltage to current - a mathematical quantity and nothing more; later, when you're very familiar, you can add analogies.
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Re: Resistance

Postby Naazima on April 25th, 2014, 1:55 pm 

CanadysPeak » April 25th, 2014, 6:05 am wrote:
But, you gotta be careful. The chemical reaction occurs, not as a step function, but more like an exponential curve that is truncated. Integrating the complete It product is more than a little difficult.

You can also get a pretty fair indication by weighing the battery (with a constant that differs for each chemical reaction). I'll leave this one for you to puzzle over for a while.


Please explain the answer. I'm curious to know.
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Re: Resistance

Postby CanadysPeak on April 25th, 2014, 2:26 pm 

Naazima » Fri Apr 25, 2014 1:55 pm wrote:
CanadysPeak » April 25th, 2014, 6:05 am wrote:
But, you gotta be careful. The chemical reaction occurs, not as a step function, but more like an exponential curve that is truncated. Integrating the complete It product is more than a little difficult.

You can also get a pretty fair indication by weighing the battery (with a constant that differs for each chemical reaction). I'll leave this one for you to puzzle over for a while.


Please explain the answer. I'm curious to know.


Take a AA battery, a AAA battery, A C battery, and a D battery, all the same kind, say alkaline. Weigh them. Look up their amp-hours or milliamp-hours on line. Make a graph of weight vs Amp-hours. I'm not interested in telling you the answer; I want you to learn on your own, even if it takes a lot of work.
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Re: Resistance

Postby Dave_Oblad on April 25th, 2014, 8:31 pm 

Hi all,

neuro wrote:I don't want to deny that your atomic/probabilistic model is nicer than the above, but what you have written here is actually incorrect:

Not incorrect. The question was why is series resistance additive. I said the water pipe analogy doesn't work for adding series resistance, as demonstrated below:

Water.jpg
Additive restrictions to water flow

Tank (A) will drain very fast.
Tank (B) will take much longer to drain because we have added a restriction.
Tank (C) will take just as long to drain as Tank (B).... Why?
Given all pipes are already filled with Water as in all wires are already filled with Electrons and all Restrictions are equal in diameter (length doesn't matter).

Then given 4 Restrictions as in (C)... the pressure in Pipe sections X,Y,Z are all equal.
Thus the 3 extra added restrictions have not reduced the flow of water volume more than a single restriction.

This is why using the Water Pipe analogy has limited accuracy and why I said the Original Poster was very astute in realizing this and questioning it.

Of course, we are told Series Resistance is additive as Rt=R1+R2+R3 but the real question is why? I seldom like the typical answer: "Shut up and Calculate". We can see the answer in the math, but what Model best represents the real world for the sake of deep understanding?

If you can think of a better Model than Probability Functions (Real World).. please express it.

A hot wire has greater Chaos in Electron Flow and a Super-Conductor has almost zero Chaos in Electron Flow. Make sense?

Best wishes all,
Dave :^)
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Re: Resistance

Postby CanadysPeak on April 25th, 2014, 9:03 pm 

Dave,

Nice example, and useful as a starting point, but C is wrong. The pressure is not the same in X, Y, and Z. Keep in mind that the water there is flowing. Make Y an orifice plate flowmeter, and you should see the problem.
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Re: Resistance

Postby Dave_Oblad on April 25th, 2014, 9:44 pm 

Hi Canady,

The flow through all three chambers of X,Y,Z will be equal. Given the restrictions are all equal in diameter.

If we were talking about a compressible medium like gas, then I might agree with you. But this is not the case. If I made a complex network of interconnecting pipes of various diameters with a pressure source at one end and nowhere for the Water to go, the pressure would be equal in all points of the water network. Correct? (I am also ignoring gravity effects somewhat, despite the fact I'm getting my initial pressure from Gravity...lol. So replace the Tanks with a precise Pressure Generator if you like)

If I allow a single restricted exit for the water at one end (anywhere), nothing changes in the distributed water pressure of each pipe, other than a general pressure drop equal everywhere because there now exists a pressure release point somewhere in the system.

I could be wrong.. but that's common sense to me. Because water is incompressible (basically) and thus the pressure is instantly distributed to a whole system.

Also, if you point out that my graphic depicts water shooting out further to Bucket (C) than Bucket (B), I'm gonna smack ya...lol.

Best Regards,
Dave :^)
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Re: Resistance

Postby CanadysPeak on April 26th, 2014, 5:42 am 

Dave_Oblad » Fri Apr 25, 2014 9:44 pm wrote:Hi Canady,

The flow through all three chambers of X,Y,Z will be equal. Given the restrictions are all equal in diameter.

If we were talking about a compressible medium like gas, then I might agree with you. But this is not the case. If I made a complex network of interconnecting pipes of various diameters with a pressure source at one end and nowhere for the Water to go, the pressure would be equal in all points of the water network. Correct? (I am also ignoring gravity effects somewhat, despite the fact I'm getting my initial pressure from Gravity...lol. So replace the Tanks with a precise Pressure Generator if you like)

If I allow a single restricted exit for the water at one end (anywhere), nothing changes in the distributed water pressure of each pipe, other than a general pressure drop equal everywhere because there now exists a pressure release point somewhere in the system.

I could be wrong.. but that's common sense to me. Because water is incompressible (basically) and thus the pressure is instantly distributed to a whole system.

Also, if you point out that my graphic depicts water shooting out further to Bucket (C) than Bucket (B), I'm gonna smack ya...lol.

Best Regards,
Dave :^)


You may be right. I may be making a mountain out of a molehill. I shall have to think about what you say. Nonetheless, I still have difficulty seeing water move without a pressure gradient, even if very small. Yet, I see that you have carefully eliminated h from the example, so perhaps there is no work being done? mmm??

This is why flow in pipes is a bad analogy. Too hard as compared to simply using Ohms Law. Why bother to use an analogy that is much more complicated than the original situation.
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Re: Resistance

Postby neuro on April 26th, 2014, 2:32 pm 

I am no physicist and I probably understand much less than the two of you engineers.

But if x liters/s flow in the restriction between Z and the bucket, then there must be a pressure drop there.
If the same flux (x l/s) occurs between C and X, X and Y and Y and Z, and the restrictions are equal, then you must have three further pressure drops (all equal).

Pressure gradients within chambers X,Y,Z, along the longitudinal axis, are almost negligible w.r.t. pressure drop across the restrictions.
I admit - as Canady underlies - that laminar flow introduces nonlinearity between pipe caliber and 1/"resistance" (almost 4th power) and that vorticosity occurs at the narrowings, which makes everything more complex.

However, you can draw, below your cartoon, a graph of pressure, and if X, Y and Z chambers are equal and the restrictions are equal you are bound to observe a continuously declining pressure along the pipe, with steepening of the gradient at the restrictions (notice that longitudinal speed correspondingly increasest there). To a first approximation, each restriction "feels" 1/4 the pressure drop in "B".

Obviously, as Canadys points out, this is not Ohmic: the "resistance" (defined as pressure/flux) changes with the speed of the flux, so it is not a proportionality "constant".
To a first approximation, however, this thing holds.

In few words, Dave, my impression is you do not consider the simple fact that any flux of anything from point A to point B implies there must be a "potential" (work per unit particle) difference between the two points A and B.

This potential is in some cases (Ohm) perfectly proportional to flow (current), whereas in other cases the general law J = E/R (flux = deltaEnergy/resistance) holds only approximately (as Canadys points out): this is true for blood flow along the aorta, the arteries, arterioles (which introduce active restrictions), capillaries and veins; this is true for air-flow into and out of the lungs (look at why asthma produces difficulty in expiring rather than inspiring); it is true for ion flow across cellular membrane channels (proteins with an internal hydrophilic pore)...
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Re: Resistance

Postby CanadysPeak on April 26th, 2014, 2:48 pm 

Neuro,

You are correct in saying that the model generally works to a first order approximation. My point, likely pedantic, is that such an analogy does not lead to better understanding, and the first purpose of any analogy should be toward that end. In that vein, I would urge folks to use only open channel flow as a way of understanding electricity (if, indeed we use flow at all).
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Re: Resistance

Postby neuro on April 26th, 2014, 3:31 pm 

Got it, Canadys.

I do have a deformed perspective, but in my defense I beg you to consider the fact that all "channels" in the human body (which is my field) are closed pipes.... :°)
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Re: Resistance

Postby Dave_Oblad on April 26th, 2014, 4:08 pm 

Hi all,

Canady wrote:I still have difficulty seeing water move without a pressure gradient, even if very small.

The funny thing Sir, is that you are very correct in your choice of words. There is a pressure Gradient in the system and it is very small. It's extremely local to the exit hole and somewhat bubble shaped with a diameter roughly equal to the diameter of the exit hole.

Not sure if anyone will get this reference, but the water moves through the system at the "Speed of Stick".

Best wishes,
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Re: Resistance

Postby CanadysPeak on April 26th, 2014, 4:49 pm 

The living Gawd that made me blessed me with a low IQ, so I have to puzzle things out without doing much thinking. So, here's my question:

Suppose I go up to the Lowes store and get a piece of 6" PVC, say about 6 feet long. I glue a cap on one end, and then put a small hose fitting (you can get those in those little boxes of parts they keep in the hardware section). I then go over to the Advance Auto and get me 10 feet of rubber vacuum hose. I put the hose on that little fitting, and then I take my magic marker and make a line about 137 mm down from the top of the open end of the PVC. I stand the whole thing upright, stretch the hose out on the ground, and stick my ole No. 2 Ticonderoga in the end of the hose to seal it. I fill the PVC tube up to the top with water, remembering to get a tall swivel chair (with wheels) to stand on, sos I can see in the PVC pipe. I then git out my smartphone, set it on stopwatch function, pull the pencil outta the end of the hose, and time how long it takes for the water level to drop to that black mark.

Then I set up the experiment again, but this time I cut that 10 foot vacuum hose smack dab in half. I get ready, pull the plug and time the same water drop. Does it now take half as long?

Iffn you don't know, you should try it. Some of you will understand exactly what happens, but the thing that happens DOES NOT happen in an electric circuit.

Just sayin'. . .
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Re: Resistance

Postby Dave_Oblad on April 26th, 2014, 7:41 pm 

Hi Canady,

It sounds like you are setting up something that resembles my Tank (A) in my Graphic. Thus the length of the hose could be your 10 feet, or 5 feet or 1 inch, but the tank will take the same amount of time to drain. I am assuming the hose is made of Teflon and doesn't offer resistance to flow along its contact surface.

If you had a 2 foot deep children's wading pool and needed to drain it, most people would use a siphon technique and stick a garden hose in the pool, touching the bottom and put the other end of the hose on the ground. Would it matter if the Hose was 10 feet long or 30 feet? Nope. I suppose a tiny difference might be measured but it has everything to do with friction along the hose and possibly laminar flow issues.

I actually had such an issue long past when I worked for a company that measured corrosion in water medium heat exchangers for processing oil. I thought I could just make a "U" shaped pipe and stick in into the main flow pipe. Then my rusting resistance element is located at the bend in the "U" shaped pipe. There was no flow through the stupid thing.

I had to put the resistance element (which looked like a shorted spark plug) directly into the flow of the main pipe. I realized there wasn't enough pressure difference at the ends of the "U" shaped pipe to get flow through it. If I made my "U" pipe large enough and spread the ends far enough apart, it would work better but all I am doing is providing an alternate pipe for sharing the water flow. Like using two wires to share the current flow in an electrical circuit from a power source to a load. But now we are entering the domain of parallel resistance issues.

Anyway, I think we all agree.. water pipes suck for modeling electrical circuits in series resistance circuits. It's not additive, the smallest aperture (Highest Resistance) is virtually the only controlling factor.

Warm Regards,
Dave :^)
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