This how I think of voltage / resistance / current

[Ratch]

Just because two physical quantities have the same units does not mean they are equivalent.  For instance, torque has units of newton-meters, which is the same as units of translational energy. No one would say that torque is energy just because it has the same physical units, would they?  Identical physical quantities must have the same units, but two quantities having the same units does not necessarily mean they are equivalent.

To be exact, torque has units of Newton × meters. It is a derived unit that is only meaningful when applied to vectors, and you cannot derive this unit from N·m because exchanging a dot for a cross product is not an allowed algebraic manipulation.
The derivation of N/m² from N·m/m³ is an allowed manipulation because a dot product is a scalar quantity.

Aren't you doing the same thing as above?  You are saying that energy/volume, which are both scalar quantities with no direction, is equivalent to pressure, which is a vector quantity, and thus has a direction.

Ratch
Hopelessly Pedantic

[Ratch]

Discussion about the basic principles always seem to attract a bunch of "my way is right, everyone else is wrong" arguments. I think there's a bit of bikeshedding going on. The other questions which have had similar effects here are "Which way does the current flow?" and "How does a transistor work?"

Sure, why not?  We can't all be right, can we?  Only by presenting informative facts for evaluation can we sort out which one of us has the correct perspective.

Current flow is an oxymoron.  I will be happy to explain that if your are interested.

To which type of transistor are you referring?  FETs, BJTs?  Each type works in different ways.

Ratch
Hopelessly Pedantic

[vulturebetrayer]

Wow guys.  Thanks for all the discussion on this topic!
I have learned a lot about the holes in my thinking and am going to go back to the drawing board before publishing anything.
The blow up on this thread is precisely why I posted here.
I don't want to publish anything that sets someone on the wrong track and from what I've seen here, I certainly would have.
Please keep the discussion going, it's very interesting.
 

[IanB]

Ratch, you have so many gaps in your knowledge and misunderstandings about basic concepts that you really should not be trying to teach others.

It is one thing to lack knowledge and be trying to learn. It is quite another to be ignorant and to be insisting you are correct as you are doing here. Claiming that you know better than physics texts and technical literature is the height of foolishness.

If you continue in the same way there are many knowledgeable people here who will rapidly lose patience with you.

[Ratch]

Ratch, you have so many gaps in your knowledge and misunderstandings about basic concepts that you really should not be trying to teach others.

With respect to knowledge of electronics and physics, I did not fall off the turnip truck yesterday.

It is one thing to lack knowledge and be trying to learn. It is quite another to be ignorant and to be insisting you are correct as you are doing here. Claiming that you know better than physics texts and technical literature is the height of foolishness.

I asked you before to show and prove that the knowledge I disseminated is incorrect.  Thus far, you have not done so.

If you continue in the same way there are many knowledgeable people here who will rapidly lose patience with you.

I would hope they would instead set me straight.

Ratch
Hopelessly Pedantic

[Ratch]

Wow guys.  Thanks for all the discussion on this topic!
I have learned a lot about the holes in my thinking and am going to go back to the drawing board before publishing anything.
The blow up on this thread is precisely why I posted here.
I don't want to publish anything that sets someone on the wrong track and from what I've seen here, I certainly would have.
Please keep the discussion going, it's very interesting.
 

It is good to see that you have a positive mental attitude.  Most folks would be intimidated and drop out from the discussion.  I applaud you.

Ratch
Hopelessly Pedantic

[Ratch]

Quote from Batch:
It was humorous seeing a disembodied head and hand doing the talking and gesturing.  I think analogs are OK for illustrating and explaining a particular point, but are confusing when coupled to another physics discipline.   Do hydraulic engineers learn their craft by studying electrical technology?  If not, then why should electrical students study hydraulics? 

The analogy is meant to foster understanding by drawing analogies with systems that are more commonly familiar to the student, i.e less abstract than the ideas of electric fields etc. If hydraulics confuse you then ignore it.     Analogs are also useful tools for solving real world engineering problems. Sometimes mechanical problems, for example, can be more easily solved by setting up an analogous electrical circuit and solving the circuit equations or merely observing the circuits behaviour.  In fact this was often done physically with analog computers back in the mid 20th century before digital computers were prevalent.

To each his own method.  But, it is one thing to use certain methods to solve problems, and another thing to teach and understand the science involved.

The professor seemed to imply that positive charges were present in a electrical circuit.  That, of course is not true.  Only negative charges are present in wires.

Of course there are positive charges in electrical circuits or there would be a humongous net negative charge.  The positive charge carriers are called protons.

Tilt!  There are no positive charge carriers in metals.  The negative charge carriers (electrons) in metals are loosely held to the metal atoms, are mobile, and contribute to the conductivity of the metal.  The positive ions of the metal are stationary, and do not transport any charge.


Why do the analog folks equate pressure (force per unit area) with voltage.  I would think they would choose force for voltage.   After all, the electric field does exert a force on the conducting electrons.

Voltage is not a force.  A force has direction, voltage does not have a direction.  Voltage is measure of potential energy difference.  Electric fields are the source of force on a charge.

Correct, but I was suggesting the analogy folks use force for voltage instead of pressure.  I never said that voltage was force.


The prof equated physical height with voltage, which I thought was not necessary.  Voltage is really an energy density, and is not a difficult concept to understand.

Voltage is not an energy density.  It is a measure of potential energy per unit of charge (Joules/Coulomb), or the amount of energy it takes to move 1C of charge against a 1 N/C E-field.  It is a scalar quantity, i.e. it has no direction.  This can be understood by remembering that Work (or energy) = force * distance, so for a coulomb of charge in an E-field the change in potential energy of the charge = E-field * distance moved, or in SI units:   V = (Newtons/Coulomb)*meters = N*m/C = J/C = Volts.

I said that voltage was the energy density per unit charge.  See reply #4 of this thread.  I agree with, and have known the facts in the above paragraph since the year one.

In the water pipe analogy:  (I use ? to represent the nabla symbol, i.e. the upside down delta)

Yes, the symbol for the gradient in vector calculus.
 

Pressure Gradient (?P) is in units of Newtons/meters^3;  it is analogous to E-field  which is in Newtons per unit of charge. These are a vector quantities.

Use whatever analogy you want.  Agreed, the E-field is a vector quantity.  Its field strength is measured in units of newtons/charge or volts/meter.  That is old news to me.
 

Pressure (P) is a measure of potential energy per unit volume of a fluid (joules/meter^3), or the amount of energy it takes to move 1 cubic meter of fluid against a pressure gradient of 1N/m^3. This is a scalar quantity just like the analogous Volts. Again, Energy = Force * distance, so for a cubic meter of fluid we have P = ?P * distance, or in SI units  (N/m^3)*m = N/m^2 = N*m/m^3 = J/m^3. (in the second term you see the more familiar definition of pressure as force per unit area: a Pascal in SI units).

If the "fluid" were a gas, I would agree with you.  But a liquid is incompressible, so so no work is done.  It is like applying a large force against a immovable object.  No work is done in that case, either.  The energy transfer to compress/expand a gas is well known in stoichiometic chemistry as PV.  No change in volume, no energy exchanged.
 

Summary of the analogy:
 
 E-field (N/C)  => Pressure Gradient (N/m^3)
 Volts  (J/C)    => Pressure                (J/m^3)

Say what you want about those analogies, I never had any use for them.

Ratch
Hopelessly Pedantic

[IanB]

Care is needed because there is no analog of inertia in the electrical world.

I'm curious why you do not think inductance counts?

Consider a short hydraulic pipe compared to a long hydraulic pipe. Increasing the flow rate in the long pipe requires work to be done to overcome the inertia of the fluid. Once the flow is established it resists change due to the stored kinetic energy (and this gives rise to water hammer effects with large pressure spikes if you try to interrupt the flow). The inertial effects are proportional to the length of the pipe.

Compare this with a long, straight wire. Increasing the current in the long wire requires work to be done to establish the magnetic field around the wire. Once the magnetic field is established it resists change due to the stored magnetic energy (and this gives rise to a back EMF with large voltage spikes if you try to interrupt the current). The inductance is proportional to the length of the wire.

The similarity here seems very direct. The main difference is that you need a very long straight wire to have a readily observable effect (unless you coil the wire around a magnetic core--but you can coil up a pipe too).

[BobsURuncle]

I said that voltage was the energy density per unit charge.  See reply #4 of this thread.  I agree with, and have known the facts in the above paragraph since the year one

Your definition is still wrong. Energy per unit charge is not the same as energy density or energy density per unit of charge.  I have two 12V batteries, which by definition each have 12V of potential energy across their terminals , but my 12V lithium ion battery has a much higher energy density than my 12V lead acid battery.

If the "fluid" were a gas, I would agree with you.  But a liquid is incompressible, so so no work is done.  It is like applying a large force against a immovable object.  No work is done in that case, either.  The energy transfer to compress/expand a gas is well known in stoichiometic chemistry as PV.  No change in volume, no energy exchanged.

A fluid can be a gas or a liquid.  You are confused between potential and kinetic energy: static pressure which has potential energy and a pressure differential which clearly can do work moving a compress-able or in-compress-able fluid through a pipe.

[LvW]

Claiming that you know better than physics texts and technical literature is the height of foolishness.

IanB - surely, I do not want to jump into your discussioin with Ratch. However, I cannot resist to comment your above quoted sentence.
Because within the last 30 years I have seen so many errors and false explanations in the techical literature, I only can warn anybody to blindly rely on statements and claims to be found in printed form or in the internet. 

[IanB]

Claiming that you know better than physics texts and technical literature is the height of foolishness.

IanB - surely, I do not want to jump into your discussioin with Ratch. However, I cannot resist to comment your above quoted sentence.
Because within the last 30 years I have seen so many errors and false explanations in the techical literature, I only can warn anybody to blindly rely on statements and claims to be found in printed form or in the internet.

I understand that mistakes can be made and that corrections are sometimes published, but broad and general errors across a body of literature are surely rather rare when describing fundamental concepts, especially in university textbooks?

I do agree with you however that blind and uncritical acceptance of anything written is something to be avoided.

[Ratch]

I said that voltage was the energy density per unit charge.  See reply #4 of this thread.  I agree with, and have known the facts in the above paragraph since the year one

Your definition is still wrong. Energy per unit charge is not the same as energy density or energy density per unit of charge.  I have two 12V batteries, which by definition each have 12V of potential energy across their terminals , but my 12V lithium ion battery has a much higher energy density than my 12V lead acid battery.

Lets get this cleared up.  Voltage is electrical energy per charge, which is an energy density.  Your example does not prove me wrong.  A Li-ion might have a larger energy capacity, but all that means it can sustain an voltage at a particular current for a longer period of time.  One does not define energy storage of a battery by its output voltage.  Voltage is still the energy density of the charge.

If the "fluid" were a gas, I would agree with you.  But a liquid is incompressible, so so no work is done.  It is like applying a large force against a immovable object.  No work is done in that case, either.  The energy transfer to compress/expand a gas is well known in stoichiometic chemistry as PV.  No change in volume, no energy exchanged.

A fluid can be a gas or a liquid.  You are confused between potential and kinetic energy: static pressure which has potential energy and a pressure differential which clearly can do work moving a compress-able or in-compress-able fluid through a pipe.

No, you did not explain clearly what you were talkiing about.  If you are moving a mass of gas or liquid through a pipeline, then yes, that takes energy.  If you are only compressing a gas, that takes energy, too.  If you try to just compress a liquid, then no work is done.  The confusion does not come from the definition of potiential or kinetic energy.  It comes from not knowing what you mean.

Ratch
Hopelessly Pedantic

[amyk]

Someone just registers and 15 of their 16 posts are in this thread, claiming everyone else is wrong and they're right. Is anyone else thinking what I'm thinking...? 

[helius]

The derivation of N/m² from N·m/m³ is an allowed manipulation because a dot product is a scalar quantity.

Aren't you doing the same thing as above?  You are saying that energy/volume, which are both scalar quantities with no direction, is equivalent to pressure, which is a vector quantity, and thus has a direction.

I don't think so. What is the direction of a balloon inflated to 50 psi?
The force vector is integrated over a surface, conceptually taking the dot product of the surface normal at each patch.
\$ \displaystyle \iint_{s} \mathbf F \cdot d \mathbf \Sigma \, = \iint_{s} ( \mathbf F \cdot \mathbf n ) \, d \Sigma \$
The result is scalar.

[Ratch]

The derivation of N/m² from N·m/m³ is an allowed manipulation because a dot product is a scalar quantity.

Aren't you doing the same thing as above?  You are saying that energy/volume, which are both scalar quantities with no direction, is equivalent to pressure, which is a vector quantity, and thus has a direction.

I don't think so. What is the direction of a balloon inflated to 50 psi?
The force vector is integrated over a surface, conceptually taking the dot product of the surface normal at each patch.
\$ \displaystyle \iint_{s} \mathbf F \cdot d \mathbf \Sigma \, = \iint_{s} ( \mathbf F \cdot \mathbf n ) \, d \Sigma \$
The result is scalar.

You are calculating the divergence of an enclosed surface using vector calculus.  The result will be 50 times the area of the balloon surface in square inches.  How does that turn energy/volume into force/area?

Ratch
Hopelessly Pedantic

[Ratch]

Someone just registers and 15 of their 16 posts are in this thread, claiming everyone else is wrong and they're right. Is anyone else thinking what I'm thinking...? 

They probably think I haven't been on this forum very long.  Isn't that the logical conclusion?  Thus far, only a very few folks have challenged my propositions.  After all, I don't submit something without giving a logical reason or proof.

Ratch
Hopelessly Pedantic

[onlooker]

How does that turn energy/volume into force/area

The discussion about this is all over internet. Just google it. say wiki or any .edu sites. For example,
from http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

"Pressure in a fluid can be seen to be a measure of energy per unit volume by means of the definition of work."

Or from wikipedia,

"Energy per unit volume has the same physical units as pressure, and in many circumstances is a synonym"

[Ratch]

How does that turn energy/volume into force/area

The discussion about this is all over internet. Just google it. say wiki or any .edu sites. For example,
from http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

"Pressure in a fluid can be seen to be a measure of energy per unit volume by means of the definition of work."

Yes, and an even more detailed answer at http://physics.stackexchange.com/questions/216342/what-is-pressure-energy.  Both those treatments of the topic involve fluid dynamics and kinematics with Bernoulli's equation thrown in.  I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one.  Will keep looking.

Ratch
Hopelessly Pedantic

[onlooker]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one

Just as in most physical systems, if not all, static case is just a special case or limiting case of the general/non-static system and concepts apply equally.

[Ratch]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one

Just as in most physical systems, if not all, static case is just a special case or limiting case of the general/non-static system and concepts apply equally.

That observation does not help much.  For example, if a have quantity of water in a closed container under pressure, what will be the energy/volume?  If the tank is part of a tube or pipeline where the water is flowing, then the kinetic and potential energy can be calculated.  But what is the energy/volume it when the tank is isolated by itself?  I can easily determine the pressure of the tank, and it is supposed to be the same as the energy/volume.  Or is it for static systems?

By the way, I checked a physics book and discovered that pressure is considered a scalar quantity and not a vector quantity like I first thought.  That is because the pressure is always considered at right angles to the surface, so no direction variation is permitted.

Ratch
Hopelessly Pedantic

[IanB]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one.

You have been shown it. It comes from the dimensional identity that force-per-unit-area is the same as energy-per-unit-volume. This occurs because work done on a system causes a change in energy of the system by the same amount, and work equals force times distance.

Thus (in SI units), if I move 1 m³ of an incompressible fluid through a pressure differential of 1 N/m², I do work on the fluid equal to 1 m³ x 1 N/m² = 1 Nm = 1 J. I have therefore done 1 J of work on 1 m³ of fluid. By moving the fluid volume through a pressure differential of 1 N/m² I have increased its potential to do work by 1 J/m³. The two statements are equivalent.

Note that I was careful to stipulate an (ideal) incompressible fluid above, since the thermodynamics of compressible fluids are more complex and the analogy with electricity breaks down.

[IanB]

By the way, I checked a physics book and discovered that pressure is considered a scalar quantity and not a vector quantity like I first thought.  That is because the pressure is always considered at right angles to the surface, so no direction variation is permitted.

That is correct. I'm glad you discovered that by your own research.

[Ratch]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one.

You have been shown it. It comes from the dimensional identity that force-per-unit-area is the same as energy-per-unit-volume. This occurs because work done on a system causes a change in energy of the system by the same amount, and work equals force times distance.

Thus (in SI units), if I move 1 m³ of an incompressible fluid through a pressure differential of 1 N/m², I do work on the fluid equal to 1 m³ x 1 N/m² = 1 Nm = 1 J. I have therefore done 1 J of work on 1 m³ of fluid. By moving the fluid volume through a pressure differential of 1 N/m² I have increased its potential to do work by 1 J/m³. The two statements are equivalent.

Note that I was careful to stipulate an (ideal) incompressible fluid above, since the thermodynamics of compressible fluids are more complex and the analogy with electricity breaks down.

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Ratch
Hopelessly Pedantic

[IanB]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Energy is relative, not absolute, so the correct question to ask is "How much energy does it take to fill the tank with fluid?"

If, for example, the tank were at the top of a tower (a water tower), then the water in the tank would have potential energy according to the formula E = Mgh, where M is the mass of liquid, g is the gravitational acceleration and h is the elevation above ground. It would take that much energy to pump the water up into the tower (plus friction losses), and you could get that much energy back again (minus friction losses) by letting the water down.

[IanB]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Let's suppose you are thinking of a closed metal tank completely full of water at atmospheric pressure, and now suppose you want to add a small additional volume of water into the tank. The work required to do this is equal to the volume of water added times the pressure difference, which initially is zero. But after you have pumped in a bit of water the pressure in the tank has gone up (let's say the walls are elastic and have stretched a bit). So the next bit of water you want to pump in will require some work as the pressure difference is no longer zero. The next bit of water after that will require more work, and the next more work still. The work to keep pumping in more water will keep increasing as the pressure in the tank goes up and up.

Hopefully you will begin to see the similarity here with pumping charge into a capacitor.

The energy you can get back out of the tank after you have pumped water into it is essentially equal to the work you did pumping water into it. The water itself doesn't have energy, but the tank plus water system has stored some energy, just like the capacitor plus charge system has stored some energy when you charge it up.

This, and the recognition that both pressure and voltage are scalar fields of potential, goes further towards explaining why voltage is "electrical pressure".