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.
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?"
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.
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.
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.
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.
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.
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.
In the water pipe analogy: (I use ? to represent the nabla symbol, i.e. the upside down delta)
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.
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).
Summary of the analogy:
E-field (N/C) => Pressure Gradient (N/m^3)
Volts (J/C) => Pressure (J/m^3)
Care is needed because there is no analog of inertia in the electrical world.
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
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.
Claiming that you know better than physics texts and technical literature is the height of foolishness.
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.
Quote from: RatchI 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 oneQuoteYour 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.Quote from: RatchIf 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.QuoteA 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
QuoteThe 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.
QuoteThe 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.
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...?
How does that turn energy/volume into force/area
QuoteHow does that turn energy/volume into force/areaThe 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."
I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one
QuoteI 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.
I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one.
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.
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?
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?