- #1

- 50

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter rdx
- Start date

- #1

- 50

- 0

- #2

- 2,985

- 15

- #3

- 2,226

- 9

Probably none,

[tex] PV = nRT = m \hat{R} T [/tex]

where [tex] \ \ \ \hat{R} = \frac{R}{m/n} [/tex]

and [itex]m/n[/itex] is the mass per mole or the mean molecular weight of the molecules in the gas.

now

[tex] P = \frac{m}{V} \hat{R} T = \rho \hat{R} T[/tex]

where [itex]\rho[/itex] is the density of the gas.

we also know that

[tex] U = m C_v \ T [/tex]

where [itex] U [/itex] is the internal energy of the gas. so the energy density is

[tex] \frac{U}{m} = C_v \ T = \frac{C_v}{\rho \hat{R}} P [/tex]

looks to me that pressure and energy density are proportional. so now the question to ask is what is the units or dimension of

[tex]\frac{C_v}{\rho \hat{R}} [/tex]

? (it should be dimensionless.) (edit: i now know it is not - i was thinking the wrong definition of energy density.)

is that a dimensionless quantity? i'm too lazy to check, but you might want to.

the same way moment [n-m] and energy [n-m] have the same units but nothing to do with each other.

uhhh, a radian is a natural unit of twist. as natural as an entire rotation ([itex]2 \pi[/itex] radians). twist a shaft with some known torque (nt-m) by one radian. how much energy did it take to do that?

Last edited:

- #4

- 2,985

- 15

I was not thinking of pressure of an Ideal gas, but pressure in a mechanical sense:

[tex] \sigma_{xx} = \frac{P}{A} [/tex]

What?

Edit: Sorry, I took your post -"how much energy did it take to do that? "-the wrong way and got snappy. I thought you were being sarcastic.

The point is, looking at units alone is not meaningful in every situation because you are implying "if" there is a rotation. The units alone make no such claim.

[tex] \sigma_{xx} = \frac{P}{A} [/tex]

uhhh, a radian is a natural unit of twist. as natural as and entire rotation ( radians). twist a shaft with some known torque (nt-m) by one radian. how much energy did it take to do that?

What?

Analytical Dynamics said:The dimensions of work, kinetic energy, and potential energy is a force times a distance. In the US customary system the commonly used unit is the ft-lb. Do not confuse the unit of energy with the unit of a moment, which has the same dimension as energy.

Edit: Sorry, I took your post -"how much energy did it take to do that? "-the wrong way and got snappy. I thought you were being sarcastic.

The point is, looking at units alone is not meaningful in every situation because you are implying "if" there is a rotation. The units alone make no such claim.

Last edited:

- #5

- 2,985

- 15

Actually, I was wrong to say probably not. They are related even in a mechanical sense.

[tex] U = \frac{1}{2} \sigma \epsilon [/tex]

Where [tex]\sigma[/tex] is the pressure.

and [tex]\epsilon[/tex] is a dimensionless fraction (called the strain), or ratio, of the change in the length of a bar under load to its original length.

Again, sorry I took what you wrote the wrong way rdj...

[tex] U = \frac{1}{2} \sigma \epsilon [/tex]

Where [tex]\sigma[/tex] is the pressure.

and [tex]\epsilon[/tex] is a dimensionless fraction (called the strain), or ratio, of the change in the length of a bar under load to its original length.

Again, sorry I took what you wrote the wrong way rdj...

Last edited:

- #6

Q_Goest

Science Advisor

Homework Helper

Gold Member

- 2,974

- 39

I'd agree with cyrus' when he said, "probably none". What's the definition of "energy density" as it relates to pressure? Is it the energy that is released assuming an isentropic expansion to 0 psia? That might make the most convenient answer, but it really is a miserable definition because that's just the conversion of internal energy to work and there's still plenty of internal energy left after the pressure reaches 0 psia! The fluid may be in liquid or even solid form upon expansion but the temperature is not likely to be anywhere close to absolute zero. There's still energy left. How much? How does the energy relate to pressure? I think I forgot the question already...

I'm afraid the question makes no sense. Regarding rbj's post, those equations may all be well and good, I'm not saying s/he's made a mistake, but those equations don't correlate to the concept of pressure and "energy density". If you want to define energy density in some way and pressure energy in some way, then there might be a correlation, but without some more usefull definition there is none.

Note that:

[tex] \frac{U}{m} = C_v \ T = \frac{C_v}{\rho \hat{R}} P [/tex]

reduces to:

[tex] \frac{U}{m} = 1 \ T = \frac{1}{\rho \hat{R}} P [/tex]

This reduces to the ideal gas equation.

- #7

- 2,226

- 9

I'm afraid the question makes no sense. Regarding rbj's post, those equations may all be well and good, I'm not saying s/he's made a mistake, but those equations don't correlate to the concept of pressure and "energy density". If you want to define energy density in some way and pressure energy in some way, then there might be a correlation, but without some more useful definition there is none.

Note that:

[tex] \frac{U}{m} = C_v \ T = \frac{C_v}{\rho \hat{R}} P [/tex]

reduces to:

[tex] \frac{U}{m} = 1 \ T = \frac{1}{\rho \hat{R}} P [/tex]

This reduces to the ideal gas equation.

of course it does. it was derived from it. (except, what happened to [itex]C_v[/itex]?)

i think the question makes perfect sense (but is not very specific). does the energy density of a

i think, for the case that the

ooops. i just realized that "energy density" of the OP was energy per unit volume, not per mass. that can also be done by mulitplying both sides by the regular mass/volume density ([itex]\rho[/itex])

[tex] \frac{U}{V} = \frac{C_v}{\hat{R}} P [/tex]

and [itex] \frac{C_v}{\hat{R}} [/itex] is clearly dimensionless. so that energy density

Last edited:

- #8

- 50

- 0

I'd agree with cyrus' when he said, "probably none". What's the definition of "energy density" as it relates to pressure? Is it the energy that is released assuming an isentropic expansion to 0 psia? That might make the most convenient answer, but it really is a miserable definition because that's just the conversion of internal energy to work and there's still plenty of internal energy left after the pressure reaches 0 psia! The fluid may be in liquid or even solid form upon expansion but the temperature is not likely to be anywhere close to absolute zero. There's still energy left. How much? How does the energy relate to pressure? I think I forgot the question already...

I'm afraid the question makes no sense. Regarding rbj's post, those equations may all be well and good, I'm not saying s/he's made a mistake, but those equations don't correlate to the concept of pressure and "energy density". If you want to define energy density in some way and pressure energy in some way, then there might be a correlation, but without some more usefull definition there is none.

Note that:

[tex] \frac{U}{m} = C_v \ T = \frac{C_v}{\rho \hat{R}} P [/tex]

reduces to:

[tex] \frac{U}{m} = 1 \ T = \frac{1}{\rho \hat{R}} P [/tex]

This reduces to the ideal gas equation.

"What's the definition of "energy density" as it relates to pressure? " I'm confused. energy density means the energy contained in a volume. Not U/m, as was suggested. It is more in the sense of the Stress-energy tensor.

- #9

- #10

- 50

- 0

This is not homework, this is just a question of general interest. I am learning about what goes where. Sorry.

- #11

- 50

- 0

pV=nRT, right? Note that the units of nRT are joules. Thus

p = joules/vol. = energy density. Doesn't this make sense?

- #12

- 2,226

- 9

pV=nRT, right? Note that the units of nRT are joules. Thus

p = joules/vol. = energy density. Doesn't this make sense?

it does dimensionally, but it doesn't derive it. it's sorta like that torque vs. energy thing above. just because they are of the same dimension of physical quantity does not mean they

[tex] PV = nRT = m \hat{R} T [/tex]

where

[tex] \hat{R} = \frac{R}{m/n} [/tex]

and [itex]m/n[/itex] is the mass per mole or the mean molecular weight of the molecules in the gas.

now

[tex] P = \frac{m}{V} \hat{R} T = \rho \hat{R} T[/tex]

where [itex]\rho[/itex] is the density of the gas.

then

[tex] U = m C_v \ T = (\rho V) C_v \ T[/tex]

where [itex] U [/itex] is the internal energy of the gas.

so the energy density, (energy per unit volume) is

[tex] \frac{U}{V} = \rho C_v \ T = \frac{C_v}{\hat{R}} P [/tex]

or

[tex] \frac{U}{V} \frac{\hat{R}}{C_v} = P [/tex]

that is why there is a relationship. otherwize the dimensional consistency might just be coincidental.

Last edited:

- #13

Q_Goest

Science Advisor

Homework Helper

Gold Member

- 2,974

- 39

Hi rbj & rdx,

[tex] \frac{U}{V} \frac{\hat{R}}{C_v} = P [/tex]

Of course, the question as it relates to thermodynamics is exactly as you've concluded, "…otherwize the dimensional consistency might just be coincidental."

I guess I can agree with you on all those things.

~

Let's consider this. Let's take for example, 3 different processes:

- Adiabatic and reversible change in pressure from P1 to P2. (ie: isentropic process)

- Adiabatic and isenthalpic change in pressure from P1 to P2. (isenthalpic process)

- Isochoric process which results in a change in pressure from P1 to P2.

Each process is the same in that they all change in pressure the exact same amount. But they are different in respect to the energy gained or lost by the fluid. So how does "energy density" as a function of pressure address these various processes? Is there an important relationship?

I guess I objected to the original question because there is no single relationship between thermodynamic energy and pressure energy because they are related only by what process the fluid undergoes.

Regardless, there is actually a relationship, but it isn't a thermodynamic relationship (ie: a relationship governed by the laws of thermodynamics). It has nothing to do with any of the processes I listed above. Bernoulli's equation uses pressure as one of the energy terms along side kinetic and potential energy. But of course, the process by which a change in fluid pressure which meets the criteria for Bernoulli's equation isn't thermodynamic, it's strictly mechanical, and unless you maintain all the assumptions used by the Bernoulli equation (ie: incompressible fluid, no work done, along a streamline, etc…) there won't be any relationship to any process unless one defines the assumptions and presents a specific relationship between energy and pressure that is valid for only those assumptions just like the Bernoulli equation does.

See also this link in http://hyperphysics.phy-astr.gsu.edu/hbase/press.html" [Broken] under "Pressure as Energy Density" half way down the page.

I'd agree 100% that there can be an equation just as you've found, that might somehow give energy per unit volume as a function of pressure in the thermodynamic sense.rbj wrote: does the energy density of a something (a fluid, perhaps, what else can we be talking about regarding pressure and energy density?) have anything to do with the pressure of the something?

i think, for the case that the something is an ideal gas, the answer is "yes".

[tex] \frac{U}{V} \frac{\hat{R}}{C_v} = P [/tex]

Of course, the question as it relates to thermodynamics is exactly as you've concluded, "…otherwize the dimensional consistency might just be coincidental."

I guess I can agree with you on all those things.

~

Let's consider this. Let's take for example, 3 different processes:

- Adiabatic and reversible change in pressure from P1 to P2. (ie: isentropic process)

- Adiabatic and isenthalpic change in pressure from P1 to P2. (isenthalpic process)

- Isochoric process which results in a change in pressure from P1 to P2.

Each process is the same in that they all change in pressure the exact same amount. But they are different in respect to the energy gained or lost by the fluid. So how does "energy density" as a function of pressure address these various processes? Is there an important relationship?

I guess I objected to the original question because there is no single relationship between thermodynamic energy and pressure energy because they are related only by what process the fluid undergoes.

Regardless, there is actually a relationship, but it isn't a thermodynamic relationship (ie: a relationship governed by the laws of thermodynamics). It has nothing to do with any of the processes I listed above. Bernoulli's equation uses pressure as one of the energy terms along side kinetic and potential energy. But of course, the process by which a change in fluid pressure which meets the criteria for Bernoulli's equation isn't thermodynamic, it's strictly mechanical, and unless you maintain all the assumptions used by the Bernoulli equation (ie: incompressible fluid, no work done, along a streamline, etc…) there won't be any relationship to any process unless one defines the assumptions and presents a specific relationship between energy and pressure that is valid for only those assumptions just like the Bernoulli equation does.

See also this link in http://hyperphysics.phy-astr.gsu.edu/hbase/press.html" [Broken] under "Pressure as Energy Density" half way down the page.

Last edited by a moderator:

- #14

- 50

- 0

So, according to the reference, energy density relates to pressure in mechanical ways. Unfortunately I was looking for something more definitive, such as a thermodynamic relationship but that seems unlikely, from your response. The problem came up originally when I was trying to model explosions mathematically. I released some amount of energy in some amount of space and it would be convenient to say the pressure was determined by that. But I assumed adiabatic expansion until enough cooling occurred for the steam to change state. Anyway, messy problem.

- #15

Q_Goest

Science Advisor

Homework Helper

Gold Member

- 2,974

- 39

Share: