A sample of ideal gas has an internal energy U and is then compressed to one half of its original volume while the temperature stays the same. What is the new internal energy of the ideal gas?

Abdii Diroo
2022-07-25

A sample of ideal gas has an internal energy U and is then compressed to one half of its original volume while the temperature stays the same. What is the new internal energy of the ideal gas?

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asked 2022-05-10

Pressure changes in Continuity equations and Poiseuille's Law?

Continuity says Q=AV, and we know that velocity and pressure are inversely related. So if we are in a closed system, like vasculature for example, Q is constant and any decrease in vessel radius would be expected to raise velocity, which would result in lower pressure.

If we look at Poiseuille's Law, on the other hand, we see the opposite! If Q is constant, then a decrease in radius/cross sectional area we should expect pressure to be raised!

What's going on?

Continuity says Q=AV, and we know that velocity and pressure are inversely related. So if we are in a closed system, like vasculature for example, Q is constant and any decrease in vessel radius would be expected to raise velocity, which would result in lower pressure.

If we look at Poiseuille's Law, on the other hand, we see the opposite! If Q is constant, then a decrease in radius/cross sectional area we should expect pressure to be raised!

What's going on?

asked 2022-05-07

Difference between Pressure and Pressure Energy in fluids

For a fluid with viscosity to flow through a pipe that has the same cross-sectional area at both ends, at a constant velocity, there has to be a pressure difference according to Poiseuille's Law. Why exactly is there a change in pressure required to keep the velocity constant?

Is it because according to Bernoulli's principle that Pressure or Pressure-Energy gets converted to Kinetic Energy to speed up the fluid so the mass flow rate at both ends of the pipe stays the same?

So if that's the case in light of Bernoulli's principle, that means the change in pressure in Poiseullie's Law is there so that pressure energy gets converted to Kinetic Energy to fight off the viscosity of the fluid to keep the velocity of the fluid constant?

And

What exactly is Pressure? I know it is the Force divided by the area. I understand that concept, but I've seen the terms Pressure and Pressure Energy used interchangeably when talking about fluids, which creates for some amount of confusion. Aren't Pressure and Pressure Energy different? But when we talk about fluids in light of Bernoulli's principles, it seems as if Pressure and Pressure Energy are the same, which is pretty confusing.

For a fluid with viscosity to flow through a pipe that has the same cross-sectional area at both ends, at a constant velocity, there has to be a pressure difference according to Poiseuille's Law. Why exactly is there a change in pressure required to keep the velocity constant?

Is it because according to Bernoulli's principle that Pressure or Pressure-Energy gets converted to Kinetic Energy to speed up the fluid so the mass flow rate at both ends of the pipe stays the same?

So if that's the case in light of Bernoulli's principle, that means the change in pressure in Poiseullie's Law is there so that pressure energy gets converted to Kinetic Energy to fight off the viscosity of the fluid to keep the velocity of the fluid constant?

And

What exactly is Pressure? I know it is the Force divided by the area. I understand that concept, but I've seen the terms Pressure and Pressure Energy used interchangeably when talking about fluids, which creates for some amount of confusion. Aren't Pressure and Pressure Energy different? But when we talk about fluids in light of Bernoulli's principles, it seems as if Pressure and Pressure Energy are the same, which is pretty confusing.

asked 2022-05-13

Do you need to increase pressure to pump a fluid from a large pipe through a small pipe?

The application of this question is towards clogged arteries and blood flow. I am wondering why people with clogged arteries have high blood pressure if a smaller artery (assuming it is smooth and cylindrical) means less pressure exerted by the blood according to Bernoulli's Equation. Is this because the heart has to pump the blood harder to be able to travel through a smaller arterial volume?

The application of this question is towards clogged arteries and blood flow. I am wondering why people with clogged arteries have high blood pressure if a smaller artery (assuming it is smooth and cylindrical) means less pressure exerted by the blood according to Bernoulli's Equation. Is this because the heart has to pump the blood harder to be able to travel through a smaller arterial volume?

asked 2022-05-20

How does Newtonian viscosity not depend on depth?

I am currently trying to self-study basic Classical Mechanics and right now I am trying to understand fluid dynamics. I have read about Newtonian viscosity: as my book says, the force exerted on a viscous fluid depends on the coefficient of viscosity, the cross-section and the velocity gradient, but, to my amazement, not on the fluid's height, or in any other way on its total amount.

Usually I'd be more than willing to trust such a result even if not proven, and the way it is used in deriving something like Poiseuille's law, for example, almost makes sense to me. But I still, to my best efforts, fail to understand or even get an intuition on what should be the most basic case of application, i.e. a fluid that can be approximated as many rectangular plates flowing on top of each other. If I try to apply the same method I would use for a cylindrical tube, I get that considering a section of any height $h$, if the velocity gradient in the rectangle is constant at every depth, the force I need to apply to that section is the same no matter the value of $h$. In particular, dividing the rectangle in an arbitrarily large number of sections of arbitrarily small value of $h$, keeping the gradient the same I get that the total force I need to apply to maintain it is arbitrarily large.

Also, it sounds unintuitive that keeping the same gradient and halving the amount of fluid I would need the same force (and double the pressure). In addition to this, even with the tube, if instead of applying the viscosity law to a full cylinder I try applying it to a cylindrical ring, I get the same problems.

Can someone explain? Am I doing this wrong (probably)? Does the viscosity law somehow only work for the "top" section of the liquid? If so, what does top mean? How can the fluid "know" what is the top and what isn't? Are different pressures really needed depending on the amount of fluid to maintain the same gradient? What's an intuitive explanantion of this? Is there something else I'm missing that would make the explanation work?

I am currently trying to self-study basic Classical Mechanics and right now I am trying to understand fluid dynamics. I have read about Newtonian viscosity: as my book says, the force exerted on a viscous fluid depends on the coefficient of viscosity, the cross-section and the velocity gradient, but, to my amazement, not on the fluid's height, or in any other way on its total amount.

Usually I'd be more than willing to trust such a result even if not proven, and the way it is used in deriving something like Poiseuille's law, for example, almost makes sense to me. But I still, to my best efforts, fail to understand or even get an intuition on what should be the most basic case of application, i.e. a fluid that can be approximated as many rectangular plates flowing on top of each other. If I try to apply the same method I would use for a cylindrical tube, I get that considering a section of any height $h$, if the velocity gradient in the rectangle is constant at every depth, the force I need to apply to that section is the same no matter the value of $h$. In particular, dividing the rectangle in an arbitrarily large number of sections of arbitrarily small value of $h$, keeping the gradient the same I get that the total force I need to apply to maintain it is arbitrarily large.

Also, it sounds unintuitive that keeping the same gradient and halving the amount of fluid I would need the same force (and double the pressure). In addition to this, even with the tube, if instead of applying the viscosity law to a full cylinder I try applying it to a cylindrical ring, I get the same problems.

Can someone explain? Am I doing this wrong (probably)? Does the viscosity law somehow only work for the "top" section of the liquid? If so, what does top mean? How can the fluid "know" what is the top and what isn't? Are different pressures really needed depending on the amount of fluid to maintain the same gradient? What's an intuitive explanantion of this? Is there something else I'm missing that would make the explanation work?

asked 2022-05-19

How to convert this derivation of Poiseuille's law into the standard one?

I am trying to derive Poiseuille's law. I have reached a point in the derivation where I have:

$V=\frac{(p1-p2)({R}^{2})}{4lu}$

Where $l$ is length, u is viscosity, p is pressure, v is flow velocity and R is radius. What I am stuck on is shifting this to the volumetric flow rate:

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

However this is incorrect as Poiseuille's law is divided by 8. I know that I am probably missing something obvious, but I can't think of a reason as to why the whole equation needs to be halved. Any help understanding why (or whether my original derivation for V was inaccurate) would be appreciated.

I am trying to derive Poiseuille's law. I have reached a point in the derivation where I have:

$V=\frac{(p1-p2)({R}^{2})}{4lu}$

Where $l$ is length, u is viscosity, p is pressure, v is flow velocity and R is radius. What I am stuck on is shifting this to the volumetric flow rate:

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

However this is incorrect as Poiseuille's law is divided by 8. I know that I am probably missing something obvious, but I can't think of a reason as to why the whole equation needs to be halved. Any help understanding why (or whether my original derivation for V was inaccurate) would be appreciated.

asked 2022-05-15

Flow rate of a syringe

Suppose a syringe (placed horizontally) contains a liquid with the density of water, composed of a barrel and a needle component. The barrel of the syringe has a cross-sectional area of $\alpha \text{}{m}^{2}$, and the pressure everywhere is $\beta $ atm, when no force is applied.

The needle has a pressure which remains equal to $\beta $ atm (regardless of force applied). If we push on the needle, applying a force of magnitude $\mu \text{}N$, is it possible to determine the medicine's flow speed through the needle?

Suppose a syringe (placed horizontally) contains a liquid with the density of water, composed of a barrel and a needle component. The barrel of the syringe has a cross-sectional area of $\alpha \text{}{m}^{2}$, and the pressure everywhere is $\beta $ atm, when no force is applied.

The needle has a pressure which remains equal to $\beta $ atm (regardless of force applied). If we push on the needle, applying a force of magnitude $\mu \text{}N$, is it possible to determine the medicine's flow speed through the needle?

asked 2022-05-15

Is there a contradiction between the continuity equation and Poiseuilles Law?

The continuity equation states that flow rate should be conserved in different areas of a pipe:

$Q={v}_{1}{A}_{1}={v}_{2}{A}_{2}=v\pi {r}^{2}$

We can see from this equation that velocity and pipe radius are inversely proportional. If radius is doubled, velocity of flow is quartered.

Another way I was taught to describe flow rate is through Poiseuilles Law:

$Q=\frac{\pi {r}^{4}\mathrm{\Delta}P}{8\eta L}$

So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:

$vA=v\pi {r}^{2}=\frac{\pi {r}^{4}\mathrm{\Delta}P}{8\eta L}$

Therefore:

$v=\frac{{r}^{2}\mathrm{\Delta}P}{8\eta L}$

Now in this case, the velocity is proportional to the radius of the pipe. If the radius is doubled, then velocity is qaudrupled.

What am I misunderstanding here? I would prefer a conceptual explanation because I feel that these equations are probably used with different assumptions/in different contexts.

The continuity equation states that flow rate should be conserved in different areas of a pipe:

$Q={v}_{1}{A}_{1}={v}_{2}{A}_{2}=v\pi {r}^{2}$

We can see from this equation that velocity and pipe radius are inversely proportional. If radius is doubled, velocity of flow is quartered.

Another way I was taught to describe flow rate is through Poiseuilles Law:

$Q=\frac{\pi {r}^{4}\mathrm{\Delta}P}{8\eta L}$

So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:

$vA=v\pi {r}^{2}=\frac{\pi {r}^{4}\mathrm{\Delta}P}{8\eta L}$

Therefore:

$v=\frac{{r}^{2}\mathrm{\Delta}P}{8\eta L}$

Now in this case, the velocity is proportional to the radius of the pipe. If the radius is doubled, then velocity is qaudrupled.

What am I misunderstanding here? I would prefer a conceptual explanation because I feel that these equations are probably used with different assumptions/in different contexts.