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?

Jeffrey Jordon

Expert2022-11-07Added 2605 answers

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. The new internal energy of the ideal gas in terms of U is U.

Note: In an isothermal process, the temperature is constant. Due to this, the internal energy of the system is constant. So the change in the internal energy is zero. $\u2206U$

The density of seawater is $1024kg/{m}^{3}$. The pressure at a depth of 5 m below the surface of the sea is ___ $N/{m}^{2}$.

The radius of the aorta is about 1.1 cm , and the blood passing through it has a speed of about 34 cm/s . Coefficient of viscosity for the whole blood (37∘) is η = 4×10−3Pa⋅s Calculate the pressure drop per cm along the aorta.

What is the pressure in atmospheres at $11$ km depth?

Theoretical question from the book: The pressure exerted by a liquid depends on...?

Please show me in deti

For the electrical resistance of a conductor, we have

$R=\rho \frac{l}{A}$

Noting the structural similarity between the Hagen-Poiseuille law and Ohm's law, we can define a similar quantity for laminar flow through a long cylindrical pipe:

${R}_{V}=8\eta \frac{l}{A{r}^{2}}$

So there's a structural difference of a factor of ${r}^{2}$ between the two. What's the intuition behind this?Theoretical vs Experimental flow rate of fluid coming out of a water bottle with a hole in it

Say there is a water bottle that is filled with 300 mL of water and has a circular hole with a radius of 2 mm. In this bottle, the water sits 7.8cm above the top of the hole (which has been drilled 1.5cm above the bottom of the bottle).

According to Bernoulli's law the velocity $v$ of the water flowing out is equal to $\sqrt{2gh}$

Using this, the flow rate can be calculated as $Q\text{}=\text{}Av\text{}=\text{}\pi (0.002\text{}m{)}^{2}\ast 1.24\text{}m/s=0.000016\text{}{m}^{3}/s=16\text{}mL/s$

This doesn't seem accurate considering that the experimental flow rate is equal to 8 mL/s (40 mL over 5 seconds). However I understand that it ignores viscosity (and other things?)

I'm wondering a few things, firstly, does the theoretical math here apply to the situation I'm describing? The hole in the bottle isn't exactly a pipe and the only examples I've seen with water flow involve pipes.

Secondly, can Poiseuille's Law be used to determine the flow rate instead, with a more accurate result? (From what I understand Q=πPR^4/8nl, however I don't understand what P is, seeing as in Bernoulli's law pressure cancels and as aforementioned this isn't a typical pipe example.)

Thirdly, I assume the theoretical flow rate will still be different from the experimental flow rate, what factors cause this?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?Why is pressure gradient assumed to be constant with respect to radius in the derivation of Poiseuille's Law?

Poiseuille's Law relies on the fact that velocity is not constant throughout a cross-section of the pipe (it is zero at the boundary due to the no-slip condition and maximum in the center). By Bernoulli's Law, this means that pressure is maximum at the boundary and minimum at the center. But in the book I have it is assumed that the pressure gradient is independent of radius (distance from the center of the pipe), and the pressure gradient is thus extricated from a radius-integral. Can anyone justify this?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.Calculate flow rate of air through a pressurized hole

I was wondering about this:

If there is a pressurized container, like a tank of compressed air at some pressure that is greater than the ambient air pressure, and this tank of air has a hole in it, what is the velocity of the escaping air through the hole? Is there a formula for this?Hydraulic dynamics question

We know the length of a pipe, say L, and the starting point pressure p1, ending point p2, cross-sectional area A. What else do we need to compute the mean velocity of flow in this pipe?

My full question actually runs as follows. Given the mean cardiac output(volume flux) be 5.5L per min, the radius of the aorta is about 1.1cm. In the systemic circulatory system, the mean radius of a capillary is about 3µm. The mean pressure at the arterial end of the capillary bed (beginning of the capillary system) is estimated to be about 30mmHg, and about 15mmHg at the venous end (end of the capillary system). The length of capillary is 0.75mm.

Calculate the mean velocity in the aorta and in a capillary.

My idea is to use the volume flux 5.5L/min and the radius of aorta, divide the flux by the crossectional area to find the mean velocity in the aorta. However, I have no idea how to deal with the velocity in a capillary. I tried to use Poiseulle's law, but given that we only know the difference of pressure and length of the capillary, it is still unsolvable. Could anyone tell me how to deal with it?How does a hole's size affect the distance that water will squirt

I took a bucket, drilled 2 different sized holes on the side near the bottom and filled it with water. The stream of water the proceeded from the larger hole traveled further than the stream from the smaller one. How does the size of the hole affect the distance that the water travels?Proving that the water leaving a vertical pipe is exponential (decay)

How can I prove that the rate of which water leaves a vertical cylindrical container (through a hole at the bottom) is exponential of the form :

$A{e}^{kx}$

I know that Torricelli's law is:

$\sqrt{2gh}$

But this only proves a square root relationship. I have data points every 10 seconds and graphed it suggests a decay function. I know the distance between the pipe is 1.5M and the internal diameter is 5cm. The hole diameter is 0.25cm, if this helps. I need to prove that the water leaving the pipe is exponentially decaying.Flow through a cylinder with pores

I'm building a model to study the flow of fluid through a cylinder with pores on its surface.

For flow through a cylinder, the velocity of the fluid flow is given by the Hagen-Poiseuille equation.

I would like to ask for suggestions on references from which I can look at derivations for a cylinder with porous walls. By porous, I mean openings on the surface of the cylinder. Example, the pores present in the fenestrated capillary.