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1

This is the pressure-gradient term integrated over all volume, converted to a surface integral and using Gauss' theorem. Note that physicists prefer the differential form of such equations (see also this Wikipedia article), when the corresponding equation becomes $$ \frac{\text{d}\boldsymbol{u}}{\text{d} t} = \frac{\partial\boldsymbol{u}}{\partial t} + ...


1

Why converging nozzle is used in subsonic flow and diverging nozzle is used in supersonic flows? Nozzle area velocity relation in differential form is. $$\frac{dA}{A} = (M^2 -1)\frac{dV}{V} $$ Here $A$ is cross sectional area, $M$ is local Mach number, $V$ is local velocity of fluid. For subsonic flow $M$ $<$ 1, so $(M^2 -1)$ is negative so ...


1

Solution based on wind energy and cost aspects: Typical design range 10-20 m/s: Wind Turbines are designed to produce maximum power under somewhat above normal mean wind speeds such that overall energy output per total cost of ownership is maximised. This typically results in optimum operating velocities in the 10-20 m/s range. Wind Turbines already ...


1

While I'm not willing to spend the money to get access the paper, one issue jumps out at a casual reading of the abstract - turbine design. Honeste_vivere's answer mentions the possibility of destroying a farm, and the abstract includes "The reduction in wind speed due to large arrays increases the probability of survival of even present turbine designs." ...


1

I think you need to be careful here. The total power contained within a hurricane ranges from $1 \times 10^{12}$ to $6 \times 10^{14}$ Watts or 1 to 600 TW. The world energy consumption in 2008 was 20,279 TWh. There are 8760 hours/year, thus our consumption rate was ~2.31 TW (1 TW = $10^{12}$ W). The point being, the energy you would need to dissipate ...


1

Assuming that the pressure change across the cone is small (e.g., no significant density changes for the flowing gas), use the continuity equation. With constant density, this simplifies to $A_1 \cdot v_1 = A_2 \cdot v_2$, where $A$ is the cross sectional area of the flow stream and $v$ is the velocity of the flow stream. If you additionally need the ...


0

You have to overcome a difference in pressure of 10.00 atm (based on your stated numbers). Regarding the weight of the hydraulic fluid, this fluid is oil, which is less dense than water. The pressure at the bottom of your hydraulic line is guaranteed to be lower than the ambient pressure at the bottom of the line, so the liquid head from the column of ...


1

I think the main point you are missing is that at supersonic speeds you can no longer treat your gas as incompressible and so must also consider the change in density. If we do this we can derive the can in speed as the nozzle expands. This gives a relation of: $$\frac{dV}{dA}= \frac{A}{V(M^2-1)}$$ where $V$ is the flow speed, $A$ is the nozzle cross ...


2

Intuitive explanation Imagine you're stuck in a traffic jam on a nice 6 lane highway. As it turns out, up ahead they closed all but one lane so everyone has to merge. As you approach and the number of lanes goes down everyone goes a little faster as the total flow rate of cars must be constant. Now once you pass the narrowest point, instead of everyone ...


0

It involves conversion of heat energy to kinetic energy. The temperature of the outflowing gas is lesser than the incoming gas, so it would not give rise to a perpetual motion ramjet. https://en.wikipedia.org/wiki/De_Laval_nozzle


1

If $A_2>A_1$, due to continuity of flow, fluid will decelerate and thus pressures will obey $p_2>p_1$. It is not possible to set up stationary flow for every pair of pressure values.


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This is actually not a completely settled question, so don't feel too bad if you don't understand it in every detail! It's also possible that the cause is not the same in all situations, and be aware that my explanation is not universally accepted, but here's my interpretation: There are actually two interacting phenomena going on: the formation of the ...


0

So the vortex street appears as the stream goes faster and faster: the stream obeys Newton's laws, "objects in motion tend to stay in motion unless acted on by a net force." The force that keeps the stream laminar around the cylinder is fluid pressure formed by the viscous forces of the fluid, which means there is a pressure gradient and a region of low ...


0

I would use the traditional drag equation: F = 0.5 * p * v^2 * Cd * A Reasoning: Stokes' law is for very small Reynolds number flows only (much less that 1), and I don't believe the more common drag model has any such restriction. The problem is that you will need to know the drag coefficient of the ball in question to use this force model. Sources: I ...


1

Depends on the Reynolds number. Stokes can be used for purely laminar flow. Complete answer to be found here.


0

I also do not fully understand the notions of weak and strong turbulence. What is clear to me is that under idealized conditions like homogeneity, isotropy and high Reynolds numbers we may differentiate three regions in the fluctuation wavenumber spectra: the lowest part up to the energy-containing wavenumber, $0 < k < k_L$; then the Kolmogorov ...


3

Some dimensions I was able to dig up (mostly from Wikipedia). Draft of the Allure of the Seas: 31 ft (10 m) Length: 1181 ft (360 m) Beam at waterline: 47 m Height: 72 m above waterline Let's just draw the section based on these simple numbers: Now if the center of gravity were in the middle of the ship (31 m above the water line), it would indeed not be ...


2

I doubt it is all fluid dynamics. The have to stay upright even with dead engines. If the integral of the lever below the water line is bigger than above then it should stay upright. Ballast at the bottom goes a long way as it has a long lever. Stuff like engines below deck tends to be heavy anyway. Weight is not a big deal as they are not going up ...


4

The reason that the speed of sound is a well-defined quantity is that, for small pertubations, the equations which govern the fluid dynamics can be linearised. In that linearised form, the solution boils down to a simple wave ansatz with linear dispersion relation, i.e. constant velocity.Those are the sound waves. It so happens that in air, this linear ...


0

Oddly enough, supercavitation can help increase max speed. According to the not-necessarily-correct PopSci, the Navy has achieved Mach 1: supercavitating torpedo . A link in wikipedia claims the Germans have achieved 800 km/h . Other than the Soviet Shkval, I can't find any updates on such torpedos, so either the Navy programs have gone deep black ...


4

Speed of sound in water at 20 degrees Celsius is 1482 m/s., (2881 knots), just for comparison to current claimed achievable speeds. Small related fact: The pistol shrimp can create sonoluminescent cavitation bubbles that reach up to 5,000 K (4,700 °C) which are as loud as 218 decibels, breaking the sound barrier in water. Says Wikipedia YouTube video of ...


0

Some years later... I am reviewing this problem mostly for my own benefit, But it could be useful if somebody is still wanting to discuss the answers, particularly without any braking system. let me start with an alternative wrong solution, aka a variation: allow the system to drop the water without horizontal velocity, for instance using a periodic ...


1

If you are above the water you will get accelerated down until the weight of the water you disperse is equal to your own weight (calling this level $x$). As soon as you are completely submerged the gravitational force downwards will be $\rho Vg$ and by Archimedes principle the force upwards will be $\rho_w V g$, where $\rho$ is your density, $V$ is your ...


1

According to the wiki page the 747 has a max takeoff "weight" of about 350,000 kg (depending on the model) and a wing surface of about 500 m2. That means that a force of 3.5 MN must be carried by 5 million square cm, or 0.7 N per square cm. If you can somehow split this evenly between the top and bottom surface, then you need to come up with a paper surface ...


1

In the end this is physics, so maybe you should try it less formal? (and not confuse eulerian and langrangian methods) Initially the mass element at $a$ had mass ( I use $\delta$ to make clear that we really should consider finite differences and then perform a limit at the end) $$m = \rho(a,0) \delta a$$ now, we follow its motion and after some time t we ...


3

The method you are looking for is called corner transport upwind (CTU) (Google search). Typically, we get the finite difference schemes via Taylor expansion, keeping 1st order terms and ignoring the rest (as either small, or naturally via subtraction a la central difference). The CTU scheme keeps these 2nd order terms, so the expansion is \begin{align} ...


1

We know the general cause for turbulence: it is that inertial effects (mass wanting to keep going in the direction that it's going) grow so large that viscous effects cannot contain the system in the laminar flow regime anymore. When those viscous effects cannot slow down a whole chunk of fluid, they are forces acting off-center on a mass: hence they create ...


0

The basic point has already been mentioned, but I would like to give my version of the answer, and point out a subtlety. The basic equations of fluid dynamics are the conservation of mass, momentum, and energy $$ \frac{\partial \rho}{\partial t} - \vec{\nabla}\cdot\vec{\jmath}_\rho \\ \frac{\partial \pi_i}{\partial t} = - \nabla_j\Pi_{ij}, \\ ...


2

Cavitation inception occurs when the local pressure is less than the vapour pressure of the liquid. In effect the liquid locally boils to form a bubble of vapour. But producing the pressure gradients required to boil the liquid require a non-zero stress, or more precisely the viscous stress tensor has to be non-zero. In a superfluid the stress is always ...


-1

Answering on the third part of your question bubbles rely on the temporary breaking of a fluid state, so it would not be possible


0

Navier-Stokes equations describe fluid in an approximate way which neglects diffusion of molecules altogether. In ordinary case, the velocity field is supposed to be smooth. "Diffusion of momentum" mentioned in connection to the term proportional to $\Delta \mathbf u$ is not really a diffusion in the molecular sense. It is rather kind of metaphor to ...


0

Put very simply: An airfoil produces lift because of angle of attack; the leading edge of the wing is higher than the trailing edge. This imparts momentum to the air through which the wing passes and also changes the potential energy of that parcel of air.John Denker's website See How it Flies probably has one of the best and most straightforward explanation ...


1

I left this as a comment but I'll expand it here since it provides another viewpoint. Imagine you have a box of gas molecules bouncing around in them. Every molecule is identical so they have the same mass, temperature and pressure. Let's also say this box has a diaphragm in the middle separating the box into two. You now remove the diaphragm and start ...


1

So it's actually a really simple reason, but you're going to have to think a little bit about what's going on. The transport equation states that everything which is a "stuff" can be viewed in this way: "A small box flows downstream; the time rate of change of the stuff inside of the box is equal to the flow of stuff through the boundary of the box, plus ...


2

You are right, the two explanations are not mutually contradictory. There are two ways of caluclation the lift on an airplane. You can look at the total mass of air being deflected downward by the wing, and equate the rate of change of momentum of the air to the upward force on the airplane. This is exactly what is going on, and for my money it is the real ...


1

On the water-splash, here's a video that I think explains what happened. https://www.youtube.com/watch?v=2UHS883_P60 (the 3 balls, not the double bounce on the trampoline). The Newton law of conservation of energy says that energy is concerved, so a share of the energy of the drop transfers to the energy of the splash. Figure you drop it from 1/2 ...


5

For a single-component fluid, the conservation of mass follows $$ \left(\begin{array}{c}\text{mass of fluid } \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of fluid } \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{sources or} \\ \text{sinks in }\Delta V\end{array}\right) $$ In terms of a ...


1

Nice argument, Why opening windows in early morning reduces temperature? Every human body need energy to survive, That energy is covered into heat while doing work or for the operation of internal organs. For example carbohydrate in our body burns with oxygen and give heat and carbon-DI-oxide. If we do not open the windows , that heat will stagnate ...


2

If the window is open just a crack, and then you open it wider, you expect this to make a difference. As the window gets larger the difference becomes less significant - and depends on whether it is diffusion or convection that is cause the cooling of air. Diffusion (no wind) depends directly on surface area: twice the area, twice the diffusion. For ...


1

First of all, put the fan in one window, set to force air to exit. This is by far the most efficient way to enhance airflow. (if there's a breeze already, make sure the fan's in the window the breeze is blowing outwards :-) ). If there's no wind outside and no fans running, then the size of the window opening is relatively unimportant, as only some ...


0

You can integrate the velocity values over $r$, the radius, using whatever integration technique you would like (trapezoidal rule, Simpson's rule, etc) and use the mean-value theorem to get the average velocity for each radial line. You then have to account for all $2\pi$ radial lines. Assuming the flow is incompressible, you can then find kinetic energy by ...


1

Surface tension is the result of inter-molecular cohesive bonding among the molecules of a liquid. At an interface between the liquid and a gas, the molecules are more attracted to themselves than they are to the gas, so they form a well-defined surface film, which is more stable than the body of liquid beneath it. Inside the body of the liquid, molecules ...


0

I'm appreciating your thinking and question. This question will be even more challenging question if the working fluid is air instead of water and fluid is moving. In this answer I'm considering bottle is moving and fluid is moving in opposite direction of bottle. (because stationary fluid is sub case of moving fluid where $v_1$=0) Let us assume the the ...


2

An interesting limitation aside from the obvious practical ones being overlooked, is that eventually you would be moving quickly enough to bond your carbon atoms with oxygen in the air. If we put a lower limit of around 750 degrees F or about 673 Kelvin to start that fire, then you would have to move at about 1,300 meters per second.


1

The opening sees water at stagnation pressure. Water will flow in until the air in the bottle is squeezed to that pressure.


0

If it is water with a specific gravity of 1 you will lose 0.0361 psi per inch in the container. Example- If the level of the container drops 12 inches you would take 12 x .0361 and you would have a pressure loss of 0.4335 psi. If it is a material with a different specific gravity you would have to make adjustments, such as mercury would be .491 psi per inch. ...


0

If u want to find exit pressure of control volume you have consider, you can use Bernoulli equation since you know exit velocity profile. Since boundary layer is thin and no flow separation, this flow is more or less irrotaional flow and I'm assuming the flow is steady. $$ p_1 +\rho \frac{v_1^2}{2} =p_2 +\rho \frac{v_2^2}{2}$$ Here $p_1$, $v_1$ are inlet ...


0

Preliminary: Pressure is the driving (or source) term in Naiver-Stokes equation that governs fluid flows in continuum region, if we know pressure at inlet and outlet we can find the velocity for simple low speed flows. This process is isentopic if there is no heat flow. That problem can be solved by Bernoulli equation if the flow is irrotational, and the ...


0

Application that you have considered is pipe flow. Pipe flow can be assumes as isentropic flow. Isentropic equations are non-linear. (Non-linear i'm using here is algebraic non-linearity. Please note the governing equation of fluid mechanics is also non-linear PDE). So resulting gas equation will be non-linear in nature. We know ideal gas relation is ...


3

Q: When we write that, do we suppose a collisionless or collisional nature of the fluids? A: It's the energy-momentum tensor for a perfect fluid Chapter 2.26 Q: If this description corresponds to collisional fluids, why cosmological simulations are N-body simulations (collisionless) and are not simply based on hydrodynamics? A: Cosmological simulations are ...



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