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What you're after is thrust, which is force, which is momentum per second. To get the same force, which is $F = ma$, you can either use a large mass with a small acceleration, or a small mass with a large acceleration. Which requires more energy? (The small mass with the large acceleration, because the imparted velocity is higher. if $v = at$, $2ma = m2a$, ...

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Dispersive mass transfer, in fluid dynamics, is the spreading of mass from highly concentrated areas to less concentrated areas. It is one form of mass transfer. "Dispersive mass flux is analogous to diffusion, and it can also be described using Fick's first law: $$J=-E\frac{dc}{dx}$$ where c is mass concentration of the species being dispersed, E is the ...

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Here is a quote by Ernst Mach (from a public lecture): "If the projectile moves faster than sound, the air ahead of it cannot recede from it quickly enough. The air is condensed and warmed, and thereupon, as all know, the velocity of sound is augmented until the head-wave travels forward as rapidly as the projectile itself, so that there is no need whatever ...

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You are right that the air can't get out of the way in a continuous fashion. When you move an object through a fluid at a speed that exceeds the fluid's sound speed, then discontinuities may form known as shocks. The sonic boom associated with superonic jets is one consequence of the formation of these shocks. Supersonic fluid motion isn't always associated ...

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A vortex is defined as a region within a fluid where the flow is mostly a spinning motion about an imaginary axis, straight or curved. Mathematically, this is defined as the curl of the velocity field, $\mathbf{v}$: $$\boldsymbol{\omega}=\nabla\times\mathbf{v}$$ If we consider the strength of the vortex tube as $$\Gamma=\int\boldsymbol\omega\cdot ... 1 In short if pipe A is much shorter than B, and both A and B have the same size and outlet pressure (aka back pressure), then pipe A and B will have identical pressure drop and pipe A will have a higher flow rate. Given a fixed inlet and outlet pressure, the pressure drop is also fixed (\Delta P = P_{inlet} - P_{outlet}) and therefore the flow in the two ... 2 I am not an aerodynamics specialist, so the following is almost certainly a huge oversimplification (or maybe downright wrong), but I think it might help with intuition. Suppose you have an amount of energy E available to spend, and you are trying to accelerate an object of mass M. Suppose you can impart the energy in the form of kinetic energy to an ... 0 You could check the works of Emilio del Giudice, Giuliano Praparata et al. The second one in particular is been involved in the study of coherence effects in quantum field theory, in particular in quantum chromo dynamics and quantum electrodynamics. In this article ... -2 It has to do with the pressure from the atmosphere. Milkshakes are thicker than water and they are also stickier causing them to stick more to the straw. 1 Stresslet coefficient is an important part of Faxen's laws - laws "relating a sphere's velocity U and angular velocity \Omega to the forces, torque, stresslet and flow it experiences under low Reynolds number (creeping flow) conditions." Forces, torque and flow seems familiar, but stresslet? So, stresslet is "a function that represents the symmetric ... 0 Like many things in electrostatics, this class of problem will come out with an unintuitive answer. Would the liquid touch the object? The thinking of the question is that the fluid is positively charged, so the fluid is attracted to the object. For the bulk fluid, this would be true. By that logic, it would be somewhat safe to say that the electric ... 0 When the liquid flows through a pipe, whether it is discharged from a pump or not, it's pressure changes. If the liquid flows in a horizontal pipe, upwards in a vertical or inclined pipe - the pressure will definitely drop. If it flows downwards, whether the pressure will rise or drop, it depends on the ratio of the friction, momentum and gravity terms in ... 1 This is a interesting question. This is how I reasoned it out. I think there are three situations here (assuming rotationally invariant): 1) The object is very deep where it cannot overcome the electrostatic force from all the positive charge above it, with only its Buoyant force. 2) The object is close enough to surface to where the buoyant force and the ... 0 The most basic definition of a fluid is a fluid is a substance that continually deforms (flows) under an applied shear stress. To model a fluid using the Euler equations, you need to satisfy the condition that the mean free path of a particle, \ell, is significantly smaller than the typical size of the domain, L (and also that viscosity and heat ... 0 How does this varying pressure conform with the constant pressure/density obtained from the equation of state? It doesn't conform. It is a contradiction to reality: you cannot have an incompressible material. However, that doesn't mean the approximation isn't a useful one under many circumstances. For example, consider water which as a density of ... 0 Continuity is just the principle of conservation of mass in differential form. The full continuity equation is (in index notation): \frac{\partial \rho}{\partial t} = -\frac{\partial }{\partial x_i}(\rho u_i) For example, consider an infinitesimal control volume (CV). The equation says that the local \rho (inside the CV) will decrease in time if the ... 0 Question number 1: A soap bubble is black when it bursts, why? Answer: When a soap bubble bursts, it looks black due to destructive interference. When a bubble bursts, Its thickness becomes negligible, that is, thickness ≈ 0, and The condition for destructive interference is satisfied.$$ 2nt = m\lambda,\tag1$$Where n is the refractive index of ... 0 Any system in which hydrostatic pressure is not the only pressure to consider could sustain a water level difference between connected containers. To give an example: a system that isn't at a static, but rather a dynamic steady state can have unequal water levels. Think of two containers with one (tank 1) elevated above the other and a two hoses connecting ... 1 Whenever it seems like two water levels should be equal but aren't, either there is a physical restriction preventing flow (like a dam keeping upstream waters higher than downstream, or surface tension causing meniscuses or capillary action), or there is energy being expended to put water back upstream as fast as gravity is pulling water downstream. In a ... 0 If the water is frozen - it won't equal out. If the containers are not connected - it won't equal out. If water can't reach the connection equally easily from both containers - e.g. if the connection is high or low and the water is vapor - it won't equal out. 0 The sink demonstration is not an optimal one to show the effect of Coriolis force, mainly because the water in the sink will never be perfectly still enough to start with in order for that force to be the dominant one to determine the direction it will swirl (you could force it to spin either direction. Better demonstrations are easy, however. One ... 0 What produces lift is circulation, which causes the airflow to be deflected in one direction, causing an equal reaction in the other direction. If you want to think in terms of Bernoulli on your soccer ball, the air on the left side is being slowed, while that on the right side is being accelerated by the spin of the ball. 2 The book derives the equation of continuity, which states that the cross-sectional area times the velocity of a flow is always constant. But nowhere in the derivation does the textbook explicitly assumes that the flow is laminar. So, does the equation hold for turbulent flows too? That is only a special case of the equation of continuity for situation ... 0 In order to have such a relation, your flow needs to be be stationary, which is never the case for turbulent flows. The conservation of the mass gives you the local continuity equation.$$\partial_t \rho+ \nabla . (\rho \vec{v})=0 $$For a stationary problem without sources, Ostrogradsky's theorem allows you to reach:$$ \oint_S \vec{v}.d\vec{S}=0 $$But ... 1 The teapot effect is flow attachment of liquid to solid due to the solid's hydrophilic or hydrophobic property, the same thing that controls water beading. This is different from what makes air follow a surface. John S. Denker has something to say about this: It appears that surface tension plays two very important roles: At the water/air ... 1 Using Bernoulli you get:$$ \frac{P_1}{\rho} + \frac{1}{2}v_1^2 = \frac{P_2}{\rho} + \frac{1}{2}v_2^2 Using your formula: \displaystyle{c_s^2 + \frac{1}{2}v_1^2 = c_s^2 + \frac{1}{2}v_2^2} and this implies: v_1 = v_2 From mass continuity: v_1 \times A_1 = v1 \times A_2, so A_1=A_2, which is a false. There is clearly a misinterpretation. ... 1 Your intuition is effectively correct. Basic corroboration can be found in the book Viscous fluid flow, second edition by Frank M. White. In section 3-9.2 he exams low Mach and Reynolds number flow, known as Stokes flow, over a sphere. Surprisingly, the analytic solution for the problem turns out to be independent of the fluid's viscosity. He then draws ... 0 Both temperature and pressure variation with altitude is given here You can use the ideal gas law to get the volume with altitude from these. 2 I say yes, things can sink in ice. Here's why: there's a thin layer of liquid on the surface of ice that is why ice is slippery. When you put a body on the surface of ice, it'll keep displacing the layer of liquid, getting deeper and deeper, and will eventually sink completely, 5 If the article you are referring to is this one, then the planes didn't sink into the ice but were buried by blizzards. Ice does exhibit ductile flow at stress of around 1 to 10MPa, but this pressure is equivalent to around 100 to 1000 tons per square metre and this is far above the stresses normally produced by objects resting on ice. Glaciers flow because ... 3 The formula that you write out is just a consequence of linearity without any additional requirements. Suppose you had a vector \mathbf{D} that was linearly dependent on another vector \mathbf{E}. Then, one would write D_i =\sum_j \epsilon_{ij}\ E_j. With no further conditions, \epsilon_{ij} would have 9=3\times3 coefficients. In your formula, the ... 1 USim can do vacuum + gas http://www.txcorp.com/home/usim/usim-overview. Not technically true vacuum, but 9 orders of magnitude density jumps. It doesn't support an incompressible fluid (liquid) though. -1 We start from point (5) above where things started to go wrong: \frac {dV_c}{ dt}= - \frac {dV_b}{ dt} \tag{5} \frac {dV_c} {dt} = m_c / (\frac {d\rho_c}{dt} ) = - \frac {m_c } {0.1} \tag{6} \begin{align} \int_{V_c,0}^{V_c,1} {dV_c} = - \frac {m_c } {0.1} \int_0^t dt \tag{7} \end{align} \begin{align} V_{c,1}-V_{c,0} = - \frac {m_c } ... 1 In your 5th "Let" statement "\frac {d\rho_c}{ dt} = 0.1kg/m^3/sec the rate pressure changes in the large container", "pressure" should be "density" and there is a sign error. Your equations 1-4 are correct, but it is important to determine what is known and what is unknown and analyze how many independent equations and unknowns you have. Also, mass of ... 1 Yup, I wrote something like that in the wikipedia article... The correspondence [between the Navier-Stokes equation and the convection-diffusion equation] is clearest in the case of an incompressible Newtonian fluid, in which case the Navier–Stokes equation is:\frac{\partial \mathbf{M}}{\partial t} = \frac{\mu}{\rho} \nabla^2\mathbf{M} -\mathbf{v} ...

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To answer Your question: yes, it is possible to extract nitrogen from the air via a centrifuge, or rather - it is possible to increase the concentration of nitrogen in a gas mixture, there are always going to be some impurities. Whether anybody does it? Probably not, since it's not the most efficient way, as it was noted. In nuclear applications it is used ...

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Consider that $$\mbox{Force} \propto \mbox{Velocity Gradient}$$ Equal force means, the same velocity gradient, i.e. linear distribution of velocities across the flow. The flow near the boundary has zero velocity and so velocity increases linearly the further away you go from the boundary.

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You can calculate the pressure drop with the following formula: $p_1 - p_2 = \frac {W^2}{2 rho}$*$(\frac {1}{A_2^2} - \frac {1}{A_1^2})$ where $W$ is the mass flow in $[\frac {kg}{s}]$, $rho$ is the density in $[\frac {kg}{m^3}]$, $A_1$ and $A_2$ are the cross-sectional areas before and after the reduction. To answer Your second question: it shouldn't ...

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A conservation law typically has the form $\frac{\partial}{\partial t} (\text{Volume Density of a quantity}) + \mathrm{div}\,(\text{Flux per unit area of that quantity}) = 0$, in the local / differential version. The integral version is more familiar: $\frac{\partial}{\partial t} (\text{Quantity inside volume}) = - \text{Total flux of quantity through ... 0 It is helpful to think of this problem in terms of timescales. Look at the navier stokes equation in 1D:$\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} = - \frac{1}{\rho} \frac{\partial p}{\partial x} + \nu \frac{\partial^2 u}{\partial x^2}$Considering timescales on the order of diffusion for a domain of length,$L, the diffusion ... 5 I think that you're basically right when you make the following suggestion: Does part of the input force (or pressure) get wasted in increasing the temperature of the gas and therefore making it less effective? Recall from the First Law of thermodynamics, \begin{align} \Delta E = Q-W \end{align} where\Delta E$is the change in internal energy of ... 3 Pretty much no. The problem is that you are not (pardon me :-) ) a rigid body, so you're going to feel a certain amount of force from the wind regardless of what sort of weights you're carrying. What can help is walking with your feet farther apart, which gives you a more stable base to work from, and to learn to turn your body sideways to the wind as ... 3 If the flow rate is dominated by the viscosity, so inertial effects can be neglected, the flow rate is given by the Hagen-Poiseuille equation: $$\Delta P = \rho gh = \frac{8 \mu L Q}{\pi r^4}$$ The pressure difference is just the pressure at the orifice or$\rho gh$as given in your first equation.$\mu$is the viscosity of the liquid,$L$and$r\$ are ...

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It's just a proportionality constant.

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Consider a ball of water floating in zero G, as demonstrated on the ISS. Ignoring for the moment the surface tension of the water (I'll come back to that) the pressure inside the water is the same as the pressure of the air around it. This is simply because without any forces, like gravity, acting on the water there is nothing to cause a pressure gradient. ...

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It's not that the random motion decreases when the flow rate increases. It is only that the random motion stays the same but the coherent motion dominates. If the diffusion velocity in a gas is 1 and the convective velocity of the flow is 1000 (units don't matter), then the diffusive action can be pretty safely ignored. The important thing to remember is ...

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