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"preserved following the flow" means preserved as you follow a fluid particle during its motion. Suppose that a fluid particle is located at position $(x,y,z)$ at time $t$ and has velocity $(u,v,w)$. …

The textbook is trying to show you the effect of varying only $Re$ while keeping everything else the same while you scale your system. Therefore you need to keep geometric parameters (among others) of …

You see laminar flow becomes turbulent by growth of disturbances introduced into it either by uncontrollable vibrations from the walls (which are ever present) or deliberately by using a trip wire. … Beyond a Reynolds number smallest disturbances get amplified (in time and/or space) and laminar flow becomes turbulent. This state of the flow is analogous to a pencil standing on its pointed end. …

In other words while mean flow may be 2-dimensional, fluctuations of flow exist in all 3 dimensions. … This means that if $(u,v,w)$ is the total flow velocity field, then it is to be decomposed as $(\bar{u}+u',v',w')$ because $\bar{v}=\bar{w}=0$. So Reynolds stress will be present. …

...volume integral is to be performed over the region where ... the fluid is not actually present in the volume where the divergence of its stress tensor is being integrated. You are correct, but …

$\mathbf{u}^\infty$ is the flow in the absence of the particle. … It is obtained by solving Navier-Stokes equation (or Stokes flow equation, if you are working in that regime) and continuity equation $\nabla\cdot\mathbf{u}^\infty=0$, when the particle is not present. …

Due to force exerted by the turbine, wind leaving the turbine has different momentum than when it came in. If you consider a control volume surrounding the turbine, then in steady state the force exer …

Also, will the velocity of water(not volumetric flow rate) just inside the syringe be zero or some value? …

If pressure difference driving the flow is constant, then it is not obvious that introducing a constriction in the flow will necessarily increase the flow speed there (compared to the flow speed before … This argument doesn't assume a compressible flow. …

Power consumed by a pump is $P=\Delta p\times Q$, so if the pressure drop $\Delta p$ that the pump must provide is constant then the power $P$ would indeed be proportional to flow rate $Q$. …

But here is something that comes close: A turbulent flow is characterised by a wide range of length and time scales, while a laminar flow is not. … flow becomes turbulent. …

So you can apply D-W equation provided flow is laminar. …

Now since flow Reynolds number is high (and therefore flow is turbulent), viscosity plays little role in determining the flow (hypothesis). In that case, $\frac{Q}{\omega d^3}\approx $constant. … Since angular speed of the fan isn't varying either I would guess that flow rate of the fan shall remain constant (to a good approximation). …

In the continuum limit the body (i.e. aggregate of blobs) seems to become a cylinder, but with increasing radius along X direction. Area of elemental circular strip at distance $x$ will be (circumfere …

More precisely, say you wish to measure pressure at a point P in the flow, where the fluid velocity is $\mathbf{v}$. … In any other reference frame the area must be oriented such that its normal is perpendicular to $\mathbf{u}$, i.e. there should be no flow across the area. …