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Incompressible flow condition intuition

As you have said it is derived from the continuity equation (which is one of Euler fluid equations) $$\frac{\partial \rho}{\partial t}+\nabla\cdot(\rho \mathbf u)=0$$ if the density of the fluid $\rho$...
Mauricio's user avatar
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Incompressible flow condition intuition

An intuitive explanation comes from the meaning of the word divergence itself. If the divergence of the velocity is zero, it means that there is no (net) inflow or outflow of fluid from a given volume....
CompassBearer's user avatar
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Incompressible flow condition intuition

You ask"How is there a relation between the density of the fluid and its speed"? There is none $\nabla u=0$ says only that the velocity does not change for example in x direction .
trula's user avatar
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Stress/forces on elements in continuum mechanics

In continuum mechanics, the stress tensor is a point function, and all 9 components exist at each point. When you do a force balance in a fluid parcel, the stress vector is evaluated locally at each ...
Chet Miller's user avatar
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Why is the strain tensor of a rod under uniform torsion the same everywhere along the rod?

Axial $z$-coordinate is homogeneous (and thus the solution doesn't depend on $z$) for: slender beams, if one excludes the region close to the extreme sections of the beam; with no distributed loads ...
basics's user avatar
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Work performed by hydrostatic pressure

With a little less math, the power of a stress distribution $\mathbf{t}_n$ over the boundary of a volume $V$ is $$P(t) = \oint_{\partial V} \mathbf{t}_n \cdot \mathbf{u} \ ,$$ being $\mathbf{u}$ the ...
basics's user avatar
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Confused about stresses in a small element in solid mechanics

So in the diagram of a stress tensor, the normal and shear stresses on opposite faces of the small cube are equal and opposite. Justified by equilibrium. This is not true. In static conditions, local ...
basics's user avatar
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Work performed by hydrostatic pressure

My intuition is that we are looking at a special case of Stokes' theorem, since we are relating surface integrals to volume changes. To this end, it seems clear enough to me that the differential ...
creillyucla's user avatar
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Rigorously proving equal tension on both ends

This isn't a rigorous proof, but moreso the intuitive answer that I use. Say you're pulling an object with a heavy rope. The end where you pull the rope will be pulling the object's mass and the rope'...
Marcus's user avatar
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