Question

find the expression for kinetic energy per unit mass

Answer 1

I/2mv^2
hope it will help you
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Answer 2

Derivation for kinetic energy flux

fluid-dynamics energy kinematics

I'm working to derive kinetic energy flux for fluids. I could not find a derivation online. I know from literature the correct answer is ϕkin=(1/2)ρv3.

The specific context is in snapshots of fluids, so its okay to assume constant acceleration.

I being with the definition of kinetic energy

Ek=12mv2.

We can assume constant acceleration between each snapshot (each time we can view the fluid). We are interested in calculating the kinetic energy flux of a system with one point, molecule or pixel and how it moves between snapshots.

We begin by placing a single particle in a box. It has energy only due to kinetic energy. As it has kinetic energy, from our previous definition of kinetic energy it must be moving.

Consider this particle in a box moving in an arbitrary direction as depicted in the figure below.



We shrink the cube so the particle must pass through it over the duration of the snapshot and measure the flux once the particle has moved through a face of the box. The box has edges which are ϵ wider then the diameter of the particle.

We describe the particle (with a vector field ) by a Dirac delta with an associated Ek scalar. The vector field therefore looks like

Fk_=EK δ3(r_−r′_)

But - this isn't actually a vector, because the Dirac delta is a scalar. But, if I include a vector, it screws up the result.

The kinetic energy flux is defined as

ϕk=∮SFk_⋅ˆn_dS

We assume the point particle does not go through an edge, so we arbitrarily take the x-y face.

ϕk=∫x+ϵx−ϵ∫y+ϵy−ϵEkδ3(r_−r′_)⋅ˆn_dxdy

Taking the normal which will get rid of one dimension. Thus our integral becomes

ϕk=∫x+ϵx−ϵ∫y+ϵy−ϵEkδ2(r_−r′_)⋅ˆn_dxdy

Again, I think the fault is in the above step--I can't just get rid of one dimension of the Dirac delta by using the dot product with the normal as an excuse. (also, I'm dotting something that isn't a vector with a vector)--but if I don't do that, I'll end up with a 3D Dirac delta in a 2D integral, which seems dodgy

We invoke the property of the Dirac delta

∫a+ϵa−ϵf(x)δ(x−a)dx=f(a)

And we are left with

ϕk=Ek∫x+ϵx−ϵ∫y+ϵy−ϵδ2(r_−r′_)⋅ˆn_dxdy=Ek

Our flux therefore is ϕk=Ek=12 m (v22−v21).

We can set the initial velocity to zero and consider the Ek in each snapshot at that instant. Thus

Ek=12 mv2

This is valid for a single particle. Extrapolating to a fluid and feeding in the density rather than mass, we find it:

Fkin=12ρv3

Which is the correct expression. (but with a flawed derivation).