Lewis Mammel wrote:
Edward Green wrote:

I have attempted to rederive Bernoulli's theorem starting with the
conversation of momentum, rather than the usual conservation of
energy,
but I can't seem to make the thing work.

From my correspondence:

But your "total force" is ill-defined. To get a change in momentum
you need to integrate dp = F dt, ( p for momentum , P for pressure )
and for a straight streamline we have

dp = F dt = dP/dx dt = 1/v dP

I don't get this step. F is not dP/dx. The units don't match.


since v = p/rho

p/rho dp = dP

so 1/2 p_2^2/rho - 1/2 p_1^2/rho = P_2 - P_1

which is consistent with Bernoulli's equation.

There is a *reason* why Bernoulli's equation can't be derived from
momentum conservation. Energy conservation and momentum conservation
are not equivalent laws.

PD

PD wrote:

Lewis Mammel wrote:
Edward Green wrote:

I have attempted to rederive Bernoulli's theorem starting with the
conversation of momentum, rather than the usual conservation of
energy,
but I can't seem to make the thing work.

From my correspondence:

But your "total force" is ill-defined. To get a change in momentum
you need to integrate dp = F dt, ( p for momentum , P for pressure )
and for a straight streamline we have

dp = F dt = dP/dx dt = 1/v dP

I don't get this step. F is not dP/dx. The units don't match.

A dP is the force on the volume A dx so dP/dx
is the force per unit volume, which I should have specified.

Also, my p is rho v, which is momentum per unit volume.
So, I just slipped into the "per unit volume" picture without
changing my nomenclature.

since v = p/rho

p/rho dp = dP

so 1/2 p_2^2/rho - 1/2 p_1^2/rho = P_2 - P_1

which is consistent with Bernoulli's equation.

There is a *reason* why Bernoulli's equation can't be derived from
momentum conservation. Energy conservation and momentum conservation
are not equivalent laws.

Right, but if you have a straight streamline, as I stipulated,
the calculation is correct. ( Note I just said it's consistent
with Bernoulli, not equivalent to it. )

Lew Mammel, Jr.

PD wrote:

There is a *reason* why Bernoulli's equation can't be derived from
momentum conservation. Energy conservation and momentum conservation
are not equivalent laws.

Right. But, my meta-reasoning was as follows:

(1) In a steady state flow, the net rate of change of momentum for a
fixed spatial volume is zero.

(2) We can apply this idea to a fixed slice of fluid flow in a pipe (or
a section of a bundle of streamlines)

(3) If we correctly identify all momentum fluxes on the boundaries of
this volume, we are bound to derive some correct relation on the flow.

(4) Under suitable assumptions, the only free variables in our fluid
flow will be pressure and velocity.

(5) Therefore, under these conditions, we are bound to derive a correct
relation between pressure and velocity.

(6) Since there are only two free variables, there is essentially only
room for one functional relation between them.

(7) This functional relation had better be Bernoulli's law.

Do you see any flaw with this program?

I would rather conjecture that there is a reason Bernoulli's law is not
usually, rather than cannot be, derived from consevation of momentum;
because, unlike the simple case starting from conservation of energy,
it seems to be rather subtle.

Edward Green wrote:

I would rather conjecture that there is a reason Bernoulli's law is not
usually, rather than cannot be, derived from consevation of momentum;
because, unlike the simple case starting from conservation of energy,
it seems to be rather subtle.

Hold everything! Granger, Fluid Dynamics ( Dover ) states that the
Bernoulli equation IS derived from momentum concepts when he compares
it to the identical-except-for-friction-term "Steady flow energy
equation" ( Section 5.5.4 )

In section 4.4.10, he derives Bernoulli's equation from the
Gromeka-Lamb form of the Navier-Stokes equation. This becomes
Lamb's equation for steady, inviscid, incompressible flow,
and Bernoulli's equation is an integral of Lamb's equation.

This is all at the end of a long section on The Differential
Form of Conservation of Momentum.

The factor of 1/2 appears in the reduction:

( del dot ) del v = 1/2 del v^2 - v X ( del X v )

So there you are.

Lew Mammel, Jr.