Fluids: Conservation of Momentum

  • #1
202
4
Ok, so for the derivation of the Navier Stokes, the conservation of momentum through a control volume (CV) is

Time rate of Momentum Change in CV = Time rate of Momentum Change into CV - Time rate of Momentum Change Out of CV + Sum of External Forces

Why is Sum of External forces in there? I know that Time rate of Momentum Change is dp/dt and represents force, so the equation is pretty much saying
force inside = force in - force out

But why are these external forces not already counted for in the Momentum Changes into and out of the CV? What is the intuition behind including the External Forces? Also, why is it a + Sum of External Forces and not
-Sum of External Forces?
 

Answers and Replies

  • #2
20,875
4,548
Ok, so for the derivation of the Navier Stokes, the conservation of momentum through a control volume (CV) is

Time rate of Momentum Change in CV = Time rate of Momentum Change into CV - Time rate of Momentum Change Out of CV + Sum of External Forces

Why is Sum of External forces in there? I know that Time rate of Momentum Change is dp/dt and represents force, so the equation is pretty much saying
force inside = force in - force out

But why are these external forces not already counted for in the Momentum Changes into and out of the CV? What is the intuition behind including the External Forces? Also, why is it a + Sum of External Forces and not
-Sum of External Forces?
It is referring to the forces acting on the boundary of the control volume like viscous stresses and pressure; also body forces, like gravity, acting on the contents of the control volume.

The complete expression Time rate of Momentum Change in CV - Rate of Momentum Entering CV+Rate of Momentum Leaving CV is just your Ma from Newton's 2nd Law. Ma must equal the sum of the external forces on the control volume.

Chet
 

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