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• Mahesh Lohano

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THE DIFFERENTIAL EQUATION OF LINEAR MOMENTUM & NAVIER STOKE EQUATION ENGR. MAHESH KUMAR

INTEGRAL RELATION FOR MOMENTUM d  F  dt  



CV V .d   CS V  (V .n)dA

Moment Flux Term denoted by Momentum

M&CS   V  (V .n)dA CS

d F   dt 

 CV V  .d     mV  out    mV  in

THE DIFFERENTIAL EQUATION OF LINEAR MOMENTUM As we Known Sides of cube are dx , dy and dz Therefore

Take component of Velocity in x,y,z direction as u,v and w

Momentum flux occur at all six faces, Picture shows mass flow in and out from all faces.

We can say

As

m here...  dx.dy.dz

Cancel row*dx.dy.dz ----------(a)

There are two types of forces act on a body force and surface force Body forces are due to external field …. i-e due to gravity and magnetism Ftotal = Fsur+ Fbody Here body force

------------------------ (1)

Stresses (sigma) are sum of hydrostatic pressure and Viscous stress (Tau) As on face P and Viscous forces both act

In X direction As we know Stresses (sigma) are sum of hydrostatic pressure and Viscous stress (Tau)

• dx.dy.dz is volume

• Similarly do for y and z direction

We can say that

-----------(2)

• Sometime above equation is expressed in divergence form

As We Know Total Force in sum of Surface and Gravitational •

---------------------------------------------

• The Value of dV/dt is given in equation (a)

----------(3)

Therefore equation (3) for x direction can be written as

Similarly for y and z direction

NAVIER STOKE EQUATION • As we know

u   x

• Put in Last Equation of linear momentum

NAVIER STOKE EQUATION

MOODY CHART