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Navier-Stokes Equation of Motion

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Applied Fluid Mechanics

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Navier-Stokes Equation of Motion

  1. 1.  The fluid element is acted upon by gravity force, pressure force and viscous force is the case of Navier- Stokes equation.  Consider an elementary small mass of fluid of size dx* dy* dz in x, y, z three directions respectively as shown in figure.
  2. 2. Pressure force on the face ABCD= p dxdy Pressure force on the face EFGH=(p+∂p/∂x dx) dy dz Hence net pressure force in x-direction is equal to = p dxdy -- (p+∂p/∂x dx) dydz = ∂p/∂x dxdydz Let R is the body force per unit mass of fluid having components X,Y,Z in the x,y,z directions respectively. Therefore the body force acting on the element of fluid in the X-direction will be Xρ dxdydz. Let tx,ty,tz are the component of shear forces in x,y,z directions Total force=mass * acceleration Xρ dxdydz---∂p/∂x dxdydz—Tx= ρ dxdydz *du/dt --------------(1)
  3. 3. Shear stress caused due to viscosity on a particular surface equal to the rate of change of velocity in a direction normal to that surface ѓ=μ dv/ dn Let us consider any two faces of parllelopiped say ABCD and EFGH The shear force acting on face ABCD = μ (dydz) du/dx The shear force acting on face EFGH = μ (dydz) ∂/ ∂x [u+ ∂u/∂x dx] = μ (dydz) *[∂u/∂x + ∂²u/∂x² dx] The net shear force along the x-axis on faces ABCD and EFGH = μ (dydz) du/dx -- μ (dydz) *[∂u/∂x + ∂²u/∂x² dx] = - μ ∂²u/∂x² dx dy dz -----------------------------------(2)
  4. 4. Similarly x-component of shear force on face BCFG = μ (dx dy) ∂u/∂z X-component of shear force on face ADEH = μ (dx dy) ∂/ ∂z [u+ ∂u/∂z dz] = μ (dx dy) *[∂u/∂z + ∂²u/∂z² dz] The net x-component of shear force on face BCFG and ADEH = μ (dx dy) ∂u/∂z --_ μ (dx dy) *[∂u/∂z + ∂²u/∂z² dz] =- μ ∂²u/∂z² dx dy dz -------------------------------(3) Similarly net x-component of shear force on face CDEF and ABGH is =- μ ∂²u/∂y² dx dy dz --------------------------------------------------(4)
  5. 5.  The total viscous resistance parallel to x-axis on the all six faces of the element is given by sum of eq (2), (3) , (4) Tx= - μ [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² ] dx dy dz Similarly for Ty and Tz are Ty = -μ [∂²v/∂x² + ∂²v/∂y² + ∂²v∂z² ] dx dy dz Tz = - μ [∂²w/∂x² + ∂²w/∂y² + ∂²w/∂z² ] dx dy dz
  6. 6. Substituting the values of Tx in eq 1 X ρ dx dy dz--- ∂p/∂x dx dy dz + μ [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² ] dx dy dz= ρ dx dy dz du/dt Dividing throughout by ρ dx dy dz , we get X - μ/ρ ∂p/∂x = du/dt –μ/ρ [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² ] X- μ/ρ ∂p/∂x = du/dt –v [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z²] [ μ/ρ=v] - ----(5) Similarly for y and z direction Y- μ/ρ ∂p/∂x = du/dt – v [∂²v/∂x² + ∂²v/∂y² + ∂²v∂z² ]--------(6) Z -μ/ρ ∂p/∂x = du/dt – v [∂²w/∂x² + ∂²w/∂y² + ∂²w/∂z² ] -----(7)
  7. 7. Navier-Stokes equations help in I. The design of aircraft and cars II. Study of blood flow III. The design of power stations IV. Analysis of pollution
  8. 8. Applications of Navier-Stokes Equation are 1. Laminar flow in circular pipes. 2. Laminar flow between two fixed plates. 3. Laminar flow between parallel plates having relative motion.
  9. 9. Thank You

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