Engr Help

Laminar Flow/Navier-Stokes Equations

θ ˗component of the Navier-Stokes Equation

1 2 3 4 5 6 7
ρ ( ∂ v_θ /∂t + v_r_, ∂ v_θ /∂r + v_θ/r ∂ v_θ /∂θ + v_r v_θ/r + v_z ∂ v_θ /∂z ) = ˗ (1/r) ∂ P /∂ θ + ρ g_θ +

8 9 10 11 12

μ [ (1/r) ∂/∂r (r ∂ v_θ /∂r) ˗ v_θ /r² + (1/r²) (∂²v_θ /∂θ²) + (2/r²) ∂ v_r /∂θ + (∂²v_θ /∂z² ]

Terms that are zero from simplifications/assumptions include:

Terms: 1 thru 5 on the left hand side and 6, 7, 11, and 12 on the right had side

The remaining terms are as follows: 0 = μ [ (1/r) ∂/∂r (r ∂ v_θ /∂r) ˗ v_θ /r²

Since v_θ = v_θ (r) this partial differential equation becomes an ordinary differential equation

As follows: d/dr (r d v_θ /dr) ˗ v_θ /r = 0

Take derivatives: d v_θ /dr + r d²v_θ /dr² ˗ v_θ /r = 0

Rearrangement gives rd²v_θ /dr² + d v_θ /dr ˗ v_θ /r = 0 (Second order d.e.

with variable coefficients)

This type of D.E. lends itself to a power law type of solution. i.e. Cr^P

1. Try the solution v1_θ = C₁ r Check to see that it satisfies the d.e.

0 + C₁ ˗ C₁ = 0 So v1_θ = C₁ r is one solution to the d.e.

2. Use the "method of reduction" to find a second solution of the differential equation.