FLOW THROUGH CLAMSHELLIN THRUST REVERSELposition

byUMA SHANKAR GANDI

Dividing the flow in two section:

1. Flow through two concentric cylinders( i.e. flow between two parallel plates in 2-D)2. Stagnation flow through an inclined plane

Assumptions:

Simplifyied flow section

Steady viscous flow

No slip conditions

2D flow analysis in Cartesian coordinates

Flow between two cylinders: (poiseulle flow)

Boundary conditions

Velocities are zero at the surfaces,

From the continuity equation

we get Vy = constant

From y- momentum equation

0u v

x y

Reduces to pressure is constant in y direction.

From the X- momentum equation

reducing with conditions we get

Let the pressure gradient along X direction be K we get

Solving the above equation with boundary conditions we get

21u u u pu v u

t x y x

2

2

u K

y

22

12

Kh yu

h

Stagnation flow on an inclined :

considering the flow directly in hit to an inclined plane with inclination

We introduce a stream function

Ψ(x, y) = V (y sin θ - x cosθ ) where u = V sinθ , v = Vy cosθ

And V is proportional to stream line velocity and proportional to characteristic body length

X- Velocity component cannot be zero ,

To satisfy the stream function viscosity is introduced in the stream function

Ψ(x,y) = V f(x)

u = -V f(x) v = V y f(x)

From X – momentum equation we get pressure is a function of

.

From Y – momentum equation assuming pressure variation constant

We get

using the dimentionless variable

We get = C which cannot be solved using analytical method

Using the velocity component in the Section 1 and substituting in section 2We can calculate flow and pressure variation along the clam shell thrust reversal

Fluent analysis of velocity variation :

Variation of velocities along Y:

Pressure variation along the simplified clam shell: