Consider a normal inviscid imcompressible fluid flow in the

Consider a normal, inviscid, imcompressible fluid flow in the two-dimensional dilation channel shown in the figure. The sloped wall at the channel is straight and the fluid the enlarged section at a velocity at V1. The width at the channel is assume to be w. Find the y-direction velocity v (boundary condition center line y = 0 to v = 0) Find the fluid acceleration ax and ay

Solution

solution:

1)here we have inviscid and incompressible flow through enlarged section means we have to just consider pressure forces acting on fluid element as euler equation.

2)inlet conditions are

width=w1

velocity is=Vo=uoi+voj

as here area is constant means there is no flow along y direction means vo=0

Vo=uoi+oj=uoi

3)consider enlarged section outlet condition to be

V2=u2i+v2j

here agian v2=0

V2=u2i

where geometry condition are for calculating angle m is

outlet widtrh=w2

length=L

tanm=(w2-w1/2L)

4)here consider another section at distance x from inlet

where velocity is V1 and pressure is P1

on applying bernoulliws equation we get that

Po/density*g+Vo^2/2*g+z1=P1/density*g+V1^2/2*g+z2

as same line z1=z2

we have equation as

Po/density*g+Vo^2/2*g=P1/density*g+V1^2/2*g

where velocity is givenas

pressure head loss as =Po/density*g-P1/density*g=H

on putting we get that

V1^2=Vo^2+2*g*H

where at point o

Vo=uo

at point 1

V1=u1i+v1j

where angle

tanm=v1/u1

v1=u1tanm

putting we get that

u1^2(1+tan^2m)=uo^2+2*g*H

we get

u1^2=uo^2+2*g*H/((1+tan^2m))

sec^2m=1+tan^2m

finally we get that

u1=cosm(uo^2+2*g*H)^.5

v1=u1tanm=sinm(uo^2+2*g*H)^.5

here velocity V1=cosm(uo^2+2*g*H)^.5i+sinm(uo^2+2*g*H)^.5j

here let calculate angle

inclined length is S

S=(x^2+y^2)^.5

sinm=y/S

cosm=x/S

let put constant A=(uo^2+2*g*H)^.5

so equation of velocity become

V1=((x/S)i+(y/S)j)A

5)hence velocity in y direction is

v1=(y/S)A=(y/S)(uo^2+2*g*H)^.5

6)where acceleration by eulerian method is

a=axi+ayj

ax=udu/dx+vdu/dy

ay=vdv/dx+vdv/dy

so on solving finally we get that

ax=(-xy^2+xy^2/(x^2+y^2)^2)A=0

ay=(x^2y-x^2y/(x^2+y^2)^2)A=0

hence resultant accelration is

a=axi+ayj=0 m/s2