The velocity distribution of a flow over a plate is parabolic with vertex 30 cm from plate, where velocity is 180 cm/s. If viscosity of fluid is 0.9 N.s/`m^2`, find velocity gradients and shear stress at distances of 0,15 and 30 cm from plate.

SOLUTION:

Given,

Distance of the vertex from the plate =30 cm= 0.3 m

Velocity at the vertex= 180 cm/s =1.8 m/s

Viscosity of the fluid=0.9 Ns/`m^2`

We know, the equation of the velocity profile, which is parabolic, is given by,

`u=ay^2+by+c`... (i)

Where, a,b,and c are constants

 The values of these constants are found from the following boundary conditions.

• At y=0, u=0

• At y=30 cm, u= 180 cm/s

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SOLUTION:

Given,

Distance of the vertex from the plate =30 cm= 0.3 m

Velocity at the vertex= 180 cm/s =1.8 m/s

Viscosity of the fluid=0.9 Ns/`m^2`

We know, the equation of the velocity profile, which is parabolic, is given by,

`u=ay^2+by+c`... (i)

Where, a,b,and c are constants

 The values of these constants are found from the following boundary conditions.

• At y=0, u=0

• At y=30 cm, u= 180 cm/s

• At y=30 cm, `(du)/(dy)=0`

From (i),

`0=0+0+c`

or, c=0

Similarly, at y=30, u=180, so

`u=ay^2+by+c`

or, `180=a*30^2+30b`

or, `180=900a+30b`... (ii)

Similarly,

`(du)/(dy)=2a y+b`

or, `2a*30+b=0`

or, `60a+b=0`... (iii)

From (ii) and (iii), we have,

`a=-0.2`

`b=12`

`c=0`

Thus, at y=0,

• Velocity gradient,`(du)/(dy)=-0.4y+12=12`per sec

• Shear stress, `tau=0.9*12=10.8`N/`m^2`

Also, at y=15 cm,

• Velocity gradient,`(du)/(dy)=-0.4y+12=6`per sec

• Shear stress, `tau=0.9*6=5.4`N/`m^2`

And, at y=30 cm,

• Velocity gradient,`(du)/(dy)=-0.4y+12=0`per sec

• Shear stress, `tau=0.9*0=0`N/`m^2`