Question

If the velocity profile of a fluid over a flat plate is parabolic with free stream velocity of 120 cm/s occurring at 20 cm from the plate. Find the velocity gradient and shear stress at a distance of 10 cm from plate. dynamic viscosity = 8.5 Poise

Answer 1

Answer:

Explanation:

Given:

Distance of vertex from plate =20cm

Velocity at vertex =120cm/sec

Dynamic viscosity = 8.5 poise

                               =8.5/10

                               =0.85 Ns/m2

Solution:

Equation of parabolic velocity profile

          U = ay2 + by + c   -----------(1)

          where a, b, c are constants and their values are determined by 3

          boundary condition

1. boundary condition first:  At y=0, U=0

   therefore equation first becomes

   0=a (0) +b (0) + c

        c=0

2. boundary condition second: At y=20cm, U=120cm/sec.

    Therefore equation first becomes

     120=a (20)2 + b(20)

     120=400a + 20b   ----------- (2)

3. Boundary conditions third: At y=20cm, dU/dy=0 (since the value is maximum  and constant therefore its differentiation is zero)

Differentiating (1) with respect to y

  dU/dy=2ay + b

  0= 2 a 20 +b

  0= 40a + b

  -40a= b    ---------------(3)

  Substituting (3) in (2)

  120= 400a + 20 (-40a)

  120=400 -800a

  120=-400a

  a= -0.3

 Now substituting the value of a in (2)

  b=12  

Therefore, substituting the values of a, b, c in equation (1)

U= -0.3y2 +12y

Velocity gradient:

differentiate U with respect to y

dU/dy = -0.3 y 2 +12  

Velocity gradient at y=10cm from the plate:

dU/dy = -0.3*2*10 +12

           = -6 + 12

           =6 /sec.    

 Shear stress:

            t= u dU/dy          --- where u = coefficient of viscosity

              = 0.85*6

              =5.1 N/m2    

Answer 2

Given:

           velocity at vertex =$120$$cm$

           distance from plate = $20cm$

            Viscosity = $8.5poise$ = $0.85Ns/m^{2}$

Formula:

         Equation for parabolic velocity profile  $U = ay^{2} + by + c$   __(1)

         Shear stress: $T=u\frac{du}{dy}$

Solution:

• boundary condition At y=20cm, U=120cm/sec:

                           $120=a (20)2 + b(20)$  (from eq 1 )

                            $120=400a + 20b$ __(2)

• Boundary conditions At y=20cm, dU/dy=0:

           * Differentiating (1) with respect to y

               [tex] dU/dy=2ay + b [/tex]

                [tex] 0= 40a + b [/tex]

                [tex] -40a= b [/tex]___(3)

• now equating equation 3 and 2 :

                     [tex]120= 400a + 20 (-40a) 120=-400a a= -0.3[/tex]

       subtituting "a" in eq 2:

                      $b=12$

• Velocity gradient at y=10cm from the plate:

                       [tex] dU/dy = -0.3*2*10 +12 =6 /sec. [/tex]

• shear force : $T=u\frac{du}{dy}$

                        [tex] = 0.85*6 =5.1 N/m2 [/tex]

hence, the shear force and velocity gradient at a distance of 10 cm is obtained .