Question 1.13: Fluid flowing over a flat plate has dynamic viscosity of 0.8......
Fluid flowing over a flat plate has dynamic viscosity of 0.8 Ns/m². Find the velocity gradient and intensity of shear stress at y = 0 m, y = 0.5 m, y = 1 m and y = 1.5 m if the velocity distribution is parabolic and parabola is having vertex at A and origin at B, as shown in the Figure 1.12.
Given,
Dynamic viscosity of fluid, μ = 0.8 Ns/m²
Velocity distribution is parabolic.
Since, velocity distribution is parabolic. Therefore, let
u = ay² + by + c (1.24)
\frac{d u}{d y}=2 a y+b (1.25)
Now, considering boundary condition, at y = 0, u = 0
c = 0
at y = 1.5, u = 5 m/s
Substituting these values in Eq. (1.24),
5 = a(1.5)² + b(1.5) + c
5 = 2.25a + 1.5b + c (1.26)
at y = 1.5, du/dy = 0
From Eq. (1.25)
0 = 2a(1.5) + b
b = —3a (1.27)
Finally, substituting the value of b and c in Eq. (1.26)
5 = 2.25a + 1.5 (-3a) + 0
5 = 2.25a — 4.5a
Therefore,
a = —2.11; b = 0.667
Velocity profile will be
u = —2.22y² + 6.667y
\frac{d u}{d y}=-4.44 y+6.667
Now, according to Newton’s law of viscosity
\begin{aligned} \tau & =\mu \frac{d u}{d y} \\ \tau & =0.8 \times(-44 y+6.667) \end{aligned}
Therefore,
at y = 0 m, τ = 0.8(0 + 6.667) = 5.336 Pa
at y = 0.5 m, τ = 0.8[-4.44(0.5) + 6.667] = 3.557 Pa
at y = 1 m, τ = 0.8[-4.44(1) + 6.667] = 1.78 Pa
at y = 1.5 m, τ = 0.8[-4.44(1.5) + 6.667] = 0 Pa