Question

The velocity profile of a fluid over a flat plate is parabolic in nature. At a distance of $15\text{cm}$ from the plate, the velocity of fluid is maximum i.e $100\text{cm/s}$. Calculate the shear stress at a distance of $10\text{cm}$ from the plate if the viscosity of the fluid is $10\text{poise}$.

• A. $4.53{\text{N/m}}^2$
• B. $2.26{\text{N/m}}^2$
• C. $2.35{\text{N/m}}^2$
• D. None of these

Answer 1

The correct option is A $4.53{\text{N/m}}^2$

Viscosity of fluid $\left(\mu \right)=\frac{10}{10}=1{\text{Ns/m}}^2$
The velocity profile given is parabolic, so the equation of velocity profile will be
$u=a{y}^2+by+c$
where $a,b\text{and}c$ are constants.
The values of $a,b\text{and}c$ will be determined from the following boundary conditions.
(a) at $y=0;u=0$
(b) at $y=15\text{cm};u=100\text{cm/s}$
(c) at $y=15\text{cm};\frac{du}{dy}=0$
From (a), we get $c=0$
From (b)
$100=a(15{)}^2+b(15)$ ..........(1)
From (c)
$\frac{du}{dy}=2ay+b$
$\Rightarrow 2a\left(15\right)+b=0$..............(2)
Solving Eq.(1) and (2), we get
$a=-0.44$ and $b=13.33$
So velocity profile become
$u=-0.44y^2+13.33y$
From formula,
shear stress $\left(\tau \right)=\mu \left(\frac{du}{dy}\right)$
At $y=10\text{cm}$,
$\tau =\mu {\left(\frac{du}{dy}\right)}_{y=10}=1\times (13.33-2\times 0.44\times 10)=4.53{\text{N/m}}^2$
Hence, option $\left(a\right)$ is the correct answer.