UPSC 2014 Maths Optional Paper 2 Q8c — Step-by-Step Solution

20 marks · Section B

Navier-Stokes equation for a viscous fluid · Mechanics & Fluid Dynamics · asked 3× in 13 yrs · Read the full method →

Question

Find Navier–Stokes equation for a steady laminar flow of a viscous incompressible fluid between two infinite parallel plates.

Technique

Reduce full Navier-Stokes by symmetries (1D flow, steady); integrate the resulting 2nd-order ODE; no-slip BCs at the plates determine the constants.

Solution

Setup. Two infinite stationary parallel plates at $y=\pm h$ . Steady incompressible viscous flow driven by a constant pressure gradient in the $x$ -direction. Assume flow is in the $x$ -direction with $\vec V=(u(y),0,0)$ — i.e., laminar parallel flow.

Step 1 — Reduce Navier-Stokes

Full Navier-Stokes:

\rho\!\left[\dfrac{\partial\vec V}{\partial t}+(\vec V\cdot\nabla)\vec V\right]=-\nabla p+\mu\nabla^{2}\vec V+\rho\vec g.

Apply assumptions:

Steady: $\partial/\partial t=0$ .
Continuity: $\nabla\cdot\vec V=\partial u/\partial x=0$ ✓ (since $u$ depends only on $y$ ).
Convective term: $(\vec V\cdot\nabla)\vec V=u\,\partial_x\vec V=0$ (no $x$ -dependence).
Gravity: assume horizontal plates, no $y$ - or $z$ -component of velocity; gravity provides only a static pressure (we’ll absorb it).

The reduced Navier-Stokes equations:

$x$ -component: $0=-\dfrac{\partial p}{\partial x}+\mu\dfrac{d^{2}u}{dy^{2}}$ .
$y$ -component: $0=-\dfrac{\partial p}{\partial y}$ , so $p=p(x,z)$ .
$z$ -component: $0=-\dfrac{\partial p}{\partial z}-\rho g$ (with $g$ along $-z$ ); gives a static pressure contribution.

Step 2 — Pressure gradient

Since $p$ has no $y$ -dependence and the dynamic part depends only on $x$ :

\dfrac{\partial p}{\partial x}=-P\quad(\text{const}),

where $P>0$ for flow in the $+x$ direction. The minus sign reflects that pressure decreases in the flow direction.

Step 3 — Velocity ODE

\mu\dfrac{d^{2}u}{dy^{2}}=-P\;\Longrightarrow\;\dfrac{d^{2}u}{dy^{2}}=-\dfrac{P}{\mu}.

Integrate twice:

\dfrac{du}{dy}=-\dfrac{Py}{\mu}+C_1,\qquad u(y)=-\dfrac{Py^{2}}{2\mu}+C_1 y+C_2.

Step 4 — Apply no-slip boundary conditions

At the plates $y=\pm h$ , the fluid is at rest: $u(\pm h)=0$ .

$u(h)=-Ph^{2}/(2\mu)+C_1 h+C_2=0$ .
$u(-h)=-Ph^{2}/(2\mu)-C_1 h+C_2=0$ .

Subtracting: $2C_1 h=0\Rightarrow C_1=0$ .

Adding: $-Ph^{2}/\mu+2C_2=0\Rightarrow C_2=Ph^{2}/(2\mu)$ .

Step 5 — Velocity profile

Answer

\boxed{\;u(y)=\dfrac{P}{2\mu}(h^{2}-y^{2}).\;}