Question 1 of 6: A1 — Laminar flow of water between parallel plates

Question 1: A1 — Laminar flow of water between parallel plates

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Fully-developed, steady laminar flow of water (constant $\rho$, $\mu$) between infinite parallel plates a distance $2\delta$ apart; $y$ measured from the mid-plane, so the walls are at $y=\pm\delta$. Flow is driven along $x$ by a constant pressure gradient $dP/dx$. For part (b), inlet temperature $T_{in}$, wall temperature $T_s$.

Find. (a) the velocity profile $u(y)$; (b) the governing PDE for $T$ together with its boundary conditions.

Part (a) — velocity profile

1. Reduce the Navier–Stokes equation. For steady, fully-developed unidirectional flow, $\vec{u}=u(y)\,\hat{\imath}$ with $u_y=u_z=0$; continuity then gives $\partial u/\partial x=0$ automatically. The $x$-momentum equation (Table A.2) loses its transient, inertial and $z$-terms, leaving a balance of pressure and viscous stress: $$0=-\frac{1}{\rho}\frac{\partial P}{\partial x}+\nu\frac{\partial^2 u}{\partial y^2}\;\Longrightarrow\;\mu\frac{d^2u}{dy^2}=\frac{dP}{dx}.$$
2. Integrate twice. With $dP/dx$ constant across the gap, $\dfrac{du}{dy}=\dfrac{1}{\mu}\dfrac{dP}{dx}\,y+C_1$ and $u=\dfrac{1}{2\mu}\dfrac{dP}{dx}\,y^2+C_1 y+C_2.$
3. Apply symmetry and no-slip. Symmetry about the mid-plane requires $du/dy=0$ at $y=0$, so $C_1=0$. No-slip at either wall, $u(\pm\delta)=0$, gives $C_2=-\dfrac{1}{2\mu}\dfrac{dP}{dx}\,\delta^2.$
4. Assemble the profile. Substituting the constants, $$u(y)=-\frac{1}{2\mu}\frac{dP}{dx}\left(\delta^2-y^2\right)=-\frac{\delta^2}{2\mu}\frac{dP}{dx}\left(1-\left(\frac{y}{\delta}\right)^2\right).$$ The centre-line value $u(0)=-\dfrac{\delta^2}{2\mu}\dfrac{dP}{dx}$ is by definition $u_{max}$ (positive, since a driving flow has $dP/dx<0$), giving the required result. $\boxed{\,u=u_{max}\left(1-\left(\dfrac{y}{\delta}\right)^2\right)\,}$

Part (b) — temperature-profile equation

The flow is now thermally developing: the fluid enters at $T_{in}$ and is heated (or cooled) toward the wall value $T_s$ as it moves downstream, so $T=T(x,y)$. Applying the energy equation for an incompressible medium (Table A.3) in rectangular coordinates and discarding the terms that vanish gives the governing PDE.

1. Reduce the energy equation. Steady state removes $\partial T/\partial t$; with $u_y=u_z=0$ only the axial convection term $u_x\,\partial T/\partial x$ survives on the left. On the right, axial conduction $\partial^2T/\partial x^2$ is negligible next to transverse conduction $\partial^2T/\partial y^2$ (large Péclet number), and there is no volumetric generation. With constant $k$: $$\rho\hat{C}_P\,u(y)\,\frac{\partial T}{\partial x}=k\,\frac{\partial^2 T}{\partial y^2}.$$
2. Insert the known velocity profile. Substituting $u(y)=u_{max}\!\left(1-(y/\delta)^2\right)$ from part (a) gives the differential equation describing the temperature field: $$\boxed{\;\rho\hat{C}_P\,u_{max}\!\left(1-\left(\frac{y}{\delta}\right)^2\right)\frac{\partial T}{\partial x}=k\,\frac{\partial^2 T}{\partial y^2}\;}$$ This is the classic Graetz (thermal-entry) problem; as instructed we stop here without solving.
3. Boundary and inlet conditions. Three conditions close the second-order-in-$y$, first-order-in-$x$ equation:
   • Inlet: $T(0,y)=T_{in}$ for all $y$.
   • Both walls isothermal: $T(x,+\delta)=T(x,-\delta)=T_s$.
   • Equivalently, by symmetry the centre-line has zero transverse gradient: $\left.\dfrac{\partial T}{\partial y}\right|_{y=0}=0$ (may replace one wall condition).