Question 2 of 6: A2 — Velocity distribution for flow between parallel plates
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. EGBC 16-CHEM-B1 Transport Phenomena, December 2017, 3 hours, open-book (one textbook permitted, no loose notes). Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section, and only the first four in the answer book are marked. All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species-continuity tables) in rectangular, cylindrical and spherical coordinates; these are quoted rather than re-derived.
Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, the transfer analogies, boundary-layer and film coefficients; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — internal-flow convection and the sphere/flat-plate correlations; F. M. White, Fluid Mechanics (McGraw-Hill) — the three-reservoir branching-pipe problem; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.
Question 2: A2 — Velocity distribution for flow between parallel plates (25 marks)
Given. Steady ($\partial/\partial t=0$), fully developed, laminar, constant-property incompressible flow between fixed parallel plates a distance $b$ apart; the only non-zero velocity is $u_x=u(y)$; $\partial P/\partial y=0$; body forces neglected; mean velocity $U_m$.
Find. The velocity profile $u(y)$ across the gap, expressed both in terms of the pressure gradient and in terms of the mean velocity $U_m$.
Fig. A2: Plane Poiseuille flow. A constant axial pressure gradient drives a parabolic velocity profile between fixed plates; the maximum (centreline) velocity is $\tfrac32$ the mean, and the wall shear is set by the profile slope at $y=0,b$.
Approach. This is a derivation: reduce the $x$-momentum (Navier–Stokes) equation using the fully-developed, unidirectional assumptions to an ordinary differential equation, integrate twice with the no-slip conditions, then evaluate the mean velocity to eliminate the pressure gradient in favour of $U_m$.
Reduce the momentum equation. With $u_y=u_z=0$ and $u_x=u(y)$ only, continuity is satisfied identically and every inertial (convective) term vanishes. The steady $x$-component of Navier–Stokes (Table A.2) collapses to a balance of pressure and viscous forces: $$0=-\frac{1}{\rho}\frac{\partial P}{\partial x}+\nu\frac{d^2u}{dy^2}\qquad\Longrightarrow\qquad \mu\frac{d^2u}{dy^2}=\frac{dP}{dx}.$$ Since $P=P(x)$ only, the right side is a constant.
Apply no-slip at both walls. With $u(0)=0$ and $u(b)=0$ (plates fixed, origin on the lower plate): $C_2=0$, and $0=\frac{1}{2\mu}\frac{dP}{dx}b^2+C_1b\Rightarrow C_1=-\frac{b}{2\mu}\frac{dP}{dx}$. Hence $$\boxed{\,u(y)=\frac{1}{2\mu}\frac{dP}{dx}\big(y^2-by\big)=-\frac{1}{2\mu}\Big(-\frac{dP}{dx}\Big)\big(by-y^2\big).}$$ Because $dP/dx<0$ for flow in $+x$, $u(y)>0$ across the gap — a downward-opening parabola pinned to zero at each wall.
Relate the pressure gradient to $U_m$. The mean velocity is $$U_m=\frac{1}{b}\int_0^b u\,dy=\frac{1}{b}\cdot\frac{1}{2\mu}\frac{dP}{dx}\int_0^b(y^2-by)\,dy=\frac{1}{2\mu b}\frac{dP}{dx}\Big(\frac{b^3}{3}-\frac{b^3}{2}\Big)=-\frac{b^2}{12\mu}\frac{dP}{dx}.$$ So $\dfrac{1}{2\mu}\dfrac{dP}{dx}=-\dfrac{6U_m}{b^2}$.
Velocity profile in terms of $U_m$. Substituting, $$\boxed{\,u(y)=6\,U_m\,\frac{y}{b}\Big(1-\frac{y}{b}\Big).}$$ The centreline value ($y=b/2$) is the maximum, $$u_{max}=u\!\left(\tfrac{b}{2}\right)=\tfrac{3}{2}U_m=\frac{b^2}{8\mu}\Big(-\frac{dP}{dx}\Big).$$
