A1 — Average velocity of laminar pipe flow: 23-Chem-B1 Transport Phenomena, May 2015, Q1

Question 1: A1 — Average velocity of laminar pipe flow (25 marks)

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. Steady, fully-developed, incompressible laminar flow of a Newtonian fluid ($\rho$, $\mu$ constant) in a horizontal circular tube of radius $R$ and length $L$; axial pressure falls from $p_0$ at inlet to $p_L$ at outlet; gravity negligible; no accumulation.

Find. The cross-sectional average (bulk) velocity $\langle u\rangle$ in terms of $p_0,\,p_L,\,R,\,\mu,\,L$.

Fig. A1: Fully-developed laminar flow in a horizontal tube. A momentum balance on the thin cylindrical annulus between $r$ and $r+dr$ links the viscous stress to the axial pressure gradient, giving the parabolic profile $u(r)$.

Approach. Balance axial momentum on a differential annular control volume to obtain the stress distribution, insert Newton’s law of viscosity to get $u(r)$, then area-average the profile.

Momentum balance on the annular control volume. For fully-developed steady flow the convective momentum in and out cancel, leaving pressure and viscous forces. On the annulus of length $L$, inner radius $r$ and thickness $dr$, force balance in $z$ gives $\dfrac{d}{dr}\big(r\,\tau_{rz}\big)=\dfrac{(p_0-p_L)}{L}\,r.$
Integrate for the stress. Integrating once, $r\,\tau_{rz}=\dfrac{(p_0-p_L)}{2L}r^2+C_1.$ The stress must be finite on the axis, so $C_1=0$ and $\tau_{rz}=\dfrac{(p_0-p_L)}{2L}\,r.$
Insert Newton’s law of viscosity. With $\tau_{rz}=-\mu\,\dfrac{du}{dr}$, $$-\mu\frac{du}{dr}=\frac{(p_0-p_L)}{2L}\,r\quad\Longrightarrow\quad \frac{du}{dr}=-\frac{(p_0-p_L)}{2\mu L}\,r.$$
Integrate and apply no-slip. $u(r)=-\dfrac{(p_0-p_L)}{4\mu L}r^2+C_2$; no-slip $u(R)=0$ fixes $C_2$, giving the parabolic profile $$u(r)=\frac{(p_0-p_L)R^2}{4\mu L}\left[1-\left(\frac{r}{R}\right)^2\right],\qquad u_{max}=\frac{(p_0-p_L)R^2}{4\mu L}.$$
Area-average the profile. The bulk velocity weights $u(r)$ by the annular area $2\pi r\,dr$: $$\langle u\rangle=\frac{\displaystyle\int_0^R u(r)\,2\pi r\,dr}{\pi R^2}=\frac{(p_0-p_L)R^2}{4\mu L}\cdot\frac{2}{R^2}\int_0^R\!\left(r-\frac{r^3}{R^2}\right)dr=\frac{(p_0-p_L)R^2}{4\mu L}\cdot\frac12.$$ Hence $$\boxed{\;\langle u\rangle=\frac{(p_0-p_L)R^2}{8\mu L}=\frac{u_{max}}{2}\;}$$ the classic Hagen–Poiseuille result: the average velocity is exactly half the centre-line maximum.

Quantity	Result
Velocity profile	$u(r)=\dfrac{(p_0-p_L)R^2}{4\mu L}\!\left[1-(r/R)^2\right]$
Maximum velocity	$u_{max}=\dfrac{(p_0-p_L)R^2}{4\mu L}$
Average velocity	$\langle u\rangle=\dfrac{(p_0-p_L)R^2}{8\mu L}=u_{max}/2$

Concept & Theory

Topic. Hagen–Poiseuille flow derived from a differential momentum balance.

Key relations. Annular force balance $\tfrac{d}{dr}(r\tau_{rz})=\tfrac{(p_0-p_L)}{L}r$; Newton’s law $\tau_{rz}=-\mu\,du/dr$; area-average $\langle u\rangle=\tfrac{1}{\pi R^2}\int u\,2\pi r\,dr$.

Why this works. Fully-developed flow makes the convective terms vanish, so the only balance is between the imposed axial pressure gradient and the radial gradient of viscous stress. That linear stress field integrates to a parabolic velocity, and the $2\pi r$ area weighting — which favours the faster fluid near the centre but multiplies by the smaller area there — halves the peak to give the mean.

Common pitfall. Forgetting the finiteness condition on the axis (setting $C_1=0$) and instead trying to impose zero stress at the wall; or number-averaging $u(r)$ over radius instead of area-averaging, which wrongly gives $u_{max}/3$ rather than $u_{max}/2$.

Source. Bird, Stewart & Lightfoot, Transport Phenomena, §2.3 (flow through a circular tube); Welty et al., Ch. 8.

Question 1 of 6: A1 — Average velocity of laminar pipe flow

Question 1: A1 — Average velocity of laminar pipe flow (25 marks)