A2 — Average-to-maximum velocity for the 1/7-power law: 23-Chem-B1 Transport Phenomena, May 2018, Q2

Question 2: A2 — Average-to-maximum velocity for the 1/7-power law (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. A fully developed turbulent profile in a round pipe of radius $R$: $V_r=V_{max}\big[(R-r)/R\big]^{1/7}$, valid to $Re\approx10^5$; incompressible (constant $\rho$). The centreline ($r=0$) carries $V_{max}$; the wall ($r=R$) carries zero.

Find. The ratio $V_{av}/V_{max}$, i.e. an expression for the flow-average velocity in terms of the peak velocity.

Fig. A2: The turbulent 1/7-power profile is far blunter than the laminar parabola — most of the cross-section moves near $V_{max}$, so the area-averaged velocity is a high fraction ($\tfrac{49}{60}\approx0.82$) of the peak, versus $\tfrac12$ for laminar flow.

Approach. The average velocity is the volumetric flow divided by the cross-sectional area; integrate the given profile over annular rings $dA=2\pi r\,dr$ and divide by $\pi R^2$, using the substitution $s=(R-r)/R$ to reduce the integral to standard powers.

Definition of the flow average. By continuity for an incompressible fluid, $$V_{av}=\frac{Q}{A}=\frac{1}{\pi R^2}\int_0^R V_r\,(2\pi r)\,dr=\frac{2V_{max}}{R^2}\int_0^R\Big(\frac{R-r}{R}\Big)^{1/7}r\,dr.$$
Substitute to standard powers. Let $s=(R-r)/R$, so $r=R(1-s)$ and $dr=-R\,ds$; the limits map $r:0\to R$ onto $s:1\to0$. Then $r\,dr=-R^2(1-s)\,ds$ and $$\int_0^R\Big(\frac{R-r}{R}\Big)^{1/7}r\,dr=R^2\int_0^1 s^{1/7}(1-s)\,ds.$$
Evaluate the beta-type integral. $$\int_0^1\!\big(s^{1/7}-s^{8/7}\big)ds=\frac{1}{\tfrac87}-\frac{1}{\tfrac{15}{7}}=\frac{7}{8}-\frac{7}{15}=\frac{49}{120}.$$
Assemble the ratio. Substituting back, the $R^2$ cancels: $$V_{av}=\frac{2V_{max}}{R^2}\cdot R^2\cdot\frac{49}{120}=\frac{98}{120}V_{max}\ \Longrightarrow\ \boxed{\dfrac{V_{av}}{V_{max}}=\dfrac{49}{60}\approx0.817.}$$

Quantity	Result
Profile integral	$\displaystyle\int_0^1 s^{1/7}(1-s)\,ds=\frac{49}{120}$
Average / maximum velocity	$V_{av}=\dfrac{49}{60}V_{max}\approx0.817\,V_{max}$
(Contrast: laminar parabola)	$V_{av}=\tfrac12 V_{max}$

Concept & Theory

Topic. Area-averaging the empirical 1/7-power turbulent velocity profile.

Key relations. Flow average $V_{av}=\frac{1}{\pi R^2}\int_0^R V_r\,2\pi r\,dr$; for a general $1/n$ law $V_{av}/V_{max}=\dfrac{2n^2}{(n+1)(2n+1)}$, which gives $\tfrac{49}{60}$ at $n=7$.

Why this works. The average is defined by equal volumetric flow, so it must be weighted by the annular area $2\pi r\,dr$ — not a simple average over the radius. Turbulent mixing flattens the core, so the weighting keeps $V_{av}$ close to $V_{max}$; only the thin near-wall region drags it below unity.

Common pitfall. Averaging $V_r$ over $r$ (giving $\tfrac{7}{8}V_{max}$) instead of over area — forgetting the $r$ in $2\pi r\,dr$. The correct area weighting yields $\tfrac{49}{60}$, not $\tfrac78$.

Source. Bird, Stewart & Lightfoot, Transport Phenomena, §5.1 (turbulent flow in tubes; the 1/7-power law); Welty et al., Ch. 12.

Question 2 of 6: A2 — Average-to-maximum velocity for the 1/7-power law

Question 2: A2 — Average-to-maximum velocity for the 1/7-power law (25 marks)