Irodov 1.334: The Parabola Already Tells the Story – Amit & Deepali Sinha Foundation

Problem Irodov 1.334

A straight tube of length $l$ and radius $R$ carries a steady, fully developed flow of a fluid of density $\rho$ and viscosity $\eta$ . At a distance $r$ from the tube’s axis the flow speed is $v(r) = v_0(1-r^2/R^2)$ , so the speed is a maximum $v_0$ on the axis and falls to zero at the wall $r=R$ . Using this profile, find (a) the volume flow rate $Q$ through any cross section, (b) the total kinetic energy $T$ of the fluid contained in the tube’s whole volume at one instant, (c) the magnitude of the friction force $F_{fr}$ the fluid exerts on the tube’s inner wall, and (d) the pressure difference $\Delta p$ between the tube’s two ends that sustains this flow.

The given parabolic speed profile $v(r)=v_0(1-r^2/R^2)$ , plotted from the axis ( $r=0$ , the maximum speed $v_0$ ) out to the wall ( $r=R$ , no slip); by symmetry the same parabola runs on the other side of the axis.

Given

$l$
$R$
$\rho$
$\eta$
$v(r) = v_0(1-r^2/R^2)$

Find

(a)

Q

; (b)

T

; (c)

F_{fr}

; (d)

\Delta p

Solution

Step 1Volume flow rate, by summing rings of the profile

A thin ring at radius $r$ , width $dr$ , carries the local speed $v(r)$ through its area $2\pi r\,dr$ ; the total flow rate is that sum over the whole cross section.

Q = \int_0^R v(r)\,2\pi r\,dr = 2\pi v_0\int_0^R\left(1-\frac{r^2}{R^2}\right)r\,dr

Expanding the integrand and integrating term by term,

\begin{align*}Q &= 2\pi v_0\int_0^R\left(r-\frac{r^3}{R^2}\right)dr \\ &= 2\pi v_0\left[\frac{r^2}{2}-\frac{r^4}{4R^2}\right]_0^R \\ &= 2\pi v_0\left(\frac{R^2}{2}-\frac{R^2}{4}\right).\end{align*}

Q = \dfrac{1}{2}\pi v_0R^2

(1)

Step 2Kinetic energy stored in the tube, by the same ring sum

Each ring’s kinetic energy is $\rho v^2/2$ per unit volume times its volume $2\pi r\,dr\,l$ ; summing over the cross section and expanding $(1-r^2/R^2)^2$ ,

T = \int_0^R \frac{1}{2}\rho v_0^2\left(1-\frac{r^2}{R^2}\right)^2 2\pi r\,l\,dr = \pi\rho l v_0^2\int_0^R\left(r-\frac{2r^3}{R^2}+\frac{r^5}{R^4}\right)dr.

Integrating term by term and evaluating at $r=R$ ,

\begin{align*}T &= \pi\rho l v_0^2\left[\frac{r^2}{2}-\frac{2r^4}{4R^2}+\frac{r^6}{6R^4}\right]_0^R \\ &= \pi\rho l v_0^2\left(\frac{R^2}{2}-\frac{R^2}{2}+\frac{R^2}{6}\right).\end{align*}

T = \dfrac{1}{6}\pi lR^2\rho v_0^2

(2)

Step 3Wall shear stress, from the velocity gradient at

r=R

The Newtonian shear stress on the fluid at the wall is $\eta$ times the speed gradient there; differentiating the given profile and evaluating at $r=R$ ,

\begin{align*}\frac{dv}{dr} &= -\frac{2v_0}{R^2}\,r \\ \left.\frac{dv}{dr}\right|_{r=R} &= -\frac{2v_0}{R}.\end{align*}

The magnitude of the shear stress on the wall is therefore $\tau_w = \eta\,(2v_0/R)$ ; by Newton’s third law the fluid pulls on the tube with this same stress, in the direction of the flow, over the whole inner surface $2\pi R l$ .

F_{fr} = \tau_w\cdot 2\pi R l = \eta\frac{2v_0}{R}\cdot 2\pi R l

F_{fr} = 4\pi\eta l v_0

(3)

Step 4Pressure difference, from the force balance on the fluid

\Delta p\cdot\pi R^2 = F_{fr} = 4\pi\eta l v_0

Dividing through by $\pi R^2$ ,

\Delta p = \dfrac{4\eta l v_0}{R^2}

(4)

Check

The mean speed implied by (1) is $Q/(\pi R^2)=v_0/2$ , exactly half the axial maximum, the standard Poiseuille fact; and the boxed $\Delta p$ is confirmed independently by a force balance on the inner core of radius $r$ alone, whose pressure push $\Delta p\,\pi r^2$ must match the shear $\eta(2v_0r/R^2)$ acting over that core’s own surface $2\pi rl$ , giving $\Delta p = 4\eta lv_0/R^2$ for every $r$ and not merely at the wall, so the given profile and this pressure gradient agree right across the tube. All four results vanish together as $v_0\to0$ , as they must for a fluid at rest.

Remark

The kinetic energy (2) has a $1/6$ where a fluid moving uniformly at the mean speed $v_0/2$ would give only a $1/8$ : a ring near the axis, moving faster than the mean, contributes more extra kinetic energy than a ring near the wall gives up, because kinetic energy grows as the square of speed. Any profile peaked at the center carries more kinetic energy than its own mean-flow estimate would suggest, for this same reason.