Irodov 1.332: The Logarithm Between Two Cylinders – Amit & Deepali Sinha Foundation

Problem Irodov 1.332

A long cylinder of radius $R_1$ slides along its own axis at a constant velocity $v_0$ inside a fixed, coaxial cylinder of radius $R_2$ , with viscous liquid filling the gap between them. Assuming the flow is laminar, find how the liquid’s velocity depends on the distance $r$ from the shared axis.

An axial section through the two coaxial cylinders. The inner cylinder of radius R_1 slides at v_0 along the dashed shared axis, inside the fixed outer cylinder of radius R_2; viscous liquid fills the shaded gap between them. — An axial section through the two coaxial cylinders. The inner cylinder of radius $R_1$ slides at $\vect v_0$ along the dashed shared axis, inside the fixed outer cylinder of radius $R_2$ ; viscous liquid fills the shaded gap between them.

Given

$R_1, R_2$ (cylinder radii)
$v_0$ (inner cylinder’s speed)

Find

Find

v(r)

, the liquid’s velocity for

R_1\le r\le R_2

Solution

Step 1The governing equation for steady axial shear flow

The flow is steady, fully developed, and driven purely by the moving inner wall (no imposed pressure gradient along the axis), so the net viscous force on every fluid annulus must vanish. For an axisymmetric velocity $v(r)$ along the axis, the viscous force on a cylindrical surface of radius $r$ and length $L$ is the shear stress $\eta\,dv/dr$ times that surface’s area,

F(r) = \eta\frac{dv}{dr}\,2\pi rL.

The annulus between $r$ and $r+dr$ feels $F(r+dr)$ on its outer face and $-F(r)$ on its inner one, and with no acceleration and no pressure push those two must cancel:

\begin{align*}F(r+dr)-F(r) &= \frac{dF}{dr}\,dr \\ &= 2\pi\eta L\,\frac{d}{dr}\left(r\frac{dv}{dr}\right)dr = 0.\end{align*}

Dividing by $2\pi\eta L\,r\,dr$ leaves the governing equation for $v(r)$ :

\frac{1}{r}\frac{d}{dr}\left(r\frac{dv}{dr}\right) = 0. \tag{1}

Step 2Integrate twice

Multiplying (1) through by $r$ and integrating once removes the outer derivative, leaving a first-order equation for $v$ ; integrating again gives $v(r)$ itself.

\begin{align*}r\frac{dv}{dr} &= C_1 \\ \frac{dv}{dr} &= \frac{C_1}{r} \\ v(r) &= C_1\ln r + C_2.\end{align*}

Step 3Apply the two boundary conditions

No-slip at each surface: the liquid takes on that surface’s own velocity, $v(R_1)=v_0$ (the inner cylinder’s own speed) and $v(R_2)=0$ (the outer cylinder is fixed):

\begin{align*}C_1\ln R_1 + C_2 &= v_0, \\ C_1\ln R_2 + C_2 &= 0.\end{align*}

Subtracting the second equation from the first eliminates $C_2$ and leaves $C_1$ alone:

C_1\ln R_1 - C_1\ln R_2 = v_0 \quad\Longrightarrow\quad C_1 = \frac{v_0}{\ln(R_1/R_2)}. \tag{2}

The second boundary equation then gives

C_2 = -C_1\ln R_2. \tag{3}

Step 4Assemble

v(r)

Substituting (3) for $C_2$ , the profile becomes

v(r) = C_1\ln r - C_1\ln R_2 = C_1\ln\frac{r}{R_2},

and substituting (2) for $C_1$ gives the velocity profile.

v = v_0\dfrac{\ln(r/R_2)}{\ln(R_1/R_2)}

(4)

The profile just assembled, plotted across the gap: $v_0$ at the inner wall $R_1$ , falling logarithmically to zero at the outer wall $R_2$ (illustrative axis values shown; the true radii are symbolic).

Check

At $r=R_1$ the ratio of logarithms is exactly $1$ , giving $v=v_0$ ; at $r=R_2$ the numerator vanishes, giving $v=0$ . Both boundary conditions are recovered directly from the closed form, as they must be.

Remark

The profile is logarithmic rather than linear for exactly the reason curved Couette flow always differs from flat Couette flow: the viscous stress on a cylindrical shell falls off as $1/r$ even though the total axial drag force it carries stays constant, so equal increments of drag are spread over ever more circumference as $r$ grows, compressing the velocity change into the region nearest the moving wall. Rotational Couette viscometers share this same two-coaxial-cylinder geometry, but not this profile: spinning one cylinder inside the other gives $v_\theta=Ar+B/r$ , not a logarithm, because in that case it is the torque, not the force, that stays the same on every cylindrical shell. The logarithmic profile found here belongs specifically to axial sliding, not to rotation.