Irodov 1.315: Streamlines Crowd on a Sharp Turn – Amit & Deepali Sinha Foundation

Problem Irodov 1.315

An ideal (inviscid, incompressible) fluid flows steadily along a flat tube of constant cross-section that lies entirely in a horizontal plane and is bent through a right angle: fluid enters along a straight run, turns the corner, and leaves along a second straight run perpendicular to the first. Point $1$ sits well up the entry run, far enough ahead of the corner that the streamlines there are still straight and parallel, a short distance from the wall that becomes the short (inner) side of the turn; point $2$ sits just past the corner in the exit run, the same short distance from that same inner wall, where the flow is still rounding the bend. Are the pressure and the flow speed at point $1$ equal to those at point $2$ ? What shape do the streamlines take through the bend?

Top view of the bent channel: fluid enters at the top and leaves at the right. The sharp corner is on the inner wall of the turn; points 1 and 2 lie equally far from that wall, before and after the bend. — Top view of the bent channel: fluid enters at the top and leaves at the right. The sharp corner is on the inner wall of the turn; points $1$ and $2$ lie equally far from that wall, before and after the bend.

Given

steady flow of an ideal fluid, constant cross-section, right-angle bend, all in one horizontal plane

Find

Compare

p_1,v_1

with

p_2,v_2

; describe the shape of the streamlines

Solution

Step 1Curved flow needs a pressure gradient across the stream

Consider a fluid element traveling along a streamline of local radius of curvature $R$ at speed $v$ . Newton’s second law applied in the direction normal to the flow, pointing away from the local center of curvature, requires a net inward force per unit volume to supply the centripetal acceleration, and only the pressure gradient can supply it in the horizontal plane (gravity has no horizontal component here). Write $n$ for the coordinate measured outward from the local center of curvature and take a thin slab of fluid of face area $S$ between $n$ and $n+dn$ . The pressure on its outer face exceeds that on its inner face, so the net inward push is the pressure difference times $S$ , and it must equal the slab’s mass $\rho S\,dn$ times the centripetal acceleration $v^2/R$ :

\bigl[p(n+dn)-p(n)\bigr]S = \frac{\partial p}{\partial n}\,S\,dn = \left(\rho S\,dn\right)\frac{v^2}{R}

Dividing by $S\,dn$ , pressure rises with $n$ and falls toward the center:

\dfrac{\partial p}{\partial n} = \dfrac{\rho v^2}{R}

(1)

Step 2Locate the two points relative to the curvature

In the straight run upstream of the bend, where point $1$ sits, the streamlines are (to a good approximation) still parallel and $R\to\infty$ : by (1) the pressure is essentially uniform across that cross-section, unaffected yet by the turn. Immediately after the bend, where point $2$ sits, the fluid has just rounded the corner: a streamline that stays close to the inner wall must turn most sharply there, so its local radius of curvature $R$ is smallest exactly on the inner side, precisely where point $2$ is measured.

Step 3Bernoulli along the near-wall streamline

Follow the streamline that runs close to the inner wall from point $1$ , through the bend, to point $2$ . The flow is steady, ideal and confined to one horizontal plane, so Bernoulli’s equation applies with no elevation term:

p_1+\frac{1}{2}\rho v_1^2 = p_2+\frac{1}{2}\rho v_2^2 \tag{2}

The small radius of curvature just past the bend, at point $2$ , makes (1) drive $\partial p/\partial n>0$ strongly outward there, which is exactly a pressure minimum on the inner side compared with the uncurved approach at point $1$ : $p_2<p_1$ . Substituting this inequality into (2) and solving for the speeds,

p_1 - p_2 = \frac{1}{2}\rho\left(v_2^2-v_1^2\right) > 0 \;\Longrightarrow\; v_2^2 > v_1^2 \;\Longrightarrow\; v_2 > v_1.

p_1 > p_2, \qquad v_1 < v_2

(3)

Step 4The streamline pattern

Three streamlines from the entry (top) to the exit (right). Evenly spaced on the approach, they leave packed tightest against the inner wall, the innermost turning most sharply.

Mass conservation within any thin streamtube ties the local flow speed to the streamtube’s cross-sectional width: a streamtube that speeds up must narrow, so streamlines that speed up must draw closer together. Since (3) says the fluid accelerates on the inner side in passing from $1$ to $2$ , the streamlines there must correspondingly crowd together on that transition, most tightly right at the corner, while they open out again toward the outer wall.

Streamlines curve through the bend, crowding toward the inner wall.

Check

Far from the bend, in either straight run, $R\to\infty$ and (1) gives $\partial p/\partial n=0$ : pressure and speed are then uniform across the width. That is exactly why point $1$ , out in the still-parallel approach, carries the undisturbed values, and it is also why the crowding drawn at the right-hand edge of the figure cannot persist: farther down the exit run than the sketch extends, the streamlines straighten and their spacing evens out again. The effect found here is tied entirely to the curvature the bend introduces, exactly as the derivation assumed.

Remark

The same curvature argument is why, in a line of constant bore whose only fittings are bends, a sharp elbow is the site of the lowest pressure and the highest local speed in the run: if that local pressure drops below the fluid’s vapor pressure the flow cavitates right at the inner corner, which is why plumbing and hydraulic fittings favor a rounded bend (a larger effective $R$ ) over a mitered right angle wherever the pressure margin is tight.