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Width equals 8/9 of minimum specific energy

Short answer

Minimizing wetted perimeter P = b + 2yc under critical flow yields b = (4/3)yc. Using yc = (2/3)Emin gives b = (8/9)Emin.

A rectangular channel section to have critical flow and at the same time the wetted perimeter is to be minimum. Show that for these two conditions simultaneously, the width of the channel must be equal to 8/9 times minimum specific energy.

Given conditions (rectangular channel, width bb, discharge QQ)

  1. Flow is critical.
  2. The wetted perimeter is minimum (for the given discharge).

Critical flow in a rectangular channel

q=Qb,yc=q2g3=(Q2gb2)1/3,Emin=32ycq=\frac Qb,\quad y_c=\sqrt[3]{\frac{q^2}{g}}=\left(\frac{Q^2}{gb^2}\right)^{1/3},\qquad E_{min}=\frac32y_c

Minimum wetted perimeter

P=b+2yc=b+2(Q2g)1/3b−2/3P=b+2y_c=b+2\left(\frac{Q^2}{g}\right)^{1/3}b^{-2/3}

Differentiate with respect to bb (QQ constant) and equate to zero:

dPdb=1−43(Q2g)1/3b−5/3=0\frac{dP}{db}=1-\frac43\left(\frac{Q^2}{g}\right)^{1/3}b^{-5/3}=0 b5/3=43(Q2g)1/3b^{5/3}=\frac43\left(\frac{Q^2}{g}\right)^{1/3}

Since (Q2g)1/3=yc b2/3\left(\dfrac{Q^2}{g}\right)^{1/3}=y_c\,b^{2/3} (from yc3=Q2/(gb2)y_c^3=Q^2/(gb^2)):

b5/3=43ycb2/3  ⇒  b=43ycb^{5/3}=\frac43y_cb^{2/3}\;\Rightarrow\;b=\frac43y_c

(The second derivative is positive, so this is a minimum.)

Relation with EminE_{min}

From Emin=32ycE_{min}=\tfrac32y_c, yc=23Eminy_c=\tfrac23E_{min}:

b=43×23Emin=89Eminb=\frac43\times\frac23E_{min}=\frac89E_{min} b=89Emin\boxed{b=\frac89E_{min}}

Hence proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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