For a rectangular channel, critical depth depends only on:

Discharge per unit width q (and g)

Why does yc stay the same over a hump?

Width and hence q do not change

The hump lowers the energy to zero

Why does yc differ in a constricted throat?

The width falls, so q = Q/b rises and yc and Ec rise

Q is reduced by the constriction

2068 Bhadra · 2 marks

Why critical depth varies with constriction only

Short answer

Critical depth y_c = (q^2/g)^(1/3) depends on discharge per unit width q = Q/b. A hump leaves the width unchanged, so q and y_c remain constant. A constriction reduces width, increasing q and thus y_c in the throat, so the critical depth differs between the approach and the contracted section.

Why does the critical depth vary for the constriction flow analysis and not vary for the hump flow analysis?

For a rectangular channel the critical depth depends only on the discharge per unit width:

y_c = \left(\frac{q^2}{g}\right)^{1/3},\qquad q = \frac{Q}{b}

Hump (bed raised): the width $b$ stays the same, so $q$ is unchanged. Hence $y_c$ is the same at every section. A hump only changes the specific energy available above its crest ( $E - \Delta z$ ). When $E - \Delta z$ falls to $E_c = 1.5\,y_c$ , the flow becomes critical on the crest with the same $y_c$ .
Constriction (width reduced): the width changes from $b_1$ to $b_2$ , so $q$ increases to $Q/b_2$ . A larger $q$ gives a larger $y_c$ and larger $E_c$ in the throat. The critical depth therefore differs from section to section.

So $y_c$ varies with a constriction because $q$ varies; it does not vary with a hump because $q$ does not.

✓ Answer checked by Claude Opus 5.5 (high effort)

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