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Hump and constriction: critical depth, 2 m channel

Short answer

For hump alone, E2 > Emin, so no critical depth; depth over hump 0.74 m. For constriction alone, E2 > Emin2, no critical depth; depth 0.96 m. With both, E2 < Emin2, critical depth 0.588 m occurs; upstream depth rises from 1.000 m to 1.010 m (10 mm increase).

A rectangular channel 2 m wide has a flow of 2.4 m³/s at a depth of 1.0 m. Determine if critical depth occurs (a) at the section where a hump of ΔZ\Delta Z = 20 cm high is installed across the bed, (b) a side wall constriction (no hump) reducing the channel width to 1.7 m, and (c) both the hump and side wall constriction combined. Will the upstream depth be affected for case (c)? If so, to what extent? Neglect head losses of the hump and constriction caused by friction, expansion and contraction.

Critical depth occurs at a section when the available specific energy there is equal to the minimum specific energy for the discharge per unit width at that section: E2=Emin=32ycE_2=E_{min}=\tfrac32y_c, where yc=q2/g3y_c=\sqrt[3]{q^2/g}.

Approach flow

Q=2.4Q=2.4 m³/s, b=2b=2 m, y1=1.0y_1=1.0 m.

q1=2.42=1.2 m2/s,V1=1.2 m/s,Fr1=1.29.81×1=0.383 (subcritical)q_1=\frac{2.4}{2}=1.2\ \text{m}^2/\text{s},\quad V_1=1.2\ \text{m/s},\quad Fr_1=\frac{1.2}{\sqrt{9.81\times1}}=0.383\ (\text{subcritical}) E1=y1+V122g=1+1.4419.62=1.0734 mE_1=y_1+\frac{V_1^2}{2g}=1+\frac{1.44}{19.62}=1.0734\ \text{m} yc1=1.229.813=0.5275 m,Emin1=1.5yc1=0.7913 my_{c1}=\sqrt[3]{\frac{1.2^2}{9.81}}=0.5275\ \text{m},\qquad E_{min1}=1.5y_{c1}=0.7913\ \text{m}

(a) Hump of 0.20 m (width unchanged)

Energy over hump E2=E1−Δz=1.0734−0.20=0.8734E_2=E_1-\Delta z=1.0734-0.20=0.8734 m >Emin1=0.7913>E_{min1}=0.7913 m.

So critical depth does not occur. The flow stays subcritical over the hump, with depth found from y+1.4419.62y2=0.8734y+\dfrac{1.44}{19.62y^2}=0.8734:

y2=0.739 m(>yc=0.528 m)y_2=0.739\ \text{m}\quad(>y_c=0.528\ \text{m})

The upstream depth remains 1.0 m.

(b) Side-wall constriction to 1.7 m (no hump)

q2=2.41.7=1.4118 m2/s,yc2=q22g3=0.5879 m,Emin2=0.8818 mq_2=\frac{2.4}{1.7}=1.4118\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=0.5879\ \text{m},\quad E_{min2}=0.8818\ \text{m}

E2=E1=1.0734>Emin2=0.8818E_2=E_1=1.0734>E_{min2}=0.8818, so critical depth does not occur. Depth in the constriction (subcritical): y2=0.964y_2=0.964 m. Upstream depth unchanged.

(c) Hump and constriction together

Available energy at the section: E2=E1−0.20=0.8734E_2=E_1-0.20=0.8734 m, while the minimum required is Emin2=0.8818E_{min2}=0.8818 m (for q2=1.412q_2=1.412).

E2<Emin2E_2<E_{min2}, so critical depth occurs (choking) over the hump in the constriction, yc=0.588y_c=0.588 m. The channel cannot pass the flow at the original upstream energy, so the upstream depth is affected: it rises until

E1′=Emin2+Δz=0.8818+0.20=1.0818 mE_1'=E_{min2}+\Delta z=0.8818+0.20=1.0818\ \text{m} y1′+1.222g y1′2=1.0818  ⇒  y1′=1.0098 my_1'+\frac{1.2^2}{2g\,y_1'^2}=1.0818\;\Rightarrow\;y_1'=1.0098\ \text{m}

Rise in upstream depth =1.0098−1.000=0.0098=1.0098-1.000=0.0098 m (about 10 mm).

Answer: (a) not critical (y=0.74y=0.74 m over hump); (b) not critical (y=0.96y=0.96 m); (c) critical depth 0.5880.588 m occurs; upstream depth rises from 1.000 m to 1.010 m, about 10 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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