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Water surface over hump with contraction

Short answer

Upstream water surface elevation remains 1.600 m. Over the 0.35 m hump and 3.5 m width, the depth is 1.158 m, giving a surface elevation of 1.508 m above the upstream bed.

Water flows at a depth of 1.6 m and velocity of 1.10 m/s in an open channel of rectangular cross section of width 4.0 m. At a certain section the width is reduced to 3.5 m and the bed is raised by 0.35 m through a smooth flat top hump. Calculate the water surface elevations at the contracted section as well as at a section upstream of it. Assume energy losses to be negligible. Show the results in specific energy diagram.

Given data

y1=1.6y_1=1.6 m, V1=1.10V_1=1.10 m/s, b1=4.0b_1=4.0 m; contracted width b2=3.5b_2=3.5 m with a hump of Δz=0.35\Delta z=0.35 m; no losses.

Q=b1y1V1=4×1.6×1.10=7.04 m3/sQ=b_1y_1V_1=4\times1.6\times1.10=7.04\ \text{m}^3/\text{s} E1=y1+V122g=1.6+1.2119.62=1.6617 m,Fr1=1.109.81×1.6=0.278E_1=y_1+\frac{V_1^2}{2g}=1.6+\frac{1.21}{19.62}=1.6617\ \text{m},\qquad Fr_1=\frac{1.10}{\sqrt{9.81\times1.6}}=0.278

Check for critical flow at the contraction

q2=7.043.5=2.0114 m2/s,yc2=q22g3=0.7444 m,Emin2=1.1165 mq_2=\frac{7.04}{3.5}=2.0114\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=0.7444\ \text{m},\quad E_{min2}=1.1165\ \text{m}

Energy available over the hump: E2=E1−Δz=1.6617−0.35=1.3117E_2=E_1-\Delta z=1.6617-0.35=1.3117 m >Emin2>E_{min2}. So the flow passes without choking, and the upstream depth is unchanged.

Depth over the hump (subcritical)

y2+q222gy22=y2+0.2062y22=1.3117  ⇒  y2=1.1579 my_2+\frac{q_2^2}{2gy_2^2}=y_2+\frac{0.2062}{y_2^2}=1.3117\;\Rightarrow\;y_2=1.1579\ \text{m}

(here q22/2g=0.2062q_2^2/2g=0.2062). Velocity V2=q2/y2=1.737V_2=q_2/y_2=1.737 m/s.

Water surface elevations (datum: upstream bed)

  • At a section upstream: 1.6001.600 m.
  • At the contracted section: 0.35+1.1579=1.50790.35+1.1579=1.5079 m.

Drop in water surface =1.600−1.508=0.092=1.600-1.508=0.092 m.

Specific energy diagram

 y (m)
1.700|                                  .:
1.600|                              1:         (1) upstream, E1 = 1.662 m
1.500|                           .:
1.400|                       .:
1.300|                    .:
1.158|               .2                        (2) over hump, E2 = 1.312 m
1.100|             . :
1.000|          . :
0.900|        . :
0.800|      .  :
0.700|     .   :                               noses: yc1 = 0.681 m, yc2 = 0.744 m
0.600|      .    :
0.500|         .       :
0.400|                   .            :
     +-----+---+------+-------------+---------> E (m)
 ticks, left to right: E_min1 = 1.021, E_min2 = 1.117, E2 = 1.312, E1 = 1.662
 "." = curve for q1 = 1.76 m²/s,   ":" = curve for q2 = 2.011 m²/s

Point (1) is on the upper (subcritical) limb of the curve for q1=1.76q_1=1.76 m²/s; point (2) is on the upper limb of the curve for q2=2.011q_2=2.011 m²/s, shifted left by Δz=0.35\Delta z=0.35 m.

Answer: depth over the hump =1.158=1.158 m, water surface elevation there =1.508=1.508 m (above the upstream bed); upstream elevation =1.600=1.600 m (unchanged).

✓ Answer checked by Claude Opus 5.5 (high effort)

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