Skip to main content

Hump check using given specific energy curve

Short answer

(i) No, available specific energy over the 0.5 m hump is 1.30 m, less than the required minimum 1.46 m. (ii) Upstream depth rises to 1.82 m; depth over the hump is critical at 0.97 m; downstream depth becomes supercritical at 0.58 m.

A rectangular channel of width 2 m is designed for irrigating land downstream. The channel is nearly horizontal, except that there is a smooth, short hump of 0.5 m in the channel bottom. The energy loss due to the hump is negligible. The point A in figure represents section A just upstream of hump. From the provided specific energy diagram below for a constant discharge of 6 m³/s, i) Determine whether the flow has adequate specific energy to sustain 6 m³/s through the hump with explanation. ii) Based on your result of i), what will be the flow profile just upstream and downstream of the hump to sustain the required discharge 6 m³/s? Show final results on the curve. [Figure: specific energy diagram of y (m) against E (m); point A on the subcritical limb at y = 1.63 m and point A' on the supercritical limb at y = 0.62 m, both at E = 1.8 m]

Given data

b=2b=2 m, Q=6Q=6 m³/s, so q=3q=3 m²/s. Hump height Δz=0.5\Delta z=0.5 m, no energy loss.

Point A (upstream, subcritical): yA=1.63y_A=1.63 m, EA=1.63+3219.62×1.632=1.8026≈1.80E_A=1.63+\dfrac{3^2}{19.62\times1.63^2}=1.8026\approx1.80 m. Its alternate depth is A' (y=0.62y=0.62 m, supercritical), with the same E=1.80E=1.80 m.

Critical state on the curve

yc=q2g3=99.813=0.972 m,Emin=32yc=1.458 my_c=\sqrt[3]{\frac{q^2}{g}}=\sqrt[3]{\frac{9}{9.81}}=0.972\ \text{m},\qquad E_{min}=\frac32y_c=1.458\ \text{m}

(i) Is the specific energy adequate?

Energy available over the hump:

Ehump=EA−Δz=1.80−0.50=1.30 mE_{hump}=E_A-\Delta z=1.80-0.50=1.30\ \text{m}

Since Ehump=1.30<Emin=1.458E_{hump}=1.30<E_{min}=1.458 m, the flow does not have adequate specific energy to pass 6 m³/s over the 0.5 m hump. (The maximum hump height the present upstream state allows is EA−Emin=1.8026−1.4575=0.345E_A-E_{min}=1.8026-1.4575=0.345 m, which is less than 0.5 m.)

(ii) What happens (profile upstream and downstream)

The hump acts as a control. Upstream, the water surface rises (backwater) until the upstream specific energy is just enough for critical flow over the hump:

E1′=Emin+Δz=1.4575+0.50=1.9575 mE_1'=E_{min}+\Delta z=1.4575+0.50=1.9575\ \text{m}

Upstream depth (subcritical root of y+919.62y2=1.9575y+\dfrac{9}{19.62y^2}=1.9575): y1′=1.819y_1'=1.819 m.

  • On the hump: critical flow, y=yc=0.972y=y_c=0.972 m.
  • Downstream of the hump: the flow passes through critical depth and becomes supercritical. After the hump the bed returns to its original level and the specific energy is again 1.95751.9575 m, so the supercritical depth is y3=0.576y_3=0.576 m. This may be followed by a hydraulic jump further downstream if the tailwater is deep enough.
  y (m)
  1.82 |            * A1 (upstream after rise, E=1.96)
  1.63 |        * A (original upstream, E=1.80)
       |
  0.97 |    * critical on hump (E=1.46)
  0.62 |        * A' (E=1.80)
  0.58 |            * downstream supercritical (E=1.96)
       +----+----+----+----+--> E (m)
          1.3  1.46 1.80 1.96

Final results on the curve: A moves up to 1.82 m (E = 1.96 m); the hump point is at the nose of the curve (0.97 m, 1.46 m); the downstream supercritical point is at 0.58 m (E = 1.96 m).

Answer: (i) No, since E=1.30E=1.30 m <Emin=1.46<E_{min}=1.46 m; (ii) the upstream depth rises to 1.82 m, the depth over the hump is critical (0.97 m), and the flow downstream is supercritical at 0.58 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Hydraulics chapter 7 questions →