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Three successive slopes: profiles and length

Short answer

Profile sequence: M2 on reach 1, S2 followed by hydraulic jump and S1 on reach 2, M1 on reach 3. The M2 drawdown profile in slope 1 is about 950 m long measured from 0.98 y_n to critical depth.

For a rectangular channel shown in fig. below, with width = 7 m, the flow rate is 8 m³/sec, S01S_{01} = 0.00032, n1n_1 = 0.011, S02S_{02} = 0.0911, n2n_2 = 0.021, S03S_{03} = 0.00020 where S0S_0 = bed slope and n = manning's coeff. i) Draw longitudinal water surface profiles for these three successive reaches. ii) Compute the length of profile in slope 1 only using direct step method or Bresse's method taking 3 steps only. [Figure: three successive reaches: slope 1 (S01S_{01}, n1n_1), slope 2 (S02S_{02}, n2n_2) and slope 3 (S03S_{03}); a depth of 1.2 m is marked at the downstream end of slope 3]

Given: b=7 mb = 7\ \text{m}, Q=8 m3/sQ = 8\ \text{m}^3/\text{s} (q=1.143 m2/sq = 1.143\ \text{m}^2/\text{s}). Slope 1: S01=0.00032S_{01} = 0.00032, n1=0.011n_1 = 0.011. Slope 2: S02=0.0911S_{02} = 0.0911, n2=0.021n_2 = 0.021. Slope 3: S03=0.0002S_{03} = 0.0002, with y=1.2y = 1.2 m at its downstream end. n3n_3 is not given, so n3=n1=0.011n_3 = n_1 = 0.011 is assumed.

Critical and normal depths

yc=(1.14329.81)1/3=0.511 my_c = \left(\frac{1.143^2}{9.81}\right)^{1/3} = 0.511\ \text{m}
Reachyny_n (m)Class
10.886mild (yn>ycy_n > y_c)
20.224steep (yn<ycy_n < y_c)
31.033mild

i) Longitudinal profiles

  • Reach 1 (mild): the flow upstream is uniform at 0.886 m. At the break to the steep slope the flow passes through critical depth, so the surface falls as an M2 curve from yny_n to ycy_c at the end of the reach.
  • Reach 2 (steep): the flow starts at ycy_c and falls as an S2 curve to the uniform supercritical depth of 0.2240.224 m.
  • Reach 3 (mild): the depth at the downstream end is 1.2 m>yn31.2\ \text{m} > y_{n3}, so an M1 backwater curve rises downstream from near yn3=1.033y_{n3} = 1.033 m to 1.2 m.
  • Jump: the sequent depth of the supercritical flow (0.224 m) is 0.9830.983 m, which is less than the depth at the head of reach 3 (≈1.033\approx 1.033 m). The tailwater is too deep for the jump to form on the mild reach, so it is pushed up onto the steep reach 2. After the jump an S1 curve rises from about 0.98 m to about 1.03 m at the break. Because S02S_{02} is very steep, this S1 curve is short.
 R1 (mild)     |  R2 (steep)          | R3 (mild)
 -- yn=0.886   |                      |            1.2 m
   \__ M2      |\ S2          J  S1   |  M1  _____---
 yc  '-.. - - -|  '-._ yn=0.224 /'''''|__---'

ii) Length of the M2 profile in reach 1

The M2 curve approaches yny_n only asymptotically, so it is taken from 0.98 yn1=0.868 m0.98\,y_{n1} = 0.868\ \text{m} to yc=0.511y_c = 0.511 m in 3 equal steps (the control is at the downstream end).

Δx=E2−E1S0−Sˉf,Sf=n2V2R4/3\Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f},\qquad S_f = \frac{n^2V^2}{R^{4/3}}
yy (m)RR (m)VV (m/s)EE (m)SfS_fSˉf\bar S_fΔx\Delta x (m)
0.86800.6951.3170.95630.00034
0.74880.6171.5260.86760.000540.00044748.6
0.62970.5341.8150.79760.000920.00073171.2
0.51060.4462.2380.76590.001780.0013530.7
L=748.6+171.2+30.7≈950 mL = 748.6 + 171.2 + 30.7 \approx 950\ \text{m}

Answer: Profile sequence: M2 (reach 1), S2, hydraulic jump and S1 (reach 2), M1 (reach 3). The M2 profile in slope 1 is about 950 m long (measured from 0.98 yn0.98\,y_n to ycy_c, in the flow direction).

✓ Answer checked by Claude Opus 5.5 (high effort)

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