Skip to main content

Muskingum routing at Chovar, Bagmati River

Short answer

Using Muskingum routing with K=10 h, x=0.2, Δt=6 h yields coefficients C0=0.0909, C1=C2=0.4545. The outflow peak is 494.70 m³/s at 24 h, an attenuation of 165.30 m³/s and a 6-h lag from the inflow peak of 660 m³/s at 18 h.

Route the following flood hydrograph at Chovar of Bagmati River using Muskingum method having coefficient k = 10 hr and x = 0.2. Following is the inflow hydrograph with outflow of Chovar at the start is 60 m3^3/s.

t (hr)06121824303642485460
Inflow (m3^3/s)60120480660460260160100906040

Method

Muskingum routing: O2=C0I2+C1I1+C2O1O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1, where

C0=−Kx+0.5ΔtK−Kx+0.5Δt,C1=Kx+0.5ΔtK−Kx+0.5Δt,C2=K−Kx−0.5ΔtK−Kx+0.5ΔtC_0 = \frac{-Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_1 = \frac{Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_2 = \frac{K - Kx - 0.5\Delta t}{K - Kx + 0.5\Delta t}

Here K=10K = 10 h, x=0.2x = 0.2, Δt=6\Delta t = 6 h, so K−Kx+0.5Δt=11K - Kx + 0.5\Delta t = 11 h.

C0=−(10)(0.2)+311=0.0909C1=(10)(0.2)+311=0.4545C2=8−311=0.4545\begin{aligned} C_0 &= \frac{-(10)(0.2) + 3}{11} = 0.0909\\ C_1 &= \frac{(10)(0.2) + 3}{11} = 0.4545\\ C_2 &= \frac{8 - 3}{11} = 0.4545\end{aligned}

Check: C0+C1+C2=1.0000C_0 + C_1 + C_2 = 1.0000.

Routing table

t (h)IC0·I2C1·I1C2·O1O (m³/s)
060.00---60.00
6120.0010.9127.2727.2765.45
12480.0043.6454.5529.75127.93
18660.0060.00218.1858.15336.33
24460.0041.82300.00152.88494.70
30260.0023.64209.09224.86457.59
36160.0014.55118.18208.00340.72
42100.009.0972.73154.87236.69
4890.008.1845.45107.59161.22
5460.005.4540.9173.28119.65
6040.003.6427.2754.3985.29

Peak attenuation and lag

  • Peak inflow = 660 m³/s at t = 18 h
  • Peak outflow = 494.70 m³/s at t = 24 h
  • Attenuation of peak = 660 − 494.70 = 165.30 m³/s (25.0 % of the inflow peak)
  • Lag of peak = 24 − 18 = 6 h (to the nearest time step)

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Engineering Hydrology chapter 7 questions →