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Muskingum routing, K=8 h, X=0.25

Short answer

Muskingum routing with K=8 h, x=0.25, Δt=6 h gives C₀=0.1111, C₁=0.5556, C₂=0.3333. Starting outflow 8 m³/s, outflows at 0,6,12,18,24,30,36,42 h: 8, 8.89, 15.19, 25.06, 27.80, 25.38, 21.24, 16.52 m³/s. Peak outflow 27.80 m³/s at 24 h; attenuation 2.20 m³/s, lag 12 h.

Route the flood hydrograph through a river reach for which Muskingum coefficient K = 8 h and X = 0.25. Initial outflow discharge from the reach is 8.0 m3^3/s.

Time (hour)06121824303642
Inflow (m3^3/s)816303025201510

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=8K = 8 h, x=0.25x = 0.25, Δt=6\Delta t = 6 h, so K−Kx+0.5Δt=9K - Kx + 0.5\Delta t = 9 h.

C0=−(8)(0.25)+39=0.1111C1=(8)(0.25)+39=0.5556C2=6−39=0.3333\begin{aligned} C_0 &= \frac{-(8)(0.25) + 3}{9} = 0.1111\\ C_1 &= \frac{(8)(0.25) + 3}{9} = 0.5556\\ C_2 &= \frac{6 - 3}{9} = 0.3333\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)
08.00---8.00
616.001.784.442.678.89
1230.003.338.892.9615.19
1830.003.3316.675.0625.06
2425.002.7816.678.3527.80
3020.002.2213.899.2725.38
3615.001.6711.118.4621.24
4210.001.118.337.0816.52

Peak attenuation and lag

  • Peak inflow = 30 m³/s at t = 12 h
  • Peak outflow = 27.80 m³/s at t = 24 h
  • Attenuation of peak = 30 − 27.80 = 2.20 m³/s (7.3 % of the inflow peak)
  • Lag of peak = 24 − 12 = 12 h (to the nearest time step)

✓ Answer checked by Claude Sonnet 5.5

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