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Muskingum routing, K=12 h, X=0.20

Short answer

Using Muskingum routing with coefficients C0=0.0476, C1=0.4286, C2=0.5238 and initial outflow 10 m³/s, peak outflow is 49.58 m³/s at 30 h. Peak inflow 60 m³/s at 18 h gives attenuation 10.42 m³/s and lag 12 h.

Route the following hydrograph through a river reach for which K = 12 h and X = 0.20. At the start of the inflow flood, the outflow discharge is 10 m3^3/s. Also find lag of peak and attenuation.

Time (h)061218243036424854
Inflow (m3^3/s)10205060554535272015

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=12K = 12 h, x=0.2x = 0.2, Δt=6\Delta t = 6 h, so K−Kx+0.5Δt=12.6K - Kx + 0.5\Delta t = 12.6 h.

C0=−(12)(0.2)+312.6=0.0476C1=(12)(0.2)+312.6=0.4286C2=9.6−312.6=0.5238\begin{aligned} C_0 &= \frac{-(12)(0.2) + 3}{12.6} = 0.0476\\ C_1 &= \frac{(12)(0.2) + 3}{12.6} = 0.4286\\ C_2 &= \frac{9.6 - 3}{12.6} = 0.5238\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)
010.00---10.00
620.000.954.295.2410.48
1250.002.388.575.4916.44
1860.002.8621.438.6132.90
2455.002.6225.7117.2345.57
3045.002.1423.5723.8749.58
3635.001.6719.2925.9746.92
4227.001.2915.0024.5840.86
4820.000.9511.5721.4133.93
5415.000.718.5717.7727.06

Peak attenuation and lag

  • Peak inflow = 60 m³/s at t = 18 h
  • Peak outflow = 49.58 m³/s at t = 30 h
  • Attenuation of peak = 60 − 49.58 = 10.42 m³/s (17.4 % of the inflow peak)
  • Lag of peak = 30 − 18 = 12 h (to the nearest time step)

✓ Answer checked by Claude Sonnet 5.5

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