Muskingum routing, k=10 h, x=0.2 (table missing)

Assumption: the inflow table is cut off in the question. The inflow ordinates are assumed to be the standard set for this problem (10, 20, 50, 60, 55, 45, 35, 27, 20, 15 m³/s at 6 h intervals from 0 to 54 h). If your table differs, use the same coefficients and the same steps with your values.

Method

Muskingum routing: $O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1$, where

$$C_0 = \frac{-Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_1 = \frac{Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_2 = \frac{K - Kx - 0.5\Delta t}{K - Kx + 0.5\Delta t}$$

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

$$\begin{aligned} C_0 &= \frac{-(10)(0.2) + 3}{11} = 0.0909\\ C_1 &= \frac{(10)(0.2) + 3}{11} = 0.4545\\ C_2 &= \frac{8 - 3}{11} = 0.4545\end{aligned}$$

Check: $C_0 + C_1 + C_2 = 1.0000$.

Routing table

Peak attenuation and lag

• Peak inflow = 60 m³/s at t = 18 h
• Peak outflow = 51.33 m³/s at t = 30 h
• Attenuation of peak = 60 − 51.33 = 8.67 m³/s (14.5 % of the inflow peak)
• Lag of peak = 30 − 18 = 12 h (to the nearest time step)