With K = 10 h, x = 0.2, dt = 5 h, which coefficients result?

C0 = 0.0909, C1 = 0.4545, C2 = 0.4545

C0 = 0.5238, C1 = 0.4286, C2 = 0.0476

What is the lag of peak?

10 h (inflow peak at 15 h, outflow peak at 25 h)

Asked 2/27
2078 Bhadra · 6+2 marks

Muskingum routing, K=10, x=0.20

Short answer

Muskingum coefficients: C0=0.0476, C1=0.4286, C2=0.5238 for K=10h, x=0.2, Δt=5h. Routing yields peak inflow 45 m³/s at 15h, peak outflow 35.00 m³/s at 25h. Attenuation = 10.00 m³/s, lag = 10 h.

Route the following hydrograph through a river reach with routing parameters K and x as 10 and 0.20, respectively. Take outflow discharge at the start of inflow flood as 5 m $^3$ /s. Also estimate attenuation and lag of peak.

Time (hr)	0	5	10	15	20	25	30	35	40
Inflow (m $^3$ /s)	5	10	30	45	40	30	18	12	8

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, $x = 0.2$ , $\Delta t = 5$ h, so $K - Kx + 0.5\Delta t = 10.5$ h.

\begin{aligned} C_0 &= \frac{-(10)(0.2) + 2.5}{10.5} = 0.0476\\ C_1 &= \frac{(10)(0.2) + 2.5}{10.5} = 0.4286\\ C_2 &= \frac{8 - 2.5}{10.5} = 0.5238\end{aligned}

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

Routing table

t (h)	I	C0·I2	C1·I1	C2·O1	O (m³/s)
0	5.00	-	-	-	5.00
5	10.00	0.48	2.14	2.62	5.24
10	30.00	1.43	4.29	2.74	8.46
15	45.00	2.14	12.86	4.43	19.43
20	40.00	1.90	19.29	10.18	31.37
25	30.00	1.43	17.14	16.43	35.00
30	18.00	0.86	12.86	18.33	32.05
35	12.00	0.57	7.71	16.79	25.07
40	8.00	0.38	5.14	13.13	18.66

Peak attenuation and lag

Peak inflow = 45 m³/s at t = 15 h
Peak outflow = 35.00 m³/s at t = 25 h
Attenuation of peak = 45 − 35.00 = 10.00 m³/s (22.2 % of the inflow peak)
Lag of peak = 25 − 15 = 10 h (to the nearest time step)

✓ Answer checked by Claude Sonnet 5.5

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Muskingum routing, K=10, x=0.20

Method

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