For K = 36 h, x = 0.15, dt = 12 h, which is correct?

C0 = 0.0164, C1 = 0.3115, C2 = 0.6721

C0 = 0.3115, C1 = 0.0164, C2 = 0.6721

C0 = 0.0164, C1 = 0.6721, C2 = 0.3115

What starting outflow is used?

Equal to the initial inflow, 42 m3/s

2080 Bhadra · 8 marks

Muskingum routing, K=36 hr, x=0.15

Short answer

Muskingum routing with K=36 h, x=0.15, Δt=12 h yields C₀=0.0164, C₁=0.3115, C₂=0.6721. Initial O=42 m³/s. Outflows at times 0–120 h (12‑h steps): 42, 42.05, 43.72, 61.26, 131.50, 199.63, 227.82, 231.12, 219.67, 200.29, 177.85 m³/s. Peak outflow 231.12 m³/s at 84 h; reduction 110.88 m³/s, lag 36 h.

The inflow hydrograph readings for a stream reach are given below for which the Muskingum coefficients of K = 36 hr and x = 0.15. Route the flood through the reach and determine the outflow hydrograph. Also, determine the reduction in peak and the time of peak of outflow. Outflow at the beginning of the flood may be taken as same as inflow.

Time (hr)	0	12	24	36	48	60	72	84	96	108	120
Inflow (cumec)	42	45	88	272	342	288	240	198	162	133	110

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 $K = 36$ h, $x = 0.15$ , $\Delta t = 12$ h, so $K - Kx + 0.5\Delta t = 36.6$ h.

\begin{aligned} C_0 &= \frac{-(36)(0.15) + 6}{36.6} = 0.0164\\ C_1 &= \frac{(36)(0.15) + 6}{36.6} = 0.3115\\ C_2 &= \frac{30.6 - 6}{36.6} = 0.6721\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	42.00	-	-	-	42.00
12	45.00	0.74	13.08	28.23	42.05
24	88.00	1.44	14.02	28.26	43.72
36	272.00	4.46	27.41	29.39	61.26
48	342.00	5.61	84.72	41.17	131.50
60	288.00	4.72	106.52	88.38	199.63
72	240.00	3.93	89.70	134.18	227.82
84	198.00	3.25	74.75	153.12	231.12
96	162.00	2.66	61.67	155.35	219.67
108	133.00	2.18	50.46	147.65	200.29
120	110.00	1.80	41.43	134.62	177.85

Peak attenuation and lag

Peak inflow = 342 m³/s at t = 48 h
Peak outflow = 231.12 m³/s at t = 84 h
Attenuation of peak = 342 − 231.12 = 110.88 m³/s (32.4 % of the inflow peak)
Lag of peak = 84 − 48 = 36 h (to the nearest time step)

✓ Answer checked by Claude Sonnet 5.5

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Muskingum routing, K=36 hr, x=0.15

Method

Routing table

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