Why is C0 negative here?

Because outflow is less than inflow

Because the inflow is increasing

Because the coefficients do not sum to 1

Which computation gives O at t = 6 h?

C0(55) + C1(35) + C2(35) = 29.94

What is true of the computed outflow at 30 h?

It is still rising, so the peak is not reached within the data

2082 Baisakh · 8 marks

Muskingum routing, x=0.28, k=38.4 hr

Short answer

Muskingum routing with K=38.4 h, x=0.28, Δt=6 h gives C₀=−0.2529, C₁=0.4487, C₂=0.8042 using O₂=C₀I₂+C₁I₁+C₂O₁. Starting from O=35 m³/s, outflow at 0,6,12,18,24,30 h is 35.00, 29.94, 25.49, 28.90, 41.10, 69.44 m³/s; peak not yet reached within given data.

The storage in the reach of a stream has been studied. The values of x and k in Muskingum equation have been identified as 0.28 and 38.4 hr. If the inflow hydrograph to reach is as given below, compute the outflow hydrograph. Assume the outflow from the reach at t = 0 is 35 m $^3$ /s.

Time (hr)	0	6	12	18	24	30
Inflow (m $^3$ /s)	35	55	92	130	160	140

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 $K = 38.4$ h, $x = 0.28$ , $\Delta t = 6$ h, so $K - Kx + 0.5\Delta t = 30.648$ h.

\begin{aligned} C_0 &= \frac{-(38.4)(0.28) + 3}{30.648} = -0.2529\\ C_1 &= \frac{(38.4)(0.28) + 3}{30.648} = 0.4487\\ C_2 &= \frac{27.648 - 3}{30.648} = 0.8042\end{aligned}

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

$C_0$ is negative because $\Delta t < 2Kx$ ( $6 < 21.5$ h). This is acceptable in the computation, but it makes the outflow dip at the start of the rise, as the table shows.

Routing table

t (h)	I	C0·I2	C1·I1	C2·O1	O (m³/s)
0	35.00	-	-	-	35.00
6	55.00	-13.91	15.70	28.15	29.94
12	92.00	-23.27	24.68	24.08	25.49
18	130.00	-32.88	41.28	20.50	28.90
24	160.00	-40.47	58.33	23.24	41.10
30	140.00	-35.41	71.79	33.06	69.44

Answer: the outflow hydrograph is 35.00, 29.94, 25.49, 28.90, 41.10 and 69.44 m³/s at t = 0, 6, 12, 18, 24 and 30 h. The outflow is still rising at 30 h (the given inflow record ends there), so the outflow peak is not reached within the data.

✓ Answer checked by Claude Sonnet 5.5

Muskingum routing, K=36 hr, x=0.15 · Engineering Hydrology, chapter 7
Reservoir routing through spillway, 6-hr inflow · Engineering Hydrology, chapter 7
Muskingum routing, K=10, x=0.20 · Engineering Hydrology, chapter 7 · asked 2×
Storage change from inflow equation I = 2t · Engineering Hydrology, chapter 1
Muskingum routing, k=10 h, x=0.2 (table missing) · Engineering Hydrology, chapter 7
Muskingum routing, K=8 h, X=0.25 · Engineering Hydrology, chapter 7
ERR from separate inflow and outflow rows · Engineering Economics, chapter 3
Storage for uniform inflow over 24 hours · Water Supply Engineering, chapter 7
RoR and PRoR from hydrograph equation (Q40) · Hydropower Engineering, chapter 3
HR trickling filter, 2.5 MLD; change if standard · Sanitary Engineering, chapter 7

All Engineering Hydrology chapter 7 questions →

Muskingum routing, x=0.28, k=38.4 hr

Method

Routing table

Related questions