Question Bank • QB_SWCE_ADV_123 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Hydrology

Subtopic: Flood routing, hydrological reservoir and channel routing

Question Statement:

In a Muskingum channel flood routing operation, the reach parameters are travel time $K = 12.0\text{ hours}$, weighting factor $x = 0.20$, and routing time step $\Delta t = 6.0\text{ hours}$. At the beginning of the interval, inflow is $I_1 = 40.0\text{ m}^3/\text{s}$ and outflow is $O_1 = 40.0\text{ m}^3/\text{s}$. At the end of the interval, inflow rises to $I_2 = 70.0\text{ m}^3/\text{s}$. The routed outflow discharge $O_2$ in $\text{m}^3/\text{s}$ is ________ (round off to one decimal place).

Answer Key:

41.4

Step-by-Step Solution & Explanation

**Method 1: Muskingum Routing Coefficients Derivation** $O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1$ 1. Common denominator $D = 2 K (1 - x) + \Delta t$: $D = 2(12.0)(1 - 0.20) + 6.0 = 2(12.0)(0.80) + 6.0 = 19.20 + 6.0 = 25.20\text{ h}$ 2. Routing coefficients: $C_0 = \frac{\Delta t - 2 K x}{D} = \frac{6.0 - 2(12.0)(0.20)}{25.20} = \frac{6.0 - 4.80}{25.20} = \frac{1.20}{25.20} = 0.047619$ $C_1 = \frac{\Delta t + 2 K x}{D} = \frac{6.0 + 4.80}{25.20} = \frac{10.80}{25.20} = 0.428571$ $C_2 = \frac{2 K(1 - x) - \Delta t}{D} = \frac{19.20 - 6.0}{25.20} = \frac{13.20}{25.20} = 0.523810$ Check sum: $C_0 + C_1 + C_2 = 0.047619 + 0.428571 + 0.523810 = 1.00000$. 3. Computing outflow $O_2$: $O_2 = (0.047619 \times 70.0) + (0.428571 \times 40.0) + (0.523810 \times 40.0)$ $O_2 = 3.3333 + 17.1428 + 20.9524 = 41.4285\text{ m}^3/\text{s} \approx 41.4\text{ m}^3/\text{s}$ **Method 2: Factoring $O_1$ and $I_1$ Group** $O_2 = C_0 I_2 + (C_1 + C_2) O_1 = 0.047619(70.0) + (0.952381)(40.0) = 3.3333 + 38.0952 = 41.43\text{ m}^3/\text{s}$
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