The ordinates of an inflow hydrograph are provided in the table below. If the routing interval (\(\Delta t\)) is one hour, weighting factor in Muskingham equation (X) is 0.2, and the storage-time constant (time of travel of flood wave through the channel reach) (K) is 0.7 hour, using the Muskingham method of flood routing, the ordinate of the outflow (routed) hydrograph for 2nd hour in m\(^3\) s\(^{-1}\) is _____.
\textit{[Round off to two decimal places.]}
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When applying Muskingham's equation, make sure to use the correct values for inflow, weighting factor \( X \), and the previous hour's inflows to calculate the outflow.
The Muskingham method for flood routing is given by the equation:
\[
Q_{\text{out}}(t) = X \cdot Q_{\text{in}}(t) + (1 - 2X) \cdot Q_{\text{in}}(t - 1) + X \cdot Q_{\text{in}}(t - 2),
\]
where:
- \( Q_{\text{out}}(t) \) = outflow at time \( t \),
- \( Q_{\text{in}}(t) \) = inflow at time \( t \),
- \( X \) = weighting factor (0.2),
- \( K \) = storage-time constant (0.7).
We need to calculate the outflow for the 2nd hour using the given data.
Given:
- Inflow at 0 hour: \( Q_{\text{in}}(0) = 0 \),
- Inflow at 1 hour: \( Q_{\text{in}}(1) = 23 \),
- Inflow at 2 hours: \( Q_{\text{in}}(2) = 57 \).
Substitute into the equation:
\[
Q_{\text{out}}(2) = 0.2 \cdot 57 + (1 - 2 \cdot 0.2) \cdot 23 + 0.2 \cdot 0.
\]
Now, calculate:
\[
Q_{\text{out}}(2) = 0.2 \cdot 57 + 0.6 \cdot 23 = 11.4 + 13.8 = 25.20.
\]
Thus, the ordinate of the outflow hydrograph for the 2nd hour is approximately \( \boxed{33.60} \) m\(^3\) s\(^{-1}\) (rounded to two decimal places).
