Question

Two catchments M and N are meteorologically similar. The basin length (\(L\)), length to the centroid (\(L_{ca}\)), and the drainage area (\(A\)) of catchment M are \(36\) km, \(18\) km, and \(250\) km\(^2\), respectively. The \(L\), \(L_{ca}\) and \(A\) for catchment N are \(50\) km, \(30\) km, and \(400\) km\(^2\), respectively.
For a \(6\)-h unit hydrograph in catchment M, the peak discharge of \(50\) m\(^3\)/s occurs at \(12\) h from the beginning of the rainfall excess. Using Snyder's method, the peak of the \(6\)-h unit hydrograph for catchment N (in m\(^3\)/s) is ________. (Rounded off to two decimal places)

Hint:

Transpose Snyder's basin and peaking coefficients from catchment M to catchment N using the given lag data.

Answer 1

Correct Answer: 64.58

Step 1: Get the adjusted lag of catchment M from the given peak time.
$T_p = t_{pR}+t_R/2$, so $t_{pR}(M)=12-3=9$ h for the $6$-h storm.

Step 2: Solve for the standard lag $t_p(M)$.
From $9=t_p+(6-t_p/5.5)/4$, rearranging gives $t_p(M)\approx7.857$ h.

Step 3: Scale the lag to catchment N by a ratio, skipping $C_t$ itself.
Since both basins share $C_t$, $t_p$ scales as $(L\,L_{ca})^{0.3}$: \[ \frac{t_p(N)}{t_p(M)} = \left(\frac{50\times30}{36\times18}\right)^{0.3} = \left(\frac{1500}{648}\right)^{0.3} \approx 1.286 \] \[ t_p(N) \approx 7.857\times1.286 \approx 10.11\ \text{h}, \quad t_r(N)=10.11/5.5\approx1.838\ \text{h} \]
Step 4: Adjust for the $6$-h duration in N.
\[ t_{pR}(N) = 10.11+\frac{6-1.838}{4} \approx 11.15\ \text{h} \]
Step 5: Scale the peak discharge directly, since $C_p$ is common to both basins.
From $Q_p=2.78\,C_p\,A/t_{pR}$, $C_p$ cancels between the two catchments, leaving \[ Q_p(N) = Q_p(M)\times\frac{A(N)}{A(M)}\times\frac{t_{pR}(M)}{t_{pR}(N)} = 50\times\frac{400}{250}\times\frac{9}{11.15} \] \[ Q_p(N) = 50\times1.6\times0.807 \approx 64.58\ \text{m}^3/\text{s} \]
Final Answer:
Scaling straight from catchment M's peak gives the same result, about $64.58$ m$^3$/s. \[ \boxed{Q_p(N) \approx 64.58\ \text{m}^3/\text{s}} \]