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: Back out the adjusted basin lag for catchment M.
Time to peak from the start of rainfall equals lag plus half the duration: \(T_p = t_{pR} + t_R/2\).
So \(t_{pR}(M) = 12 - 6/2 = 9\) h.

Step 2: Recover the standard basin lag \(t_p\) of catchment M.
Snyder's duration correction is \(t_{pR} = t_p + \dfrac{t_R - t_p/5.5}{4}\). Substituting \(t_{pR}=9\), \(t_R=6\):
\[ 9 = t_p + \frac{6-t_p/5.5}{4} \implies t_p \approx 7.857\ \text{h}, \quad t_r = t_p/5.5 \approx 1.429\ \text{h} \]
Step 3: Find catchment coefficient \(C_t\) from M.
\(t_p = C_t (L\,L_{ca})^{0.3}\); with \(L\,L_{ca} = 36\times18=648\), \((648)^{0.3}\approx6.97\), so \[ C_t = 7.857/6.97 \approx 1.127 \]
Step 4: Find peaking coefficient \(C_p\) from M's given peak.
\(Q_p = \dfrac{2.78\,C_p\,A}{t_{pR}}\), so \(50 = \dfrac{2.78\,C_p\times250}{9}\), giving \(2.78 C_p = 1.8\) (i.e. \(C_p \approx 0.6475\)).

Step 5: Transfer \(C_t\), \(C_p\) to catchment N (meteorologically similar basins share these coefficients).
\(L\,L_{ca}(N) = 50\times30=1500\), \((1500)^{0.3}\approx8.97\), so \(t_p(N) = 1.127\times8.97 \approx 10.11\) h, and \(t_r(N)=10.11/5.5\approx1.838\) h.

Step 6: Adjust the lag for the same \(6\)-h duration in N.
\[ t_{pR}(N) = 10.11 + \frac{6-1.838}{4} \approx 11.15\ \text{h} \]
Step 7: Get the peak of N's \(6\)-h unit hydrograph.
\[ Q_p(N) = \frac{2.78\,C_p\,A(N)}{t_{pR}(N)} = \frac{1.8\times400}{11.15} \approx 64.58\ \text{m}^3/\text{s} \]
Final Answer:
The transposed \(6\)-h unit hydrograph for catchment N peaks near \(64.58\) m\(^3\)/s. \[ \boxed{Q_p(N) \approx 64.58\ \text{m}^3/\text{s}} \]