Rainfall, lag time and time of concentration from hydrograph

Reading the hyetograph: rainfall is 6 cm/h in hour 1, 12 cm/h in hour 2 and 6 cm/h in hour 3 (rain from t = 0 to 3 h). The hydrograph peaks at t = 5 h and ends at about t = 12 h.

(i) Total rainfall and average intensity

$$P = 6\times 1 + 12\times 1 + 6\times 1 = 24 \text{ cm}$$ $$i_{avg} = \frac{24}{3} = 8 \text{ cm/h}$$

Answer: total rainfall = 24 cm, average intensity = 8 cm/h.

(ii) Basin lag and time of concentration

Basin lag $t_L$ = time from the centre of mass of the rainfall to the peak of the hydrograph. The rain is symmetric (6, 12, 6), so its centroid is at 1.5 h.

$$t_L = 5 - 1.5 = 3.5 \text{ h}$$

Time of concentration $t_c$ = time from the end of the rainfall to the end of direct runoff (time for the farthest water to reach the outlet):

$$t_c = 12 - 3 = 9 \text{ h}$$

Answer: basin lag = 3.5 h; time of concentration ≈ 9 h. (If $t_c$ is measured only to the point of inflection on the recession limb, at about t = 8 h, then $t_c \approx 8 - 3 = 5$ h.)

(iii) Time base of flow

The direct runoff starts at t = 0 and ends at about t = 12 h.

Answer: time base $t_b$ ≈ 12 h.