Question

The mass curve of an isolated storm in a 500 ha watershed as follows Time (hour) 0 2 4 6 8 10 12 14 16 18 Cumulative Rainfall (cm) 0 0.8 2.5 2.9 5.1 7.3 10.8 12 12.4 12.8 If the direct runoff produced by the storm is measured at the watershed as 0.340 Mm3 estimate the phi index of the storm​

Answer 1

Explanation:

The mass curve of an isolated storm in a 500 ha watershed is as follows: Time 0 2 4 6 8 10 12 14 16 18 from start (h) Cumulati 0 0.8 2.6 2.8 4.1 7.3 10.8 11.8 12.4 12.6 ve rainfall (cm) – If the direct runoff produced by the storm is measured at the outlet of the watershed as 0.340Mm", estimate the index of the storm and duration of rainfall excess.

Answer 2

Answer:

0.4 is the phi-index of the storm and the excess rainfall duration = 5*2 hours = 10 hours.

Explanation:

Runoff depth = Runoff/ Area

Compute rainfall intensities for all time pulses.

Assume all 18 hours (9 pulses) contributing to runoff.

Runoff Depth= Sum (rainfall intensities * time duration for each (i.e. 2 hours) )- phi index*18.

This gives you Phi = 0.322 means only 6 pulses with rainfall intensity > phi can contribute to runoff.

Similarly, assume 7 pulses contributing to runoff which gives phi= 0.385.

Now eliminate one more pulse and compute the value of phi using the same procedure which gives you its value as 0.4.