1. The total flow at the outlet of a catchment is given in...

1. The total flow at the outlet of a catchment is given in the table below. This flow was the result of a rainfall event with three consecutive rainfall pulses, each lasting three hours. Each of the pulse delivered the following precipitation depths: 1.25 inches, 2.2 inches, and 1.55 inches, respectively. The area of the watershed is 80 acres. a. Use the straight-line method to determine the ordinates of the DRH for this storm. b. Determine the \(\phi\)-index for infiltration c. Plot the hyetograph of the rain event. Include the \(\phi\)-index in your plot (found in step b) Time Q Time Q (hrs) (cfs) (hrs) (cfs) -12 42 30 110 -6 41 36 75 0 40 42 53 6 90 48 45 12 155 54 42 18 210 60 40 24 180 66 40

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00:01 All right, so you know that the forest vegetation that covers a 120 kilometers square wildland watershed uses an average of 4 millimeters of water per day during the summer.

00:10 You also know that the soil averages 150 centimeters in depth, and that at field capacity can store 0 .2 millimeters of water per millimeter of soil depth.

00:20 However, at the beginning of the season, the soil was quite dry, holding an average of only 0 .1 millimeter per millimeter.

00:26 So as the watershed manager, you are asked to predict how much water.

00:30 Will be carried by the streams draining the watersheds during the 90 -day summer period when 450 millimeter of precipitation falls on the area so use the water balance equation to make a rep prediction of the stream discharge as a percentage of the precipitation and in cubic meters of water so first to start out here we will find the area of wildland watershed and that was 120 square kilometers, convert this over to 1 .2 times 10 to the 14 millimeter square.

01:25 So the volume of water received during the summer, we've written as 450 times 1 .2 times 10 to the 14th.

01:53 This would be millimeters cubed, and we should get an answer of 5 .4 times 10 to the 16th cubic millimeters.

02:16 So at the beginning of the season, the water content of the soil was equal to 1 ,500 times 0 .1 millimeter depth.

02:39 And this is equal to 150 millimeter depth.

02:48 So the total content of water at the beginning of the season would be 150 times 1 .2 times 10 to the 14th.

03:24 And then we should get an answer of 1 .8 times 10 to the 16th.

03:34 This would be in millimeters cubed.

03:41 So then the highest water depth can be 1 ,500 times 0 .2...

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