- ahmad alhayek

Problem

A major storm event was recorded for Little Bear Creek. The incremental rainfall and measured hydrograph data for this storm are provided in Table in 1-hr increments. The drainage area for the basin is 3.25 mi2. Assume that the base flow for Little Bear is zero. (See Fig. 1-26.)

Table

Little Bear Creek Data

Date/Time		Incremental Precipitation (in.)	Flow (cfs)
6/8/01	1 6:00	0	0.0
6/8/01	1 7:00	0.05	0.5
6/8/01	18:00	0.04	0.5
6/8/01	1 9:00	0	0.6
6/8/01	20:00	0.08	14
6/8/01	21 :00	0.05	32
6/8/01	22:00	0.42	54
6/8/01	23:00	0.1	78
6/9/01	0:00	0.3	105
6/9/01	1 :00	0.14	158
6/9/01	2:00	0.19	201
6/9/01	3:00	0.45	228
6/9/01	4:00	0.01	258
6/9/01	5:00	0	272
6/9/01	6:00		225
6/9/01	7:00		189
6/9/01	8:00		142
6/9/01	9:00		119
6/9/01	1 0:00		98
6/9/01	1 1 :00		66
6/9/01	1 2:00		42
6/9/01	1 3:00		27
6/9/01	1 4:00		0

(a) Using storm hydrograph, estimate the vol of runoff in inches over the watershed.
(b) Estimate the vol of infiltration (in.).
(c) Estimate tp for the entire event.

Step-by-step solution

Step 1 of 12
(a)
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Step 2 of 12
Determine the total flow rate (Q) from 6/8/2001 (16:00) to 6/9/2001 (14:00):
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Step 3 of 12
Thus, the total flow rate storm event is .
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Step 4 of 12
Find the precipitation per hour:
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Step 5 of 12
Similarly, calculate and tabulate the value of for remaining time periods as in table (1).
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Step 6 of 12
Show the time increment as well as the flow rate for respective time period as in table (1).
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Step 7 of 12

Date	Time	Time increment (hrs)	Precipitation (in.)	(in./hr)	Flow (cfs)
6/8/2001	16:00	0	0	0	0
6/8/2001	17:00	1	0.05	0.05	0.5
6/8/2001	18:00	2	0.04	0.08	0.5
6/8/2001	19:00	3	0	0	0.6
6/8/2001	20:00	4	0.08	0.32	14
6/8/2001	21:00	5	0.05	0.25	32
6/8/2001	22:00	6	0.42	2.52	54
6/8/2001	23:00	7	0.1	0.7	78
6/9/2001	0:00	8	0.3	2.4	105
6/9/2001	1:00	9	0.14	1.26	158
6/9/2001	2:00	10	0.19	1.9	201

Comment

Step 8 of 12

Date	Time	Time increment (hrs)	Precipitation (in.)	(in./hr)	Flow (cfs)
6/9/2001	3:00	11	0.45	4.95	228
6/9/2001	4:00	12	0.01	0.12	258
6/9/2001	5:00	13	0	0	272
6/9/2001	6:00	-	-	-	225
6/9/2001	7:00	-	-	-	189
6/9/2001	8:00	-	-	-	142
6/9/2001	9:00	-	-	-	119
6/9/2001	10:00	-	-	-	98
6/9/2001	11:00	-	-	-	66
6/9/2001	12:00	-	-	-	42
6/9/2001	13:00	-	-	-	27
6/9/2001	14:00	-	-	-	0
SUM	1.83	14.55	2,309.6

Comment

Step 9 of 12
The volume can be determined by using the individual flows, timing it by the time interval and then adding them. The volume will be in cfs-hr. in this problem time interval is 1 hr.
Therefore, the sum flow is equal to the volume of runoff.
Convert the unit of area from square miles to acres:
Determine the volume of runoff that occurred in inches over the watershed using the formula.
Substitute 2,309.6 ac-inch for sum of volume and 2,080 acres for A.
Thus, the volume of the runoff over the watershed is .
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Step 10 of 12
(b)
The subtraction of total rainfall and amount of direct runoff gives the value of loss due to infiltration. The total rainfall can be found with the help of incremental precipitation.
Determine the volume of infiltration loss using the relation:
Substitute 1.83 inch for total precipitation and 1.11 inch for infiltration loss.
Thus, the volume of infiltration loss is .
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Step 11 of 12
(c)
Time to peak can be calculated knowing the center of mass of rainfall excess and the time to the peak of the hydrograph. The peak of the hydrograph happens at 13:00 hrs after the storm starts at 5 o’ clock.
Determine the time to canter of mass of rainfall using the formula.
Substitute 14.55 inch-hr for sum of and 1.83 inch for
Thus, the time to center of mass of rainfall is .
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Step 12 of 12
From table (1), the time to peak flow is 13 hrs.
Determine the time to peak using the formula.
Substitute 13 hrs for time to peak flow and 7.95 hrs for time to center of mass of rainfall.
Thus, the time to peak flow for this watershed for the entire storm is .
Comment