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An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.
Area of the catchment in hectares is (Answer up to the nearest integer)
Correct answer is '9720'. Can you explain this answer?
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An average rainfall of 16 cm occurs over a catchment during a period o...
To calculate the direct runoff hydrograph, we need to convolve the unit hydrograph with the rainfall excess.
Step 1: Calculate the rainfall excess
Total rainfall = 16 cm
Time period = 12 hours
Average rainfall intensity = Total rainfall / Time period = 16 cm / 12 hours = 1.33 cm/hour
Duration of the unit hydrograph = 6 hours
Time of concentration (Tc) = 0.6 x (Catchment length in km)^0.6 = assumed to be 1 hour
Time of concentration (Tc) is less than the duration of the unit hydrograph (Tc < t),="" therefore,="" the="" rainfall="" excess="" can="" be="" calculated="" />
Rainfall excess = (Rainfall intensity - Losses) x Duration
Losses = 0.2 x Square root of Tc = 0.2 x Square root of 1 = 0.2 cm
Rainfall excess = (1.33 - 0.2) x 6 = 6.78 cm
Step 2: Normalize the unit hydrograph
The area under the unit hydrograph is 1 cm x 6 hours = 6 cm-hours. To normalize the unit hydrograph, we need to divide each ordinate by the total area.
Normalized unit hydrograph = Unit hydrograph / Total area
Total area = 6 cm-hours
Normalized unit hydrograph = (0, 0), (5, 0.5/6), (10, 1/6), (15, 1.5/6), (20, 1/6), (25, 0.5/6), (30, 0)
Step 3: Convolve the rainfall excess with the normalized unit hydrograph
To convolve the rainfall excess with the normalized unit hydrograph, we need to shift the normalized unit hydrograph by the time of concentration (1 hour) and multiply each ordinate by the rainfall excess (6.78 cm).
Direct runoff hydrograph = Convolution of rainfall excess and normalized unit hydrograph
Time (hours) = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Rainfall excess (cm) = 6.78, 6.78, 6.78, 6.78, 6.78, 6.78, 0, 0, 0, 0, 0, 0, 0
Normalized unit hydrograph (cumecs) = 0, 0, 0.083, 0.25, 0.417, 0.583, 0.75, 0.583, 0.417, 0.25, 0.083, 0, 0
Direct runoff hydrograph (cumecs) = 0, 0, 0.566, 1.697, 2.825, 3.961, 4.95, 4.074, 2.998, 2.07, 1.269, 0.283, 0
Therefore, the direct runoff hydrograph peaks at 4.95 cumecs after 6 hours from the start of the rainfall excess. The total direct runoff volume is calculated by finding the
Step 1: Calculate the rainfall excess
Total rainfall = 16 cm
Time period = 12 hours
Average rainfall intensity = Total rainfall / Time period = 16 cm / 12 hours = 1.33 cm/hour
Duration of the unit hydrograph = 6 hours
Time of concentration (Tc) = 0.6 x (Catchment length in km)^0.6 = assumed to be 1 hour
Time of concentration (Tc) is less than the duration of the unit hydrograph (Tc < t),="" therefore,="" the="" rainfall="" excess="" can="" be="" calculated="" />
Rainfall excess = (Rainfall intensity - Losses) x Duration
Losses = 0.2 x Square root of Tc = 0.2 x Square root of 1 = 0.2 cm
Rainfall excess = (1.33 - 0.2) x 6 = 6.78 cm
Step 2: Normalize the unit hydrograph
The area under the unit hydrograph is 1 cm x 6 hours = 6 cm-hours. To normalize the unit hydrograph, we need to divide each ordinate by the total area.
Normalized unit hydrograph = Unit hydrograph / Total area
Total area = 6 cm-hours
Normalized unit hydrograph = (0, 0), (5, 0.5/6), (10, 1/6), (15, 1.5/6), (20, 1/6), (25, 0.5/6), (30, 0)
Step 3: Convolve the rainfall excess with the normalized unit hydrograph
To convolve the rainfall excess with the normalized unit hydrograph, we need to shift the normalized unit hydrograph by the time of concentration (1 hour) and multiply each ordinate by the rainfall excess (6.78 cm).
Direct runoff hydrograph = Convolution of rainfall excess and normalized unit hydrograph
Time (hours) = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Rainfall excess (cm) = 6.78, 6.78, 6.78, 6.78, 6.78, 6.78, 0, 0, 0, 0, 0, 0, 0
Normalized unit hydrograph (cumecs) = 0, 0, 0.083, 0.25, 0.417, 0.583, 0.75, 0.583, 0.417, 0.25, 0.083, 0, 0
Direct runoff hydrograph (cumecs) = 0, 0, 0.566, 1.697, 2.825, 3.961, 4.95, 4.074, 2.998, 2.07, 1.269, 0.283, 0
Therefore, the direct runoff hydrograph peaks at 4.95 cumecs after 6 hours from the start of the rainfall excess. The total direct runoff volume is calculated by finding the
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Community Answer
An average rainfall of 16 cm occurs over a catchment during a period o...
Area of unit hydrograph provides the volume of water due to 1 cm of ER
A x 0.01 = 1/2 x 30 x 18 x 60 x 60
A = 9720 x 104 m2
= 9720 hectare
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An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer?
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An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer?.
An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer?.
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An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer?, a detailed solution for An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer? has been provided alongside types of An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours.Φ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.Area of the catchment in hectares is (Answer up to the nearest integer)Correct answer is '9720'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.
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