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The infiltration rate f in a basin under ponding condition is given by f = 30+ 10e-2t , where, f is in mm/h and t is time in hour. Total depth of infiltration (in mm, up to one decimal place) during the last 20 minutes of a storm of 30 minutes duration is _________
Correct answer is '(11.7)'. Can you explain this answer?
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The infiltration rate f in a basin under ponding condition is given by...
Infiltration rate f (t ) = 30+ 10 e−2t
Total infiltration depth in time 10 min. to 30 min. i.e., 0.166 hour to 0.5 hour
Total infiltration depth in time 10 min. to 30 min. i.e., 0.166 hour to 0.5 hour
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The infiltration rate f in a basin under ponding condition is given by...
Given:
- Infiltration rate, f = 30 * 10^(-2t) mm/h
- Time duration of the storm, t = 30 minutes
- Total depth of infiltration during the last 20 minutes of the storm
To Find:
- Total depth of infiltration during the last 20 minutes of the storm
Calculation:
To find the total depth of infiltration during the last 20 minutes of the storm, we need to integrate the infiltration rate function over the interval (t = 10 to t = 30).
We are given that the infiltration rate, f, is in mm/h and time, t, is in hours. So, we need to convert the time from minutes to hours.
1 minute = 1/60 hour
Therefore, t = 30 minutes = 30/60 = 0.5 hour
Integration:
The integral of the infiltration rate function f = 30 * 10^(-2t) with respect to t over the interval (t = 10 to t = 30) gives us the total depth of infiltration during the last 20 minutes of the storm.
∫(10 to 30) f dt = ∫(10 to 30) 30 * 10^(-2t) dt
To integrate this function, we can use the power rule of integration, which states that for any constant k, the integral of x^k dx is (x^(k+1))/(k+1).
Applying the power rule of integration, we have:
∫(10 to 30) 30 * 10^(-2t) dt = 30 * ∫(10 to 30) 10^(-2t) dt
Now, let u = -2t and du = -2dt. We can rewrite the integral as:
∫(10 to 30) 10^(-2t) dt = 30 * (-1/2) * ∫(10 to 30) 10^u du
Simplifying further:
∫(10 to 30) 10^u du = (-1/2) * [10^u] from 10 to 30
Plugging in the limits of integration:
(-1/2) * [10^30 - 10^10]
Calculating the values:
(-1/2) * [10^30 - 10^10] ≈ -1/2 * [1e30 - 1e10] ≈ -1/2 * (1e30) ≈ -0.5e30
Since the depth of infiltration cannot be negative, we take the absolute value:
| -0.5e30 | ≈ 0.5e30
Converting the depth of infiltration to mm:
0.5e30 mm ≈ 0.5 * 10^30 mm ≈ 5 * 10^29 mm
The answer is given up to one decimal place, so rounding to one decimal place:
5 * 10^29 mm ≈ 1.2 mm
Therefore, the total depth of infiltration during the last 20 minutes of the storm is approximately 1.2 mm.
- Infiltration rate, f = 30 * 10^(-2t) mm/h
- Time duration of the storm, t = 30 minutes
- Total depth of infiltration during the last 20 minutes of the storm
To Find:
- Total depth of infiltration during the last 20 minutes of the storm
Calculation:
To find the total depth of infiltration during the last 20 minutes of the storm, we need to integrate the infiltration rate function over the interval (t = 10 to t = 30).
We are given that the infiltration rate, f, is in mm/h and time, t, is in hours. So, we need to convert the time from minutes to hours.
1 minute = 1/60 hour
Therefore, t = 30 minutes = 30/60 = 0.5 hour
Integration:
The integral of the infiltration rate function f = 30 * 10^(-2t) with respect to t over the interval (t = 10 to t = 30) gives us the total depth of infiltration during the last 20 minutes of the storm.
∫(10 to 30) f dt = ∫(10 to 30) 30 * 10^(-2t) dt
To integrate this function, we can use the power rule of integration, which states that for any constant k, the integral of x^k dx is (x^(k+1))/(k+1).
Applying the power rule of integration, we have:
∫(10 to 30) 30 * 10^(-2t) dt = 30 * ∫(10 to 30) 10^(-2t) dt
Now, let u = -2t and du = -2dt. We can rewrite the integral as:
∫(10 to 30) 10^(-2t) dt = 30 * (-1/2) * ∫(10 to 30) 10^u du
Simplifying further:
∫(10 to 30) 10^u du = (-1/2) * [10^u] from 10 to 30
Plugging in the limits of integration:
(-1/2) * [10^30 - 10^10]
Calculating the values:
(-1/2) * [10^30 - 10^10] ≈ -1/2 * [1e30 - 1e10] ≈ -1/2 * (1e30) ≈ -0.5e30
Since the depth of infiltration cannot be negative, we take the absolute value:
| -0.5e30 | ≈ 0.5e30
Converting the depth of infiltration to mm:
0.5e30 mm ≈ 0.5 * 10^30 mm ≈ 5 * 10^29 mm
The answer is given up to one decimal place, so rounding to one decimal place:
5 * 10^29 mm ≈ 1.2 mm
Therefore, the total depth of infiltration during the last 20 minutes of the storm is approximately 1.2 mm.
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The infiltration rate f in a basin under ponding condition is given by f = 30+ 10e-2t , where, f is in mm/h and t is time in hour. Total depth of infiltration (in mm, up to one decimal place) during the last 20 minutes of a storm of 30 minutes duration is _________Correct answer is '(11.7)'. Can you explain this answer?
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