For the Horten model, the infiltration rate is equal to 0 cm/hr, the final infiltration rate is equal to 13 cm/hr and the time constant is 2hr-1. Determine the infiltration capacity at t=1, t=2, t=3, t=5, and the volume of water infiltrated up to 5 hours.

For the Horten model, the infiltration rate is equal to 0 cm/hr, the final infiltration rate is equal to 13 cm/hr and the time constant is 2hr-1. Determine the infiltration capacity at t=1, t=2, t=3, t=5, and the volume of water infiltrated up to 5 hours.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Problem # 1 The infiltration capacities of an area for different intervals of time are given below: time (hr)0 0.25 0.5 1.0 1.25 1.50 0.75 1.75 2.0 I (cm/hr) 10.0 5.5 3.1 2.2 1.55 1.25 1.2 1.0 1.0 Find the equation for infiltration capacity (Horton's equation).,..
If the rain starts at 8.00PM on 10November 2015, use Horton's Equation to solve for the infiltration capacity of the soil (mm per hat 12:30AM the next day. Given the initial infiltration capacity of the soil is 0.0386 cm per min, the time constant as 0.1018 per hour, and the capacity of soit for hrial infiltration is 0.6683 cm per hour. Detailed solution step by step
The infiltration capacity in a basin is represented by Horton's equation as fp=3.0+e^-2t where fp is in cm/h and t is in hours. Assuming the infiltration to take place at capacity rates in a storm of 60 min duration, estimate the depth in infiltration in every 15 minutes in an hour.
Q3. It was rained on a basin, where the soil has a k constant of 1.7, an initial infiltration rate of 3.2 mm/hr, and an infiltration rate of 2.3 mm/hr minutes later. The precipitation depths versus time are tabulated below: t, hr 600 615 630 645 700 715 730 745 800 815 80 845 900 915 | ΣΡ, cm 0.0 0.4 2.0 3.8 5.0 5.0 5.0 UH, (m/s)/cm o 10 35 65 85 75 50 35 25 18 12 6. 2 a) Obtain Horton's infiltration equation (10 points), b) Calculate the depth of total infiltrated water, the depth of surface runoff, and the runoff coefficient (20 points), c) 30-minute unit hydrograph (UH) ordinates are also tabulated above. Obtaining 15-minute UH and neglecting the losses except for the infiltration, determine the direct runoff hydrograph. (30 point).
Paragraph Problem 1: A pipe is laid horizontally and filled with soil, with its open end being saturated. With the given data, solve for its infiltration in 36mins if the soil column had been initially placed upright with its upper surface saturated. Area of the pipe's cross section = 212.6cm2 Infiltrated by 2126cm? of water after 26mins Hydraulic conductivity is 0.001cm/s (Final answer should be in 4 decimal places.) Problem 2: With the given data, solve for the infiltration rate and cumulative infiltration after 360mins on a sandy clay soil. Assume water is ponded to a small but negligible depth on the surface. Moisture Content of the soil = 34.5% Residual Moisture Content of the soil = 26% (Final answers should be in 4 decimal places.) The table will be provided in page 2 Problem 3: Using the data from Problem 2. What is its ponding time if: a) Intensity of rainfall is 0.000126cm/s. b) Intensity of rainfall is 5.8344cm/d. (Final answers should be in 4 decimal places.) Problem 4:…
instantaneous infiltration rate can be estimated by the equation, f= 0.5(a / +0.5) + b, where 'a' and 'b' are constants and 't' is the time elapsed from the beginning of the storm. find total infiltration during 10 hours if a=5 mm/hº.5 and b = 10 mm/h.
A 15 mm rainfall is recorded over a small catchment according to the table below.Infiltration tests in the area have given the following Horton equation parameters:fo= 27 mm/hr; fc = 6 mm/hr; k= 3.5 hr.Time (hr) Rainfall intensity (mm/hr)0.00-0.25 400.25-0.50 30.50-0.75 120.75-1.00 16(a) Calculate the infiltration capacity as a function of time. It is assumed that the soil surface is always ponded with water.(b) Can the Horton equation be applied to calculate infiltration for thisrainfall event?
The initial infiltration capacity fo of a watershed is estimated as 1.5 in./hr,and the time constant is taken to be 0.35 hr -1. The equilibrium capacity fe is 0.2 in./hr. Use Horton's equation to find (a) the values of fat t= 10 min, 30 min, 1 hr, 2 hr, and 6 hr, and (b) the total volume of infiltration over the 6-hr period
The infiltration rate for excess rain on a small area was observed to be 4.5 in/hr at the beginning of the rain, and it decreased exponentially toward an equilibrium of 0.5 in/hr. A total of 30 inches of water infiltrated during a 10-hour interval. Determine the value of rate of decrease in infiltration (k) using Horton's equation.
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
Lli 3 Find the infiltration rate if the initial rate = 1.3 in/hr. and the final rate = 0.2 * ?in/hr, k= 0.35-hr. after 0.5 hr. and 1 hr. using Hortonz equation 1.29, 1.12 O 1.43, 1.29 1.43,1.12 O
If the rain starts at 8:00PM on 10November2015, use Horton's Equation to solve for the infiltration capacity of the soil (mm per hr) at 3:12AM the next day. Given the initial infiltration capacity of the soil is 0.0386 cm per min, the time constant as 0.1018 per hour, and the capacity of soil for final infiltration is 0.6683 cm per hour.