Problem 1: A catchment has the following Horton’s infiltration parameters: f0=280 mm/hr, fc=25 mm/hr and k = 2.5 hr-1 . For the
rainfall distribution is given below Interval (mins) 0-10 10-20 20-30 30-40 40-50 Rainfall (mm/hr) 25 50 80 205 170 (a) What is the cumulative infiltration rate (F), infiltration rate (fp), depression storage (S) and runoff (R) at each time step (b) What time does ponding occur?
1 answer:
Answer:
Ponding will occur in 40mins
Explanation:
We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.
Consider checking attachment for the step by step solution.
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Answer:
vertical load = 10 kN
Modulus of elasticity = 200GPa
Yield stress on the cable = 400 MPa
Safety factor = 2.0
Explanation:
Data
let L =
= 3.35 m
substituting 1.5 m for h and 3 m for the tern (a + b)
= tan⁻¹(
)
= 45⁰
substituting 1.5 for h and 3 m for (a+ b) yields:
₂ = tan⁻¹ (
)
=25.56⁰
checking all the forces, they add up to zero. This means that the system is balanced and there is no resultant force.
Answer:
Q = 125.538 W
Explanation:
Given data:
D = 30 cm
Temperature degree celcius
Heat coefficient = 12 W/m^2 K
Efficiency 80% = 0.8
Q = 125.538 W