Brainstorming is the problem-solving method engineers use most.TrueFalse
1 answer:
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Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above and also note that I derived the whole expressions in variable firm that are required to solve this question and giving reasons for each step so that you can employ these steps in solving new problems.
Answer:
The graphs are attached below
Explanation:
Ans) We know,
F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]
where K = hydraulic conductivity =0.1 cm/hr
z = capillary suction =20cm
Δ∅ = ∅ (1 - Se)
∅ = effective porosity , Se = initial moisture content
Δ∅ = 0.40(1 - 0.15)
Δ∅ = 0.34
Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]
F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]
Also, infiltration rate (f),
f = K [(zΔ∅ + F)/F]
f = 0.40 [6.8 + F]/F
Condition of ponding,
Ponding time tp = K zΔ∅ i(i-k)
where, i = rainfall rate (1cm/hr)
tp = 0.40(6.8) / 0.60
tp= 4.53 hours
Now, cumulative infiltraton at ponding Fp = i tp
Fp = 1 x 4.53 or 4.53 cm
For infiltration at time less then ponding time , infiltration rate = rainfall rate
For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm
For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm
For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm
For t = 1 hr , f = 1cm/hr ; F = 1cm/hr
For t > tp ,
Equivalent time origin(to),
to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)
to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)
to = 4.53 - 2.5 ( 4.53 - 3.44)
to = 1.82 hr
Hence,
F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]
Solving above equation for t by assuming F, and further solving equation for infiltration rate
Ans (a) Following is required curve :
O.S 1-5 2 time Chr)ホー Lt 2 time Chr)
Ans (c) Following is required table :
Ans (d) For time step t = 0.25 hrs
At t < tp ; i = f = 1 cm/hr
F = i x t
F = 1 x 0.25
F = 0.25 cm
