Answer:
δu/δx+δu/δy = 6x-6x =0
9r^2
Explanation:
The flow is obviously two-dimensional, since the stream function depends only on the x and y coordinate. We can find the x and y velocity components by using the following relations:
u =δψ/δy = 3x^2-3y^2
v =-δψ/δx = -6xy
Now, since:
δu/δx+δu/δy = 6x-6x =0
we conclude that this flow satisfies the continuity equation for a 2D incompressible flow. Therefore, the flow is indeed a two-dimensional incompressible one.
The magnitude of velocity is given by:
|V| = u^2+v^2
=(3x^2-3y^2)^2+(-6xy)^2
=9x^4+18x^2y^2+9y^2
=(3x^2+3y^2)^2
=9r^2
where r is the distance from the origin of the coordinates, and we have used that r^2 = x^2 + y^2.
The streamline ψ = 2 is given by the following equation:
3x^2y — y^3 = 2,
which is most easily plotted by solving it for x:
x =±√2-y^3/y
Plot of the streamline is given in the graph below.
Explanation for the plot: the two x(y) functions (with minus and plus signs) given in the equation above were plotted as functions of y, after which the graph was rotated to obtain a standard coordinate diagram. The "+" and "-" parts are given in different colors, but keep in mind that these are actually "parts" of the same streamline.
Answer:
a) ≈ 30 mins
b) 8 vpm
Explanation:
<u>a) Determine how long after the first vehicle arrival will the queue dissipate</u>
The time after the arrival of the first vehicle for the queue to dissipate
= 29.9 mins ≈ 30 mins
<u>b) Determine the average service rate at the parking lot gate </u>
U = A / t
where : A = 240 vehicles , t = 30
U = 240 / 30 = 8 Vpm
attached below is a detailed solution of the given problems above
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
Answer:
The answer is 2.715 In
Explanation:
An estimated 60% of annual precipitation in a watershed (drainage area = 20000 acres) is evaporated. If the average annual river flow at the outlet of the basin has been observed to be 2.5 cfs, determine the annual precipitation (inches) in the basin
The annual precipitation in inches in the basin is 2.715 inches.
The solution and steps is explained in the attachment.
I hope i have been able to help.
Answer:
Explanation:
Given data:
mass = 2.00 kg
slope angle = 38.0
From figure
balancing force
.....1
Balancing torque
......2
for pure rolling
from 1 and 2nd equation
N =normal force
solving for coefficent of friction we get
