Discuss the ethics of the circumstances that resulted in the Columbia shuttle disaster. Considering the predictions that were ma
1 answer:
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Answer:
the volume flow rate per unit depth is:
the ratio is :
Explanation:
From the question; the equations of the velocities profile in the system are:
----- equation (1)
The above boundary condition can now be written as :
At y= 0; u =0 ----- (a)
At y = h; u =0 -----(b)
At y = ; u =
------(c)
where ;
A,B and C are constant
h = distance between two plates
u = velocity
= maximum velocity
y = measured distance upward from the lower plate
Replacing the boundary condition in (a) into equation (1) ; we have:
Replacing the boundary condition (b) in equation (1); we have:
Replacing the boundary condition (c) in equation (1); we have:
replacing for A in (d); we get:
replacing the values of A, B and C into the velocity profile expression; we have:
To determine the volume flow rate; we have:
Replacing
Thus; the volume flow rate per unit depth is:
Consider the discharge ;
Q = VA
where :
A = bh
Q = Vbh
Also;
Then;
Thus; the ratio is :
Answer:
138.68 kN
Explanation:
I assume the figure is the one included in my answer.
Let's say r is the distance from the hinge A. For a narrow section of the gate at this position, the length is dr, the width is w, and the area is dA.
dA = w dr
The pressure is:
P = ρgh
Using geometry, we can write h in terms of r.
(OA + r)² = h² + h²
(5√2 − 1.2 + r)² = 2h²
5√2 − 1.2 + r = √2 h
h = (5√2 − 1.2 + r) / √2
So the pressure at position r is:
P = ρg (5√2 − 1.2 + r) / √2
The force at position r is:
dF = P dA
dF = ρgw (5√2 − 1.2 + r) / √2 dr
The moment about hinge A caused by this force is:
dM = dF r
dM = (ρgw/√2) ((5√2 − 1.2) r + r²) dr
The total torque caused by the pressure is is:
M = ∫ dM
M = (ρgw/√2) ∫ ((5√2 − 1.2) r + r²) dr
M = (ρgw/√2) (½ (5√2 − 1.2) r² + ⅓ r³) [from r=0 to r=1.2]
M = (ρgw/√2) (½ (5√2 − 1.2) (1.2)² + ⅓ (1.2)³)
Sum of the moments on the gate:
∑τ = Iα
F(1.2) − M = 0
F = M / 1.2
F = (ρgw/√2) (½ (5√2 − 1.2) (1.2) + ⅓ (1.2)²)
Given that ρ = 1000 kg/m³, g = 9.81 m/s², and w = 5 m:
F = (1000 kg/m³ × 9.8 m/s² × 5 m /√2) (½ (5√2 − 1.2) (1.2) + ⅓ (1.2)²)
F = 138.68 kN
Round as needed.
