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Answer:
The height of the water is 1.25 m
Explanation:
copper properties are:
Kc=385 W/mK
D=20x10^-3 m
gc=8960 kg/m^3
Cp=385 J/kg*K
R=10x10^-3 m
Water properties at 280 K
pw=1000 kg/m^3
Kw=0.582
v=0.1247x10^-6 m^2/s
The drag force is:
The bouyancy force is:
The weight is:
Laminar flow:
Reynold number:
Not flow region
For Newton flow region:
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.
