Hinged Valve Submerged in Water

1. The problem statement, all variables and given/known data

An automatic valve consists of a 225 X 225 mm square plate that is pivoted about a horizontal axis through A located at a distance h = 90 mm above the lower edge. Determine the depth (d) of the water for which the valve will open." Given: h = 90 mm, y = 225 mm
*Answer given in the back of the book = 300 mm*

Diagram: http://ift.tt/1hq3Rv9

2. Relevant equations

F = Pressure(P) * Area(A)

F = ∫∫A P * dA

Pressure P = ρgh, where
g = 9.81 m/s^2
h = depth in meters
ρ = density = 10^3 kg/m^3

3. The attempt at a solution

My theory: If the force on the valve above the hinge equals the force on the valve below the hinge, the valve will begin to open.

Force on Top: FTOP = [itex]\int ^{225}_{0}\int ^{225}_{90} \rho g(d-y) dydx[/itex]

Simplified: FTOP = 255[itex]\int ^{225}_{90} \rho g(d-y) dy[/itex]

Force on bottom: FBOT = [itex]\int ^{225}_{0}\int ^{90}_{0} \rho g(d-y) dydx[/itex]

Simplified: FBOT = 255[itex]\int ^{90}_{0} \rho g(d-y) dy[/itex]

[itex]\Rightarrow[/itex] FTOP = FBOT

Equate: 255[itex]\int ^{225}_{90} \rho g(d-y) dy[/itex] = 255[itex]\int ^{90}_{0} \rho g(d-y) dy[/itex]

Simplify: [itex]\int ^{225}_{90} (d-y) dy[/itex] = [itex]\int ^{90}_{0} (d-y) dy[/itex]

Integrate: (dy – [itex]\frac{y^2}{2}[/itex]|[itex]\stackrel{225}{90}[/itex]) = (dy – [itex]\frac{y^2}{2}[/itex]|[itex]\stackrel{90}{0}[/itex])

[itex]\Rightarrow[/itex] 135d – 21262.5 = 90d – 4050

[itex]\Rightarrow[/itex] 45d = 17212.5

[itex]\Rightarrow[/itex] d = 382.5 mm

Answer given in book = 300 mm

4. Notes

I know there are many ways to solve this problem. I would like to see a solution involving calculus because I believe that is where all other solution methods are derived from. Thanks in advance.

Captain1024

http://ift.tt/1gXVewI

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