There is a thin layer of 10W-30 oil (?=0.17 Ns/m2 ) of thickness w=5 mm on each

Question

There is a thin

layer of 10W-30 oil (?=0.17 Ns/m2) of thickness w=5 mm on each of the interior walls of the duct. (Oil layer has a constant thickness and does not mix with the water). The remainder of the duct is filled with a plug of length L=5m, and mass m=900

kg, that also has a square cross-section. The plug is initially located at x=0, as shown

Duct is used to connect two water reservoirs. h1=50 m, and h2=30 m, and D=3m. Assume that the reservoirs are so big that their depths are not dependent on the position of the plug. a. What force is required to hold the plug in place at x=0? Assume for convenience that the plug

There is a thin layer of 10W-30 oil (?=0.17 Ns/m2) of thickness w=5 mm on each of the interior walls of the duct. (Oil layer has a constant thickness and does not mix with the water). The remainder of the duct is filled with a plug of length L=5m, and mass m=900 kg, that also has a square cross-section. The plug is initially located at x=0, as shown Duct is used to connect two water reservoirs. h1=50 m, and h2=30 m, and D=3m. Assume that the reservoirs are so big that their depths are not dependent on the position of the plug. a. What force is required to hold the plug in place at x=0? Assume for convenience that the plug's cross-section also has side length D. b. Now assume that the plug is allowed to slide freely in the duct. In which direction will it move? Find the differential equation that governs its motion. Ignore any forces related to the motion of the water. c. What's the speed of the plug when it reaches to a steady state (t??)? Is this realistic? Which assumption do you think is not appropriate and cause this unrealistic answer? What will be the new depth immersed in water if the mooring cord is cut?

Explanation / Answer

a)

Avg pressure on left side of plug = rho*g*(h1 - D/2)

Avg pressure on right side of plug = rho*g*(h2 - D/2)

Area = D^2

Average force on left side = rho*g*(h1 - D/2)*D^2

Average force on rightt side = rho*g*(h2 - D/2)*D^2

Net force due to fluid = rho*g*D^2*(h1 - h2) = Force required to hold the plug

Force required to hold the plug = 1000*9.81*(50 - 30)*3^2 = 1765800 N = 1765.8 kN

b)

shear stress on each side due viscosity = -mu*(du / dw)

Shear force = -mu*(du / dw)*D^2

Shear force on 4 walls = -4*mu*(du / dw)

Assuming linear velocity variation, du / dw = u/w

Shear force on 4 walls = -4*mu*(u / w)

Also, u = x' = dx / dt

Using Newton's 2nd law,

mx'' = 1765800 - 4*0.17*(x' / 0.005)

900*x'' = 1765800 - 136*x'

x'' = 1962 - 0.1511 x'

c)

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