California State University, Northridge



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|College of Engineering and Computer Science

Mechanical Engineering Department

Mechanical Engineering 390

Fluid Mechanics | |

| |Spring 2008 Number: 11971 Instructor: Larry Caretto |

Exercise Two Solutions – Forces on Submerged Objects and Buoyancy

1. A rectangular gate that is 2 m wide is located in the vertical wall of a tank containing water as shown in the figure at the right copied from the text. It is desired to have the gate open automatically when the depth of the water above the top of the gate reaches 10 m. (a) At what distance, d, should the frictionless horizontal shaft be located? (b) What is the magnitude of the force on the gate when it opens? (Problem 2.63 in text.)

In this problem the horizontal shaft is located in the middle of the gate. Part of the gate is above the shaft and part is below it. If the net pressure below the location of the shaft is greater than that above the shaft the gate will remain closed because of the block on the right side of the bottom of the gate. However, if the net pressure above the shaft is greater than the net pressure below the shaft, the gate will open because there is no blockage in the gate.

The figure at the left shows the gate with the coordinates of the centroid and the the center of pressure. (In this figure yCP is denoted as yR.) If the hinge is located at the center of pressure, yR, the net pressure forces on the top and bottom of the shaft will be exactly the same. Any increase in depth will cause the gate to open.

In the figure as shown the coordinate of the centroid of the gate, yc = 10 m + 2 m = 12 m relative to the surface of the liquid. The area of the gate is (4 m)(2 m) = 8 m2. The 4 m side is in the vertical direction that is used in the formula for the moment of inertia about the centroid: Ixc = (horizontal side) (vertical side) / 12. Using the equation for the center of pressure gives the following result.

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From the diagram we see that the location of yR equals d + 10 m so that d = yR – 10 m; d = 2.11 m

To determine the force we can use our usual equation for the resultant force with an angle of 90o and the specific weight of water.

[pic] FR = 941 kN

2. The closed vessel shown at the right (copied from the text) contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout that is closed by a 6-inch circular diameter gate that is hinged along one side as illustrated. (Note that the slope of the gate is that of the hypotenuse of a right triangle with sides of 3 and 4 as shown.) The horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum force that must be applied at the hinge to hold the gate shut. Neglect the weight of the gate and the friction at the hinge. (Problem 2.67 in text.)

Here we have the usual computation of the forces on a submerged surface with an additional force, the air pressure at the top of the water. When we analyzed the force on a submerged object in class, we computed the force as the integral (dF = (PdA = ((p0 + γysinθ)dA with p0 = 0. In this case, p0 = 10 psi so it must be considered in the integral. The result is simple: (p0dA = p0A.

Furthermore, when we consider the point where this equivalent force due to p0 acts, it will be exactly at the centroid, not at the centre of pressure. Thus we have to consider two separate forces on the gate: F1 is the force of the pressure at the surface of the liquid, p0, which acts at the centroid, and F2 is the usual force from the pressure of the fluid, which acts at the center of pressure. This is shown in the diagram at the left.

The force acting on the gate due to the pressure at the top of the water, p0 is simply p0 times the area, A, of the gate.

[pic]

The pressure force from the liquid is found by the usual formula, FR = γhcA = γycAsinθ; in this case θ = tan-1(3/4) = 0.6435 so sinθ = sin(0.6453) = 0.6. The value of hc is the 10 ft from the water surface to the top of the liquid, plus sinθ times the (6 in)/2 = 0.25 ft distance from the top of the circular gate to the centroid in the center of the circular gate. This gives hc = 10 ft + (0.5 ft)sinθ = 10 ft + (0.25 ft)(0.6) = 10.15 ft. So, the pressure force due to the water is found to be

[pic]

To compute the location of the resultant pressure force we need the distance yc from the top of the water to the centroid along the slanted line. We can find this as yc = hc/sinθ = (10.15 ft)/(0.6) = 16.916667 ft. For a circle of radius R, the moment of inertia about the centroid is πR4/4. The location of the center of pressure is then found as follows.

[pic]

The net moment about the hinge is the sum of the two moments from the two forces. The value of yR found above is essentially the same as the centroid so we could assume that both forces act at the same point, the centroid of the circular gate, 3 in = 0.25 ft from the hinge. The solution below illustrates the general approach where we have to determine the distance from the hinge, which is equal to 16.91759 ft – (10 ft)/sinθ, to get the correct moment arm. (This calculation gives a result of 0.2509 ft; so, using the centroid coordinate, 0.25, would give an error of only 0.4%.) Setting the sum of the moments about the hinge to zero gives the following equation for the resulting hinge moment, C.

[pic]

C = 102 ft▪lbf

3. A plate of negligible weight closes a 1-ft diameter hole in a tank containing air and water as shown in the figure at the right (copied from the text.) A block of concrete (specific weight = 150 lbf/ft3), having a volume of 1.5 ft3, is suspended from the plate and is completely immersed in the water. As the air pressure is increased the differential reading, Δh, on the inclined-tube mercury manometer increases. Determine Δh just before the plate starts to lift off the hole. The weight of the air has a negligible effect on the manometer reading. (Problem 2.88 in text.)

When the plate is just about to lift off the hole, the pressure force on the plate is just equal to the weight of the concrete less the buoyancy force on the concrete. (See the free body diagram at the left.) In equation form this gives pA = W – FB. (Note that the use of a gage pressure gives the net resultant force of the pressure inside the tank acting on the bottom of the plate and the atmospheric pressure outside the tank acting on the top of the plate.) The weight of the block is simply the volume of the block times the specific weight of the concrete:

W = γconcreteV = [pic]

The buoyancy force is the weight of the displaced water which is γwaterV = [pic]

The area of the plate on which the pressure acts is A = πD2/4 = π(1 ft)2/4 = 0.7854 ft2.

Applying these values in the force balance gives 225 lbf = 93.6 lbr + p(0.7854 ft2) so that p = (225 lbf – 93.6 lbr) / (0.7854 ft2) = 167 lbf/ft2. This air pressure causes the manometer displacement of Δh so that the vertical height difference of the mercury column is Δh sin30o. This elevation difference is caused by the difference between the (gage) air pressure inside the tank and the open side of the manometer where the gage pressure is zero. Applying the manometer equation gives

[pic]

Δh = 0.394 ft

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