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Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062

College of Engineering and Computer ScienceMechanical Engineering Department

Mechanic al Engin eering 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 verticalwall of a tank containing water as shown in the figure at theright copied from the text. It is desired to have the gate openautomatically when the depth of the water above the top of the gate reaches 10 m. (a) At what distance, d, should thefrictionless horizontal shaft be located? (b) What is themagnitude of the force on the gate when it opens? (Problem2.63 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 thanthat above the shaft the gate will remain closed because of theblock on the right side of the bottom of the gate. However, if thenet pressure above the shaft is greater than the net pressurebelow the shaft, the gate will open because there is no blockagein the gate.

The figure at the left shows the gate with thecoordinates of the centroid and the thecenter of pressure. (In this figure yCP isdenoted as yR.) If the hinge is located at thecenter of pressure, yR, the net pressureforces on the top and bottom of the shaft willbe exactly the same. Any increase in depthwill cause the gate to open.

In the figure as shown the coordinate of thecentroid of the gate, yc = 10 m + 2 m = 12 mrelative to the surface of the liquid. The areaof the gate is (4 m)(2 m) = 8 m

2. The 4 m

side is in the vertical direction that is used inthe formula for the moment of inertia aboutthe centroid: Ixc = (horizontal side) (verticalside) / 12. Using the equation for the center

of pressure gives the following result.

mmmm

mm y

A y

I y c

c

xc R 11.1212

812

12

42

2

3

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.

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Exercise two solutions ME 390, L. S. Caretto, Spring 2008

o

c R mmm

kN A y F 90sin812

80.9sin

2

3 F R = 941 kN

2. The closed vessel shown at the right(copied from the text) contains water withan air pressure of 10 psi at the water

surface. One side of the vessel containsa spout that is closed by a 6-inch circular diameter gate that is hinged along oneside as illustrated. (Note that the slope of the gate is that of the hypotenuse of aright triangle with sides of 3 and 4 asshown.) The horizontal axis of the hingeis located 10 ft below the water surface.Determine the minimum force that mustbe applied at the hinge to hold the gate shut. Neglect the weight of the gate and thefriction at the hinge. (Problem 2.67 in text.)

Here we have the usual computation of the forces on a submerged surface with an additionalforce, 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 + ysin)dA with p0 = 0.

In this case, p0 = 10 psi so it must be considered in the integral. The result is simple: p0dA = p0 A.

Furthermore, when we consider the point where this equivalent force due to p0 acts, it will beexactly at the centroid,not at the centre of pressure. Thus we haveto consider two separateforces on the gate: F1 isthe force of the pressureat the surface of theliquid, p0, which acts at

the centroid, and F2 isthe usual force from thepressure of the fluid,which acts at the center of pressure. This isshown in the diagram atthe left.

The force acting on the gate due to the pressure at the top of the water, p 0 is simply p0 times thearea, A, of the gate.

f

f lbin

in

lb A p F 2836

4

102

201

The pressure force from the liquid is found by the usual formula, FR = hc A = yc Asin; in this case = tan

-1(3/4) = 0.6435 so sin= sin(0.6453) = 0.6. The value of hc is the 10 ft from the water

surface to the top of the liquid, plus sin 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)sin =10 ft + (0.25 ft)(0.6) = 10.15 ft. So, the pressure force due to the water is found to be

f

f

c lbin

ft in f t

f t

lb Ah F 124

1446

415.10

4.62

2

2

2

32

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To compute the location of the resultant pressure force we need the distance y c from the top of

the water to the centroid along the slanted line. We can find this as yc = hc/sin = (10.15 ft)/(0.6)

= 16.916667 ft. For a circle of radius R, the moment of inertia about the centroid is R4/4. The

location of the center of pressure is then found as follows.

ft ft

ft ft

ft

y A y

I y c

c

xc R 91759.16916667.16

25.0916667.16

25.0

4 2

4

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 atthe same point, the centroid of the circular gate, 3 in = 0.25 ft from the hinge. The solution belowillustrates the general approach where we have to determine the distance from the hinge, which

is equal to 16.91759 ft – (10 ft)/sin, to get the correct moment arm. (This calculation gives aresult 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 theresulting hinge moment, C.

6.0

1091759.1612425.0238

2211

ft ft lb ft lb F F C f f

C = 102 ft▪lbf

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3. A plate of negligibleweight closes a 1-ftdiameter hole in a tankcontaining air and water as shown in the figure atthe right (copied from thetext.) A block of concrete(specific weight = 150lbf /ft

3), having a volume of

1.5 ft3, is suspended from

the plate and is completelyimmersed in the water. Asthe air pressure isincreased the differential

reading, h, on the

inclined-tube mercury manometer increases. Determine h just before the plate starts tolift 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 thebuoyancy force on the concrete. (See the free body diagram atthe left.) In equation form this gives pA = W – FB. (Note that the

use of a gage pressure gives the net resultant force of thepressure inside the tank acting on the bottom of the plate and theatmospheric pressure outside the tank acting on the top of theplate.) The weight of the block is simply the volume of the blocktimes the specific weight of the concrete:

W = concreteV = f

f lb f t

f t

lb2255.1

1503

3

The buoyancy force is the weight of the displaced water which is water V = f

f lb ft

f t

lb6.935.1

4.62 3

3

The area of the plate on which the pressure acts is A = D2/4 = (1 ft)

2/4 = 0.7854 ft

2.

Applying these values in the force balance gives 225 lb f = 93.6 lbr + p(0.7854 ft2) so that p = (225

lbf – 93.6 lbr ) / (0.7854 ft2) = 167 lbf /ft

2. This air pressure causes the manometer displacement of

h so that the vertical height difference of the mercury column is h sin30o. This elevation

difference is caused by the difference between the (gage) air pressure inside the tank and theopen side of the manometer where the gage pressure is zero. Applying the manometer equationgives

3

2

4.626.13

0167

f t

lb f t

lb

SG p p p phh p p

f

f

water Hg

openair

Hg

openair Hg openair

h = 0.394 ft

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