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FLUID MECHANICS

Fluid Statics

BUOYANCY

When a body is either wholly or partially immersed in a fluid, the hydrostatic

lift due to the net vertical component of the hydrostatic pressure forces experiencedby the body is called the Buoyant Force and the phenomenon is called Buoyancy.

Fig. Buoyancy

The Buoyancy is an upward force exerted by the fluid on the body when thebody is immersed in a fluid or floating on a fluid. This upward force is equal to theweight of the fluid displaced by the body.

CENTER OF BUOYANCY

(a) Floating Body (b) Submerged Body

Fig. Center of Buoyancy

Center of Buoyancy is a point through which the force of buoyancy is

supposed to act. As the force of buoyancy is a vertical force and is equal to theweight of the fluid displaced by the body, the Center of Buoyancy will be the centerof the fluid displaced.

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LOCATION OF CENTER OF BUOYANCY

Consider a solid body of arbitrary shape immersed in a homogeneous fluid.Hydrostatic pressure forces act on the entire surface of the body. Resultanthorizontal forces for a closed surface are zero.

Fig. Location of Center of Buoyancy

The body is considered to be divided into a number of vertical elementaryprisms of cross section d(Az). Consider vertical forces dF1 and dF2acting on the two

ends of the prism.

dF1 = (Patm +gz1)d(Az)dF

2= (Patm +gz2) d(Az)

The buoyant force acting on the element:dF

B= dF

2- dF1=g(z2z1)d(Az) = g(dv) where dv = volume of the element.

The buoyant force on the entire submerged body (FB) = vg(dv) = gV;Where V = Total volume of the submerged body or the volume of the displacedliquid.

LINE OF ACTION OF BUOYANT FORCE

To find the line of action of the Buoyant Force FB, take moments about z-axis,

XBFB = xdFBBut FB = gV and dFB = g(dv)Substituting we get, XB = (1/V) v xdv where XB = Centroid of the Displaced

Volume.

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ARCHIMEDES PRINCIPLE

The Buoyant Force (FB) is equal to the weight of the liquid displaced by the

submerged body and acts vertically upwards through the centroid of the displacedvolume.

Net weight of the submerged body = Actual weight Buoyant force.

The buoyant force on a partially immersed body is also equal to the weight of thedisplaced liquid. The buoyant force depends upon the density of the fluid andsubmerged volume of the body. For a floating body in static equilibrium, the buoyantforce is equal to the weight of the body.

Problem 1Find the volume of the water displaced and the position of Center of Buoyancy for awooden block of width 2.0m and depth 1.5m when it floats horizontally in water.

Density of wooden block is 650kg/m3 and its length is 4.0m.

Volume of the block, V = 12m3

Weight of the block, gV= 76518NVolume of water displaced =76,518 / (1000 9.81)= 7.8m3

Depth of immersion, h =7.8 / (24) = 0.975m.The Center of Buoyancy is at 0.4875m from the base.

Problem-2A block of steel (specific gravity= 7.85) floats at the mercury-water interface asshown in figure. What is the ratio (a / b) for this condition?(Specific gravity of Mercury = 13.57)

Let A = Cross sectional area of the blockWeight of the body = Total buoyancy forces

A(a+b) 7850 g=A(b 13.57 +a) g 1000

7.85 (a+b) = 13.57 b + a(a / b) = 0.835

Problem 3

A body having the dimensions of 1.5m 1.0m 3.0m weighs 1962N in water. Find itsweight in air. What will be its specific gravity?

Volume of the body = 4.5m3 = Volume of water displaced.

Weight of water displaced = 1000 9.81 4.5 = 44145NFor equilibrium,Weight of body in air Weight of water displaced = Weight in waterWair = 44145N + 1962N = 46107N

Mass of body = (46107 / 9.81) = 4700kg.

Density = (4700 / 4.5) = 1044.4 kg/m3

Specific gravity =1.044

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STABILITY OF UN-CONSTRAINED SUBMERGED BODIES

IN A FLUID

When a body is submerged in a liquid (or a fluid), the equilibrium requires thatthe weight of the body acting through its Center of Gravity should be co-linear withthe Buoyancy Force acting through the Center of Buoyancy. If the Body is NotHomogeneous in its distribution of mass over the entire volume, the location ofCenter of Gravity (G) does not coincide with the Center of Volume (B). Dependingupon the relative locations of (G) and (B), the submerged body attains differentstates of equilibrium: Stable, Unstable and Neutral.

STABLE, UNSTABLE AND NEUTRAL EQUILIBRIUM

Stable Equilibrium: (G) is located below (B). A body being given a smallangular displacement and then released, returns to its original position by retainingthe original vertical axis as vertical because of the restoring couple produced by theaction of the Buoyant Force and the Weight.

Fig. Stable Equilibrium

Unstable Equilibrium: (G) is located above (B). Any disturbance from theequilibrium position will create a destroying couple that will turn the body away fromthe original position

Fig. Unstable Equilibrium

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Neutral Equilibrium: (G) and (B) coincide. The body will always assume thesame position in which it is placed. A body having a small displacement and thenreleased, neither returns to the original position nor increases its displacement- Itwill simply adapt to the new position.

Fig. Neutral Equilibrium

A submerged body will be in stable, unstable or neutral equilibrium if theCenter of Gravity (G) is below, above or coincident with the Center of Buoyancy (B)respectively.

Fig. Stable, Unstable and Neutral Equilibrium

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STABILITY OF FLOATING BODIES

Stable conditions of the floating body can be achieved, under certainconditions even though (G) is above (B). When a floating body undergoes angulardisplacement about the horizontal position, the shape of the immersed volumechanges and so, the Center of Buoyancy moves relative to the body.

META CENTER

Fig. Meta Center

Fig. (a) shows equilibrium position; (G) is above (B), FBand W are co-linear.

Fig. (b) shows the situation after the body has undergone a small angulardisplacement () with respect to the vertical axis. (G) remains unchanged relative tothe body. (B) is the Center of Buoyancy (Centroid of the Immersed Volume) and it

moves towards the right to the new position [B1]. The new line of action of thebuoyant force through [B1] which is always vertical intersects the axis BG (old

vertical line through [B] and [G]) at [M]. For small angles of (), point [M] is practicallyconstant and is known as Meta Center.

Meta Center [M] is a point of intersection of the lines of action of BuoyantForce before and after heel. The distance between Center of Gravity and MetaCenter (GM) is called Meta-Centric Height. The distance [BM] is known as Meta-Centric Radius.

In Fig. (b), [M] is above [G], the Restoring Couple acts on the body in its

displaced position and tends to turn the body to the original position - Floating bodyis in stable equilibrium.

If [M] were below [G], the couple would be an Over-turning Couple and thebody would be in Unstable Equilibrium.

If [M] coincides with [G], the body will assume a new position without anyfurther movement and thus will be in Neutral Equilibrium.

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For a floating body, stability is determined not simply by the relative positionsof [B] and [G]. The stability is determined by the relative positions of [M] and [G]. Thedistance of the Meta-Center [M] above [G] along the line [BG] is known as theMeta-Centric height (GM).

GM=BM-BGGM>0, [M] above [G]------- Stable Equilibrium

GM=0, [M] coinciding with [G]------Neutral EquilibriumGM

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There exists a buoyant force dFB upwards on the wedge (ODD1) and dFB

downwards on the wedge (OAA1) each at a distance of (2/3)(b/2)=(b/3) from thecenter.The two forces are equal and opposite and constitute a couple of magnitude,

dM= dFB (2/3)b =[(wb2 l tan )/8](2/3)b =w(lb3/12)tan =wIYY tan Where, IYY is

the moment of inertiaof the floating object about the longitudinal axis.This moment is equal to the moment caused by the movement of buoyant force from(B) to (B1).

W(BM) tan =w(IYY

)tan ; Since W=wV, where V=volume of liquid displaced by the

object, wV(BM) tan =w IYY

tan Therefore,BM= (I

YY/V) and GM = BM-BG= (IYY /V) - BG

Where BM = [Second moment of the area of the plane of flotation about thecentroidal axis perpendicular to the plane of rotation / Immersed Volume]

EXPERIMENTAL METHOD OF DETERMINATION OF

META-CENTRIC HEIGHT

Let w1= known weight placed over the center of the vessel as shown in Fig. (a) and

vessel is floating. Let W=Weight of the vessel including (w1)

G=Center of gravity of the vesselB=Center of buoyancy of the vessel

Fig. Experimental method for determination of Meta-centric height.

Move weight (w1) across the vessel towards right by a distance (x) as shown in Fig.

(b). The angle of heel can be measured by means of a plumb line. The new Centerof Gravity of the vessel will shift to (G1) and the Center of Buoyancy will change to

B1. Under equilibrium, the moment caused by the movement of the load (w1)through a distance (x)=Moment caused by the shift of center of gravity from (G) to(G1).

Moment due to the change of G = W (GG1) = W (GM tan )

Moment due to movement of w1=w1(x) = W (GM. tan )Therefore, GM= [(w

1x) / (Wtan )]

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Problem - 1A block of wood (specific gravity=0.7) floats in water. Determine the meta-centric

height if its size is 1m1m 0.8m.Let the depth of immersion=hWeight of the wooden block = 5494 N

Weight of water displaced = 1000 9.81 11 h= 5494N

Therefore, h=0.56 m; AB =0.28 m; AG=0.4 m; BG=0.12 m;GM=(Iyy/V) - BG

Iyy=(1 / 12) m4, V=0.56m3

GM = 1 / (12 0.56) - 0.12 = 0.0288m(The body is Stable)

Problem - 2A rectangular barge of width b and a submerged depth H has its center of gravityat the water line. Find the meta-centric height in terms of (b/H) and show that forstable equilibrium of the barge, (b/ H) >6

OB =(H / 2) and OG = HL= Length of the bargeBG = (H / 2);

BM = (Iyy / V)= [(Lb3 / 12) /(LbH)] = (b2 /12H)

GM=BM-BG= (b2 / 12H) - (H/ 2)

=(H/2)[{(b/H)2 / 6}- 1];For stable equilibrium, GM > 0;

Therefore, [b /H] >6

Problem 3

A wooden cylinder having a specific gravity of 0.6 is required to float in an oil ofspecific gravity 0.8. If the diameter of cylinder is d and length is L, show that Lcannot exceed 0.817d for the cylinder to float with its longitudinal axis vertical.

Weight of the cylinder = Weight of oil displaced

(d2 /4) L600g = (d2/4) H800g;Therefore, H=0.75LOG=(L/2); OB=(H/2) = (3L/8)BG= OG - OB = (L/8)

BM=(Iyy /V); Iyy= (d4/64); V=(d2H /4)

BM= (d2 / 12L)

GM=BM -BG = (d2 / 12L) - (L /8);For Stable equilibrium,GM > 0; 0.817d > L or L< 0.817d