www.getmyuni.com 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 experienced by 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 the body is immersed in a fluid or floating on a fluid. This upward force is equal to the weight 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 the weight of the fluid displaced by the body, the Center of Buoyancy will be the center of the fluid displaced.

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www.getmyuni.com 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. Resultant horizontal 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 elementary

prisms of cross section d(A z). Consider vertical forces dF 1 and dF 2 acting on the two ends of the prism.

dF 1 = (P atm +ρgz 1)d(A z)

dF 2 = (P atm +ρgz 2) d(A z) The buoyant force acting on the element:

dF B = dF 2 - dF 1 =ρg(z 2–z1)d(A z) = ρg(dv) where dv = volume of the element. ∫ ∫ ∫ The buoyant force on the entire submerged body (F B) = v ρg(dv) = ρgV; Where V = Total volume of the submerged body or the volume of the displaced liquid.

LINE OF ACTION OF BUOYANT FORCE

To find the line of action of the Buoyant Force F B, take moments about z-axis, XBFB = ∫xdF B But F B = ρgV and dF B = ρg(dv) Substituting we get, X B = (1/V) v∫ ∫ ∫ xdv where X B = Centroid of the Displaced Volume.

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www.getmyuni.com ARCHIMEDES PRINCIPLE

The Buoyant Force (F B) is equal to the weight of the liquid displaced by the submerged body and acts vertically upwards through the centroid of the displaced volume. 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 the displaced liquid. The buoyant force depends upon the density of the fluid and submerged volume of the body. For a floating body in static equilibrium, the buoyant force is equal to the weight of the body.

Problem –1 Find the volume of the water displaced and the position of Center of Buoyancy for a wooden block of width 2.0m and depth 1.5m when it floats horizontally in water. Density of wooden block is 650kg/m 3 and its length is 4.0m.

Volume of the block, V = 12m 3 Weight of the block, ρgV= 76518N Volume of water displaced = 76,518 / (1000 × 9.81)= 7.8m 3 Depth of immersion, h =7.8 / (2 ×4) = 0.975m. The Center of Buoyancy is at 0.4875m from the base.

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

Let A = Cross sectional area of the block Weight 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 its weight in air. What will be its specific gravity? Volume of the body = 4.5m 3 = Volume of water displaced. Weight of water displaced = 1000 × 9.81 × 4.5 = 44145N For equilibrium, Weight of body in air – Weight of water displaced = Weight in water Wair = 44145N + 1962N = 46107N Mass of body = (46107 / 9.81) = 4700kg. Density = (4700 / 4.5) = 1044.4 kg/m 3 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 that the weight of the body acting through its Center of Gravity should be co-linear with the Buoyancy Force acting through the Center of Buoyancy. If the Body is Not Homogeneous in its distribution of mass over the entire volume, the location of Center of Gravity (G) does not coincide with the Center of Volume (B). Depending upon the relative locations of (G) and (B), the submerged body attains different states of equilibrium: Stable, Unstable and Neutral.

STABLE, UNSTABLE AND NEUTRAL EQUILIBRIUM

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

Fig. Stable Equilibrium

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

Fig. Unstable Equilibrium

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www.getmyuni.comNeutral Equilibrium: (G) and (B) coincide. The body will always assume the same position in which it is placed. A body having a small displacement and then released, neither returns to the original position nor increases it’s displacement- It will simply adapt to the new position.

Fig. Neutral Equilibrium

A submerged body will be in stable, unstable or neutral equilibrium if the Center 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 certain conditions even though (G) is above (B). When a floating body undergoes angular displacement about the horizontal position, the shape of the immersed volume changes 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), F B and W are co-linear. Fig. (b) shows the situation after the body has undergone a small angular displacement ( θ) with respect to the vertical axis. (G) remains unchanged relative to the body. (B) is the Center of Buoyancy (Centroid of the Immersed Volume) and it

moves towards the right to the new position [B 1]. The new line of action of the buoyant force through [B 1] which is always vertical intersects the axis BG (old vertical line through [B] and [G]) at [M]. For small angles of ( θ), point [M] is practically constant and is known as Meta Center.

Meta Center [M] is a point of intersection of the lines of action of Buoyant Force before and after heel. The distance between Center of Gravity and Meta Center (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 body is in stable equilibrium.

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

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

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For a floating body, stability is determined not simply by the relative positions of [B] and [G]. The stability is determined by the relative positions of [M] and [G]. The distance of the Meta-Center [M] above [G] along the line [BG] is known as the Meta-Centric height (GM). GM=BM-BG GM>0, [M] above [G]------Stable Equilibrium GM=0, [M] coinciding with [G]------Neutral Equilibrium GM<0, [M] below [G]------Unstable Equilibrium.

DETERMINATION OF META-CENTRIC HEIGHT

Consider a floating object as shown. It is given a small tilt angle( θ) from the initial state. Increase in the volume of displacement on the right hand side displaces the Center of Buoyancy from (B) to (B 1)

Fig. Determination of Meta-centric Height.

The shift in the center of Buoyancy results in the Restoring Couple = W (BM tan θ); Since F B= W; W=Weight of the body= Buoyant force= F B This is the moment caused by the movement of Center of Buoyancy from (B) to (B 1) Volume of the liquid displaced by the object remains same. Area AOA 1=Area DOD 1 Weight of the wedge AOA 1(which emerges out)=Weight of the wedge DOD 1(that was submerged) Let (l) and (b) be the length and breadth of the object. . Weight of each wedge shaped portion of the liquid = dFB = (w/2)(b/2)(b/2)(tan θ)(l) =[(wb 2 l tan θ)/8] w= ρg=specific weight of the liquid.

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There exists a buoyant force dFB upwards on the wedge (ODD 1) and dFB downwards on the wedge (OAA 1) each at a distance of (2/3)(b/2)=(b/3) from the center. The two forces are equal and opposite and constitute a couple of magnitude, 2 3/ dM= dFB (2/3)b =[(wb l tan θ)/8](2/3)b =w(lb 12)tan θ=wI YY tan θ Where, I YY is the moment of inertia of the floating object about the longitudinal axis. This moment is equal to the moment caused by the movement of buoyant force from (B) to (B 1).

W(BM) tan θ=w(I YY ) tan θ; Since W=wV, where V=volume of liquid displaced by the

object, wV(BM) tan θ=w I YY tan θ Therefore,

BM= (I YY / V) and GM = BM -BG= (I YY / V) - BG Where BM = [Second moment of the area of the plane of flotation about the centroidal axis perpendicular to the plane of rotation / Immersed Volume]

EXPERIMENTAL METHOD OF DETERMINATION OF META-CENTRIC HEIGHT

Let w 1= 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 (w 1) G=Center of gravity of the vessel B=Center of buoyancy of the vessel

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

Move weight (w 1) 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 Center of Gravity of the vessel will shift to (G 1) and the Center of Buoyancy will change to B1. Under equilibrium, the moment caused by the movement of the load (w 1) through a distance (x)=Moment caused by the shift of center of gravity from (G) to (G 1). Moment due to the change of G = W (GG 1) = W (GM tan θ)

Moment due to movement of w 1=w 1(x) = W (GM. tan θ)

Therefore, GM= [(w 1 x) / (Wtan θ)]

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Problem - 1 A block of wood (specific gravity=0.7) floats in water. Determine the meta-centric height if it’s size is 1m ×1m × 0.8m. Let the depth of immersion=h Weight of the wooden block = 5494 N Weight of water displaced = 1000 × 9.81 × 1×1 × h= 5494N Therefore, h=0.56 m; AB =0.28 m; AG=0.4 m; BG=0.12 m; GM=(I yy/ V) - BG Iyy =(1 / 12) m 4, V=0.56m 3 GM = 1 / (12 × 0.56) - 0.12 = 0.0288m (The body is Stable)

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

OB =(H / 2) and OG = H L= Length of the barge BG = (H / 2); BM = (I yy / V)= [(Lb 3 / 12) /(LbH)] = (b 2 /12H) GM=BM -BG= (b 2 / 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 of specific gravity 0.8. If the diameter of cylinder is ‘d’ and length is ‘L’, show that ‘L’ cannot exceed 0.817d for the cylinder to float with its longitudinal axis vertical.

Weight of the cylinder = Weight of oil displaced (πd2 /4) × L×600 ×g = ( πd2/ 4) × H×800 ×g; Therefore, H=0.75L OG=(L/2); OB=(H/2) = (3L/8) BG= OG - OB = (L/8) BM=(I yy /V); I yy = ( πd4/ 64); V=( πd2H /4) BM= (d 2 / 12L) GM=BM -BG = (d 2 / 12L) - (L /8); For Stable equilibrium, GM > 0; 0.817d > L or L< 0.817d

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