Buoyancy.pptx

Buoyancy and Buoyancy Force

1. Sanjivani Rural Education Society’s Sanjivani College of Engineering, Kopargaon, 423603. An Autonomous Institute Affiliated to Savitribai Phule Pune University, Pune Subject: Fluid Mechanics Buoyancy and Buoyancy Force By Prof. V. S. Chaudhari (Assistant Professor) Department of Civil Engineering Sanjivani College of Engineering, Kopargaon-423603

2. Buo y ancy o r F o r ce o f B u o y ancy Wh e n a n y objec t is i m me r s ed in l iquid, the l i qui d e x ert some f o r ce o n th a t objec t . Th e v erti c a l f o r ce e x er t e d b y the liquid i s c alled as B u o y ancy o r F o r ce o f Bu o y an c y . Thi s f o r ce o f bu o y an c y is e qua l t o the w e i g h t o f the liquid displa c ed b y the bod y . F o r equilibrium c ondition , w eig h t o f the bo d y is e qua l t o the f o r ce e x er t ed b y the liqui d . C e n t r e o f Bu o y ancy : It i s t h e p oi n t th r ou g h whi c h f o r ce of bu o y ancy act s . It will act a t the c e n t e r o f g ra vity o f w eig h t o f li q ui d d i s p laced b y the bod y .

3. What are the laws of Floatation? Floatation depends upon the density. If an object has density less than the density of water, it floats. Principle of floatation is stated by the Archimedes. Archimedes,  the Ancient Greek scientist first stated the principle of floatation. According to him, all the objects placed in a liquid experience an upward force which allows the body to float if it displaces water with the weight equal to the weight of the body. This upward force is known as  buoyant force  and the law is known as  the law of buoyancy.  

4. Buo y ancy o r F o r ce o f B u o y ancy

5. Stability of Submerged or Floating Body For submerged of floating body stability means the tendency for the body to return to the original upright position after it has been displaced slightly When body is completely submerged in a fluid, the position of C.G and C.B are fixed The various relative position of C.G and C.B determines the stability of submerged body There are two types of displacement Linear displacement in horizontal or Vertical direction Angular displacement (Tilt)

6. Condition of S t abi lity o f subme r g ed Body S t abili t y o f s u bme r g ed bod y i s d et erm i n e d b y the positi o n of ce n t r e o f bu o y ancy with r espect t o ce n t r e o f g ra vit y . Wh e n c e n t r e o f bu o y a nc y i s ab o v e c e n t r e o f g ra vit y , th e n the subme r g ed bod y r emains in s t able equilibrium . Wh e n c e n t r e o f bu o y ancy is b el o w ce n t r e o f g ra vit y , then the subme r g ed bod y r emains in un s t able equilibrium .

7. Unstable equilibrium:  B below G. if the body is given a slight tilt, the couple of forces W and FB also in the same direction. Neutral equilibrium:  B coincide G. There is no torque couple formed by W and FB.

8. M e t a c e n t r e and M e t ac e n tric Heig h t Me t a c e n t r e : It i s d e f i n e d as a p oi n t with r e sp ect t o which a bod y oscill at e s in a liquid, w h en a bod y is til te d th r o ug h a smal l angle. It c an also b e d e fi ne d a s the i n t e r s ecting poi n t b et w e e n n e ut r al a x i s li n e o f t h e bod y an d li n e o f a ction o f f o r ce of bu o y an c y . Me t a c e n tric Heig h t : Di s t an c e be t w ee n the m e t a c e n t r e and ce n t er o f g ra vity is kn o wn a s m e t ace n tric heig h t.

9. S t abi lity o f flo a ting body A floating or submerged body is said to be stable if it comes back to its original position after giving a slight disturbance The condition of stability of these bodies depends on the position of its  metacentre, centre of buoyancy and centre of gravity . 

10. Conditions of Stability of floating body .

11. STABLE EQUILLIBRIUM UNSTABLE EQUILLIBRIUM NEUTRAL EQUILLIBRIUM M is above G M is Below G M and G Coincides GM is Positive GM is Negative GM=O Metacentric height is Positive Metacentric height is Negative -- BM>BG BM<BG BM=BG

12. S t abi lity o f flo a ting body S t abili t y o f flo a ting b od y is de t ermi n ed b y the posit ion of m e t a ce n t r e with r espect t o ce n t r e o f g ra vit y . Wh e n m e t a ce n t r e i s ab o v e c e n t r e o f g ra vit y , th e n t h e body r emains in s t able equilibriu m . Wh e n m e t a ce n t r e is bel o w ce n t r e o f g ra vi t y , th e n the body r emains in un s t able equilibrium .

13. Analyti c al m e thod o f d e t ermining M e t ac e n tric Heig h t In equilibrium situation, centre of gravity G and centre of buoyancy B will lie on same axis which is displayed here in following figure with a vertical line. Let us assume that we have given an angular displacement to the body in clockwise direction as displayed here in above figure. Centre of buoyancy will be shifted now towards right side from neutral axis and let us assume that it is now B1 When the ship is given small angular displacement in clockwise direction, after that effect of heeling of ship are Centre of buoyancy shifted to B to B1. Immersed portion of ship changed from ‘ABCD’ to ‘A1B1C1D1’ So By taking in account above two effect it is possible to find metacentric height of ship

14. 1. Centre of buoyancy shifted from B to B1  θ M B B1 2. Immersed portion of ship changed from ‘ABCD’ to ‘A1B1C1D1’ To evaluate overturning couple due to due to wedge consider two small elementary prism or strip of wedge at distance ‘x’ from longitudinal axis O-O Let us consider one small strip of thickness dx at a distance x from the centre O, at right side of the axis, as displayed in above figure dx A A1 O  θ x Area of the strip = x θ dx Volume of the strip = x θ dx L Where L is the length of the floating body Weight of the strip = ρ g (x θ) L dx Prism or strip of length ‘L’ Eq.1

15. Similarly, we will consider one small strip of thickness dx at a distance x from the centre O, at left side of the axis and we will have the weight of the strip ρ g x θ L dx. Above two forces i.e. weights are acting in the opposite direction and therefore there will be developed one couple. Eq.2 But these two couple are same hence equate eq. 1 and 2

16. Notes ɣ = W/V W = ɣ Volume W = ρ g Volume

17. Wt of block = ɣ x Volume W = 0.6*9.81*(Π/4 *4 2 )*3 = 221.89 KN Wt of water displaced = ɣwater * Volume of wood submerged 9.81 * *(Π/4 *4 2 )*h = 123.27 h

18. Analyti c al m e thod o f d e t ermining M e t ac e n tric Heig h t

19. Experime n t al m e thod o f d e t ermining M e t ac e n tric Heig h t

20. T hank Y ou