Fluid Mechanics PDF

60 76 Fluid Mechanics Fluid is the name given to a substance which begins to flow when external force is applied E3 on it. Liquids and gases are fluids. Fluids do not have their own shape but take the shape of the containing vessel. The branch of physics which deals with the study of fluids at rest...

60 76 Fluid Mechanics Fluid is the name given to a substance which begins to flow when external force is applied E3 on it. Liquids and gases are fluids. Fluids do not have their own shape but take the shape of the containing vessel. The branch of physics which deals with the study of fluids at rest is called hydrostatics and the branch which deals with the study of fluids in motion is called hydrodynamics. ID 11.1 Pressure. The normal force exerted by liquid at rest on a given surface in contact with it is called thrust of liquid on that surface. U The normal force (or thrust) exerted by liquid at rest per unit area of the surface in contact with it is called pressure of liquid or hydrostatic pressure. D YG If F be the normal force acting on a surface of area A in contact with liquid, then pressure exerted by liquid on this surface is P  F / A (1) Units : N / m 2 or Pascal (S.I.) and Dyne/cm2 (C.G.S.) (2) Dimension : [P]  [F] [MLT 2 ]   [ML1 T  2 ] 2 [ A] [L ] (3) At a point pressure acts in all directions and a definite direction is not associated with it. So U pressure is a tensor quantity. ST (4) Atmospheric pressure : The gaseous envelope surrounding the earth is called the earth's atmosphere and the pressure exerted by the atmosphere is called atmospheric pressure. Its value on the surface of the earth at sea level is nearly 1.013  10 5 N / m 2 or Pascal in S.I. other practical units of pressure are atmosphere, bar and torr (mm of Hg) 1atm  1.01  10 5 Pa  1.01 bar  760 torr The atmospheric pressure is maximum at the surface of earth and goes on decreasing as we move up into the earth's atmosphere. (5) If P0 is the atmospheric pressure then for a point at depth h below the surface of a liquid of density  , hydrostatic pressure P is given by P  P0  h  g P0  h P Fluid Mechanics 77 60 (6) Hydrostatic pressure depends on the depth of the point below the surface (h), nature of liquid (  ) and acceleration due to gravity (g) while it is independent of the amount of liquid, ID E3 shape of the container or cross-sectional area considered. So if a given liquid is filled in vessels of different shapes to same height, the pressure at the base in each vessel's will be the same, though the volume or weight of the liquid in different vessels will be different. (A) (B) (C) U PA  PB  PC but W A  W B  WC U D YG (7) In a liquid at same level, the pressure will be same at all points, if not, due to pressure difference the liquid cannot be at rest. This is why the height of liquid is the same in vessels of different shapes containing different amounts of the same liquid at rest when they are in communication with each other. ST (8) Gauge pressure : The pressure difference between hydrostatic pressure P and atmospheric pressure P0 is called gauge pressure. P  P0  h g Problem 1. Sample problems based on Pressure If pressure at half the depth of a lake is equal to 2/3 pressure at the bottom of the lake then what is the depth of the lake (a) 10 m (b) 20 m (c) 60 m (d) 30 m Solution : (b) Pressure at bottom of the lake = P0  hg and pressure ay half the depth of a lake  P0  h g 2 According h to given 2 P0 2  10 5   20 m. g 10 3  10 condition P0  1 2 h g  (P0  h g) 2 3  1 1 P0  hg 3 6  78 Fluid Mechanics Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm3. If the mass of the other is 48 g, its density in g / cm3 is [CBSE 1994] 4 3 (a) (b) 3 2 (c) 3 m Solution : (c) Apparent weight  V (   )g   (   )g (d) 5 60 Problem 2. where m  mass of the body,   density of the body and   density of water If two bodies are in equilibrium then their apparent weight must be equal. 1 (1   )g  m2 2 ( 2   )g  36 48 (9  1)  ( 2  1)g. By solving we get 2  3. 9 2 An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm3) (a) 350 cm3 ID Problem 3. m1 E3  (b) 300 cm3 (c) 250 cm3 (d) 22 cm3 Solution : (b) According to Boyle's law, pressure and volume are inversely proportional to each other i.e. 1 V  P1V1  P2 V2  h D YG  (P0  h w g)V1  P0 V2 P 2V 2 U P  h w g  V1  V2   1  P0    47.6  10 2  1  1000  V2   1   70  13.6  1000    V1   (P1 V1) [As P2  P0  70 cm of Hg  70  13.6  1000 ]  V2  (1  5)50 cm 3  300 cm 3. A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on U Problem 4. ST the right contains mercury (density 13.6 g/cm3). The level of mercury in the narrow limb is at a distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water Water (a) 1.2 cm (b) 2.35 cm Mercur y (c) 0.56 cm (d) 0.8 cm Solution : (c) If the rise of level in the right limb be x cm. the fall of level of mercury in left limb be 4x cm because the area of cross section of right limb 4 times as that of left limb.  Level of water in left limb is (36 + 4x) cm. Now equating pressure at interface of mercury A B A 4x A' x B B' Fluid Mechanics 79 and water (at A' B') (36  4 x )  1  g  5 x  13.6  g By solving we get x = 0.56 cm. Problem 5. A uniformly tapering vessel is filled with a liquid of density 900 kg/m3. The force that acts on the base of the vessel due to the liquid is (g  10 ms 2 ) 60 Area = 10– 3 2 m (a) 3.6 N (b) 7.2 N 0.4 m (c) 9.0 N E3 Area=2 × 10–3m2 (d) 14.4 N Solution : (b) Force acting on the base F  P  A  hdgA  0.4  900  10  2  10 3  7.2 N Problem 6. A tank 5 m high is half filled with water and then is filled to the top with oil of density 0.85 (a) 1.85 g/cm2 ID g/cm3. The pressure at the bottom of the tank, due to these liquids is (b) 89.25 g/cm2 Problem 7. (d) 500 g/cm2 g g g  [250  1  250  0.85 ]  250 [1.85 ]  462.5 2 cm 2 cm cm 2 U Solution : (c) Pressure at the bottom P  (h1d1  h2 d 2 ) (c) 462.5 g/cm2 A siphon in use is demonstrated in the following figure. The density of the liquid flowing in D YG siphon is 1.5 gm/cc. The pressure difference between the point P and S will be Q 5 10 cm (a) 10 N/m (b) 2 × 105 N/m R 20 cm P S (c) Zero (d) Infinity U Solution : (c) As the both points are at the surface of liquid and these points are in the open atmosphere. So both point possess similar pressure and equal to 1 atm. Hence the pressure difference will be zero. The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio ST Problem 8. of density of mercury to that of air is 104. The height of the hill is (a) 250 m (b) 2.5 km (c) 1.25 km (d) 750 m Solution : (b) Difference of pressure between sea level and the top of hill P  (h1  h2 )   Hg  g  (75  50 )  10 2   Hg  g......(i) and pressure difference due to h meter of air P = h   air  g......(ii) By equating (i) and (ii) we get h  air  g  (75  50 )  10 2   Hg  g   Hg    25  10 2  10 4  2500 m  Height of the hill = 2.5 km.  h  25  10  2    air  11.2 Density. 80 Fluid Mechanics m dm  V 0 V dV (1) In case of homogenous isotropic substance, it has no directional properties, so is a scalar. In a fluid, at a point, density  is defined as:   lim 60 (2) It has dimensions [ML3 ] and S.I. unit kg/m3 while C.G.S. unit g/cc with 1 g / cc  10 3 kg / m 3 (3) Density of substance means the ratio of mass of substance to the volume occupied by the substance while density of a body means the ratio of mass of a body to the volume of the body. So for a solid body. Density of body = Density of substance E3 While for a hollow body, density of body is lesser than that of substance [As Vbody  Vsub. ] ID (4) When immiscible liquids of different densities are poured in a container the liquid of highest density will be at the bottom while that of lowest density at the top and interfaces will be plane. (5) Sometimes instead of density we use the term relative density or specific gravity which is defined as : Density of body Density of water U RD  (6) If m 1 mass of liquid of density  1 and m 2 mass of density  2 are mixed, then as D YG m  m 1  m 2 and V  (m 1 /  1 )  (m 2 /  2 )  If m 1  m 2  [As V  m /  ] m m1  m 2  mi   V (m1 / 1 )  (m 2 /  2 ) (m i / pi ) 2 1  2  Harmonic mean 1   2 (7) If V1 volume of liquid of density  1 and V2 volume of liquid of density  2 are mixed, then as: U m  1 V1   2 V2 and V  V1  V2 If V1  V2  V [As   m / V ]   (1   2 ) / 2 = Arithmetic Mean ST (8) With rise in temperature due to thermal expansion of a given body, volume will increase while mass will remain unchanged, so density will decrease, i.e.,  (m / V ) V0 V0    0 (m / V0 ) V V0 (1   ) or  [As V  V0 (1   ) ] 0 ~ –  0 (1   ) (1   ) (9) With increase in pressure due to decrease in volume, density will increase, i.e.,  (m / V ) V0   0 (m / V0 ) V But as by definition of bulk-modulus [As   m ] V Fluid Mechanics 81 B  V0 p  p  i.e., V  V0 1  B  V       0 1  So p   B  1 ~  0  1  p  B   Problem 9. 60 Sample problems based on Density A homogeneous solid cylinder of length L (L  H / 2). Cross-sectional area A / 5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L / 4 in the 5 d 4 (b) 4 d 5 (c) Ad d H/ 2 2d d 5 3L/4 L U (d) H/ 2 ID (a) E3 denser liquid as shown in the fig. The lower density liquid is open to atmosphere having pressure P0. Then density D of solid is given by [IIT-JEE1995] Solution : (a) Weight of cylinder = upthrust due to both liquids D YG ALd g D d A  A 3   A L V  D  g  . L   d  g  .   2d  g  . L  D. g    4 5 4 5 4  5 4 5  D  5 d 4 Problem 10. Density of ice is  and that of water is . What will be the decrease in volume when a mass M of ice melts (a) M   U Solution : (c) Volume of ice  (b) M    M , volume of water  M  1 1  (c) M        Change in volume  (d) M   1 1 1     M    1 1  M       M Problem 11. Equal masses of water and a liquid of density 2 are mixed together, then the mixture has a ST density of (a) 2/3 (b) 4/3 (c) 3/2 (d) 3 Solution : (b) If two liquid of equal masses and different densities are mixed together then density of mixture  2 1  2 2 1  2 4   12 3 1   2 Problem 12. Two substances of densities 1 and  2 are mixed in equal volume and the relative density of mixture is 4. When they are mixed in equal masses, the relative density of the mixture is 3. The values of  l and  2 are (a) 1  6 and  2  2 (b) 1  3 and  2  5 (c) 1  12 and  2  4 (d) 82 Fluid Mechanics Solution : (a) When substances are mixed in equal volume then density  1   2 2  1   2  8 4.......(i) When substances are mixed in equal masses then density  2 1  2 3 1   2  2 1  2  3(1   2 ) By solving (i) and (ii) we get  1  6 and 2  2. 60.......(ii) Problem 13. A body of density d1 is counterpoised by M g of weights of density d 2 in air of density d. Then the true mass of the body is Solution : (d) Let M 0  mass of body in vacuum.  d  (c) M 1   d1   M (1  d / d 2 ) (1  d / d1 ) E3  d   (b) M  1  d 2   (a) M (d) ID Apparent weight of the body in air = Apparent weight of standard weights in air  Actual weight – upthrust due to displaced air = Actual weight displaced air  d  M 1    d 2  dg  M 0  .   d  1   d1   U  M  dg  Mg     d2 D YG  M M 0 g   0  d1 – upthrust due to 11.3 Pascal's Law. It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same. or U The increase in pressure at one point of the enclosed liquid in equilibrium of rest is transmitted equally to all other points of the liquid and also to the walls of the container, provided the effect of gravity is neglected. ST Example : Hydraulic lift, hydraulic press and hydraulic brakes Working of hydraulic lift : It is used to lift the heavy loads. If a small force f is applied on piston of C then the pressure exerted on the liquid P  f /a [a = Area of cross section of the piston in C] This pressure is transmitted equally to piston of cylinder D. f Load C D F Hence the upward force acting on piston of cylinder D. FPA f A a  A f  a As A  a , therefore F  f. So heavy load placed on the larger piston is easily lifted upwards by applying a small force. Fluid Mechanics 83 11.4 Archimedes Principle. Accidentally Archimedes discovered that when a body is immersed partly or wholly in a fluid, in rest it is buoyed up with a force equal to the weight of the fluid displaced by the body. This principle is called Archimedes principle and is a necessary consequence of the laws of fluid 60 statics. When a body is partly or wholly dipped in a fluid, the fluid exerts force on the body due to hydrostatic pressure. At any small portion of the surface of the body, the force exerted by the fluid is perpendicular to the surface and is equal to the pressure at that point multiplied by the E3 area. The resultant of all these constant forces is called upthrust or buoyancy. To determine the magnitude and direction of this force consider a body immersed in a fluid of density  as shown in fig. The forces on the vertical sides of the body will cancel each other. The top surface of the body will experience a downward force. [As P  h g  P0 ] ID F1  AP1  A(h1g  P0 ) F2  AP2  A(h2g  P0 ) h1 h2 F2 U While the lower face of the body will experience an upward force. F1 As h2  h1 , F2 will be greater than F1 , so the body will experience a net upward force D YG F  F2  F1  A g(h2  h1 ) If L is the vertical height of the body F  A gL  Vg [As V  AL  A(h 2  h1 )] ] i.e., F = Weight of fluid displaced by the body. This force is called upthrust or buoyancy and acts vertically upwards (opposite to the weight of the body) through the centre of gravity of displaced fluid (called centre of buoyancy). U Though we have derived this result for a body fully submerged in a fluid, it can be shown to hold good for partly submerged bodies or a body in more than one fluid also. (1) Upthrust is independent of all factors of the body such as its mass, size, density etc. except the volume of the body inside the fluid. ST (2) Upthrust depends upon the nature of displaced fluid. This is why upthrust on a fully submerged body is more in sea water than in fresh water because its density is more than fresh water. (3) Apparent weight of the body of density (  ) when immersed in a liquid of density ( ).   Apparent weight = Actual weight – Upthrust  W  Fup  Vg  Vg  V (   )g  Vg1        W APP  W 1     (4) If a body of volume V is immersed in a liquid of density  then its weight reduces. 84 Fluid Mechanics W1 = Weight of the body in air, W2 = Weight of the body in water Then apparent (loss of weight) W1  W2  Vg  V  Relative density of a body (R.D.)= density of body Weight of body  = density of water Weight of equal volu me of water 60 (5) W1  W2 g Weight of body Water thru st W1 Weight of body in air Weight of body = = Weight in air – weight in water Loss of weight in water W1  W2 E3  (6) If the loss of weight of a body in water is 'a ' while in liquid is 'b' ID  L Upthrust on body in liquid a Wair  Wliquid Loss of weight in liquid      W Upthrust on body in water Loss of weight in water b Wair  W water  11.5 Floatation. (1) Translatory equilibrium : When a body of density  and volume V is immersed in a U liquid of density  , the forces acting on the body are Weight of body W  mg  Vg, acting vertically downwards through centre of gravity of the D YG body. Upthrust force = Vg acting vertically upwards through the centre of gravity of the displaced liquid i.e., centre of buoyancy. If density of body is equal to that of liquid    If density of body is lesser than that of liquid    Weight Weight will be less than upthrust so the body will move upwards and in equilibrium will float partially immersed in the liquid Such that, W  Vin g  V g  Vin g ST U If density of body is greater than that of liquid    Weight will be more than upthrust so the body will sink will be equal to upthrust so the body will float fully submerged in neutral equilibrium anywhere in the liquid. V  Vin Where Vin is the volume of body in the liquid Important points (i) A body will float in liquid only and only if    Fluid Mechanics 85 (ii) In case of floating as weight of body = upthrust So W App = Actual weight – upthrust = 0 (iii) In case of floating Vg  Vin g 60 So the equilibrium of floating bodies is unaffected by variations in g though both thrust and weight depend on g. (2) Rotatory Equilibrium : When a floating body is slightly tilted from equilibrium position, the centre of buoyancy B shifts. The vertical line passing through the new centre of ID E3 buoyancy B and initial vertical line meet at a point M called meta-centre. If the meta-centre M is above the centre of gravity the couple due to forces at G (weight of body W) and at B  (upthrust) tends to bring the body back to its original position. So for rotational equilibrium of floating body the meta-centre must always be higher than the centre of gravity of the body. Th M G B' G B B U (B) (C) D YG (A) W Th G M W B B' However, if meta-centre goes below CG, the couple due to forces at G and B  tends to topple the floating body. That is why a wooden log cannot be made to float vertical in water or a boat is likely to capsize if the sitting passengers stand on it. In these situations CG becomes higher than MC and so the body will topple if slightly tilted. U (3) Application of floatation (i) When a body floats then the weight of body = Upthrust ST  Vg  Ving  Vin    V      Vout  V  Vin  1   V   i.e., Fraction of volume outside the liquid fout  (ii) For floatation V  Vin    Vout    1   V   Vin   fin V If two different bodies A and B are floating in the same liquid then  A ( fin ) A   B ( fin )B (iii) If the same body is made to float in different liquids of densities  A and  B respectively. 86 Fluid Mechanics V  (Vin ) A  A  (Vin ) B  B   A (Vin ) B   B (Vin ) A (iv) If a platform of mass M and cross-section A is floating in a liquid of density  with its height h inside the liquid Mg  hA g 60......(i) Now if a body of mass m is placed on it and the platform sinks by y then (M  m )g  (y  h) A g......(ii)......(iii) E3 Subtracting equation (i) and (ii), mg  A  y g , i.e., W  y So we can determine the weight of a body by placing it on a floating platform and noting the depression of the platform in the liquid by it. ID Sample problems based on Archimedes principle Problem 14. A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance l U and h are shown there. After some time the coin falls into the water. Then Coin (a) l decreases and h increases l D YG (b) l increases and h decreases h (c) Both l and h increase (d) Both l and h decrease Solution : (d) As the block moves up with the fall of coin, l decreases, similarly h will also decrease because when the coin is in water, it displaces water equal to its own volume only. Problem 15. A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 10 3kg/m3. If U outer diameter and the density of the bowl are 1 m and 2 × 104 kg/m3 respectively, then the inner diameter of the bowl will be (a) 0.94 m (b) 0.97 m ST Solution : (c) Weight of the bowl = mg = Vg  (c) 0.98 m (d) 0.99 m 3 3 4  D   d          g 3  2   2   where D is the outer diameter , d is the inner diameter and  is the density of bowl 3 Weight of the liquid displaced by the bowl  Vg  4  D    g 3 2 where  is the density of the liquid. For the flotation 3 3 3 3  1 3  d 3  4 D 4  D   d   1    g         g     1.2  10 3        2  10 4 3 2 3  2   2   2  2    2  By solving we get d = 0.98 m. Fluid Mechanics 87 Problem 16. In making an alloy, a substance of specific gravity s1 and mass m1 is mixed with another substance of specific gravity s 2 and mass m 2 ; then the specific gravity of the alloy is Solution : (c) Specific gravity of alloy   m1  m 2  m1 m 2   p2  1     w       (c) m1  m 2  m1 m 2     s 2   s1 60  s1 s 2 (b)   m1  m 2     Density of alloy Mass of alloy  Volume of alloy  density of water Density of water m1  m 2 m  m2  1 m1 m2 m1 m 2   s1 s2 1 / w  2 / w  density of substance   As specific gravity of substance  density of water    E3  m  m2 (a)  1  s1  s 2  m1 m 2     s 2   s1 (d) m1  m 2 Problem 17. A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The (a) 8 ID specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be [AIIMS 1995] (b) 4 (c) 3 (d) Zero U Solution : (b) Let specific gravities of concrete and saw dust are 1 and  2 respectively. According to principle of floatation weight of whole sphere = upthrust on the sphere  D YG 4 4 4  (R 3  r 3 )1 g  r 3  2 g  R 3  1  g  R 3 1  r3 1  r3 2  R 3 3 3 3 R 3 (1  1)  r3 (1  2 )  R 3 1   2  1  1 r3  R 3  r 3 1  2  1  1   1  1 r3 (R 3  r 3 )1  1   2  1    r3 2  1  1   2 Mass of concrete  1  0.3  2.4  4  Mass of saw dust  2. 4  1  0.3 U  Problem 18. A vessel contains oil (density = 0.8 gm/cm3) over mercury (density = 13.6 gm/cm3). A ST homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in gm/cm3 is (a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8 Solution : (c) As the sphere floats in the liquid. Therefore its weight will be equal to the upthrust force on it Weight of sphere  4 R 3 g 3 Upthrust due to oil and mercury  Equating (i) and (ii)...... (i) 2 3 2 R   oilg  R 3 Hg g......(ii) 3 3 Oil Mercur y 88 Fluid Mechanics 4 2 2 R 3 g  R 3 0.8 g  R 3  13.6 g  2   0.8  13.6  14.4    7.2 3 3 3 Problem 19. A body floats in a liquid contained in a beaker. The whole system as shown falls freely under gravity. The upthrust on the body due to the liquid is 60 (a) Zero (b) Equal to the weight of the liquid displaced (c) Equal to the weight of the body in air E3 (d) Equal to the weight of the immersed position of the body Solution : (a) Upthrust  Vliquid (g  a) ; where, a = downward acceleration, V = volume of liquid displaced  Upthrust = 0 But for free fall a = g Problem 20. A metallic block of density 5 gm cm–3 and having dimensions 5 cm × 5 cm × 5 cm is weighed ID in water. Its apparent weight will be (a) 5 × 5 × 5 × 5 gf (b) 4 × 4 × 4 × 4 gf (c) 5 × 4 × 4 × 4 gf (d) 4 × 5 × 5 × 5 gf Solution : (d) Apparent weight  V (   )g  l  b  h  (5  1)  g  5  5  5  4  g Dyne or 4  5  5  5 gf. U Problem 21. A wooden block of volume 1000 cm3 is suspended from a spring balance. It weighs 12 N in air. It is suspended in water such that half of the block is below the surface of water. The reading of the spring balance is (b) 9 N D YG (a) 10 N (c) 8 N (d) 7 N Solution : (d) Reading of the spring balance = Apparent weight of the block = Actual weight – upthrust  12  Ving  12  500  10 6  10 3  10  12  5  7 N. Problem 22. An iceberg is floating in sea water. The density of ice is 0.92 gm/cm3 and that of sea water is 1.03g/cm3. What percentage of the iceberg will be below the surface of water (a) 3% (b) 11% U Solution : (c) For the floatation of ice-berg, (d) 92% Weight of ice = upthrust due to displaced water   0. 92  V    V  0. 89 V  1. 03    Vg  Vin g  Vin   (c) 89%  Vin  0.89 V or 89%. ST Problem 23. A cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with acceleration of g/3, the fraction of volume immersed in the liquid will be 1 2 3 (b) 8 2 (c) 3 3 (d) 4 (a) g 3 Fluid Mechanics 89  Solution : (a) Fraction of volume immersed in the liquid Vin   V i.e. it depends upon the densities of the   block and liquid. So there will be no change in it if system moves upward or downward with constant velocity or some acceleration. 60 Problem 24. A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8. The relative density of silver is 10.5. The tension in the string in kgwt is (b) 1.94 Solution : (b) Apparent weight  V (   )g  M  (c) 3.1 (d) 5.25   0.8   (   )g  M 1   g  2. 1 1   g  1.94 g Newton = 1.94 10.5      E3 (a) 1.6 Kg-wt Problem 25. A sample of metal weighs 210 gm in air, 180 gm in water and 120 gm in liquid. Then (a) Metal is 3 ID relative density (RD) of (b) Metal is 7 (c) Liquid is 3 (d) Liquid is 1 3 U Let the density of metal is  and density of liquid is . Solution : (b, c) If V is the volume of sample then according to problem........(i) 180  V (  1)g........(ii) 120  V (   )g........(iii) D YG 210  Vg By solving (i), (ii) and (iii) we get   7 and   3. Problem 26. Two solids A and B float in water. It is observed that A floats with half its volume immersed and B floats with 2/3 of its volume immersed. Compare the densities of A and B (a) 4 : 3 (b) 2 : 3 (c) 3 : 4 U Solution : (c) If two different bodies A and B are floating in the same liquid then (d) 1 : 3  A ( fin ) A 1 / 2 3     B ( fin ) B 2/3 4 ST Problem 27. The fraction of a floating object of volume V0 and density d 0 above the surface of a liquid of density d will be (a) d0 d (b) dd 0 d  d0 Solution : (c) For the floatation V0 d0 g  Vin d g  Vout  V0  Vin  V0  V0  Vin  V0 (c) d  d0 d (d) dd 0 d  d0 d0 d V d  d0 d0  d  d0   V0 .  out   V0 d d  d  Problem 28. A vessel with water is placed on a weighing pan and reads 600 g. Now a ball of 40 g and density 0.80 g/cc is sunk into the water with a pin as shown in fig. keeping it sunk. The weighing pan will show a reading 90 Fluid Mechanics (a) 600 g (b) 550 g (c) 650 g (d) 632 g 60 Solution : (c) Upthrust on ball = weight of displaced water m  40  V  g     g   1  g  50 g Dyne or 50 gm 0.8  E3 As the ball is sunk into the water with a pin by applying downward force equal to upthrust on it. So reading of weighing pan = weight of water + downward force against upthrust = 600 + 50 = 650 gm. ID 11.6 Some Conceptual Questions. Que.1 Why a small iron needle sinks in water while a large iron ship floats For floatation, the density of body must be lesser or equal to that of liquid. In case of iron needle, the density of needle, i.e., iron is more than that of water, so it will sink. However, the density of a ship due to its large volume is lesser than that of water, so it will float. U Ans. Que.2 A man is sitting in a boat which is floating in a pond. If the man drinks some water from Ans. D YG the pond, what will happen to the level of water in the pond If the man drinks m g of water from the pond, the weight of (boat + man) system will increase by mg and so the system will displace mg more water for floating. So due to removal of water from pond, the water level in pond will fall but due to water displaced by the floating system the water level in the pond will rise and so the water removed from the pond is equal to the water displaced by the system; the level of water in the pond will remain unchanged. Que.3 A boy is carrying a fish in one hand and a bucket full of water in the other hand. He then No, when he places the fish in water in the bucket, no doubt the weight of fish is reduced due to upthrust, but the weight of (water + bucket) system is increased by the same amount, so that the total weight carried by him remains unchanged. ST Ans. U places the fish in the bucket thinking that in accordance with Archimedes' principle he is now carrying less weight as the weight of the fish will reduce due to upthrust. Is he right Que. 4 A bucket of water is suspended from a spring balance. Does the reading of balance change (a) when a piece of stone suspended from a string is immersed in the water without touching the bucket? (b) when a piece of iron or cork is put in the water in the bucket? Ans. (a) Yes, the reading of the balance will increase but the increase in weight will be equal to the loss in weight of the stone (Vg) and not the weight of stone (Vg) [ Vg as    ]. (b) Yes, the reading of the balance will increase but the increase in weight will be equal to the weight of iron or cork piece. Fluid Mechanics 91 Que. 5 Why a soft plastic bag weighs the same when empty or when filled with air at atmospheric pressure? Would the weight be the same if measured in vacuum Ans. If the weight of empty bag is W0 and the volume of bag is V, when the bag is filled with air of density thrust of air Vg will decrease its weight; so W  W0  Vg  Vg  W0 60  at NTP, its weights will increase by Vg. Now when the bag filled with air is weighed in air, the i.e., the weight of the bag remains unchanged when it is filled with air at NTP and weighed in air. However if the bag is weighed in vacuum will be W0 when empty and (W0  Vg) when filled with air E3 (as there is no upthrust), i.e., in vacuum an air-filled bag will weigh more than an empty bag. Que.6 Why does a uniform wooden stick or log float horizontally? If enough iron is added to one end, it will float vertically; explain this also. When a wooden stick is made to float vertically, its rotational equilibrium will be unstable as its meta-centre will be lower than its CG and with Th Th a slight tilt it will rotate under the action of the couple formed by thrust and weight in the G G direction of tilt, till it becomes horizontal. B M B ID Ans. B W D YG U G However, due to loading at the bottom, the CG G of the stick (or log) will be lowered and so may W be lower than the meta-centre. In this situation the equilibrium will be stable and if the stick (or log) is tilted, it will come back to its initial vertical position. Que.7 A boat containing some pieces of material is floating in a pond. What will happen to the level of water in the pond if on unloading the pieces in the pond, the piece (a) floats (b) sinks? Ans. If M is the mass of boat and m of pieces in it, then initially as the system is floating M  m  VD W U i.e., the system displaces water VD  M W  m......(i) W When the pieces are dropped in the pond, the boat will still float, so it displaces water M  V1 W , ST i.e, V1  (M /  W ) (a) Now if the unloaded pieces floats in the pond, the water displaced by them m  V2 W , i.e, V2  (m /  W ) So the total water displaced by the boat and the floating pieces V1  V2  M W  m W.....(ii) Which is same as the water displaced by the floating system initially (eqn. 1); so the level of water in the pond will remain unchanged. (b) Now if the unloaded pieces sink the water displaced by them will be equal to their own volume, i.e, 92 Fluid Mechanics V2  m  as   V    m  and so in this situation the total volume of water displaced by boat and sinking pieces will be  M m V1  V2      W  .....(iii) 60 Now as the pieces are sinking    W , so this volume will be lesser than initial water displaced by the floating system (eq. 1); so the level of water in the pond will go down (or fall) 11.7 Streamline, Laminar and Turbulent Flow. E3 In this problem if the pieces (either sinking or floating) are unloaded on the ground, the water displaced after unloading, V2  M /  W , will be lesser than before unloading. V  (M  m ) /  W ; so the level of water in the pond will fall. ID (1) Stream line flow : Stream line flow of a liquid is that flow in which each element of the liquid passing through a point travels along the same path and with the same velocity as the preceeding element passes through that point. A streamline may be defined as the path, straight or curved, the tangent to which at any point gives the direction of the flow of liquid at that point. U The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the greater is the velocity of liquid particles at that place. A v1 B C v2 v3 D YG Path ABC is streamline as shown in the figure and v 1 , v 2 and v 3 are the velocities of the liquid particles at A, B and C point respectively. (2) Laminar flow : If a liquid is flowing over a horizontal surface with a steady flow and moves in the form of layers of different velocities which do not mix with each other, then the flow of liquid is called laminar flow. U In this flow the velocity of liquid flow is always less than the critical velocity of the liquid. The laminar flow is generally used synonymously with streamlined flow. (3) Turbulent flow : When a liquid moves with a velocity ST greater than its critical velocity, the motion of the particles of liquid becomes disordered or irregular. Such a flow is called a turbulent flow. In a turbulent flow, the path and the velocity of the particles Pilla r of the liquid change continuously and haphazardly with time from point to point. In a turbulent flow, most of the external energy maintaining the flow is spent in producing eddies in the liquid and only a small fraction of energy is available for forward flow. For example, eddies are seen by the sides of the pillars of a river bridge. 11.8 Critical Velocity and Reynold's Number. Fluid Mechanics 93 The critical velocity is that velocity of liquid flow upto which its flow is streamlined and above which its flow becomes turbulent. Reynold's number is a pure number which determines the nature of flow of liquid through a pipe. for a flowing fluid. NR  Inertial force per unit area Viscous force per unit area 60 It is defined as the ratio of the inertial force per unit area to the viscous force per unit area of liquid flowing through the tube per second E3 If a liquid of density  is flowing through a tube of radius r and cross section A then mass dm  volume flowing per second × density = dt  Inertial force per unit area = ID Av   dp / dt v(dm / dt) v Av  = = v2  A A A v U Viscous force per unit area F / A  r D YG So by the definition of Reynolds number N R  v2 v r Inertial force per unit area    Viscous force per unit area v / r If the value of Reynold's number (i) Lies between 0 to 2000, the flow of liquid is streamline or laminar. (ii) Lies between 2000 to 3000, the flow of liquid is unstable changing from streamline to turbulent flow. (iii) Above 3000, the flow of liquid is definitely turbulent. U Sample problems based on Stream lined and Turbulent flow ST Problem 29. In which one of the following cases will the liquid flow in a pipe be most streamlined (a) Liquid of high viscosity and high density flowing through a pipe of small radius (b) Liquid of high viscosity and low density flowing through a pipe of small radius (c) Liquid of low viscosity and low density flowing through a pipe of large radius (d) Liquid of low viscosity and high density flowing through a pipe of large radius Solution : (b) For streamline flow Reynold's number N R  r  should be less. For less value of N R , radius and density should be small and viscosity should be high. 94 Fluid Mechanics Problem 30. Two different liquids are flowing in two tubes of equal radius. The ratio of coefficients of viscosity of liquids is 52:49 and the ratio of their densities is 13:1, then the ratio of their critical velocities will be (b) 49 : 4 Solution : (a) Critical velocity v  N R (c) 2 : 7 (d) 7 : 2 v    52 1 4  1  1  2 .   r v 2  2 1 49 13 49 60 (a) 4 : 49 11.9 Equation of Continuity. E3 The equation of continuity is derived from the principle of conservation of mass. A non-viscous liquid in streamline flow passes through a tube AB of varying cross section. Let the cross sectional area of the pipe at points A and B be a1 and a 2 respectively. Let the liquid enter with normal velocity v 1 at A and leave with v2 B a1 v1 ID velocity v 2 at B. Let  1 and  2 be the densities of the liquid at point A and B respectively. a2 A a1v1 1  a2 v 2  2 U Mass of the liquid entering per second at A = Mass of the liquid leaving per second at B [If the liquid is incompressible (1   2 ) ] D YG a1v1  a 2 v 2 or av  constant or a 1 v v a This expression is called the equation of continuity for the steady flow of an incompressible and U non-viscous liquid. (1) The velocity of flow is independent of the liquid (assuming the liquid to be non-viscous) ST (2) The velocity of flow will increase if cross-section decreases and vice-versa. That is why : (a) In hilly region, where the river is narrow and shallow (i.e., small cross-section) the water current will be faster, while in plains where the river is wide and deep (i.e., large cross-section) the current will be slower, and so deep water will appear to be still. (b) When water falls from a tap, the velocity of falling water A1 v1 A2 v2 under the action of gravity will increase with distance from the tap (i.e, v 2  v1 ). So in accordance with continuity equation the cross section of the water stream will decrease (i.e., A2  A1 ), i.e., the falling stream of water becomes narrower. Sample problems based on Equation of continuity Fluid Mechanics 95 Problem 31. Two water pipes of diameters 2 cm and 4 cm are connected with the main supply line. The velocity of flow of water in the pipe of 2 cm diameter is (a) 4 times that in the other pipe (b) 1 4 times that in the (d) 1 2 times that in the (c) 2 times that in the other pipe 60 other pipe other pipe Solution : (a) d A  2 cm and d B  4 cm  rA  1 cm and rB  2 cm vA a  (rB ) 2  2   B      v A  4v B vB aA  (rA ) 2  1  av = constant  E3 2 From equation of continuity Problem 32. An incompressible liquid flows through a horizontal tube as shown in the following fig. Then the velocity v of the fluid is v2 = 1.5 m/s v1 = 3 m/s A ID A 1.5 A (b) 1.5 m/s D YG (a) 3.0 m/s U v (c) 1.0 m/s (d) 2.25 m/s Solution : (c) If the liquid is incompressible then mass of liquid entering through left end, should be equal to mass of liquid coming out from the right end.  M  m1  m 2  Av1  Av 2  1.5 A. v  A  3  A  1.5  1.5 A. v  v  1 m / s Problem 33. Water enters through end A with speed v1 and leaves through end B with speed v 2 of a cylindrical tube AB. The tube is always completely filled with water. In case I tube is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end B upwards. We have v1  v 2 for U (a) Case I (b) Case II (c) Case III (d) Each case ST Solution : (d) This happens in accordance with equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, either tube remain horizontal or vertical. Problem 34. Water is moving with a speed of 5.18 ms–1 through a pipe with a cross-sectional area of 4.20 cm2. The water gradually descends 9.66 m as the pipe increase in area to 7.60 cm2. The speed of flow at the lower level is (a) 3.0 ms–1 (b) 5.7 ms–1 (c) 3.82 ms–1 (d) 2.86 ms–1 Solution : (d) a 1 v 1  a 2 v 2  4.20  5.18  7.60  v 2  v 2  2.86 m / s 11.10 Energy of a Flowing Fluid. A flowing fluid in motion possesses the following three types of energy Pressure Energy Potential energy Kinetic energy 96 Fluid Mechanics It is the energy possessed by liquid by virtue of its height or position above the surface of earth or any reference level taken as zero level. It is the energy possessed by a liquid by virtue of its motion or velocity. Pressure energy of the liquid PV Potential energy of the liquid mgh Kinetic energy of the liquid Potential energy per unit mass of the liquid gh Kinetic energy per unit mass of Potential energy per volume of the liquid gh Kinetic energy per unit volume P  Pressure energy per volume of the liquid P unit unit the liquid ID of the liquid 1 mv 2 2 E3 Pressure energy per unit mass 60 It is the energy possessed by a liquid by virtue of its pressure. It is the measure of work done in pushing the liquid against pressure without imparting any velocity to it. of the liquid 1 v 2 2 U 11.11 Bernoulli's Theorem. 1 2 v 2 D YG According to this theorem the total energy (pressure energy, potential energy and kinetic energy) per unit volume or mass of an incompressible and C P2 v2 non-viscous fluid in steady flow through a pipe remains A2 constant throughout the flow, provided there is no source or P1 B h2 v1 sink of the fluid along the length of the pipe. Mathematically for unit volume of liquid flowing through a pipe. h1 1 2 v  constant 2 U P  gh  A1 To prove it consider a liquid flowing steadily through a tube of non-uniform area of crosssection as shown in fig. If P1 and P2 are the pressures at the two ends of the tube respectively, ST work done in pushing the volume V of incompressible fluid from point B to C through the tube will be W  P1 V  P2 V  (P1  P2 )V......(i) This work is used by the fluid in two ways. (i) In changing the potential energy of mass m (in the volume V ) from mgh1 to mgh2, i.e., U  mg (h 2  h1 ) (ii) In changing the kinetic energy from......(ii) 1 1 1 mv 12 to mv 22 , i.e., K  m(v 22  v12 ) 2 2 2 Now as the fluid is non-viscous, by conservation of mechanical energy W  U  K......(iii) Fluid Mechanics 97 (P1  P2 ) V  mg (h2  h1 )  P1  P2  g(h 2  h1 )  or or P1  gh1  or P  gh  1 m(v 22  v12 ) 2 1 (v 22  v 12 ) 2 [As   m / V ] 1 2 1 v 1  P2  gh2  v 22 2 2 60 i.e., 1 2 v  constant 2 E3 This equation is the so called Bernoulli's equation and represents conservation of mechanical energy in case of moving fluids. (i) Bernoulli's theorem for unit mass of liquid flowing through a pipe can also be written P   gh  1 2 v  constant 2 v2 P is called pressure head, h is called gravitational head and is called velocity 2g g D YG Here P v2 h = constant g 2g U (ii) Dividing above equation by g we get ID as: head. From this equation Bernoulli's theorem can be stated as. In stream line flow of an ideal liquid, the sum of pressure head, gravitational head and velocity head of every cross section of the liquid is constant. 11.12 Applications of Bernoulli's Theorem. U (i) Attraction between two closely parallel moving boats (or buses) : When two boats or buses move side by side in the same direction, the water (or air) in the region between them moves faster than that on the remote sides. Consequently in ST accordance with Bernoulli's principle the pressure between them is reduced and hence due to pressure difference they are pulled towards each other creating the so called attraction. P0 B1 P