Fluid Dinamics

MECHANICAL PROPERTIES OF MATTER When a fluid moves in such a way that there are relative motions among the fluid particles, the fluid is said to be flowing. TYPES OF FLUID FLOW : Fluid flow can be classified as : l Steady and Unsteady Flow Steady flow is defined as that type of flow in which the fluid characteristics like velocity, pressure and density at a point do not change with time. In an unsteady flow, the velocity, pressure and density at a point in the flow varies with time. 2 Streamline Flow In steady flow all the particles passing through a given point follow the same path and hence a unique line of flow. This line or path is called a streamline. Streamlines do not intersect each other because if they intersect each other the particle can move in either direction at the point of intersection and flow cannot be steady. 3 Laminar and Turbulent Flow Laminar flow is the flow in which the fluid particles move along well–defined streamlines which are straight and parallel. In laminar flow the velocities at different points in the fluid may have different magnitudes, but there directions are parallel. Thus the particles move in laminar or layers gliding smoothly over the adjacent layer. Turbulent flow is an irregular flow in which the particles can move in zig–zag way due to which eddies formation take place which are responsible for high energy losses. 4 Compressible and Incompressible Flow In compressible flow the density of fluid varies from point to point i.e. the density is not constant for the fluid whereas in incompressible flow the density of the fluid remains constant throughout. Liquids are generally incompressible while gases are compressible. 5 Rotational and Irrotational Flow Rotational flow is the flow in which the fluid particles while flowing along path–lines also rotate about their own axis. In irrotational flow particles do not rotate about their axis. So they have no net angular velocity. EQUATION OF CONTINUITY The continuity equation is the mathematical expression of the law of conservation of mass in fluid dynamics. In the steady flow the mass of fluid entering into a tube of flow in a particular time interval is equal to the mass of fluid leaving the tube. m1 m 2  1A1v1 =2A2v2 ( 1 = 2)  A1v1 = A2v2  Avv = constant t t (Here = density of fluid, v= velocity of fluid, A = Area of cross–section of tube) Therefore the velocity of liquid is smaller in the wider parts of a tube and larger in the narrower parts. BERNOULLI'S THEOREM Bernoulli's equation is mathematical expression of the law of mechanical energy conservation in fluid dynamics. Bernoullis theorem is applied to the ideal fluids. Characteristics of an ideal fluid are : (i) The fluid is incompressible. (ii) The fluid is non–viscous. (iii) The fluid flow is steady. (iv) The fluid flow is irrotational. Every point in an ideal fluid flow is associated with three kinds of energies : Kinetic Energy If a liquid of mass (m) and volume (V) is flowing with velocity (v) then Kinetic Energy= per unit volume = Potential Energy Kinetic Energy 1 m 2 1 2  v = v 2 volume 2 V 1 mv2and kinetic energy 2 If a liquid of mass (m) and volume (V) is at height (h) from the surface of the earth then its Potential Energy = mgh and potential energy per unit volume = Potential Energy m = gh = gh volume V Pressure Energy If P is the pressure on area A of a liquid and the liquid moves through a distance () due to this pressure then Pressure energy = Work done = force x displacement = pressure x area x displacement = PA = PV [ A = volume V] Pressure energy per unit volume = Pressure energy =P volume THEOREM According to Bernoulli's Theorem , in case of steady flow of incompressible and non–viscous fluid through a tube of non–uniform cross–section, the sum of the pressure, the potential energy per unit volume and the kinetic energy per unit volume is same at every point in the tube, i.e., P + gh  1 2 v  constant. 2 Consider a liquid flowing steadily through a tube of non–uniform area of cross–section as shown in figure. If P1 and P2 are the pressures at the two ends of the tube respectively, work done in pushing the volume V of incompressible liquid from point B to C through the tube W = P (V) = (P1–P2)V ...(i) This work is used by the liquid in two ways : (i) In changing the potential energy of mass m (in the volume V) from mgh 1 to mgh2 i.e., U=mg (h2–h1) ...(ii) MECHANICAL PROPERTIES OF MATTER (ii) 1 1 1 mv12 to mv22 , i.e. K  m v22  v12 2 2 2 Now as the liquid is non–viscous, by conservation of mechanical energy,  In changing the kinetic energy from 1 2 2 W = U+K i.e.,  P1  P2  V=mg  h 2  h 1  + m v2  v1 2  P1–P2 = g(h2–h1) +   1  v22  v12 [as   m / V ] 2   1 1 2 1 v1  P2  gh 2  v22  P + gh+ v2= constant 2 2 2 This equation is the Bernoulli's equation and represents conservation of mechanical energy in case of moving fluids. P1 + gh 1  VENTURIMETER It is a gauge put on a flow pipe to measure the speed of flow of a liquid shown in figure. Let the liquid of density  be flowing through a pipe of area of cross section A1. Let A2 be the area of cross section at the throat and a manometer is attached as shown in the figure. Let v1 and P1 be the velocity of the flow and pressure at point A, v2 and P2 be the corresponding quantities at point B. Using Bernoulli’s theorem : P1 P 1 1  gh 1  v12  2  gh 2  v22  2  2 we get 1 P1 P 1 1  gh  v12  2  gh  v22 (Since h1 = h2 = h) or (P1–P2)= (v22 – v12) 2  2  2 ....(i)  A1  According to continuity equation, A1v1 = A2v2 or v2   A  v1  2 Substituting the value of v2 in equation (i) we have 1  P1  P2   2   Since A1 > A2, therefore,  A  2 1  1  v12  v12  v12  A 2   2 P1 > P2 or v12  2(P1  P2 )  A  2    1   1   A 2    A  2   1   1   A 2    2 A 22 (P1  P2 )  (A12  A 22 ) where (P1 – P2) = mgh and h is the difference in heights of the liquid levels in the two tubes. v1  2 m gh  A  2    1   1   A 2   The flow rate (R) i.e., the volume of the liquid flowing per second is given by R = v1A1. 36 • Torricelli's Law of Efflux (Fluid Outflow) As shown in the figure since the area of cross–section at A is very large as compared to that at orifice B, speed at A i.e. vA 0. Also the two fluid particles at A and B are at same pressure P0 (atmospheric pressure). Applying Bernoulli's theorem at A and B. P0  gH  1 2 1 1 v A  P0  g(H  h )  v2B  v2B = gh vB = 2 2 2 2gh Equation is same as that of freely falling body after falling through h height and is known as Torricelli's law. Writing equation of uniformly accelerated motion in vertical direction H–h=0+ 1 2 2 (H  h ) 1 gt (from sy = uyt + ayt2)  t  2 g 2 Horizontal range R = vxt = 2gh × 2(H  h) = 2 h(H  h) g Range R will be maximum when R2 is maximum. i.e., d d H R2 = 0  4 (Hh–h2) = 0  H – 2h = 0, i.e., h = dh dh 2    Rmaximum = 2 Ex. H 2 H  H  2  = H   A cylindrical tank 1 m in radius rests on a platform 5 m high. Initially the tank is filled with water upto a height of 5 m. A plug whose area is 10–4 m2 is removed from an orifice on the side of the tank at the bottom. Calculate (a) initial speed with which the water flows from the orifice (b) initial speed with which the water strikes the ground Sol. (a) Speed of efflux vH  2 gh  = 2  1 0  5  10 m/s (b) As initial vertical velocity of water is zero, so its vertical velocity when it hits the ground vv= 2 gh = 2  1 0  5  10 m/s So the initial speed with which water strikes the ground, v = v2H  v2V =102 = 14.1 m/s MECHANICAL PROPERTIES OF MATTER Ex. In a given arrangement B (a) Find out velocity of water coming out of ‘C’ container of area A (b) Find out pressure at A, B and C. Sol. (a) h h1 A h3 Applying Bernoulli’s equation between liquid surface and point ‘C’. liquid 1 2 1 2 pa + v1 = pa – gh3 + v2 2 2 through continuity equation Av1 = av2 , v1 = C 2 gh 3 , v2 = v = a2 1 2 A B 2 gh 3 a2 1 2 A h2 h Pressure at A just outside the tube pA = patm + gh1 For pressure at B : pA + 0 + 0 = pB + gh2 + pB = PA – gh2 – Pressure at C : area of cross section a av2 1 2 1 a2 2   2 v2 = –gh3 + v2 A 2 2 A 2 2 (b) h2 h1 v1 A h3 1 v 2 2 B 1  2 gh 3  2  a2  1  2 A C v2    pC = patm Ex. A tank is filled with a liquid upto a height H. A small hole is made at the bottom of this tank. Let t1 be the time taken t1 to empty first half of the tank and t2 time taken to empty rest half of the tank, then find t . 2 Sol. Let at some instant of time the level of liquid in the tank is y. Velocity of efflux at this instant of time v = 2 gy . Now,,  dV1  at this instant volume of liquid coming out the hole per second is   dt   dV2  Volume of liquid coming down in the tank per second is  dt    dy   dy  dV1 dV2 = av=A  =A    dt   a 2 gy dt  dt dt (Here area of cross-section of hole and tank are respectively a and A) ...(i) Substituting the proper limits in equation (i),  t1 0 dt   Similarly, A a 2g  H/ 2 H  t2 y 1 / 2 dy  t1 = dt   A  H 2A   A  y H =   = 2  a a 2g   H / 2 a 2 g  2A 0 H/ 2 y 1 / 2 dy  t2 = a 2g t1 From equations (ii) and (iii), t = 2–1 = 0.414 2 0 A a H g H g   2  1 ...(ii) ...(iii) Ex. A fixed container of height ' H ' with large cross-sectional area ' A ' is completely filled with water. Two small orifice of cross-sectional area ' a ' are made, one at the bottom and the other on the vertical side of the container at a distance H/2 from the top of the container. Find the time taken by the water level to reach a height of H/2 from the bottom of the container. Sol. v1 = 2 g (h  H / 2 ) ; v2 =  By continuity equation 2 gh  dh  dh A   d t  = a (v1 + v2)  A   d t  = a or  A a 2g H/ 2  H   dt  t = hH/ 2  H / 2)  2 g h  0 2A 3a  2 1  v1 H/2 H/2 v2 t dh h   a h a H g Ex. A cylindrical vessel filled with water upto a height of 2 m stands on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. Find the minimum diameter of the hole so that the vessel begins to move on the floor if the plug is removed. The coefficient of friction between the bottom of the vessel and the plane is 0.4 and total mass of water plus vessel is 100 kg. Sol. From Torricelli's theorem, velocity of efflux = 2 gh Momentum per second carried by water stream = density × volume coming out per second × velocity =  × av × v = av2 Hence force on cylindrical vessel = a2gh Cylinder starts to move when reaction force is just equal to maximum force of friction. i.e., Mg=a2gh  a = Area of circular hole = M 0 .4  10 0 = = 0.01 m2 2 h 2  103  2 d 2 = 0.01 m2  d = 4 0 .01  4 = 0.113m  EXAMPLES OF BERNOULLI'S THEOREM l. Magnus Effect (Spinning Ball) Tennis and cricket players usually experience that when a ball is thrown spinning it moves along a curved path. This is called swing of the ball. This is due to the air which is being dragged round by the spinning ball. When the ball spins, the layer of the air around it also moves with the ball. So, as shown in figure the resultant velocity of air increases on the upper side and reduces on the lower side. force on ball speed of air flow increases \pressure reduced spin speed of air flow decreases\pressure increased motion of a spin ball Hence according to Bernoulli's theorem the pressure on the upper side becomes lower than that on the lower side. This pressure difference exerts a force on the ball due to which it moves along a curved path. This effect is known as Magnus–effect. MECHANICAL PROPERTIES OF MATTER 2. Motion of the Ping–Pong Ball When a ping–pong ball is placed on a vertical stream of water– fountain, it rises upto a certain height above the nozzle of the fountain and spins about its axis. The reason for this is that the streams of water rise up from the fountain with very large velocity air so that the air–pressure in them decreases. Therefore, whenever the ball goes out from the streams, the outer air which is at atmospheric pressure pushes it back into the streams ( in the region of low pressure). Thus the ball remains in stable equilibrium on the fountain. If we blow air at one end of a narrow tube, the air emerges from the other end at high speed and so the pressure falls there. If a ping–pong ball is left free slightly below this end, it does not fall down due to the large pressure (atmospheric) below the ball. Similarly, if we blow air in between two ping–pong balls suspended by light threads near each other, the balls come close to each other due to the decrease of air pressure between them. Same is the reason that when air is blown below a pan of a physical balance the pan is depressed down. 3. Aerofoil This is a device which is shaped in such a way so that the relative motion between it and a fluid produces a force perpendicular to the flow. As shown in the figure the shape of the aerofoil section causes the fluid to flow faster over the top surface then over the bottom i.e. the streamlines are closer above than below the aerofoil. By Bernoullis theorem the pressure above is reduced and that underneath is increased. high speed, reduced pressure aerofoil lift low speed, increased pressure principle of an aerofoil Thus a resultant upward force is created normal to the flow and it is this force which provides most of lift upward force for an aeroplane. Examples of aerofoils are aircraft wings, turbine blades and propellers. 4. Pull–in or Attraction Force by Fast Moving Trains If we are standing on a platform and a train passes through the platform with very high speed we are pulled towards the train. This is because as the train comes at high speed, the pressure between us and the train decreases. Thus the air behind us which is still at atmospheric pressure pushes us towards the train. The reason behind flying–off of small papers, straws and other light objects towards the train is also the same. 5. Sprayer or Atomizer This is an instrument used to spray a liquid in the form of small droplets (fine spray). It consists of a vertical tube whose lower end is dipped in the liquid to be sprayed, filled in a vessel. The upper end opens in a horizontal tube. At one end of the horizontal tube there is a rubber bulb and at the other end there is a fine bore (hole). When the rubber bulb is squeezed, air rushes out through the horizontal tube with very high velocity and thus the pressure reduces (according to Bernoulli's theorem). So the liquid rises and comes out through narrow end in form of droplets. It is used in spray gun, perfumes, deodorant and etc. spray Rubber bulb air flowing out with high velocity 6. Bunsen's Burner (Jet) It is based on the working of jet. It consists of a long tube having a fine nozzle O at the bottom. The burning gas enters the tube through O and burns in a flame at the top of the tube. flame gas +air To produce a non–luminous flame, the air of the atmosphere is mixed with the gas. Since the nozzle O is fine, the gas enters with a large velocity and so the pressure inside the tube is lowered than the outer atmospheric pressure. Therefore, air from outside gas rushes into the tube through a hole and is mixed with the burning gas. 7. Filter Pump (Nozzle) It is based on the working of nozzle. It consists of a wide tube M N in the upper part of which there is another tube A. The upper end of A is connected to a water tank, while its lower end has a fine bore through which water comes out in the form of a jet. The vessel which is to be evacuated is connected to the tube M N as shown. The velocity of the emerging water jet is very large. Therefore, the pressure of the air near the jet becomes less than the pressure in the vessel. Hence air from the vessel rushes into the tube M N and is carried out along with the water jet. Thus partial vacuum is created in the vessel. 8. Blowing–off of Tin Roof Tops in Wind Storm air nosel water M A vessel air N water jet water and air When wind blows with a high velocity above a tin roof, it causes lowering of pressure above the roof, while the pressure below the roof is still atmospheric. Due to this pressure–difference the roof is lifted up. P wind P0 v large So P< P0