Hydrodynamics Lecture 3 PDF

Lecture 3 HYDRODYNAMICS  HYDRODYNAMICS When a fluid is in motion, its flow can be characterized in one of two ways. The flow is said to be streamline, or laminar, if every particle that passes a particular point moves along exactly the same smooth path followed by...

Lecture 3 HYDRODYNAMICS  HYDRODYNAMICS When a fluid is in motion, its flow can be characterized in one of two ways. The flow is said to be streamline, or laminar, if every particle that passes a particular point moves along exactly the same smooth path followed by previous particles passing that point. This path is called a streamline. Different streamlines can’t cross each other under this steady-flow condition, and the streamline at any point coincides with the direction of the velocity of the fluid at that point. In contrast, the flow of a fluid becomes irregular, or turbulent, above a certain velocity or under any conditions that can cause abrupt changes in velocity. Irregular motions of the fluid, called eddy currents, are characteristic in turbulent flow.  HYDRODYNAMICS In discussions of fluid flow, the term viscosity is used for the degree of internal friction in the fluid. This internal friction is associated with the resistance between two adjacent layers of the fluid moving relative to each other. A fluid such as kerosene has a lower viscosity than does crude oil or molasses. Many features of fluid motion can be understood by considering the behavior of an ideal fluid, which satisfies the following conditions: 1- The fluid is non-viscous which means there is no internal friction force between adjacent layers 2- The fluid is incompressible which means its density is constant 3- The fluid motion is steady meaning that the velocity, density, and pressure at each point in the fluid don’t change with time 4-The fluid moves without turbulence This implies that each element of the fluid has zero angular velocity about its center, so there can’t be any eddy currents present in the moving fluid  HYDRODYNAMICS Turbulent flow Streamline Flow The velocity of liquid is high and unsteady. The velocity of liquid is low and steady. Velocity of liquid is greater than the critical velocity. Velocity of liquid is below its critical velocity.  Steady flow of liquids a fluid moving with streamline flow through a pipe of varying cross- sectional area. The volume of fluid flowing through A1 in a time interval Δt must equal the volume flowing through A2 in the same time interval. Since the liquid is incompressible and there are no places in the pipe in which the liquid can be stored, the volume of liquid which flows through any cross- sectional area perpendicular to the stream lines in any interval of time must be the same everywhere in the pipe.  Steady flow of liquids Consider two typical areas A1 and A2 perpendicular to the stream lines as shown in Figure 2. In a small time interval Δt the liquid at the cross- section of area A1 moves a distance Δx1 = u1Δt, where u1 is the velocity through the cross- section of area A1. Then the mass of liquid flows is: Δ𝑀1 = 𝜌𝐴1Δ𝑋1 = 𝜌𝐴1 𝑢1 Δ𝑡 where ρ is the density of the liquid. Similarly, the liquid that passes through A 2 and moving a distance Δx2 in the same time interval Δt has the mass ΔM2, as: Δ𝑀2 = 𝜌𝐴2𝑢2Δ𝑡 where u2 is the liquid velocity through A2. Since the mass of the liquid is conserved and the flow is steady, ΔM 1 must equal ΔM2. Therefore, 𝜌𝐴1 𝑢1 Δ𝑡 = 𝜌𝐴2 𝑢2Δ𝑡 or 𝐴1 𝑢1 = 𝐴2 𝑢2 𝑢1 𝐴2 where ‘Av’ is the rate of flow, which is defined that the volume, V, = 𝑢2 𝐴1 (1) passing through the area, A, in unit time  Steady flow of liquids 𝐴𝑢 = Δ𝑢/Δ𝑡 = 𝑐𝑜𝑛𝑠 tan 𝑡 (2) This expression is called the equation of continuity. From this result, we see that :- “The product of the cross-sectional area of the pipe and the fluid speed at that cross section is a constant”. Therefore, the speed is high where the tube is constricted and low where the tube has a larger diameter. The product Au, which has dimensions of volume per unit time, is called the flow rate. The condition Au = constant is equivalent to the fact that :- The volume of fluid that enters one end of the tube in a given time interval equals the volume of fluid leaving the tube in the same interval, assuming that the fluid is and there are no leaks  Kinetic energy of a liquid When a fluid is in A quantity of liquid of mass ‘m’ moving with velocity ‘v’ 1 possesses kinetic energy 2 𝑚𝑣 2 It acquires this energy as a result of work done on it by external forces. For example, consider the case of a liquid flowing out of a tank which is filled to a height ‘h’ and open to the atmosphere. Suppose that the liquid flows out of the bottom of the tank through a horizontal pipe of cross- sectional area ‘A’. Let us focus our attention on a small volume of liquid ‘ΔV’ which has just entered this pipe. There are two forces acting on this small cylinder of liquid, one to the right, owing to the pressure ‘P1’ of the liquid in the tank, and the other to the left, because of the pressure ‘P2’ of the liquid in the pipe. Since each of these pressures acts on an area ‘A, the resultant force ‘F’ acting on this cylinder is given by:- P = F \ A ⸫ F= PA 𝐹 = (𝑃1 − 𝑃2 )𝐴 (1) 𝑊 = 𝐹𝑙 = 𝑃1 − 𝑃2 𝑙𝐴 (2)  Kinetic energy of a liquid But ‘A ℓ‘ is the volume ‘ΔV’ of the liquid. Hence, 𝑊 = (𝑃1 − 𝑃2)Δ𝑉 the work done in forcing the liquid into the pipe is the difference of pressure multiplied by the volume of the liquid. It is this work which gives the liquid its kinetic energy; hence we may write: 1 (𝑃1 − 𝑃2)Δ𝑉 = 𝑚𝑣 2 ** 2 the pressures can be evaluated. The pressure ‘P2’ is due to pressure of the atmosphere ‘B’ on the open end of the pipe, while the pressure ‘P1’ is due to the atmosphere ‘B’ plus that due to the height ‘h’ of the liquid above the pipe. Therefore, the difference between these two pressures is 𝑃1 − 𝑃2 = ℎ𝜌𝑔 where ‘ρ’ is the density of the liquid m= ρ ΔV 1 (ℎ𝜌𝑔)Δ𝑉 = 𝑚𝑣 2 2  Kinetic energy of a liquid Hence, 𝑣 2 = 2(ℎ𝑔) v = (2hg)1/2 the speed of the liquid coming out of a tank filled to a height ‘h’ above the opening is exactly the same as if the liquid had fallen through the same height’  Bernoullis Equation As a fluid moves through a pipe of varying cross section and elevation, the pressure changes along the pipe. In 1738, Daniel Bernoulli (1700–1782) derived an expression that relates the pressure of a fluid to its speed and elevation. Bernoulli’s equation is not a freestanding law of physics; rather, it’s a consequence of energy conservation as applied to an ideal fluid. The motion of the liquid will be taken as sufficiently slow to permit streamline flow (laminar flow). As it moves from a cross- sectional area A1 to A2 along the streamline, its velocity increases from v1 to v2, the elevation as h1 and h2 and the pressure of the liquid as P1 and P2. In a small time interval Δt the liquid through A1 moves a distance Δx1=v1Δt, and the work done through this distance is w1=(P1A1)v1Δt, where F1=P1A1 and w = F1 Δx1. According to Equation 𝐴𝑣 = Δ𝑣/Δ𝑡 = 𝑐𝑜𝑛𝑠 tan 𝑡 the work w1 per unit time becomes w1=P1ΔV. Similarly, w2= P2 ΔV through A2 in the same time interval Δt.  Bernoullis Equation The net work done on the liquid between the two sections A1 and A2 is given by: Δ𝑊 = 𝑊1 − 𝑊2 = (𝑃1 − 𝑃2 )Δ𝑣 This work done by the pressure difference produces the change in kinetic energy and potential energy of the liquid between the two sections A1 and A2. if m is the mass of the small volume ΔV of the liquid, its kinetic energy at A1 is ½ mv12 and at A2 is ½ mv22, so that the net change in kinetic energy per unit time between the two 1 Δ𝐾 = 𝑚(𝑣22 − 𝑣12) 2 1 Δ𝐾 = 𝜌𝛻𝑉(𝑣22 − 𝑣1 2) 2 Also, the net change in potential energy of the liquid between the two sections is Δ𝑈 = 𝑚𝑔(ℎ2 − ℎ1)  Bernoullis Equation Or Δ𝑈 = 𝜌𝑔Δ𝑉(ℎ2 − ℎ1 ) As we have Δ𝑊 = Δ𝐾 + Δ𝑈 1 (𝑃1 − 𝑃2 )Δ𝑉 = 𝜌Δ𝑉(𝑣2 2 − 𝑣1 2 ) + 𝜌𝑔Δ𝑉(ℎ2 − ℎ1 ) 2 SO 1 1 𝜌𝑣12 + 𝜌𝑔ℎ1 + 𝑃1 = 𝜌𝑉22 + 𝜌𝑔ℎ2 + 𝑃2 2 2 Therefore 1 𝜌𝑉2 2 + 𝜌𝑔ℎ1 + 𝑃 = 𝑐𝑜𝑛𝑠 tan 𝑡 2 It is known as Bernoulli’s equation. It means that the pressure, kinetic energy per unit volume and potential energy per unit volume has the same value at any point along the streamline.  Bernoullis Equation Note that last Equation isn’t strictly true for gases because they aren’t incompressible. The qualitative behavior is the same, however: As the speed of the gas increases, its pressure decreases. Dividing this Equation by ρ we get: 𝑣2 𝑃 ( ) + (ℎ) + ( ) = 𝑐𝑜𝑛𝑠 tan 𝑡 2𝑔 𝜌𝑔 pressure head velocity head elevation head The sum of these three heads is called total head. Also, P is called the static pressure and ½ ρ v2 is called the dynamic pressure. If the tube is horizontal, then h1= h2 and ( h1-h2 ) = 0. Therefore, the net work done 1 1 𝜌𝑣 2 + 𝑃1 = 𝜌𝑉2 2 + 𝑃2 2 1 2  Bernoullis Equation OR 1 2 𝜌𝑣 + 𝑃 = 𝑐𝑜𝑛𝑠 tan 𝑡 2 for a flow in an open channel, where P1=P2, then 𝑣1 2 𝑣2 2 ℎ1 + = ℎ2 + 2𝑔 2𝑔 Or 𝑣 2 ℎ+ = 𝑐𝑜𝑛𝑠 tan 𝑡 2𝑔