14-1. Fluids in Motion There are two types of fluid motion called laminar flow and turbulent flow.

Transcription

1 Fluid Dynamics Sections Covered in the Text: Chapter 15, except 15.6 To complete our study of fluids we now examine fluids in motion. For the most part the study of fluids in motion was put into an organized state by scientists working a generation after Pascal and Torricelli. We shall see here that much of the physics of fluids is encapsulated in the statement of Bernoulli s principle. The study of fluid flow was driven by the demands of the industrial revolution. It was vital for progress and profit that the movement of water, steam and oil through pipes, and the movement of aerofoils through the air, were understood mathematically. Today, the physics of fluids in motion is of special interest in the various branches of the environmental and life sciences. Flow Rate When a fluid occupies a pipe of cross sectional area A and flows with average speed v, the rate of flow Q is given by Q = Av. [14-1] The units of Q are m 3.s 1. These are units of volume.s 1, so flow rate is the same as volume rate. The paths of particles in a fluid moving with laminar flow are called streamlines. Streamlines never cross one another (Figure 14-1). Fluids in Motion There are two types of fluid motion called laminar flow and turbulent flow. A fluid will execute laminar flow when it is moving at low velocity. The particles of the fluid follow smooth paths that do not cross, and the rate of fluid flow remains constant in time. This is the easiest type of flow to describe mathematically. A fluid will execute turbulent flow when it is moving above a certain critical velocity. Strings of vortices form in the fluid, resulting in highly irregular motion. A white water rapid is a good example. A system moving irregularly is difficult to describe mathematically, so we shall not be concerned with turbulent flow here. An Ideal Fluid Since a fluid is in general a complicated medium to describe mathematically, even in laminar flow, we shall assume for simplicity that the fluid is ideal. By ideal we mean 1 The fluid is moving in laminar flow; viscous forces between adjacent layers are negligible. The flow is steady, that is, the flow rate does not change with time. 3 The fluid has a uniform density and is thus incompressible. 4 The flow is irrotational, that is, the angular momentum about any point is zero; in common parlance the fluid does not swirl. Figure Particles in a fluid moving with laminar flow follow streamlines that do not cross. The Equation of Continuity A fluid moving in laminar flow in a flow tube (that may be a pipe) can be shown to satisfy a simple relationship. Consider an ideal fluid flowing through a tube of variable cross-section (Figure 14-). In an elapsed time t, a volume A 1 v 1 t of fluid crosses area A 1, and a volume A v t crosses area A. Since the fluid is incompressible and the streamlines do not cross the volume of fluid crossing A 1 must equal the volume of fluid crossing A, so A 1 v 1 t = A v t, from which it follows that A 1 v 1 = A v or Q 1 = Q. This means that the flow rate Q = Av = const. [14-] An important consequence of eq[14-] is that if the cross sectional area of the flow tube is reduced at some point, then the flow speed increases. Eq[14-] is 14-1

2 known as the equation of continuity. 1 Example Problem 14-1 Speed of Blood Flow in Capillaries The radius of the aorta is about 1.0 cm and the blood flowing through it has a speed of about 30.0 cm.s 1. Calculate the average speed of the blood in the capillaries using the fact that although each capillary has a diameter of about 8.0 x 10 4 cm, there are literally billions of them so that their total cross section is about 000 cm. From the equation of continuity the speed of blood in the capillaries is v = v A 1 1 = 0.30(m.s 1 ) 3.14 (0.010) (m ) A (m ) = 5.0 x 10 4 m.s 1 or about 0.5 mm.s 1. This is a very low speed. Figure 14-. Illustration of the equation of continuity. The flow tube of a fluid is shown at two positions 1 and. At no time does fluid enter or leave the flow tube. The equation of continuity can be applied to explain the various rates of blood flow in the body. Blood flows from the heart into the aorta from which it passes into the major arteries; these branch into the small arteries (arterioles), which in turn branch into myriads of tiny capillaries. The blood then returns to the heart via the veins. Blood flow is fast in the aorta, but quite slow in capillaries. When you cut a finger (capillary) the blood oozes, or flows very slowly. We can show this by means of a numerical example. 1 The equation of continuity can be thought of as a statement of the conservation of fluid. As the fluid flows through the pipe the volume of fluid remains constant; it neither increases nor decreases. 14- Bernoulli s Equation Daniel Bernoulli ( ), a Swiss mathematician and scientist, lived a generation after Pascal and Torricelli. The equation he derived is a more general statement of the laws and principles of fluids we have examined thus far. Bernoulli allowed for the flow tube to undergo a possible change in height (Figure 14-3). Consider points 1 and. Let point 1 be at a height y 1 and let v 1, A 1 and p 1 be the speed of the fluid, cross sectional area of the tube and pressure of the fluid at that point. Similarly let v, A and p be the same variables at point. The actual system is the volume of fluid in the flow tube. In an elapsed time t the amount of fluid crossing A 1 is V 1 = A 1 v 1 t and the amount of fluid crossing A is V = A v t. But from the equation of continuity, A 1 v 1 = A v. So the volume of fluid crossing either area is the same; let us simply write it as ΔV = AvΔt. Fluid is moved in the flow tube as the result of the work done on the fluid by the surrounding fluid (the environment). The net work W done on the fluid in the elapsed time t is

3 p v + ρgy = const. Note 14 [14-6] Many of the principles and laws we have seen can be shown to be special cases of Bernoulli s equation. We shall consider a number of them. Many homes and buildings in the colder climates are heated by the circulation of hot water in pipes. Even if they are not explicitly aware of it, architects must ensure that their designs conform to Bernoulli s principle in order to avoid system failure. Let us consider an example. Figure Illustration of Bernoulli s equation. W = ( F F 1 )Δx = ( F F 1 ) ΔV A = ( p p 1)ΔV. This work goes into achieving two things: 1 changing the kinetic energy of the fluid between the two points by the amount: ΔK = ρδv v v 1 ( ). [14-3] changing the gravitational potential energy of the fluid between the two points by the amount mg h or ΔU = ρδvg( y y 1 ). [14-4] Since W = ( p p 1 )ΔV = ΔK + ΔU, we have, by substituting eqs[14-3] and [14-4]: ( p 1 p )ΔV = ρδv v v 1 ( ) + ρδvg y y 1 ( ). Dividing through by V we obtain the general form of Bernoulli s equation: p 1 v + ρgy = p + ρ 1 1 v + ρgy. We can put this equation into the simpler form: [14-5] Example Problem 14- Application of Bernoulli s Principle Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50 m.s 1 through a pipe of diameter 4.0 cm in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a pipe of diameter.6 cm on the second floor 5.0 m above? Let the basement be level 1 and the second floor be level. We can obtain the flow speed by applying the equation of continuity, eq[14-]: v = v 1 A 1 A = 0.50(m.s 1 ) π(0.00m) π(0.013m) = 1. m.s 1. To find the pressure p we use Bernoulli s equation, eq[14-5]: p = p 1 + ρg( y 1 y ) + 1 ρ v ( 1 v ). Substituting p 1 = 3.0 x 10 5 Pa, ρ water = 1000 kg.m 3, g = 9.8 m.s, y 1 = 0, y = 5.0 m, v 1 = 0.50 m.s 1, v = 1. m.s 1 we get p =.5 x 10 5 Pa. Thus p < p 1. Obviously, the pipe on the second floor of the house must be able to withstand less pressure than the pipe in the basement. Eqs[14-5] and [14-6] have the look of conservation of energy expressions including work, because that, in effect, is what they are. 14-3

4 1 Torricelli s Theorem Consider the container filled with fluid in Figure A distance h below the surface of the fluid a small hole allows fluid to escape. What is the velocity of the outflowing fluid? h Example Problem 14-3 Applying Torricelli s Theorem A glass container of height 1.0 m is full of water. A small hole appears on the side of the container at the bottom (as shown in Figure 14-4). What is the speed of the water flowing out the hole? This is a straightforward application of Torricelli s theorem and eq[14-7]. The speed of the water is Figure A container of fluid with a small hole a distance h below the surface allowing fluid to escape. Applying Bernoulli s equation we have p v + ρgy = const. Assuming a container of normal laboratory size, the atmospheric pressure p is essentially the same at its top and bottom, so we can cancel p and write to a good approximation ρ v + ρgy = const. But if the container is not vanishingly small then the fluid velocity is essentially zero at the top (at y = 0). So the velocity v of the outflowing fluid (at y = h) is given by ρ v + ρg( h) = 0, from which it follows that v = gh. [14-7] This is what is known as Torricelli s theorem. Note that the velocity is independent of the fluid density. Note too that our treatment here neglects the effect of fluid viscosity (that would otherwise reduce the speed). v = gh = 9.80(m.s 1 ) 1.0(m) = 4.43 m.s 1 The value measured would be less than this if the fluid (for example, shampoo) had a significant viscosity. A Fluid at Rest For a fluid at rest (static fluid) the speed v in eq[14-6] is zero. Bernoulli s equation then reduces to p + ρgy = const. This is just Pascal s law; see eq[13-3] in Note 13. The point to be made here is that Bernoulli s equation is a more general statement of the physics of fluids than is Pascal s law. 3 Pressure in a Flowing Fluid We studied hydrostatic pressure in Note 13. We can show that in a moving fluid the pressure is dependent on velocity. A fluid flowing along a horizontal level experiences a constant gravitational potential (y = constant). Bernoulli s equation therefore becomes p v = const. [14-8] This means that as the speed v of the fluid increases the pressure p must fall. This result forms the principle of operation of many practical devices. One is the Venturi tube, which we consider next. Eq[14-5] is the same expression as obtained for the final velocity of an object dropped from rest at a height h (Note 09). 14-4

5 The Venturi Tube The Venturi tube (Figure 14-5) is a device that is used to measure the flow rate Q of a fluid. It consists of two sections of different cross sectional areas A 1 and A that are known with good precision. The difference in pressure in the fluid in the two sections (p 1 p ) is measured with a built-in manometer. Example Problem 14-4 Using a Venturi Tube Note 14 A Venturi tube is used to measure the flow of water. It has a main diameter of 3.0 cm tapering down to a throat diameter of 1.0 cm. The pressure difference p 1 p is measured to be 18 mm Hg. Calculate the velocity v 1 of the fluid input and the flow rate Q. The pressure difference in Pa is, using the conversion expression eq[13-4]: p 1 p = 18 mm Hg x 10 5 (Pa.m 1 )x18x10 3 (m) = 4.0 x 10 Pa. Therefore from eq[14-9] the speed of the input fluid is ( ) p v 1 = A 1 p ρ A 1 A ( ) = 4.6 cm.s 1. Multiplying by A 1 we obtain the flow rate Q: Q = π(1.5cm) 4.6(cm.s 1 ) =174 cm 3.s 1 Figure The Venturi tube. From Bernoulli s equation with y = y 1 we have p 1 p = ρ v v 1 ( ). Substituting the equation of continuity to eliminate v we can rearrange and solve for v 1. The flow rate Q is therefore given by ( ) p Q = A 1 v 1 = A 1 A 1 p ρ A 1 A ( ). [14-9] Since the density of the fluid ρ is also known, as are the areas A 1 and A, Q can be calculated once (p 1 p ) is measured with the manometer. These results, too, would be quite in error if the fluid had a non-negligible viscosity. The viscosity of water has a negligible effect in this example. The Aerofoil Bernoulli s equation helps to explain why an airplane is equipped with wings to help it stay in the air. An airplane wing is an example of an aerofoil (Figure 14-6). In a moving stream of air, the air travels more quickly over the top surface of an aerofoil than over the bottom surface. According to Bernoulli s principle the pressure on the top surface is less than the pressure on the bottom surface, contributing to a net upward force called aerodynamic lift. When an airplane is flying at constant altitude and speed, the upward aerodynamic lift balances the downward gravitational force and prevents the plane from falling. This is sometimes called the Bernoulli effect. A glider moving on an air track is a kind of aerofoil. On the top surface of a glider the pressure is atmospheric pressure. In the space between the bottom 14-5

6 surface and the surface of the air track the pressure of the air is higher than atmospheric. The difference in pressure accounts for the aerodynamic lift that keeps the glider from contacting the air track surface and grinding to a halt. wing. From eq[14-5] we can write p above air v above = p below air v below [14-10] where above and below refer to the airplane wing. The difference in pressure accounts for the aerodynamic lift, that is, the difference in pressure is equal to the resultant force per unit area on the wing. The magnitude of force must just equal the weight of the airplane. Thus p below p above = F A = (kg) 9.80(m.s ) 100(m ) = 16,300 Pa. Rearranging eq[14-10] we can write Figure Streamlines around an aerofoil. Example Problem 14-5 Aerodynamic Lift An airplane has a mass of.0 x 10 6 kg and the air flows past the lower surface of the wings at 100 m.s 1. If the wings have a surface area of 100 m, how fast must the air flow over the upper surface of the wing if the plane is to stay in the air? Consider only the Bernoulli effect. The wing is a surface moving horizontally through the air. The effect is the same as if the wing were stationary and the air were flowing horizontally over the v above = p p below above + v below. ρ / Taking the density of air as the value at the Earth s surface, i.e., 1.9 kg.m 3 we have v above = (1.9 /) Evaluating and taking the square root we have finally v above = 190 m.s 1. Thus the air must flow over the upper surface of the wing nearly twice as fast as past the lower surface. 14-6

7 To Be Mastered Note 14 Definitions: laminar flow, turbulent flow, equation of continuity Physics of: Bernoulli s Equation p 1 v + ρgy = p + ρ 1 1 v + ρgy Physics of: the Venturi tube Physics of: the aerofoil Typical Quiz/Test/Exam Questions 1. State Benoulli s law relating pressure p, velocity v and density ρ in a moving fluid.. Sketch a Venturi tube, labelling the important features. 3. Explain briefly what a Venturi tube is used for, and how it is used