14-1 Note 14 Fluid Dynamics

Note 14 Fluid Dynamics Sections Covered in the Text: Chapter 15, except 15.6

To complete our study of fluids we now examine Flow Rate fluids in motion. For the most part the study of fluids When a fluid occupies a pipe of cross sectional area A in motion was put into an organized state by scientists and flows with average speed v, the rate of flow Q is working a generation after Pascal and Torricelli. We given by shall see here that much of the physics of fluids is Q = Av . …[14-1] encapsulated in the statement of Bernoulli’s principle. The study of fluid flow was driven by the demands The units of Q are m3.s–1. These are units of volume.s–1, of the industrial revolution. It was vital for progress so flow rate is the same as volume rate. and profit that the movement of water, steam and oil The paths of€ particles in a fluid moving with laminar through pipes, and the movement of aerofoils through flow are called streamlines. Streamlines never cross the air, were understood mathematically. Today, the one another (Figure 14-1). physics of fluids in motion is of special interest in the various branches of the environmental and life sciences.

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 Figure 14-1. Particles in a fluid moving with laminar flow form in the fluid, resulting in highly irregular motion. follow streamlines that do not cross. 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 The Equation of Continuity flow here. A fluid moving in laminar flow in a flow tube (that may be a pipe) can be shown to satisfy a simple rela- An Ideal Fluid tionship. Consider an ideal fluid flowing through a Since a fluid is in general a complicated medium to tube of variable cross-section (Figure 14-2). In an describe mathematically, even in laminar flow, we elapsed time ∆t, a volume A1v1∆t of fluid crosses area shall assume for simplicity that the fluid is ideal. By A1, and a volume A 2v2∆t crosses area A2. Since the ideal we mean fluid is incompressible and the streamlines do not

cross the volume of fluid crossing A1 must equal the 1 The fluid is moving in laminar flow; viscous forces volume of fluid crossing A2, so A1v1∆t = A2v2∆t, from between adjacent layers are negligible. which it follows that A 1v1 = A2v2 or Q 1 = Q2. This 2 The flow is steady, that is, the flow rate does not means that the flow rate change with time. 3 The fluid has a uniform density and is thus incom- Q = Av = const . …[14-2] pressible. 4 The flow is irrotational, that is, the angular momen- An important consequence of eq[14-2] is that if the tum about any point is zero; in common parlance cross sectional area of the flow tube is reduced at the fluid does not “swirl”. some point,€ then the flow speed increases. Eq[14-2] is 14-1 Note 14 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 2000 cm2.

Solution: From the equation of continuity the speed of blood in the capillaries is

−1 2 2 v1A1 0.30(m.s ) × 3.14 × (0.010) (m ) v2 = = −1 2 A2 2.0 ×10 (m )

= 5.0 x 10–4 m.s–1

€ or about 0.5 mm.s–1. This is a very low speed.

Bernoulli’s Equation Daniel Bernoulli (1700-1782), 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 Figure 14-2. Illustration of the equation of continuity. The points 1 and 2. Let point 1 be at a height y1 and let v1, flow tube of a fluid is shown at two positions 1 and 2. At no A1 and p1 be the speed of the fluid, cross sectional area time does fluid enter or leave the flow tube. of the tube and pressure of the fluid at that point.

Similarly let v2, A2 and p 2 be the same variables at point 2. The actual system is the volume of fluid in the The equation of continuity can be applied to explain flow tube. the various rates of blood flow in the body. Blood In an elapsed time ∆t the amount of fluid crossing A1 flows from the heart into the aorta from which it is ∆V1 = A1v1∆t and the amount of fluid crossing A2 is passes into the major arteries; these branch into the ∆V2 = A2v2∆t. But from the equation of continuity, A1v1 small arteries (arterioles), which in turn branch into = A2v2. So the volume of fluid crossing either area is myriads of tiny capillaries. The blood then returns to the same; let us simply write it as the heart via the veins. Blood flow is fast in the aorta, but quite slow in capillaries. When you cut a finger ΔV = AvΔt . (capillary) the blood oozes, or flows very slowly. We can show this by means of a numerical example. 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 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. 14-2 Note 14

 ρ 2 p +  v + ρgy = const . …[14-6]  2 

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 14-3. Illustration of Bernoulli’s equation. Example Problem 14-2 Application of Bernoulli’s Principle V Δ Water circulates throughout a house in a hot water W = (F2 − F1)Δx = (F2 − F1) = (p2 − p1)ΔV . A 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 This work goes into achieving two things: basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a pipe of diameter 2.6 € 1 changing the kinetic energy of the fluid between cm on the second floor 5.0 m above? the two points by the amount: Solution:  ρΔV  2 2 Let the basement be level 1 and the second floor be ΔK =   v − v . …[14-3]  2 ( 2 1 ) level 2. We can obtain the flow speed by applying the equation of continuity, eq[14-2]:

2 changing the gravitational potential energy of the 2 fluid between the two points by the amount mg∆h v1A1 −1 π(0.020m) v2 = = 0.50(m.s ) 2 or € A2 π(0.013m) ΔU = ρΔVg(y2 − y1). …[14-4] = 1.2 m.s–1.

Since W = (p2 − p1)ΔV = ΔK + ΔU , € To find the pressure p2 we use Bernoulli’s equation, we have,€ by substituting eqs[14-3] and [14-4]: eq[14-5]: 1 2 2 p2 = p1 + ρg(y1 − y2 ) + ρ(v1 − v2 ). € (p1 − p2 )ΔV 2

5 –3  ρΔV  Substituting p1 = 3.0 x 10 Pa, ρwater = 1000 kg.m , g = 2 2 9.8 m.s –2, y = 0, y = 5.0 m, v = 0.50 m.s–1, v = 1.2 m.s–1 =  ( v2 − v1 ) + ρΔVg(y2 − y1) . 1 2 1 2  2  € we get € 5 p2 = 2.5 x 10 Pa. Dividing through by ∆V we obtain the general form of

Bernoulli’s equation: Thus p2 < p1. Obviously, the pipe on the second floor € of the house must be able to withstand less pressure  ρ 2  ρ 2 than the pipe in the basement. p +  v + ρgy = p +  v + ρgy . 1  2  1 1 2  2  2 2 …[14-5] Eqs[14-5] and [14-6] have the look of conservation of energy expressions including work, because that, in We can put this equation into the simpler form: effect, is what they are. € 14-3 Note 14 1 Torricelli’s Theorem Consider the container filled with fluid in Figure 14-4. Example Problem 14-3 A distance h below the surface of the fluid a small Applying Torricelli’s Theorem hole allows fluid to escape. What is the velocity of the outflowing fluid? 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?

Solution: h This is a straightforward application of Torricelli’s theorem and eq[14-7]. The speed of the water is

1 Figure 14-4. A container of fluid with a small hole a distance v = 2gh = 2 × 9.80(m.s− ) ×1.0(m) = 4.43 m.s–1 h below the surface allowing fluid to escape. The value measured would be less than this if the fluid (for example, shampoo) had a significant Applying Bernoulli’s equation we have € viscosity.

 ρ 2 p +  v + ρgy = const .  2  2 A Fluid at Rest Assuming a container of normal laboratory size, the For a fluid at rest (static fluid) the speed v in eq[14-6] atmospheric pressure p is essentially the same at its is zero. Bernoulli’s equation then reduces to top€ and bottom, so we can cancel p and write to a good approximation p + ρgy = const .

 ρ 2 This is just Pascal’s law; see eq[13-3] in Note 13. The  v + ρgy = const .  2  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. But if the container is not vanishingly small then the fluid velocity is essentially zero at the top (at y = 0). So 3 Pressure in a Flowing Fluid the velocity v of the outflowing fluid (at y = –h) is € We studied hydrostatic pressure in Note 13. We can given by show that in a moving fluid the pressure is depen-  ρ 2  v + ρg(−h) = 0 , dent on velocity.  2  A fluid flowing along a horizontal level experiences a constant gravitational potential (y = constant). Ber- from which it follows that noulli’s equation therefore becomes

 ρ 2 € v = 2gh . …[14-7] p +  v = const . …[14-8]  2  This is what is known as Torricelli’s theorem. Note that the velocity is independent of the fluid density. Note This means that as the speed v of the fluid increases too that our€ treatment here neglects the effect of fluid the pressure p must fall. This result forms the viscosity (that would otherwise reduce the speed). 2 principle€ of operation of many practical devices. One is the Venturi tube, which we consider next.

2 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 Note 14 The Venturi Tube The Venturi tube (Figure 14-5) is a device that is used Example Problem 14-4 to measure the flow rate Q of a fluid. It consists of two Using a Venturi Tube sections of different cross sectional areas A1 and A 2 that are known with good precision. The difference in A Venturi tube is used to measure the flow of water. It pressure in the fluid in the two sections (p1 – p2) is has a main diameter of 3.0 cm tapering down to a measured with a built-in manometer. throat diameter of 1.0 cm. The pressure difference p1 – p2 is measured to be 18 mm Hg. Calculate the velocity v1 of the fluid input and the flow rate Q.

Solution: The pressure difference in Pa is, using the conversion expression eq[13-4]:

5 –1 –3 p1 – p2 = 18 mm Hg ≡ 1.333 x 10 (Pa.m )x18x10 (m)

= 24.0 x 102 Pa.

Therefore from eq[14-9] the speed of the input fluid is

2(p1 − p2 ) v1 = A2 2 2 ρ(A1 − A2 )

= 24.6 cm.s–1.

Multiplying€ by A1 we obtain the flow rate Q:

2 1 Q = π(1.5cm) × 24.6(cm.s− ) =174 cm3.s–1

Figure 14-5. The Venturi tube. 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. From Bernoulli’s equation with y2 = y1 we have

 ρ 2 2 p − p =   v − v . 1 2  2 ( 2 1 ) The Aerofoil Bernoulli’s equation helps to explain why an airplane is equipped with wings to help it stay in the air. An Substituting the equation of continuity to eliminate v 2 airplane wing is an example of an aerofoil (Figure 14- we can rearrange and solve for v . The flow rate Q is 1 6). In a moving stream of air, the air travels more therefore given by € quickly over the top surface of an aerofoil than over the bottom surface. According to Bernoulli’s principle the 2(p1 − p2 ) pressure on the top surface is less than the pressure on Q = A1v1 = A1A2 . …[14-9] ρ A2 − A2 the bottom surface, contributing to a net upward force ( 1 2 ) called aerodynamic lift. When an airplane is flying at constant altitude and speed, the upward aerodynamic Since the density of the fluid ρ is also known, as are lift balances the downward gravitational force and the areas A1 and A2, Q can be calculated once (p1 – p2) prevents the plane from falling. This is sometimes is€ measured with the manometer. 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 Note 14 surface and the surface of the air track the pressure of wing. From eq[14-5] we can write the air is higher than atmospheric. The difference in pressure accounts for the aerodynamic lift that keeps  ρ   ρ  p +  air v 2 = p +  air v 2 the glider from contacting the air track surface and above  2  above below  2  below grinding to a halt. …[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

F 2.0 ×106 (kg) × 9.80(m.s−2) p − p = = below above A 1200(m2 )

= 16,300 Pa.

€ Rearranging eq[14-10] we can write   Figure 14-6. Streamlines around an aerofoil. 2 pbelow − pabove 2 vabove =   + vbelow .  ρ /2 

Taking the density of air as the value at the Earth’s surface, i.e., 1.29 kg.m–3 we have Example Problem 14-5 € Aerodynamic Lift 2 16300 4 vabove = +10 . An airplane has a mass of 2.0 x 106 kg and the air (1.29/2) flows past the lower surface of the wings at 100 m.s–1. If the wings have a surface area of 1200 m2, how fast Evaluating and taking the square root we have finally must the air flow over the upper surface of the wing if –1 the plane is to stay in the air? Consider only the € vabove = 190 m.s . Bernoulli effect. Thus the air must flow over the upper surface of the Solution: wing nearly twice as fast as past the lower surface. The wing is a surface moving horizontally through the air. The effect is the same as if the wing were stat- ionary and the air were flowing horizontally over the

14-6 Note 14 To Be Mastered

• Definitions: laminar flow, turbulent flow, equation of continuity

 ρ 2  ρ 2 • Physics of: Bernoulli’s Equation p +  v + ρgy = p +  v + ρgy 1  2  1 1 2  2  2 2 • 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.

2. Sketch a Venturi tube, labelling the important features.

3. Explain briefly what a Venturi tube is used for, and how it is used.

5. 6.

14-7