Chapter 5 — Fluids

1. Chapter 5 — Fluids Hydrostatics and Hydrodynamics

2. Fluids • A fluid is a material that has the ability to flow

3. Hydrostatics and Hydrodynamics • Hydrostatics studies fluids that are not moving • Hydrodynamics studies fluids in motion

4. Pressure in a Fluid P = density x acceleration due to gravity x height P = rgh (to calculate the pressure at any depth in any fluid, if density of fluid is known) If you omit g, you get non-standard (but fairly common) “pressure” units in terms of mass per area • g/cm2 is an example • N/m2 is the standard metric unit of pressure • N/m2 = Pa (Pascal) • Pa = Kg m.s2 N = kg.m s2 • P: Pressure of a fluid • g: gravity = 9.8m/s2 (constant) • h: height of fluid column _ r: (Greek letter “rho”) is the density of a fluid (density of water is a constant: 1000kg/m3)

5. Pressure: KiloPascals (kPa) versus Pounds per Square Inch (psi) • 1,000 Pa = 1 kPa = 0.15 psi • 6.90 kPa = 1 psi

6. Pressure in a Fluid • Imagine a beaker filled with water. • The beaker measures 10 cm in diameter and the height of the water is 20 cm. • How much pressure does the water exert on the bottom of the beaker? Pbottom=? P = rgh P=(1000kg)(0.20m)(9.8m) m3 s2 P=1,960=1.96x103kg =1.96x103 Pa (Pascals) m.s2 P top P bottom

7. Pressure in a Fluid • Imagine a beaker filled with water. • The beaker measures 10 cm in diameter and the height of the water is 20 cm. • How much pressure does the water exert at a depth of 5 cm? Pat 5cm=? P = rgh P=(1000kg)(0.05m)(9.8m) m3 s2 P=490=4.90x102kg =4.90x102 Pa (Pascals) m.s2

8. Pressure in a Fluid • The pressure at any depth in a fluid can be calculated if you know the pressure at some other depth and the density of the fluid: General expression for pressure as a function of depth in a fluid, where point 1 is above point 2 by a height h.

9. Pressure in a Fluid • In addition, if one has an open container of fluid, then the pressure at a depth h becomes Patm = 1 atm = 101,325 Pa What is the pressure at a depth of 1,000m below the surface of the ocean (P2), assuming that ocean water has a density of 1.03x103 kg/m3. P2 = 101,325 Pa + (1.03x103 kg/m3)(9.8m/s2)(1000m)= 10,195,325 Pa P2 = 1.02x107 Pa

10. Pressure at the same depth • At the same depth in a fluid, the fluid exerts the same pressure and in all directions This “dot” represents a “theoretical” point particle which is suspended in a fluid (occupies no space or volume)

11. Pressure Is Independent of Container Shape Height of the fluid is the same in all 4 containers (equal height = equal pressure) http://www.enasco.com/product/SB16537M

12. Pascal’s Principle • When an external pressure is applied to a confined fluid, it is transmitted unchanged to every point within the fluid • The pressure applied to a confined fluid increases the pressure throughout the fluid by the same amount • P2 = P1 + rgh = (Patm + external pressure) + rgh • For example, if you increase the pressure on top of a confined fluid by 5 Pa, the pressure everywhere in the fluid goes up by 5 Pa. • P2 = P1 + rgh = (Patm + 5 Pa) + rgh

13. Question • You’re applying a pressure of 3 psi on the plunger ofa syringe with a plugged up needle. • How does the pressure in the barrel compare to the pressure in the needle? • It will be the same throughout because the needle is plugged up

14. Buoyancy • Does wood float in water? Why? • Yes, because the density of many types of wood is less than the density of water. • How does the weight of an object immersed in water compare to the weight of the object in air? • The object weighs less when immersed in water • Fluids exert a buoyant force on objects immersed in them • Buoyant force = difference between the upward directed force and the downward directed force

15. Floating and Density • If the density of the object is greater than that of the fluid, it will sink (buoyant force is less than the true weight of the object). • If the density of the object is less than the density of the fluid, it will float (buoyant force is greater than the true weight of the object). • If the object sinks, the buoyant force is less than the weight of the object. • If the object floats, the buoyant force is greater than the weight of the object.

16. F1 F2 Buoyant Force P2 = rgh2 P2 = F2/A F2 =Argh2 P1 = rgh1 P1 = F1/A F1 =Argh1 h1 h2 Since h1 < h2 F1 < F2 Buoyant Force Equation Fb =rgV Wapparent = Wtrue-Fb

17. Principle of Archimedes • An object floating on a fluid will displace a volume of fluid that has a weight equal to the weight of the object. • An object immersed in a fluid displaces a volume of fluid equal to the volume of the object • An object in a fluid (floating or immersed) feels a buoyant force equal to the weight of fluid displaced. Archimedes (our committees) http://www.merriam-webster.com/cgi-bin/audio-medlineplus.pl?bixarc02.wav=Archimedes

18. Calculating a Bouyant Force • The buoyant force equals the weight of fluid displaced by an object. • How much will a 700 N person weigh under water, if they have a volume of 65 L? Buoyant Force Equation Fb =rgV Wapparent = Wtrue-Fb First, convert Volume given in L to m3 V = 65 L (1m3) = 65 m3 = 0.065 m3 = 6.5 x 10-2m3 1000 L 1000 Fb =rgV = (1000 kg)(9.8 m)(6.5 x 10-2m3) = 637 N m3 s2 Wapparent = Wtrue-Fb = 700N – 637N = 63N

19. Hydrometers • A hydrometer measures density (A hydrometer will sink until it displaces an amount of fluid exactly equal to its weight). • If fluid is dense, the hydrometer will displace only a small amount of fluid (it will not sink very deep)

20. Moving Fluids (HYDRODYNAMICS) • Laminar flow (smooth flow) means fluids move without internal turbulence • Pattern where adjacent layers of fluid smoothly slide past each other (smooth and orderly) • Turbulent flow means there are eddies • Continuously varying pattern of flow (chaotic and abrupt) • We will consider only laminar flow

21. Flow Rate (R=V/t)(volume of fluid passing a particular point per unit time) • Imagine a fluid moving through a cylinder • The flow rate of gas can be expressed as volume per time, e.g., mL/minute • Other examples (non-gases): gallons/min, L/hr

22. Velocity (Speed) and Diameter • As the diameter of the tube decreases, the velocity of the fluid increases (e.g., thumb over the edge of garden hose, blood moving from arteries to arterioles) • If there are no leaks, the flow rate has to be the same throughout the system (volume of the 2 “blue sections” are the same) • Thus, the fluid has to flow faster through a narrow tube

23. Velocity (Vel) and Diameter length Area Big Side: Rate = Vbig/t Volume = A•l so, Rate= Abig• lbig/t Rate = Abig•Velbig Small Side: R = Vsmall/t Volume = A•l so, Rate= Asmall•lsmall/t Rate = Asmall •Velsmall Abig • velbig = Asmall • velsmall

24. Principle of Equivalence* length Area Abig • velbig = Asmall • velsmall A1v1 = A2v2 Remember that Atube = pr2 Flow Rate = Av = pr2v or A1v1 = A2v2 *This is also known as the Equation of Continuity

25. Example • Blood is moving at 15 cm3/second (flow rate) through the aorta. If the diameter of the aorta is 1.0 cm (radius = 0.5cm), what is the velocity of the blood? Flow rate = (Area)(Velocity) Velocity = Flow rate = 15 cm3/s = 15 cm3/s = 19.0986 = 19 cm/s Area p (0.5 cm)2 0.7854cm2

26. Example • The barrel of a syringe has a diameter of 1 cm. The diameter of the needle is 0.02 cm. If you apply a pressure on the plunger so that the medicine moves at 1 cm/sec through the barrel, how fast does it move through the needle? • A1v1 = A2v2 (equation of continuity) • Assign “1” to the Barrel. Assign “2” to the needle (V2 = ?) • V2 = A1V1 = (p (0.5 cm)2)(1cm/s) = 0.7854cm2/s = 2.5 x 103 cm/s A2 (p (0.010 cm)2) 3.14159 x 10-4 cm2

27. Bernoulli’s Principle • The faster a fluid flows, the less pressure it exerts. • This phenomenon is rooted in conservation of energy(and the equivalence of work and energy) • The Bernoulli effect: • Provides the lift for airplanes • Makes the shower curtains to get sucked towards you when you first turn on the shower • Provides the basis for a Venturi flowmeter • Let’s start by imagining the horizontal laminar flow of a sample of fluid moving down the wide section, and this causes an equal volume of fluid to move in the narrow section

28. d1(thickness) d2 (thickness) F2=P2A2 F1=P1A1 Wtotal = W1 + W2 = F1 • d1 – F2 • d2 But F = P • A and V = A • d Wtotal = P1 • A1 • d1 – P2 • A2 • d2 = P1 • V1 – P2 • V2 W = ΔKE ½ m2 (v2) 2 – ½ m1 (v1) 2 = P1 • V1 – P2 • V2 m = rV ½ rV2 (v2) 2 – ½ rV1 (v1) 2 = P1 • V1 – P2 • V2 ½ r(v2) 2 – ½ r(v1) 2 = P1 – P2 or P1 + ½ r (v1) 2 = P2 + ½ r (v2) 2 Work flows into the system from the left and out of the system on the right…thus W1 > 0 and W2 < 0

29. Bernoulli’s Equation • ½ r(v2) 2 – ½ r(v1) 2 = P1 – P2 or • P1 + ½ r (v1) 2 = P2 + ½ r (v2) 2 Pressure Difference = P1 – P2 = ½ rΔ(v 2) Caution: Watch the units!! I will give problems where the units match When in doubt, use the standard MKS (Meter/Kg/Sec) units Pressure is in pascals (N/m2) velocity is in m/s density is in kg/m3

30. Example • What is the pressure differential across a flat roof having an area of 240 m2 when the wind blows at 25 m/s? The density of air “r ” is 1.29kg/m3. P1 – P2 = ½ r(v2) 2 – ½ r(v1) 2 • Let’s assume that the speed of air inside the house is 0m/s. • Let’s assign “2” to the outside of the house and a “1” to inside of the house; therefore v2 = 25m/s and v1 = 0m/s P1 – P2 = ½ r(v2) 2 – ½ r(v1) 2 = ½ (1.29kg/m3)(25m/s)2 – ½ (1.29kg/m3)(0m/s)2 = 403 Pa If we multiply this pressure times the area of the roof, we find the force of the roof from the air inside the house amounts to 96,700 N or 10.9 tons (403 Pa)(240m2) = 96,700 N 1 ton = 8,896.4 N  96,700 N x 1 ton = 10.9 tons 8,896.4 N

31. Venturi Tube Flowmeter (devices to measure fluid speeds in tubes) • If the liquid in the tube is mercury, the pressure has units of cm Hg flow flow ΔP (given by Δh) Open to U-tube U-tube (manometer): contains fluid of known density to measure ΔP

32. Venturi Example(Venturi tube flowmeter equation)P1 – P2 = ½ r(v1)2{(A1/A2)2 – 1} • Cyclopropane has a density (r) of 0.001816 g/cm3 • What pressure difference (P1 – P2) is generated by a flow rate of 50 cm3/second through a Venturi tube having radii of 0.50 cm and 0.030 cm? • Let’s call the big tube “1” and the small tube “2” • First calculate A1 and A2 using:Atube = pr2 and convert to SI (MKS) units • A1 = pr2= p(0.50cm)2= 0.7854 cm2x (1m/100cm)2= 7.854 x 10-5 m2 • A2 = pr2= p(0.030cm)2= 2.827 x 10-3 cm2x (1m/100cm)2= 2.827 x 10-7 m2 • We know the flow rate (50 cm3/s) but we need the speed (V1). flow rate = A1V1 so V1 = flow rate A1

33. Venturi Example From Previous Page(Venturi tube flowmeter equation) A1= 7.854 x 10-5 m2 P1 – P2 = ½ r(v1)2{(A1/A2)2 – 1} A2= 2.827 x 10-7 m2 V1 = flow rate A1 V1 = 50cm3/s = 63.662 cm/s x 1m = 0.63662 m/s 0.7854cm2100cm Convert cyclopropane density to SI units r = 0.001816 g x 1kg x (100cm/1m)3 = 1.816 kg/m3 cm3 1000g P1 – P2 = ½ r(v1)2{(A1/A2)2 – 1} = ½(1.816 kg/m3)(0.63662 m/s)2{(7.854 x 10-5 m2/ 2.827 x 10-7 m2)2-1} = ½(1.816 kg/m3)(0.63662 m/s)2{7.71845 x 104 – 1} = 2.8 x 104 Pa

34. Viscosity • Viscosity is a measure of a fluid’s resistance to flow • Think of it as resulting from friction between molecules • Ideal fluids have no loss of energy due to friction • No interactions between fluid molecules and: • the pipe, tubing, or container • The older literature measures viscosity in poise • The standard unit of viscosity is the pascal-second 1 Pa • s = 10 poise

35. Measuring Viscosity • his the viscosity • F is force required to push a plate having area A at a velocity v over a layer of liquid having a thickness l

36. Poiseuille’s Theorem • The flow rate for a fluid through a tube depends on: • The radius of the tube • The length of the tube • The pressure difference • The viscosity of the fluid applies only to laminar flow

37. Example • Blood has a viscosity of 0.0015 pascal-seconds (1.5x10-3 Pa.s). If a pressure of 100 mm Hg (or 13,000 Pa) (1.3x104 Pa) is applied to the aorta (r = 0.010 m; L = 1 m), what flow results? Flow = (1.3x104 Pa) p(0.010m)4 = 3.4 x 10-2 m3/s 8 (1.5x10-3 Pa.s)(1m)

38. Example • An IV bottle hangs 2 meters above a patient. The tube leads to a needle having a length of 0.04 m and a radius of 4.0 x 10 -4 m. This configuration results in a pressure (P1-P2) of 1000 Pa (1.0 x 103 Pa) above the patient’s blood pressure.If the viscosity of the liquid is 0.0015 Pa-s (1.5 x 10-3 Pa.s), what flow rate (in m3/second) results? Flow = (1.0x103 Pa) p(4 x 10-4 m)4 = 1.7 x 10-7 m3/s 8 (1.5x10-3 Pa)(0.04m)

39. Thank you!