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Chapter 13

Chapter 13. The mechanics of nonviscous fluids. Steady/Unsteady flow. In steady flow the velocity of particles is constant with time

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Chapter 13

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  1. Chapter 13 The mechanics of nonviscous fluids T . Norah Ali Al moneef

  2. Steady/Unsteady flow • In steady flow the velocity of particles is constant with time • Unsteady flow occurs when the velocity at a point changes with time. Turbulence is extreme unsteady flow where the velocity vector at a point changes quickly with time e.g. water rapids or waterfall. • When the flow is steady, streamlines are used to represent the direction of flow. • Steady flow is sometimes called streamline flow • Streamlines never cross. A set of streamlines can define a tube of flow, the borders of which the fluid does not cross Turbulent Flow Turbulent flow is an extreme kind of unsteady flow and occurs when there are sharp obstacles or bends in the path of a fast-moving fluid. In turbulent flow, the velocity at a point changes erratically from moment to moment, both in magnitude and direction. T . Norah Ali Al moneef

  3. Viscous/Non viscous flow • A viscous fluid such as honey does not flow readily, it has a large viscosity. • Water has a low viscosity and flows easily. • A viscous flow requires energy dissipation. • Zero viscosity – requires no energy. with no dissipation of energy.Some liquids can be taken to have zero viscosity e.g. water. • An incompressible, non viscous fluid is said to be an ideal fluid T . Norah Ali Al moneef

  4. 13. 2 Stream flow T . Norah Ali Al moneef T . Norah Ali Al moneef

  5. The volume flow rate, Q is defined as the volume of fluid that flows past an imaginary (or real) interface. A v represents the volume of fluid per second that passes through the tube and is called the volume flow rate Q T . Norah Ali Al moneef

  6. If the fluid is incompressible, the density remains constant throughout • Av represents the volume of fluid per second that passes through the tube and is called the volume flow rate Q • This just means that the amount of fluid moving in any “section of pipe” must remain constant. • If the area is reduced the fluid must speed up! • Speed is high where the pipe is narrow and speed is low where the pipe has a large diameter T . Norah Ali Al moneef

  7. The product of the area and the fluid speed at all points along a pipe is constant for an incompressible fluid. T . Norah Ali Al moneef

  8. Example: Oil is flowing at a speed of 1.22 m/s through a pipeline with a radius of 0.305 m. How much oil flows in 1 day? Q: Have you ever used your thumb to control the water flowing from the end of a hose? A: When the end of a hose is partially closed off, thus reducing its cross-sectional area, the fluid velocity increases. This kind of fluid behavior is described by the equation of continuity. T . Norah Ali Al moneef

  9. example example • A river flows in a channel that is 40. m wide and 2.2 m deep with a speed of 4.5 m/s. • The river enters a gorge that is 3.7 m wide with a speed of 6.0 m/s. • How deep is the water in the gorge? • The area is width times depth. A1 = w1d1 • Use the continuity equation. v1A1 = v2A2 • v1w1d1 = v2w2d2 • Solve for the unknown d2. • d2 = v1w1d1 / v2w2 =(4.5 m/s)(40. m)(2.2 m) / (3.7m)(6.0 m/s) = 18 m T . Norah Ali Al moneef

  10. example T . Norah Ali Al moneef

  11. example If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter 40mm takes 30% of total discharge and pipe 3 diameter 60mm. What are the values of discharge and mean velocity in each pipe? T . Norah Ali Al moneef

  12. example T . Norah Ali Al moneef

  13. example example T . Norah Ali Al moneef

  14. Example: The volume rate of flow in an artery supplying the brain is 3.6x10-6 m3/s. If the radius of the artery is 5.2 mm, A) determine the average blood speed. B) Find the average blood speed if a constriction reduces the radius of the artery by a factor of 3 (without reducing the flow rate). T . Norah Ali Al moneef

  15. Example: Example,decreasein area of stream of water coming from tap. So v2 > v1 P2< p1 T . Norah Ali Al moneef

  16. Example if the cross-section area, A, is 1.2 x 10-3m2 and the discharge, Q is 24 l / s , then the mean velocity, Example if the area A 1= 10 x10-3 m2 and A 2= 3 x10-3m 2 and the upstream mean velocity, 1 = 2.1 m / s , then the downstream mean velocity T . Norah Ali Al moneef

  17. Example: • The figure shows 3 straight pipes through which water flows. The figure gives the speed of the water in each pipe. • Rank them according to the volume of water that passes through the cross-sectional area per minute, greatest first. 6 same Example: • Water flows smoothly through the pipe shown in the figure, descending in the process. Rank the four numbered sections of pipe according to • (a) the volume flow rate Q through them • (b) the flow speed v through them • (c) the water pressure p within them, greatest first. a)Same b) 1,2,3,4 c ) 4,3,2,1 T . Norah Ali Al moneef

  18. 1) 2) 3) T . Norah Ali Al moneef

  19. A = 2 cm2 1.5 kg Pressure Pressure is the ratio of a force F to the area A over which it is applied: P = 73,500 N/m2 The greater the force, the greater the pressure is. The greater the area, the smaller the force is. example a) Calculate the total force of the atmosphere acting on the top of a table that measures (b) What is the total force acting upward on the underside of the table? (a) The total force of the atmosphere on the table will be the air pressure times the area of the table. (b) Since the atmospheric pressure is the same on the underside of the table (the height difference is minimal), the upward force of air pressure is the same as the downward force of air on the top of the table,

  20. example T . Norah Ali Al moneef

  21. Example The mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm deep. a) Find the weight of the water in the mattress. The volume density of water at the normal condition (0oC and 1 atm) is 1000kg/m3. So the total mass of the water in the mattress is Therefore the weight of the water in the mattress is b) Find the pressure exerted by the water on the floor when the bed rests in its normal position, assuming the entire lower surface of the mattress makes contact with the floor. Since the surface area of the mattress is 4.00 m2, the pressure exerted on the floor is T . Norah Ali Al moneef T . Norah Ali Al moneef

  22. 13.3 Bernoulli’sEquation • Relates pressure to fluid speed and elevation • Bernoulli’s equation is a consequence of Work Energy Relation applied to an ideal fluid • Assumes the fluid is incompressible and nonviscous, and flows in a nonturbulent, steady-state manner T . Norah Ali Al moneef T . Norah Ali Al moneef

  23. This work is negative because the force on the segment of fluid is to the left and the displacement is to the right. Thus, the net work done on the segment by these forces in the time interval T . Norah Ali Al moneef

  24. T . Norah Ali Al moneef

  25. States that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline T . Norah Ali Al moneef

  26. 13.4 static consequences of Bernoulli's equation • Fluid At Rest In a Container : • Pressure in a continuously distributed uniform static fluid varies only with • vertical distance and is independent of the shape of the container. • • The pressure is the same at all points on a given horizontal plane in a fluid. • • For liquids, which are incompressible, we have: • pb + ρgh1 = patm + ρgh2 • pb = patm + ρg (h2 - h1) = patm + ρgd Y a = y b = y c = yd y A = y B = y C = y D d h2 h1 T . Norah Ali Al moneef T . Norah Ali Al moneef

  27. Absolute Pressure and Gauge Pressure The pressure P is called the absolute pressure Remember, P = Patm + ρ g h P – Patm = ρ g h ( soρ g h is the gauge pressure) , because it is the actual value of the system’s pressure. Gauge pressure: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere • Usual pressure gauges record gauge pressure. To calculate absolute pressure: P = P atm + P gauge As a fluid moves along if it changes speed or elevation then the pressure changes and vice versa. Bernoulli’s equation is really just conservation of energy for the fluid • Bernoulli’s equation. • P + ½ ρv2 + ρg h = constant • Bernoulli’s equation shows that as the velocity of a fluid speeds up it’s pressure goes down…..this idea used in airplane wings and frisbees (difference in pressure leads to upward force we call Lift) T . Norah Ali Al moneef

  28. PA>PB PA=PC PA – Patm = ρ g h PA = PB = PC = PD • P atm is atmosphericpressure=1.013 x 105 Pa • The pressure does not depend upon the shape of the container T . Norah Ali Al moneef

  29. Variation of Pressure with Depth • If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium • All points at the same depth must be at the same pressure – Otherwise, the fluid would not be in equilibrium • Pressure changes with elevation • The pressure gradient in the vertical direction is negative • The pressure decreases as we move upward in a fluid at rest • Pressure in a liquid does not change due to the shape of the container • The fluid would flow from the higher pressure region to the lower pressure region • the pressure at a point in a fluid depends only on density, gravity and depth T . Norah Ali Al moneef

  30. example T . Norah Ali Al moneef

  31. Pressure Measurements: The manometer • Manometers are devices in which one or more columns of a liquid are used to determine the pressure difference between two points. • U-tube manometer • One end of the U-shaped tube is open to the atmosphere • The other end is connected to the pressure to be measured T . Norah Ali Al moneef

  32. Pressure at B is P=Patm+ρgh Pressure at A > Patm Pressure in a continuous static fluid is the same at any horizontal level so, Pressure at A = Pressure at B P A = PB = Patm+ ρgh T . Norah Ali Al moneef T . Norah Ali Al moneef T . Norah Ali Al moneef

  33. Blood Pressure measurements by cannulation Pressure at c =pressure at D PC = PD PC = PA + ρsgh PB= Patm+ ρgh PA+ ρsgh= Patm + ρgh PA= Patm + ρg h – ρsgh Pblood =PA = Patm + ρ g h – ρsgh T . Norah Ali Al moneef

  34. Blood Pressure Sphygmometer • Blood pressure is measured with a special type of manometer called a sphygmomano-meter • Pressure is measured in mm of mercury T . Norah Ali Al moneef

  35. Pressure with depth Estimate the amount by which blood pressure changes in an actuary in the foot P2 and in the aorta P1 when the person is lying down and standing up Take density of blood = 1060kg/m3 T . Norah Ali Al moneef

  36. 26.8 K P a T . Norah Ali Al moneef

  37. example A fluid of constant density ρ = 960 kg / m3 is flowing steadily through the above tube. The diameters at the sections are d 1 =100 mm and d 2 = 80 mm . The gauge pressure at 1 is p1 = 200 k N/ m2 ,and the velocity here is u 1 = 5 m /s . We want to know the gauge pressure at section 2. The tube is horizontal, with y1 = y2 so Bernoulli gives us the following equation for pressure at section 2: P2 = 2 x10 5 + 960 { 52 – ( 7.8 )2} /2 = 182796.8 pa T . Norah Ali Al moneef

  38. An example of the U-Tube manometer Using a u-tube manometer to measure gauge pressure of fluid density ρ = 700 kg/m3, and the manometric fluid is mercury, with a relative density of 13.6. What is the gauge pressure if: h1 = 0.4m and h2 = 0.9m? pB = pC pB = pA + ρgh1 pB = pAtmospheric + ρman gh2 We are measuring gauge pressure so patmospheric= 0 pA = ρman gh2 - ρgh1 a) pA = 13.6 x 103x9.8 x 0.9 - 700 x9.8 x 0.4= 117 327 N, b) pA = 13.6 x 103x9.8 x (-0.1) - 700 x 9.8x 0.4= -16 088.4 N, The negative sign indicates that the pressure isbelow atmospheric T . Norah Ali Al moneef

  39. Example: Fluid is flowing from left to right through the pipe. Points A and B are at the same height, but the cross-sectional areas of the pipe differ. Points B and C are at different heights, but the cross-sectional areas are the same. Rank the pressures at the three points, from highest to lowest. A) A and B (a tie), C B) C, A and B (a tie) C) B, C, A D) C, B, A E) A, B, C E) PA > PB > PC T . Norah Ali Al moneef

  40. Pressure Measurements: • A long closed tube is filled with mercury and inverted in a dish of mercury • Measures atmospheric pressure as ρ g h • Since the closed end is at vacuum, it does not exert any force. p=0 Vacuum! Vacuum pressure = 0 patm h For mercury, h = 760 mm. How high will water rise? No more than h = patm/g = 10.3m T . Norah Ali Al moneef T . Norah Ali Al moneef

  41. h v2 example We can set (assume the hole is on the ground or is where we measure height from). We also have 1 atm. So we have T . Norah Ali Al moneef

  42. example T . Norah Ali Al moneef

  43. T . Norah Ali Al moneef

  44. example P2 =p1 + ρ (v12 –v22 ) / 2 = 180X103 + 103X(22 – 182) = 20X 103 pa T . Norah Ali Al moneef

  45. example T . Norah Ali Al moneef

  46. If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium • If the height doesn’t change much, Bernoulli becomes: y1 = y2 • Where speed is higher, pressure is lower. • Speed is higher on the long surface of the wing – creating a net force of lift. FL T . Norah Ali Al moneef

  47. example T . Norah Ali Al moneef

  48. example T . Norah Ali Al moneef

  49. 1.5m example Water flows through the pipe as shown at a rate of .015 m3/s. If water enters the lower end of the pipe at 3.0 m/s, what is the pressure difference between the two ends? A2 = 20 cm2 T . Norah Ali Al moneef

  50. Example Estimate the force exerted on your eardrum due to the water above when you are swimming at the bottom of the pool with a depth 5.0 m. We first need to find out the pressure difference that is being exerted on the eardrum. Then estimate the area of the eardrum to find out the force exerted on the eardrum. Since the outward pressure in the middle of the eardrum is the same as normal air pressure Estimating the surface area of the eardrum at 1.0cm2=1.0x10-4 m2, we obtain T . Norah Ali Al moneef

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