# Fluids in motion.ppt - Fluid Motion Paul E Tippens The...

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This preview shows page 1 out of 29 pages. Unformatted text preview: Fluid Motion Paul E. Tippens The lower falls at Yellowstone National Park: the water at the top of the falls passes through a narrow slot, causing the velocity to increase at that point. In this chapter, we will study the physics of fluids in motion. Objectives: After completing this module, you should be able to: • Define the rate of flow for a fluid and solve problems using velocity and crosssection. • Write and apply Bernoulli’s equation for the general case and apply for (a) a fluid at rest, (b) a fluid at constant pressure, and (c) flow through a horizontal pipe. Fluids in Motion All fluids are assumed in this treatment to exhibit streamline flow. •• Streamline Streamline flow flow isis the the motion motion of of aafluid fluid in in which which every every particle particle in in the the fluid fluid follows follows the the same same path path past past aaparticular particular point point as as that that followed followed by by previous previous particles. particles. Assumptions for Fluid Flow: •• All All fluids fluids move move with with streamline streamline flow. flow. •• The The fluids fluids are are incompressible. incompressible. •• There There isis no no internal internal friction. friction. Streamline flow Turbulent flow Rate of Flow The rate of flow R is defined as the volume V of a fluid that passes a certain cross-section A per unit of time t. The volume V of fluid is given by the product of area A and vt: A VV Avt Avt vt Volume = A(vt) Avt R vA t Rate of flow = velocity x area Constant Rate of Flow For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases: R v1 A1 v2 A2 A1 22 11 11 22 22 22 vv dd vv dd R = A1v1 = A2v2 A2 v1 v2 v2 Example 1: Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What must be the diameter of the nozzle in order that the water emerge at 16 m/s? The area is proportional to the square of diameter, so: v1d12 v2 d 22 2 1 1 vd (4 m/s)(2 cm) d v2 (20 cm) 2 2 2 2 dd22 == 0.894 0.894 cm cm Example 1 (Cont.): Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What is the rate of flow in m3/min? R v1 A1 v2 A2 d12 R v1 A1 ; A1 4 2 2 d1 (4 m/s) (0.02 m) R1 v1 4 4 m3 1 min R1 0.00126 min 60 s R1 = 0.00126 m3/s 3 RR11 ==0.0754 0.0754 m m3/min /min Problem Strategy for Rate of Flow: •• Read, Read, draw, draw, and and label label given given information. information. •• The The rate rate of of flow flow RR is is volume volume per per unit unit time. time. •• When When cross-section cross-section changes, changes, RR is is constant. constant. R v1 A1 v2 A2 •• Be Be sure sure to to use use consistent consistent units units for for area area and and velocity. velocity. Problem Strategy (Continued): •• Since Since the the area area AA of of aa pipe pipe is is proportional proportional to to its its diameter diameter dd,, aa more more useful useful equation equation is: is: 2 1 1 v d v2 d 2 2 •• The The units units of of area, area, velocity, velocity, or or diameter diameter chosen chosen for for one one section section of of pipe pipe must must be be consistent consistent with with those those used used for for any any other other section section of of pipe. pipe. The Venturi Meter h A B C The higher velocity in the constriction B causes a difference of pressure between points A and B. PPAA -- PPBB == gh gh Demonstrations of the Venturi Principle Examples of the Venturi Effect The increase in air velocity produces a difference of pressure that exerts the forces shown. Work in Moving a Volume of Fluid A2 P2 A1 P1 Volum F1 P1 ; F1 P1 A1 e V A1 A1 P1 F1 P2 Note differences in pressure P and area A F2 A2 ; F2 P2 A2 A2 P 2 , F2 h Fluid is raised to a Work on a Fluid (Cont.) v2 F1 = P1A1 h1 s1 F2 = P2A2 A2 v1 A1 s2 h2 Net Net work work done done on on fluid fluid isis sum sum of of work work done done by by input input force force FFi iless less the the work work done done by by resisting resisting force force FF22,, as as shown shown in in figure. figure. Net Work = P1V - P2V = (P1 - P2) V Conservation of Energy v2 Kinetic Energy K: K ½ mv22 ½ mv12 Potential Energy U: F1 = P1A1 h1 U mgh2 mgh1 Net Work = K + U F2 = P2A2 A2 v1 A1 s2 h2 s1 also Net Work = (P1 - P2)V 2 2 2 1 ( P1 P2 )V (½ mv ½ mv ) (mgh2 mgh2 ) Conservation of Energy 2 2 2 1 ( P1 P2 )V (½ mv ½ mv ) (mgh2 mgh2 ) Divide by V, recall that density m/V, then simplify: P1 gh1 ½ v12 P2 gh2 ½ v22 v2 v1 Bernoulli’s Theorem: 2 1 h2 P1 gh1 ½ v Const h1 Bernoulli’s Theorem (Horizontal Pipe): P1 gh1 ½ v12 P2 gh2 ½ v22 Horizontal Pipe (h1 = h2) P1 P2 ½ v22 ½ v12 v1 h v2 h 1 = h2 Now, since the difference in pressure P = gh, Horizontal Pipe P gh ½ v22 ½ v12 Example 3: Water flowing at 4 m/s passes through a Venturi tube as shown. If h = 12 cm, what is the velocity of the water in the constriction? Bernoulli’s Equation (h1 = h2) 2 2 2 1 P gh ½ v ½ v h v1 = 4 m/s Cancel , then clear fractions: v2 h = 6 cm 2gh = v22 - v12 v2 2 gh v12 2(9.8 m/s 2 )(0.12 m) (4 m/s) 2 vv22 == 4.28 4.28 m/s m/s Note that density is not a factor. Bernoulli’s Theorem for Fluids at Rest. For many situations, the fluid remains at rest so that v1 and v2 are zero. In such cases we have: P1 gh1 ½ v12 P2 gh2 ½ v22 P1 - P2 = gh2 - gh1 This is the same relation seen earlier for finding the pressure P at a given depth h = (h2 - h1) in a fluid. P P = = g(h g(h22-hh11)) h = 1000 kg/m3 Torricelli’s Theorem When there is no change of pressure, P1 = P2. 2 1 P1 gh1 ½ v P2 gh2 ½ v Consider right figure. If surface v2 and P1= P2 and v1 = v we have: Torricelli’s theorem: v 2 gh 2 2 v2 h2 h h1 v 2 gh Interesting Example of Torricelli’s Theorem: Torricelli’s theorem: v 2 gh • Discharge velocity increases with depth. v v v • Maximum range is in the middle. • Holes equidistant above and below midpoint will have same horizontal range. Example 4: A dam springs a leak at a point 20 m below the surface. What is the emergent velocity? Torricelli’s theorem: h v 2 gh Given: h = 20 m g = 9.8 m/s2 v 2(9.8 m/s 2 )(20 m) 2 vv == 19.8 19.8 m/s m/s2 v 2 gh Strategies for Bernoulli’s Equation: •• Read, Read, draw, draw, and and label label aa rough rough sketch sketch with with givens. givens. •• The The height height hh of of aa fluid fluid isis from from aa common common reference reference point point to to the the center center of of mass mass of of the the fluid. fluid. •• In In Bernoulli’s Bernoulli’s equation, equation, the thedensity density is is mass mass 33 density and the appropriate units are kg/m density and the appropriate units are kg/m .. •• Write Write Bernoulli’s Bernoulli’s equation equation for for the the problem problem and and simplify simplify by by eliminating eliminating those those factors factors that that do do not not change. change. P1 gh1 ½ v12 P2 gh2 ½ v22 Strategies (Continued) P1 gh1 ½ v12 P2 gh2 ½ v22 • For a stationary fluid, v1 = v2 and we have: = 1000 P P = = g(h g(h22-h kg/m3 hh11)) • For a horizontal pipe, h1 = h2 and we obtain: P1 P2 ½ v22 ½ v12 Strategies (Continued) P1 gh1 ½ v12 P2 gh2 ½ v22 • For no change in pressure, P1 = P2 and we have: Torricelli’s Theorem v 2 gh General Example: Water flows through the pipe at the rate of 30 L/s. The absolute pressure at point A is 200 kPa, and the point B is 8 m higher than point A. The lower section of pipe has a diameter of 16 cm and the upper section narrows to a diameter of 10 cm. Find the velocities of the stream at points A and B. B R = 30 L/s = 0.030 m3/s A R 2 ; D R 2 AA = (0.08 m)2 = 0.0201 m3 AB = (0.05 m)2 = 0.00785 m3 R 0.030 m 3 /s vA 1.49 m/s; 2 AA 0.0201 m vA = 1.49 m/s R=30 L/s 8m A R 0.030 m 3 /s v2 3.82 m/s 2 A2 0.00785 m vB = 3.82 m/s General Example (Cont.): Next find the absolute pressure at Point B. Given: vA = 1.49 m/s vB = 3.82 m/s PA = 200 kPa h B - hA = 8 m B R=30 L/s 8m A Consider the height hA = 0 for reference purposes. 0 PA + ghA +½vA2 = PB + ghB + ½vB2 PB = PA + ½vA2 - ghB - ½vB2 PB = 200,000 Pa + 1113 3 –78,400 Pa PB = 200,000 Pa + ½1000 kg/mPa )(1.49 m/s)2 – 7296 Pa – (1000 kg/m3)(9.8 m/s2)(8 m) - ½1000 kg/m3)(3.82 m/s)2 PPBB == 115 115 kPa kPa Summary Streamline Fluid Flow in Pipe: v1d12 v2 d 22 R v1 A1 v2 A2 Fluid at Rest: PA - PB = gh Horizontal Pipe (h1 = h2) P1 P2 ½ v22 ½ v12 Bernoulli’s Theorem: P1 gh1 ½ v12 Constant Torricelli’s theorem: v 2 gh Summary: Bernoulli’s Theorem •• Read, Read, draw, draw, and and label label aa rough rough sketch sketch with with givens. givens. •• The The height height hh of of aa fluid fluid isis from from aa common common reference reference point point to to the the center center of of mass mass of of the the fluid. fluid. •• In In Bernoulli’s Bernoulli’s equation, equation, the thedensity density rr isis mass mass 33 density and the appropriate units are kg/m density and the appropriate units are kg/m .. •• Write Write Bernoulli’s Bernoulli’s equation equation for for the the problem problem and and simplify simplify by by eliminating eliminating those those factors factors that that do do not not change. change. P1 gh1 ½ v12 P2 gh2 ½ v22 ...
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