Introduction to Mechanical Engineering Chapter 1 The Mechanical Engineering Profession Chapter Problem-Solving and Communication Skills Chapter 3 Forces in Structures and Machines Chapter 4 Materials and Stresses Chapter 5 Fluids Engineering Chapter 6 Thermal and Energy Systems Chapter 7 Motion and Power Transmission Chapter 8 Mechanical Design 1
Hydrodynamics Aerodynamics Steam Water recycling Biomedical Engineering Buoyancy, drag and lift by fluid 5-
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Properties of Fluids 1. Not equilibrium on shear stress. Liquid(incompressible) 3. Gas(compressible) 4
Properties of Fluids Continuous motion (flow) by shear stress No-Slip Condition at microscopic level; only several molecules thick, adheres to solid surface 5-5
Properties of Fluids F A Viscosity( ); resistance of a fluid to flow and also measure of stickness,friction v ; flow velocity v (Newtonian fluid) h units of viscosity; kg/m.s mass/(length- time) 6
special unit for viscosity ; P, cp kg 1P( Poise) 0.1 m. s 1cP 0.01P 7
Example 5.1 Machine Tool Guideways oil viscosity ; 40 cp width of ways ; 8 cm length of ways ; 40 cm force ; 90 N velocity ; 15 cm/s 8
Example 5.1 Machine Tool Guideways oil viscosity ; 40 cp width of ways ; 8 cm length of ways ; 40 cm force ; 90 N velocity ; 15 cm/s What is thickness of oil film? F v A h Av h F ()(0.08)(0.4)( m )(0.4 kg/ m. s)(0.15 90( kg. m / s ) m / s).56x10 5 m 5.6 m 9
Pressure and Buoyancy Force P Po gh 1 w gv ;density of fluid V ;volume of fluid 10
Buoyancy and pressure liquid mass, m Ah equilibriu m force balance p p 1 A p p 0 A ( Ah) g gh 1 0 0 11
Units 1 pascal = 1 Pa(N/m ) psi = lb/in psf = lb/ft 1 atm = 1.013 x 10 5 Pa 1
Pressure and Buoyancy Force F B fluid gv object 13
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Example 5. Aircraft s Fuel Capacity Fuel capacity 90,000 L Density ; 840 kg/m 3 What is the weight of fuel? w mg Vg (840 kg 3 m m )(90,000L)(0.001 L 3 )(9.81 m s ) 7.416x10 5 kg. m ( ) s 741.6 kn 15
Example 5.3 Deep Submergence Rescue Vehicle Max. dive depth ; 5000 ft What is water pressure in psi? p 1 p p p 0 1 gh p 0 gh slug (1.99 )(3. 3 ft ft s )(5000 ft) 3.04x10 5 ( slug ft. s ) 3.04x10 5 slug. ( s ft )( 1 ft ) 3.04x10 5 ( lb ft )( 1 ft x 1 in ) 5 psi 16
Example 5.4 Great White Shark Attack 55 gal barrel weighing 35 lb What force Shark to overcome? F B T T w 0 F B w fluid slugs (1.99 )(3. 3 ft gv object slug. ft 471.( ) 35( lb) s 35( lb) ft )(55gal)(0.1337 s 436. lb 3 ft ) 35( lb) gal 17
5.4 Laminar and Turbulent Fluid Flows Reynolds number Re vl ( density, speed, viscosity, characteristic length ) 18
Example 5.5 Reynolds Number Reynolds number vl Re Winchester. speed : v 70m / s d 7.6mm R e 1.( kg/ m^3)(70m / s)(7.6x10^3m) 1.8x10^5kg/ m. s ) 3.679x10 5 Re for water flowing in the pipe R e 3 (1000kg/ m )(0.5m / s)(0.01m) 3 1.0x10 kg/( m. s) 5000 19
Example 5.5 Reynolds Number Reynolds number vl Re 3) SAE 30 oil. instead. of. water 3 (917kg/ m )(0.5m / s)(0.01m) Re 0.6kg/( m. s) 4) Submarine with hull diameter of 17.63 10 m and speed of 8 m/ s Re 3 106 kg/m )( 8 m/s)( 10m) 3 1. x 10 kg/(m.s) 6. 8x10 7 0
5.5 Fluid Flows in Pipes 1
In pipe, Not with any disturbances and low enough Re number(<000), then flow is laminar v v r vmax (1 ( ) R d p 16 L max ) ( in ( in case of case of Re Re 000) 000)
000) Re ( 18 8 ) ( 000) Re ( 1 ; / 4 max max max max case special L p d v d v d v A q case special v v Av q rate flow Volumetric t x v x A V avg avg avg Poiseuille s law 3 000) Re case of ( 16 max in L p d v
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A v 1 1 A v p v gh constant 5
Bernoulli s equation p v gh constant 1 st ; Work of pressure force nd ; Kinetic energy of flow fluid 3 rd ; Gravitational potential energy of it 6
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Example 5.6 Automotive Fuel Line 1. A car speed ; 40 mph. With fuel economy rate; 8 miles/gallon 1. Fuel line inside dia.; 3/8 in. a) volumetric flow rate is car speed q fuel rate b)line flow rate is v (3.968x10 avg q A c) Re 4 40mi/ h 8mi/ gal gal s 40mi/ h x 8mi/ gal 1 3600 3 ft )(0.1337 ) 5.306x10 gal a) Volumetric flow of oil in ft3/s? b) Fuel avg. velocity in in./s? c) What Re number? 3 5 ft 5.306x10 ( ) s ft 6.917x10 1ft (0.375in.) ( ) s 4 1in. 3 vl (1.3slug / ft )(6.917x10 ft / s)(3/ 8)( in.)(1/1)( ft / in.) 6 6.1x10 slug /( ft. s) 5 ( h ( ) 3.968x10 s ft s 3 ) 4 gal s 467.8 8
Drag forces and Viscosity FD Av CD : fluid ' s density A: frontal area of object v : relative speed C D 1 : coefficien t of drag 9
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Nearly constant 31
Drag forces and Viscosity F D 3dv (Special case for sphere; Re 1) d ; sphere's diameter μ ; fluid' s viscosity(kg/m.s, pois) 3
Example 5.7 Golf Ball in Flight 1. 1.68 in. dia. golf ball speed 70 mph. Drag force? A) smooth ball B) C D =0.7 1 FD Av CD A) To knowre F Re F D D vl 1 5.10x10 8.813x10 (.33x10 D number 3 slug. ft s 4 for deciding ( very high, so slug )(1.538x10 3 ft 5.104x10 slug. ft 9.45x10 9.45x10 s B) With C 0.7 instead, lb apply C ft C lb D D ) )(10.7 ft s ) (0.5) 33
34 1) (Re ) ( 18 6) / ( 6, /, 3 0 3 3 fluid object object object object fluid B D D B ρ ρ gd v ρ d m d V mg W gv ρ πμdv, F F -w F F balance force Terminal Velocity
Drag and Lift forces in fluids 35
Wind tunnel 36
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p v gh constant Bernoulli' s equation 40
Lift force by pressure difference on the airfoil p v gh constant p If v v top constant v, then p bottom top p bottom
1 FL Av C C ; coefficien t L L of lift 4
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Summary 1. Buoyancy in fluids F B fluid gv object. Drag forces in fluids F D 3dv (Re 1) 1 D C D F Av 3. Lift forces in fluids Re vl F 1 Av C L L 46
Problem 4.36: (a) A luxury sports car has a frontal area of.4 ft and a 0.9 coefficient of drag at 60 mi/hr. What is the drag force on the vehicle at this speed? (b) A sport utility vehicle has C D = 0.45 at 60 mi/hr, and the slightly larger frontal area of 9.1 ft. What is the drag force in this case? F 1 Av D C D 47
Approach: Apply Equation using the density of air listed in Table 4.3. Convert velocity to consistent units when calculating the drag force. Solution: a) Luxury sports car Velocity: mi ft 4 hr v 60 580.77810 88 hr mi s 1 ft F D 6 s 3 3.3310 slug / ft.4 ft 88 0.9 58. lb b) Sport utility vehicle 1 ft F D 118 s 3 3.3310 slug / ft 9.1 ft 88 0.45 lb 3.3310 slug / ft ft s 3 48
Problem 4.8: A steel storage tank is filled with gasoline. The tank has partially corroded on its inside, and small particles of rust have contaminated the fuel. The rust particles are spherical, have diameter 5 µm, and density 5.3 g/cm3. (a) What is the terminal velocity of the particles as they fall through the gasoline? (b) How long would it take the particles to fall 5 m and settle out of the tank? Approach: Apply terminal velocity Equation to find the terminal velocity using the density(680 kg/m 3 ) and viscosity(.9x10-4 kg/m.s) of gasoline listed in Table 4.3. Calculate the Reynolds number to confirm that the velocity is low enough that the equation was correctly applied. Solution: a) Terminal velocity v gd ( p 18 ) 0.0054( m / s) g sphere g 5.3 cm 3 cm 100 m 6 9.81 m / s 510 m 4 18.9 10 kg/ m s 3 kg 0.001 g kg 5300 3 m kg 5300 3 m kg 680 3 m 49
Check Reynolds number, Re vd 3 6 680kg / m 0.0054m/ s 510 m.910 4 kg / m s 0.31 which is dimensionless. Since the numerical value of above Re is less than one, the velocity was properly calculated. b) Fall time 5 m 0.0054m/ s 96s 15.4 min 50
Problem 4.39: Submarines dive by opening vents which allow air to escape from ballast tanks, and water to flow in and fill them. In addition, diving planes located at the bow are angled downward to help push the boat below the surface. Calculate the diving force produced by a 0 ft hydroplane that is inclined by 3 o as the boat cruises at 15 knots (1 knot = 1.15 mi/hr). Approach: Apply Equation (4.19) using consistent units for velocity. From Figure 4.33 at an angle of attack of 3o, CL = 0.3. Use the density of sea water(1.99 slut/ft3) Velocity v 15 kt mi/ hr 1.1516 580 kt ft mi.77810 4 hr s 5.3 ft s Downward L 1 Av diving force 1 3 slug ft CL 1.9910 4088 3 ft s 0 ft 5.3 0.3 lb 51
Reference Wickert J., Lewis K., An Introduction to Mech anical Engineering, 3 rd Edition, 013. www.slideshare.com 5