Chapter 15. FLUIDS What volume does 0.4 kg of alcohol occupy? What is the weight of this volume? m m 0.4 kg. ρ = = ; ρ = 5.

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1 Chapter 15. FLUIDS Density What volume does 0.4 kg of alcohol occupy? What is the weight of this volume? m m 0.4 kg ρ = ; = = ; = 5.06 x 10-4 m ρ 790 kg/m W = D = ρg = 790 kg/m )(9.8 m/s )(5.06 x 10-4 m ); W =.9 N 15-. n unknown substance has a volume of 0 ft and weighs 70 lb. What are the weight density and the mass density? W 70 lb D = = ; D = 168 lb/ft D 168 lb/ft ρ = = ; ρ = 5.7 slugs/ft 0 ft g 9.8 m/s 15-. What volume of water has the same mass of 100 cm of lead? What is the weight density of lead? First find mass of lead: m = ρ = (11. g/cm )(100 cm ); m = 110 g Now water: m 110 g w = = = 110 cm ; w = 110 cm D = ρg ρw 1 g/cm D = (11,00 kg/m )(9.8 m/s ) = 110,740 N/m ; D = 1.11 x 10 5 N/m mL flask (1 L = 1 x 10 - m ) is filled with an unknown liquid. n electronic balance indicates that the added liquid has a mass of 176 g. What is the specific gravity of the liquid? Can you guess the identity of the liquid? = 00 ml = 0.00 L = x 10-4 m ; m = kg m kg ρ = = ; ρ = 880 kg/m, Benzene 4 x 10 m 0

2 Fluid Pressure Find the pressure in kilopascals due to a column of mercury 60 cm high. What is this pressure in lb/in. and in atmospheres? P = ρgh = (1,600 kg/m )(9.8 m/s )(0.60 m); P = 80 kpa lb/in. P = 80 kpa ; 1 kpa P = 11.6 lb/in. 80 kpa P = ; P = atm 101. kpa/atm pipe contains water under a gauge pressure of 400 kpa. If you patch a 4-mm-diameter hole in the pipe with a piece of tape, what force must the tape be able to withstand? D (0.004 m) π π = = = 1.57 x 10 m ; P = ; F F = P = (400,000 Pa)(1.57 x 10-5 m ); P = 5.0 N submarine dives to a depth of 10 ft and levels off. The interior of the submarine is maintained at atmospheric pressure. What are the pressure and the total force applied to a hatch ft wide and ft long? The weight density of sea water is around 64 lb/ft. P = Dh = (64 lb/ft )(10 ft); P = 7680 lb/ft ; P = 5. lb/in. F = P = (7680 lb/ft )( ft)( ft); F = 46,100 lb If you constructed a barometer using water as the liquid instead of mercury, what height of water would indicate a pressure of one atmosphere? P 101,00 Pa P = ρgh; h = = ; ρ g (1000 kg/m )(9.8 m/s ) h = 10. m or 4 ft! 04

3 kg piston rests on a sample of gas in a cylinder 8 cm in diameter. What is the gauge pressure on the gas? What is the absolute pressure? D (0.08 m) π π = = = 5.07 x 10 m ; F mg P = = (0 kg)(9.8 m/s ) 4 P = =.90 x 10 kpa; P = 9.0 kpa x 10 m P abs = 1 atm + P gauge = 101. kpa kpa; P abs = 140 kpa * n open U-shaped tube such as the one in fig is 1 cm in cross-section. What volume of water must be poured into the right tube to cause the mercury in the left tube to rise 1 cm above its original position? ρmhm (1.6 g/cm )(1 cm) ρ 1 g/cm ρ gh = ρ gh ; h = = = 1.6 cm m m = h = (1 cm )(1.6 cm); = 1.6 cm or 1.6 ml The gauge pressure in an automobile tire is 8 lb/in.. If the wheel supports 1000 lb, what area of the tire is in contact with the ground? F 1000 lb = = ; = 5.7 in. P 8 lb/in Two liquids that do not react chemically are placed in a bent tube like the one in Fig Show that the heights of the liquids above their surface of separation are inversely h1 ρ proportional to their densities: = h ρ 1 The gauge pressure must be the same for each column: ρ 1 gh 1 = ρ gh So that: h h ρ = ρ

4 15-1. ssume that the two liquids in the U-shaped tube in Fig are water and oil. Compute the density of the oil if the water stands 19 cm above the interface and the oil stand 4 cm above the interface. Refer to Prob h h ρ hwρ w (19 cm)(1000 kg/m ) = ; ρoil = = ; ρ oil = 79 kg/m ρ h 4 cm 1 1 oil water-pressure gauge indicates a pressure of 50 lb/in. at the foot of a building. What is the maximum height to which the water will rise in the building? P (50 lb/in. )(144 in. /ft ) P = Dh; h = = ; h = 115 ft D 6.4 lb/ft ) The Hydraulic Press The areas of the small and large pistons in a hydraulic press are 0.5 and 5 in., respectively. What is the ideal mechanical advantage of the press? What force must be exerted to lift a ton? Through what distance must the input force move, if the load is lifted a distance of 1 in.? [1 ton = 000 lb ] M I o 5 in. = = ; M I = in. i 000 lb F i = ; F i = 40.0 lb 50 s i = M I s o = (50)(1 in.); s i = 50 in force of 400 N is applied to the small piston of a hydraulic press whose diameter is 4 cm. What must be the diameter of the large piston if it is to lift a 00-kg load? F d F (00 kg)(9.8 m/s ) = = = = ; F d F o o o o ; d (4 cm) o di i i i i 400 N d o = 8.85 cm 06

5 The inlet pipe that supplies air pressure to operate a hydraulic lift is cm in diameter. The output piston is cm in diameter. What air pressure (gauge pressure) must be used to lift an 1800-kg automobile. Fo (1800 kg)(9.8 m/s ) Pi = Po ; Pi = = ; P i = 19 kpa π (0.16 m) o The area of a piston in a force pump is 10 in.. What force is required to raise water with the piston to a height of 100 ft? [ 10 in. (1 ft /144 in. ) = ft ] F = P = (Dh); F = (6.4 lb/ft )(100 ft)( ft ); F = 4 lb rchimedes Principle * g cube cm on each side is attached to a string and then totally submerged in water. What is the buoyant force and what is the tension in the rope? = (0.0 m) = 8 x 10 6 m ; m = kg; F B = ρg T F B T F B = (1000 kg/m )(9.8 m/s )(8 x 10-6 m ); F B = N mg ΣF y = 0; T + F B mg = 0; T = mg F B ; T = (0.100 kg)(9.8 m/s ) N; T = N N; T = 0.90 N *15-0. solid object weighs 8 N in air. When it is suspended from a spring scale and submerged in water, the apparent weight is only 6.5 N. What is the density of the object? W 8 N m m = = ; m = kg ρ = g 9.8 m/s To find density, we need to know the volume of the block which is the same as the volume of the water displaced. F 1.50 N F B = 8 N 6.5 N; F B = 1.5 N; F B = ρg; = B = ρ g (1000 kg/m )(9.8 m/s ) T F B 8 N T=6.5 N m kg 1.5 x 10 m -4 = 1.5 x 10 m ; ρ = = ; ρ = 5 kg/m -4 07

6 *15-1. cube of wood 5.0 cm on each edge floats in water with three-fourths of its volume submerged. (a) What is the weight of the cube? (b) What is the mass of the cube? (c) What is the specific gravity of the wood? [ = (0.05 m) = 1.5 x 10 4 m ] The volume of water displaced is ¾ of the volume of the block: D = ¾(1.5 x 10-4 m ) = 9.8 x 10-5 m F B = W F B = ρg D = (1000 kg/m )(9.8 m/s )(9.8 x 10-5 m ) = N F B W The weight is the same as the buoyant force if the block floats: W = N W N m = = ; m = kg or m = 9.8 g g 9.8 m/s Specific gravity: ρ m 4 (15 cm ) = = ; Specific gravity = 0.75 ρ m 15 cm w w * g piece of metal has a density of 4000 kg/m. It is hung in a jar of oil (1500 kg/m ) by a thin thread until it is completely submerged. What is the tension in the thread? First find volume of metal from its mass and density: ρ m m ρ 0.0 kg 4000 kg/m -6 = ; = = = 5 x 10 m ; F B = ρg T F B mg T F B = (1500 kg/m )(9.8 m/s )(5 x 10-6 m ); F B = N ΣF y = 0; T + F B mg = 0; T = mg F B ; mg = (0.0 kg)(9.8 m/s ) T = N N; T = 0.1 N *15-. The mass of a rock is found to be 9.17 g in air. When the rock is submerged in a fluid of density 87 kg/m, its apparent mass is only 7.6 g. What is the density of this rock? pparent weight = true weight buoyant force; m g = mg m f g m = m m f ; m f = m - m = 9.17 g 7.6 g = 1.91 g 08

7 *15-. (Cont.) The volume f displaced is found from the mass displaced: m f = 1.91 g f m kg f -6 = = =.19 x 10 m ; f = rock =.19 x 10-6 m ρ f 87 kg/m - mrock 9.16 x 10 kg ρ rock = = ; ρ -6 rock = 4191 kg/m.19 x 10 m rock *15-4. balloon 40 m in diameter is filled with helium. The mass of the balloon and attached basket is 18 kg. What additional mass can be lifted by this balloon? First find volume of balloon: b = (4/)πR = (4/) π (0 m) b =.51 x 10 4 m ; F B = W balloon + W Helium + W added F b = ρ air g b = (1.9 kg/m )(9.8 m/s )(.51 x 10 4 m ) = 4.4 x 10 5 N F B W He W b W L W He = ρg = (0.178 kg/m )(9.8 m/s )(.51 x 10 4 m ) = x 10 4 N; W b = m b g W add = F B W b W Helium = 4.4 x 10 5 N (18 kg)(9.8 m/s ) x 10 4 N W add =.65 x 10 5 N; m add 5 Wadd.65 x 10 N = = ; m add = 7,00 kg g 9.8 m/s The densities of air and helium were taken from Table 15-1 in the text. Fluid Flow Gasoline flows through a 1-in.-diameter hose at an average velocity of 5 ft/s. What is the rate of flow in gallons per minute? (1 ft = 7.48 gal). How much time is required to fill a 0-gal tank? [ = πr = π(0.5 in.) = in (1 ft/144 in. ) = x 10 - ft ] R = v = (5 ft/s)(5.454 x 10 - ft ) = 0.07 ft /s; R = (0.07 ft /s)(7.48 gal/ft )(60 s/min); R = 1. gal/min. 0 gal Time = ; Time = 1.6 min 1. gal/min 09

8 15-6. Water flows from a terminal cm in diameter and has an average velocity of m/s. What is the rate of flow in liters per minute? ( 1 L = m. ) How much time is required to fill a 40-L container? [ = π(0.015 m) = 7.07 x 10-4 m ; = 40 L = 0.04 m ] R = v = ( m/s)(7.07 x 10-4 m ) = 1.41 x 10 - m /s 0.04 m Time = ; Time = 8. s x 10 m /s What must be the diameter of a hose if it is to deliver 8 L of oil in 1 min with an exit velocity of m/s? 8 L m 1 min R = = 1 min 1 L 60 s x 10 m /s ; R = v 4 4(1. x 10 m /s) = π D R ; D R 5.66 x 10 m 4 = v = π v = π ( m/s) = D = 5.66 x 10 m ; D = 7.5 x 10 - m; D = 7.5 mm Water from a -in. pipe emerges horizontally at the rate of 8 gal/min. What is the emerging velocity? What is the horizontal range of the water if the pipe is 4 ft from the ground? 8 gal 1 ft 1 min ( ft) R = = ft /s; = π = ft 1 min 7.48 gal 60 s 4 1 R ft /s v = = ; v = ft/s ft To find the range x, we first find time to strike ground from: y = ½gt y (4 ft) t = = = s ; Now range is x = v x t g ft/s Range: x = vt = (0.817 ft/s)(0.5 s); x = ft or x = 4.90 in. 10

9 15-9. Water flowing at 6 m/s through a 6-cm pipe is connected to a -cm pipe. What is the velocity in the small pipe? Is the rate of flow greater in the smaller pipe? 1 v 1 = v v d v d v 6 cm = = = (6 m/s) ; v = 4.0 m/s cm The rate of flow is the same in each pipe. pplications of Bernoulli s Equation Consider the situation described by Problem If the centers of each pipe are on the same horizontal line, what is the difference in pressure between the two connecting pipes? For a horizontal pipe, h 1 = h : P 1 + ρgh 1 +½ρv 1 = P + ρgh + ½ρv P 1 - P = ½ρv - ½ρv 1 = ½ρ(v v 1 ); ρ = 1000 kg/m P 1 P = ½(1000 kg/m )[(4 m/s) (6 m/s) ] =.70 x 10 5 Pa; P = 70 kpa What is the emergent velocity of water from a crack in its container 6 m below the surface? If the area of the crack is 1. cm, at what rate of flow does water leave the container? [ The pressures are the same at top and at crack: P 1 = P ] P 1 + ρgh 1 +½ρv 1 = P + ρgh + ½ρv ; ρgh 1 +½ρv 1 = ρgh + ½ρv Notice that ρ cancels out and recall that v 1 can be considered as zero. Setting v 1 = 0, we have: ρgh 1 = ρgh + ½ρv or v = g(h 1 - h ) v = (9.8 m/s )(6 m); v = 10.8 m/s = (1. cm )(1 x 10-4 m /cm ) = 1.0 x 10-4 m ; R = v R = (10.84 m/s)(1.0 x 10-4 m ); R = 1.41 x 10 - m /s or R = 1.41 L/s 11

10 15-. -cm-diameter hole is in the side of a water tank, and it is located 5 m below the water level in the tank. What is the emergent velocity of the water from the hole? What volume of water will escape from this hole in 1 min? [ pply Torricelli s theorem ] v = gh = (9.8 m/s )(6 m) ; v = 9.90 m/s π D π (0.0 m) = = =.18 x 10 m ; R = v R = (9.90 m/s)(.18 x 10-5 m ); R =.11 x 10-4 m /s R x 10 m 60 s 1 L = ; R = 187 L/min 1 s 1 min m *15-. Water flows through a horizontal pipe at the rate of 8 ft /min. pressure gauge placed on a 6-in. cross-section of this pipe reads 16 lb/in.. What is the gauge pressure in a section of pipe where the diameter is in.? 1 = π( in.) = 8. in. = ft 1 = π(1.5 in.) = 7.07 in. = ft 6 in. 1 in. R 1 = R = 8 ft /min(1 min/60 s) = 1.7 ft /s; v 1 1 = v ; R R v = = = ft/s; = = = 7.84 ft/s 1.7 ft /s 1.7 ft /s 1 v ft ft Now find absolute pressure: P 1 = 16 lb/in lb/in. = 0.7 lb/in. ; lb 144 in. 1 P = 0.7 = 441 lb/ft in. 1 ft (bsolute pressure pipe 1) D = ρg; ρ = D 6.4 lb/ft 1.95 slugs/ft g = ft/s = Use consistent units for all terms: ft, slugs) 1

11 *15-. (Cont.) P 1 = 441 lb/ft ; v 1 = ft/s; v = 7.84 ft/s; ρ = 1.95 slugs/ft ; Since h 1 = h, P 1 +½ρv 1 = P + ½ρv and P = P 1 + ½ρv 1 - ½ρv P = 441 lb/ft + ½(1.95 slugs/ft )(6.959 ft/s) ½(1.95 slugs/ft )(7.84 ft/s) P = 441 lb/ft lb/ft lb/ft ; P = 71 lb/ft lb 1 ft P = lb/in. = ft 144 in. (bsolute pressure in pipe ) Gauge pressure P = 5.8 lb/in lb/in. ; P = 11.1 lb/in. *15-4. Water flows at the rate of 6 gal/min though an opening in the bottom of a cylindrical tank. The water in the tank is 16 ft deep. What is the rate of escape if an added pressure of 9 lb/in. is applied to the source of the water? The rate of escape is proportional to the difference in pressure between input and output. P = ρgh = Dh = (6.4 lb/ft )(16 ft); P = lb/ft = 6.9 lb/in. Thus, P of 6.9 lb/in. produces a rate of 6.0 gal/min. Now the new P is: P = 6.9 lb/in. + 9 lb/in. = 15.9 lb/in., By ratio and proportion, we have R 6 gal/min = and R = 1.8 gal/min 15.9 lb/in. 6.9 lb/in. *15-5. Water moves through a small pipe at 4 m/s under an absolute pressure of 00 kpa. The pipe narrows to one-half of its original diameter. What is the absolute pressure in the narrow part of the pipe? [ v 1 d 1 = v d ] 1 v 1 = 4 m/s; v d = = d (4 m/s) 16 m/s d 1 d P 1 +½ρv 1 = P + ½ρv and P = P 1 + ½ρv 1 - ½ρv 1

12 *15-5. (Cont.) P = P 1 + ½ρv 1 - ½ρv ; ρ = 1000 kg/m ; v 1 = 4 m/s; v = 16 m/s P = 00, 000 Pa + ½(1000 kg/m )(4 m/s) ½(1000 kg/m )(16 m/s) P = 00,000 Pa Pa 18,000 Pa = 80,000 Pa; P = 80.0 kpa *15-6. Water flows steadily through a horizontal pipe. t a point where the absolute pressure in 00 kpa, the velocity is m/s. The pipe suddenly narrows, causing the absolute pressure to drop to 100 kpa. What will be the velocity of the water in this constriction? P 1 +½ρv 1 = P + ½ρv ½ρv = P 1 +½ρv 1 P ½(1000 kg/m )v = 00,000 Pa + ½(1000 kg/m )( m/s) 100,000 Pa (500 kg/m )v = 0,000 Pa; v = 0.1 m/s 1 d 1 d 1 d Challenge Problems *15-7. Human blood of density 1050 kg/m is held a distance of 60 cm above an arm of a patient to whom it is being administered. How much higher is the pressure at this position than it would be if it were held at the same level as the arm? P = ρgh = (1050 kg/m )(9.8 m/s )(0.6 m); P = 6.17 kpa *15-8. cylindrical tank 50 ft high and 0 ft in diameter is filled with water. (a) What is the water pressure on the bottom of the tank? (b) What is the total force on the bott0m? (c) What is the pressure in a water pipe that is located 90 ft below the water level in the tank? (a) P = (6.4 lb/ft )(50 ft) = 10 lb/ft or 1.7 lb/in. (Gauge pressure) (b) F = P = (10 lb/ft )[π(10 ft) ]; F = 9.80 x 10 5 lb (c) P = (6.4 lb/ft )(90 ft) = 5616 lb/ft ; P = 9.0 lb/in. (Gauge pressure) 14

13 *15-9. block of wood weighs 16 lb in air. lead sinker with an apparent weight of 8 lb in water, is attached to the wood, and both are submerged in water. If their combined apparent weight in water is 18 lb, find the density of the wooden block? F B = 8 lb + 16 lb 18 lb = 6 lb; F B = D = F B = 6 lb = ft ; 16 lb wood D 6.4 lb/ft D = ; D = 8.4 lb/ft ft * g block of wood has a volume of 10 cm. Will it float in water? Gasoline? For water: F B = ρgh = (1000 kg/m )(9.8 m/s )(10 x 10-6 m ) = N W = mg = (0.1 kg)(9.8 m/s ) = N F B > W YES For gasoline: F B = ρgh = (680 kg/m )(9.8 m/s )(10 x 10-6 m ) = N W = mg = (0.1 kg)(9.8 m/s ) = N F B < W NO * vertical test tube has cm of oil (0.8 g/cm ) floating on 9 cm of water. What is the pressure at the bottom of the tube? [ P T = ρ ο gh o + ρ ω gh w ] P T = (800 kg/m )(9.8 m/s )(0.0 m) + (1000 kg/m )(9.8 m/s )(0.09 m) P T = 5. Pa + 88 Pa = 1117 Pa; P T = 1.1 kpa *15-4. What percentage of an iceberg will remain below the surface of seawater (ρ = 100 kg/m )? If iceberg floats, then: F B = m i g and F B = ρ ω g w F B m i g = ρ i i g; therefore, ρ i i g = ρ ω g w So that: w i ρi 90 kg/m = = = 0.89 or 89.% ρ 100 kg/m w F B Thus, 89. percent of the iceberg remains underwater, hence the expression: That s just the tip of the iceberg. 15

14 *15-4. What is the smallest area of ice 0 cm thick that will support a 90-kg man? The ice is floating in fresh water (ρ = 1000 kg/m ). [ = h so that = /h ] The question becomes: What volume of ice will support a 90-kg person? F B = W ice + W man ; F B = ρ w g w ; W ice = ρ ice g ice ; W man = mg ρ w g w = ρ ice g ice + mg; ( gravity divides out ) The smallest area occurs when the ice is level with the surface and ice = w ρ w w = ρ ice w + m or w = 1.15 m ; (1000 kg/m ) w (90 kg/m ) w = 90 kg Now since = h, we have 1.15 m = w = and =.75 m h 0.00 m * spring balance indicates a weight of 40 N when an object is hung in air. When the same object is submerged in water, the indicated weight is reduced to only 0 N. What is the density of the object? [ F B = 40 N 0 N = 10 N ] F B = ρ w g w ; w = FB 10 N ; 1.0 x 10 m w ρ g = (1000 kg/m )(9.8 m/s ) = w - The volume of the object is the same as the water displaced: = 1.0 x 10 - m W 40 N m 4.08 kg m = = = 4.08 kg; ρ = = ρ = 4000 kg/m g 9.8 m/s 1.0 x 10 m * thin-walled cup has a mass of 100 g and a total volume of 50 cm. What is the maximum number of pennies that can be placed in the cup without sinking in the water? The mass of a single penny is.11 g. F B = ρg = m c g + m p g ; m p = ρ - m c ; m p = (1 g/cm )(50 cm ) g m p = 150 g; Since each penny is.11 g, Nbr of pennies = 48 16

15 What is the absolute pressure at the bottom of a lake that is 0 m deep? P abs = 1 atm + ρgh = 101,00 Pa + (1000 kg/m )(9.8 m/s )(0 m); P = 95 kpa fluid is forced out of a 6-mm-diameter tube so that 00 ml emerges in s. What is the average velocity of the fluid in the tube? [ R = v ; 00 ml = x 10-4 m ] d (0.006 m) π π = = =.8 x 10 m ; -4 x 10 m 6.5 x 10-6 m /s R = = s -6 R 6.5 x 10 m /s v = = ; v = 0.1 m/s -5.8 x 10 m * pump of -kw output power discharges water from a cellar into a street 6 m above. t what rate in liters per second is the cellar emptied? Fs mgh m Power 000 W Power = = ; = = = 4.0 kg/s t t t gh (9.8 m/s )(6 m) Now, ρ w = (1000 kg/m )(0.001 m /L) = 1 kg/l; Therefore, 1 kg of water is one liter: Rate = 4.0 L/s horizontal pipe of diameter 10 mm has a constriction of diameter 40 mm. The velocity of water in the pipe is 60 cm/s and the pressure is 150 kpa (a) What is the velocity in the constriction? (b) What is the pressure in the constriction? d 1 10 mm v1d 1 = vd ; v = v1 = (60 cm/s) ; v = 540 cm/s d 40 mm P 1 +½ρv 1 = P + ½ρv and P = P 1 + ½ρv 1 - ½ρv P = 150, 000 Pa + ½(1000 kg/m )(0.06 m/s) ½(1000 kg/m )(5.4 m/s) P = 150,000 Pa Pa 14,580 Pa = 16,000 Pa; P = 16 kpa 17

16 * The water column in the container shown in Fig stands at a height H above the base of the container. Show that the depth h required to give the horizontal range x is given by H H x h = ± Use this relation to show that the holes equidistant above and H y h x below the midpoint will have the same horizontal range? [ y = H h ] The emergent velocity v x is: v = gh and x = v x t and y = ½gt x t x x x v gh gh = = ; t = ; ( Now substitute t into y = ½gt ) x x gx gx y = ½ g ; y = ; y = H h = gh 4h 4h Multiply both sides by h and rearrange: h - Hh + x /4 = 0 Now, b ± b 4ac h = where a = 1, b = H, and c = a x 4 H ± H 4 x /4 H H h = x ; h = ± (1) The midpoint is (H/), and the ± term indicates the distance above and below that point. * column of water stands 16 ft above the base of its container. What are two hole depths at which the emergent water will have a horizontal range of 8 ft? From Problem 15-50: H H x h = ± 16 ft (16 ft) (8 ft) h = ± = 8 ft ± 6.9 ft ; h = 1.07 ft and 14.9 ft 18

17 *15-5. Refer to Fig and Problem Show that the horizontal range is given by x = h( H h) Use this relation to show that the maximum range is equal to the height H of the water column. From Problem 15-50: H h = x /4h or x = 4hH 4h x = 4h(H h) From which we have: x = h( H h) The maximum range x max occurs when h = (H/) xmax = H H H = H 4 and x max = H *15-5. Water flows through a horizontal pipe at the rate of 60 gal/min (1 ft = 7.48 gal). What is the velocity in a narrow section of the pipe that is reduced from 6 to 1 in. in diameter? R = = min 7.48 gal 60 s gal 1 ft 1 min ft /s ; d 1 = 6 in. = 0.5 ft R 0.17 ft /s 1 1 π (0.5 ft) v = = ; v 1 = ft/s d 1 6 in. v1d 1 = vd ; v = v1 = ( ft/s) ; v = 4.5 ft/s d 1 in What must be the gauge pressure in a fire hose if the nozzle is to force water to a height of 0 m? P = ρgh = (1000 kg/m )(9.8 m/s )(0 m); P = 196 kpa 19

18 * Water flows through the pipe shown in Fig. 15- at the rate of 0 liters per second. The absolute pressure a pint is 00 kpa, and the point B is 8 m higher than pint. The lower section of pipe has a diameter of 16 cm and the upper section narrows to a diameter of 10 cm. (a) Find the velocities of the stream at points and B. (b) What is the absolute pressure at point B? [ R = 0 L/s = 0.00 m /s ] B = π(0.08 m) = m B = π(0.05 m) = m 8 m v = R 1.49 m/s;.8 m/s = = = = = 0.00 m /s R 0.00 m /s v m m The rate of flow is the same at each end, but v 1 = 1.49 m/s and v =.8 m/s To find pressure at point B, we consider the height h = 0 for reference purposes: 0 P + ρgh +½ρv = P B + ρgh B + ½ρv B ; P B = P + ½ρv - ρgh B - ½ρv B P B = 00,000 Pa + ½(1000 kg/m )(1.49 m/s) (1000 kg/m )(9.8 m/s )(8 m) - ½(1000 kg/m )(.8 m/s) P B = 00,000 Pa Pa 78,400 Pa 796 Pa; P B = 115 kpa Critical Thinking Questions living room floor has floor dimensions of 4.50 m and.0 m and a height of.40 m. The density of air is 1.9 kg/m. What does the air in the room weight? What force does the atmosphere exert on the floor of the room? = (4.50 m)(.0 m)(.40 m); m = 4.56 m ; floor = (4.5 m)(. m) = 14.4 m m ; ρ = ; W = mg = ρ = (1.9 kg/m )(4.56 m ) = 44.6 kg ; W = (44.6 kg)(9.8 m/s ); W = 47 N F = P = (101,00 Pa)(14.4 m ); F = 1,460,000 N 0

19 tin coffee can floating in water (1.00 g/cm ) has an internal volume of 180 cm and a mass of 11 g. How many grams of metal can be added to the can without causing it to sink in the water. [ The buoyant force balances total weight. ] F B = ρg = m c g + m m g ; m m = ρ - m c ; The volume of the can and the volume of displaced water are the same. m m = (1 g/cm )(180 cm ) - 11 g m m = 68.0 g * wooden block floats in water with two-thirds of its volume submerged. The same block floats in oil with nine-tenths of its volume submerged. What is the ratio of the density of the oil to the density of water (the specific gravity)? The buoyant force is same for oil and water, F B (oil) = F B (water) ρ oil g oil = ρ w g w ; ρoil ρ w w = = ; Specific gravity = oil 10 * n aircraft wing 5 ft long and 5 ft wide experiences a lifting force of 800 lb. What is the difference in pressure between the upper and lower surfaces of the wing? P 800 lb (5 ft)(5 ft) 1 P = = 6.40 lb/ft ; P = lb/in. * ssume that air (ρ = 1.9 kg/m ) flows past the top surface of an aircraft wing at 6 m/s. The air moving past the lower surface of the wing has a velocity of 7 m/s. If the wing has a weight of 700 N and an area of.5 m, what is the buoyant force on the wing? ssume h 1 = h for small wing: P 1 +½ρv 1 = P + ½ρv P 1 P = ½(1.9 kg/m )(6 m/s) - ½(1.9 kg/m )(7 m/s) ; P = 65.7 Pa F P = = 65.7 Pa; F = (65.7 Pa)(.50 m ); F = 180 N 1

20 * Seawater has a weight density of 64 lb/ft. This water is pumped through a system of pipes (see Fig. 15-) at the rate of 4 ft /min. Pipe diameters at the lower and upper ends are 4 and in., respectively. The water is discharged into the atmosphere at the upper end, a distance of 6 ft higher than the lower section. What are the velocities of flow in the upper and lower pipes? What are the pressures in the lower and upper sections? ft 1 min R = = min 60 s ft /s d = 4 in.; d B = in. 6 ft B π ( ft) π ( ft) = = = = ft ; B ft v = R ft/s;.06 ft/s = = = = = ft /s R ft /s v ft ft The rate of flow is the same at each end, but v = ft/s and v B =.06 ft/s Pressure at point B is 1 atm = 116 lb/ft ; Consider h = 0 for reference purposes. D = ρg; D 64 lb/ft ρ = = ; ρ =.00 slugs/ft g ft/s 0 P + ρgh +½ρv = P B + ρgh B + ½ρv B ; P = P B + ½ρv B + ρgh B - ½ρv P = 116 lb/ft + ½( slugs/ft )(.06 ft/s) + ( slugs/ft )( ft/s )(6 ft) - ½( slugs/ft )(0.764 ft/s) P = 116 lb/ft lb/ft + 84 lb/ft lb/ft ; P = 509 lb/ft P lb 1 ft = 09 ft 144 in. ; P = 17.4 lb/in. (absolute)

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