Hydrostatics and Fluid Dynamics Density: mass per unit volume m Density ρ = V ex. Find the mass of a block of iron 10.0 cm x 5.0 cm x 7.5 cm. The density of iron is 7.8 g/cm3. V = l w h = ( 10.0 cm )( 5.0 cm )( 7.5 cm ) = 375 cm3 m m = ρ V = ( 7.8 g/cm3 )( 375 cm3 ) ρ = V m = 2900 g

For most substances, density decreases with increased temperature ; not water Ice: ρ = 0.91 g/cm3 As ice melts, hydrogen bonds break, and volume decreases Max density at 4oC: ρ = 1.00 g/cm3

specific gravity: the ratio of a substance's density to that of water at 4oC ex. Density of aluminum = 2.7 g/cm3 specific gravity of aluminum 2.7 g/cm3 = = 2.7 1.0 g/cm3 Aluminum is 2.7 times as dense as water

So specific gravity of block is 0.67 ex.: A block of wood, 12.0 cm x 10.0 cm x 6.0 cm, is placed in water. At equilibrium, 2.0 cm of the largest face is above water. What is the specific gravity of the wood? 2 cm What fraction of the block is under water? 6 cm 12 cm 2 3 Convert to percent: 67% So specific gravity of block is 0.67 Density = 0.67 g/cm3 Convert to kg/m3

Density = 0.67 g/cm3 Convert to kg/m3 1000 g = 1 kg 100 cm = 1 m 0.67 g kg 100 cm 100 cm 100 cm cm3 1000 g m m m = 670 kg/m3

Pressure: force per unit area F P = A ex. Your foot has an approximate area of 300 cm2. If you have a mass of 65.0 kg, what is the pressure on the soles of your feet? F Wt. m g ( 65 kg )( 9.8 m/s2 ) P = = = = A A A 300 cm2 P = 2.12 N/cm2

Pressure in Liquids As you swim down into water, the pressure on your body increases The pressure is a result of the weight of the water above you * The pressure at equal depths within a uniform liquid is equal So the pressure 2 meters below the surface of the Aquatic Center pool is the same as the pressure 2 meters below the surface of Lake Michigan (but not the same as the pressure 2 meters below the surface of the Pacific Ocean, because salt water is a different liquid)

F m g P = = A A m ρ = m = ρ V V V = A h ρ V g ρ (A h) g P = = A A ρ = m = ρ V V V = A h ρ V g ρ (A h) g P = = A A ρ = density h = depth Pressure in liquids P = ρ g h

ex. What is the pressure at a depth of 150 m in the ocean? The density of seawater is 1.025 g/cm3 , or 1025 kg/m3 P = ρ g h = ( 1025 kg/m3 )( 9.8 m/s2 )( 150 m ) P = 1.51 x 106 N/m2 Another pressure unit: lb/in2, or psi 1 lb/in2 = 6895 N/m2 1.51 x 106 N/m2 1 lb/in2 = 219 lb/in2 6895 N/m2 Check this out:

Pressure in Gases Kinetic Theory of Gases: - Particles in a gas are in constant motion - Particles collide with each other and the walls of the container - Particles exert a force when they collide, and cause pressure * Pressure in gases results from collisions At room temperature, gas molecules undergo 109 collisions per second

Atmospheric Pressure We live at the bottom of an ocean of air 9 km up and you have 70% below you (height of Mt. Everest) 38 km up and you have 99% below you (height of “space jump”)

Torricelli’s Barometer Fill tube with mercury; cap and invert into dish of mercury; remove cap Weight of air pushes mercury up the tube At equilibrium, 76.0 cm of mercury would be above level in dish (Pressure will vary slightly as different weather systems move through ) Definition: 1 atmosphere = 29.92 inches Hg = 760 mm Hg = 101 300 N/m2 = 101 300 Pa = 14.7 lb/in2

Gauge Pressure Gauges measure difference between tire's inside pressure (pushing outward) and atmospheric pressure (pushing inward) Absolute pressure is sum of two: Absolute Pressure PA = atmospheric pressure PG = gauge pressure P = PA + PG

Gauge Pressure ex. Inflate a car tire to 32.0 psi. What is absolute pressure? P = PA + PG = 14.7 psi + 32.0 psi P = 46.7 psi

ex. A scuba diver dives to a depth of 15.0 m in the ocean. Determine the absolute pressure at that depth. P = ρ g h ρseawater = 1025 kg/m3 P = ( 1025 kg/m3 )( 9.8 m/s2 )( 15.0 m ) P = 150 675 N/m2 Absolute Pressure = Water pressure + Atmospheric Pressure = 150 675 N/m2 + 101 300 N/m2 Absolute Pressure = 2.52 x 105 N/m2

Pascal's Principle and Hydraulics Pressure applied to a confined fluid increases the pressure throughout by the same amount Apply force to small piston Force exerts pressure on fluid Pressure exerts force on large piston Pressure is transmitted throughout fluid

the small piston of the hydraulic system below. ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on the small piston of the hydraulic system below. (a) Find the pressure that small piston exerts on the fluid. F2 F1 = 800 N A1 = 100 cm2 A2 = 1500 cm2 F1 800 N P1 = = P1 = 8.00 N/cm2 A1 100 cm2

the small piston of the hydraulic system below. ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on the small piston of the hydraulic system below. (b) Find the force exerted on the large piston. F2 F1 = 800 N A1 = 100 cm2 A2 = 1500 cm2 P1 = 8.00 N/cm2 P2 = 8.00 N/cm2 By Pascal’s Principle, P2 = P1 = 8.00 N/cm2 F2 Then P2 = F2 = P2 A2 = ( 8.00 N/cm2 )( 1500 cm2 ) A2 F2 = 12 000 N

the small piston of the hydraulic system below. ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on the small piston of the hydraulic system below. (c) Find the mass of the car that can be lifted by this system. F2 = 12 000 N F1 = 800 N m = ? A1 = 100 cm2 A2 = 1500 cm2 P1 = 8.00 N/cm2 P2 = 8.00 N/cm2 F2 = Wt. of car = m g F2 12 000 N m = = = m = 1220 kg g 9.8 m/s2

the small piston of the hydraulic system below. ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on the small piston of the hydraulic system below. (d) If car is to be raised 2.0 m, how far must small piston move? A1 = 100 cm2 V2 h2 = 2.0 m A2 = 1500 cm2 V1 h1 = ? V1 = V2 V = A h A1 h1 = A2 h2 A2 h2 ( 1500 cm2 )( 2.0 m ) h1 = = = h1 = 30 m A1 100 cm2

Other Phenomena of Liquids Result from the interplay between cohesive and adhesive forces cohesive forces cause liquid particles to stick to each other adhesive forces cause liquid particles to stick to other things Water droplets assume a spherical shape due to cohesive forces Spheres have the largest volume to surface area ratio of any shape (cubes, tetrahedrons, etc.)

Other spheres in nature:

Cohesive forces also cause surface tension Water forms a "skin" at the surface Surfactants take away the surface tension; key ingredient in laundry detergents

Water also exhibits adhesive properties Responsible for things getting “wet” As object is pulled out of water, cohesive forces compete with adhesive forces; things get wet because adhesive forces are greater than cohesive forces

When water is put into a narrow tube, adhesive forces cause it to climb up the sides of the tube; known as capillary action A meniscus forms in a graduated cylinder due to adhesion; volume is read at bottom of meniscus

Other examples:

Could this work as a generator of energy?

Water is adhesive Mercury is cohesive

Buoyant Force and Archimedes Principle Fill a tank with water 1 kg Consider a 1-kg “piece” of the water 9.8 N buoyant force Wt. Water weighs 9.8 N Water just below the piece exerts an upward "buoyant“ force of 9.8 N to produce equilibrium

Buoyant Force and Archimedes Principle Now replace the piece with a stone of equal volume, but not of equal mass 1.5 kg Let mass of stone = 1.5 kg 9.8 N buoyant force The weight of the rock outside of the water = 14.7 N 14.7 N The water still exerts a buoyant force of 9.8 N, as if the "piece” of water was still there If the rock was released, it would accelerate to the bottom

Buoyant Force and Archimedes Principle Now replace the piece with a stone of equal volume, but not of equal mass 4.9 N 1.5 kg Let mass of stone = 1.5 kg 9.8 N buoyant force The weight of the rock outside of the water = 14.7 N 14.7 N If suspended from a spring scale, what would it read, in newtons? 14.7 N - 9.8 N = 4.9 N

Archimedes Principle: The buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid displaced by the body. ex. A block of metal weighs 15.4 N in air. When submerged in water, it weighs 9.7 N. (a) What is the buoyant force exerted by the water? FB = 15.4 N - 9.7 N = FB = 5.7 N FB 9.7 N

ex. A block of metal weighs 15.4 N in air. When submerged in water, it weighs 9.7 N. (b) What is the volume of water displaced by the metal? Wt. of water = FB = 5.7 N Wt. 5.7 N Wt. = m g m = = g 9.8 m/s2 = 0.582 kg = 582 g m ρ = ρ = 1.00 g/cm3 V V ρ = m FB m 582 g 9.7 N V = = = V = 582 cm3 ρ 1.00 g/cm3

ex. A block of metal weighs 15.4 N in air. When submerged in water, it weighs 9.7 N. (c) What is the density of the metal? Wt. in air = 15.4 N Vm = Vw = 582 cm3 m m = ? ρ = V Wt. 15.4 N m = = = 1.57 kg g 9.8 m/s2 15.4 N = 1570 g m 1570 g ρ = = = ρ = 2.7 g/cm3 V 582 cm3

Archimedes Principle: The buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid displaced by the body. Things That Float: Lower boat into water

Archimedes Principle: The buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid displaced by the body. Things That Float: Lower boat into water

Archimedes Principle: The buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid displaced by the body. Things That Float: Lower boat into water Volume of water displaced Volume of boat below water line =

Archimedes Principle: The buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid displaced by the body. Things That Float: Buoyant Force FB Lower boat into water Weight of water displaced = Buoyant Force

If the weight of the displaced water equals the weight of the object, the object will float. Things That Float: Buoyant Force FB Lower boat into water Weight of water displaced = Buoyant Force

Fluids in Motion Consider a flowing river ; enters a narrow canyon What happens to the rate of flow? Water must speed up

Fluids in Motion Volumes must be equal A1 V1 A2 V2 x2 x1 V1 = V2 V = A x A1 x1 = A2 x2 A1 v1 t = A2 v2 t x = v t Continuity Equation A1 v1 = A2 v2

Now put pipes into the flow, with the valves closed

When valves are opened, lateral pressure pushes water up pipes, but not an equal amount Bernoulli’s Principle: A fast-moving fluid exerts less lateral pressure than a slow-moving fluid

Air moving past an airfoil: Fast-moving air Slow-moving air

Air moving past an airfoil: Lift Fast-moving air; low pressure Slow-moving air; high pressure

Frisbee: fast air slow air

Spinning ball through air:

Other examples: umbrellas on a windy day; standing next to a train track as a train passes; toilets on a windy day

Bernoulli’s Equation P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2 ex: A frisbee of area 0.120 m2 has air flowing under it at 15.0 m/s and over it at 15.8 m/s. Find the lift force in newtons. Lift comes from a difference in pressure, P2 - P1 P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2 h1 (height of air at bottom of frisbee) = h2 (height at top) P1 + ½ ρ v12 = P2 + ½ ρ v22

P1 + ½ ρ v12 = P2 + ½ ρ v22 Solve for ( P2 - P1 ) - P1 - P1 ½ ρ v12 = P2 - P1 + ½ ρ v22 - ½ ρ v22 - ½ ρ v22 ½ ρ v12 - ½ ρ v22 = P2 - P1 P2 - P1 = ½ ρ v12 - ½ ρ v22 = ½ ρ ( v12 - v22 ) ρair = 1.29 kg/m3 = ½ ( 1.29 kg/m3 )( [15.0 m/s]2 - [15.8 m/s]2 ) P2 - P1 = 15.9 N/m2 F P = F = P A = ( 15.9 N/m2 )( 0.120 m2 ) A F = 1.91 N

ex: Water is pumped from the ground 35 m up to the top of a water tower. At the bottom, the pipe is 1.5 m in diameter and the water flows at 0.45 m/s under a pressure of 4.8 atm. The pipe tapers to a diameter of 0.50 m at the top. Find the flow rate and the pressure at the top. P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2 A1 v1 = A2 v2 A1 = π ( 0.75 m )2 = 1.767144 m2 A2 = π ( 0.25 m )2 = 0.196349 m2 v1 = 0.45 m/s v2 = ? P1 = 4.8 atm 101 300 N/m2 atm P1 = 486 240 N/m2 P2 = ?

ex: Water is pumped from the ground 35 m up to the top of a water tower. At the bottom, the pipe is 1.5 m in diameter and the water flows at 0.45 m/s under a pressure of 4.8 atm. The pipe tapers to a diameter of 0.50 m at the top. Find the flow rate and the pressure at the top. A1 = 1.767144 m2 A2 = 0.196349 m2 v1 = 0.45 m/s A1 v1 = A2 v2 A1 v1 v2 = A2 ( 1.767144 m2 )( 0.45 m/s ) = 0.196349 m2 v2 = 4.0 m/s

ex: Water is pumped from the ground 35 m up to the top of a water tower. At the bottom, the pipe is 1.5 m in diameter and the water flows at 0.45 m/s under a pressure of 4.8 atm. The pipe tapers to a diameter of 0.50 m at the top. Find the flow rate and the pressure at the top. P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2 v1 = 0.45 m/s v2 = 4.0 m/s P1 = 486 240 N/m2 P2 = ? Let h1 = 0 h2 = 35 m

v1 = 0.45 m/s v2 = 4.0 m/s h1 = 0 P1 = 486 240 N/m2 P2 = ? h2 = 35 m P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2 - ½ ρ v22 - ρ g h2 - ½ ρ v22 - ρ g h2 P2 = P1 + ½ ρ v12 - ½ ρ v22 - ρ g h2 ρ = 1000 kg/m3 ½ ρ v12 = ½ ( 1000 kg/m3 )( 0.45 m/s )2 = 101.25 N/m2 ½ ρ v22 = ½ ( 1000 kg/m3 )( 4.0 m/s )2 = 8201.25 N/m2 ρ g h2 = ( 1000 kg/m3 )( 9.8 m/s2 )( 35.0 m ) = 343 000 N/m2 P2 = ( 486 240 ) + ( 101.25 ) - ( 8201.25 ) - ( 343 000 ) P2 = 135 140 N/m2 atm P2 = 1.3 atm 101 300 N/m2