Fluid Mechanics Notes Archimedes' Principle The buoyant force equals the weight of the displaced water. Fluids at Rest Density is a property of the material: ρ = m V \rho = \frac{m}{V} ρ = V m Ranking the mass of objects:{mountain} > m {statue} > m_{counter} Ranking the density of objects:{statue} = \rho {counter} > \rho_{mountain} Practice Problems Problem 1: Smuggling gold in a backpack.Density of gold: ρ = 19 k g L \rho = 19 \frac{kg}{L} ρ = 19 L k g Backpack volume: V = 30 L V = 30L V = 30 L Mass of gold: m = ρ V = 19 k g L ∗ 30 L = 570 k g m = \rho V = 19 \frac{kg}{L} * 30 L = 570kg m = ρ V = 19 L k g ∗ 30 L = 570 k g Problem 2: Mass of a cube with side length 2L.Original cube: mass m m m L L L ρ \rho ρ New cube: length 2 L 2L 2 L Volume of new cube: V ′ = ( 2 L ) 3 = 8 L 3 V' = (2L)^3 = 8L^3 V ′ = ( 2 L ) 3 = 8 L 3 Mass of new cube: m ′ = ρ V ′ = ρ ( 2 L ) 3 = 8 ρ L 3 = 8 m m' = \rho V' = \rho (2L)^3 = 8 \rho L^3 = 8m m ′ = ρ V ′ = ρ ( 2 L ) 3 = 8 ρ L 3 = 8 m Density Experiments Finding the Density of Clay Procedure:Mold clay into spheres of different sizes. Measure the diameter of each sphere. Measure the mass of each sphere. Create a table of data (titles & units). Plot a linear graph using the data. Calculate the density of the clay using the graph. Finding the Density of Cleaning Liquid Procedure:Pour liquid into a measuring cup at different marked volumes. Measure the mass of the liquid at each volume interval. Create a table of data (titles & units). Plot a linear graph using the data. Calculate the density of the liquid using the graph. Homework Write a detailed report of your experimental work.Objective: To graphically find the density of playdoh/cleaning liquid. Hand in the report next class. Classwork - Density Complete the density worksheet with table group partners and justify answers. Warm Up Ranking substance densities from greatest to least:Correct order: Gold, silver, copper, iron. Specific Gravity Specific Gravity: S G = density of the substance density of water at 4°C and 1 atm SG = \frac{\text{density of the substance}}{\text{density of water at 4°C and 1 atm}} SG = density of water at 4°C and 1 atm density of the substance Water density is highest at 4°C.ρ 100 ° C = 988.35 k g m 3 \rho_{100°C} = 988.35 \frac{kg}{m^3} ρ 100° C = 988.35 m 3 k g ρ 4 ° C = 999.97 k g m 3 \rho_{4°C} = 999.97 \frac{kg}{m^3} ρ 4° C = 999.97 m 3 k g ρ 0 ° C water = 999.84 k g m 3 \rho_{0°C \text{ water}} = 999.84 \frac{kg}{m^3} ρ 0° C water = 999.84 m 3 k g ρ 0 ° C ice = 917 k g m 3 \rho_{0°C \text{ ice}} = 917 \frac{kg}{m^3} ρ 0° C ice = 917 m 3 k g ρ − 180 ° C ice = 934 k g m 3 \rho_{-180°C \text{ ice}} = 934 \frac{kg}{m^3} ρ − 180° C ice = 934 m 3 k g Anomalous Behavior of Water - Applications 4°C water is denser and sinks. Water below freezing is less dense than 4°C water, so ice forms at the top, insulating marine life below at 4°C. During winter, lakes, rivers, and seas freeze at the top, allowing marine life to survive at the bottom. Potholes on roads worsen during winter and rainy days due to water expanding when it freezes. Pressure Pressure: P = F A P = \frac{F}{A} P = A F Pressure exerted by a box on a table depends on the face in contact. Pascal's Principle Pressure at any point is the same in all directions. Force due to fluid pressure is always perpendicular to the contact surface. Pressure from a fluid at rest: P f l = ρ g h P_{fl} = \rho g h P f l = ρ g h h h h Derivation: P = F A = m g A = ρ V g A = ρ ( h A ) g A = ρ g h P = \frac{F}{A} = \frac{mg}{A} = \frac{\rho V g}{A} = \frac{\rho (hA) g}{A} = \rho g h P = A F = A m g = A ρ V g = A ρ ( h A ) g = ρ g h If there is atmospheric pressure P < e m > 0 P<em>0 P < e m > 0 P = P < / e m > 0 + ρ g h P = P</em>0 + \rho g h P = P < / e m > 0 + ρ g h Balloon Example At deeper levels, surrounding water pressure is greater, compressing the balloon and increasing its density, causing it to sink. Gauge and Absolute Pressure Gauge Pressure: P G = ρ g h P_G = \rho g h P G = ρ g h Pressure due to the weight of the fluid above the measuring point. Absolute Pressure: P < e m > a b s = P < / e m > 0 + P G P<em>{abs} = P</em>0 + P_G P < e m > ab s = P < / e m > 0 + P G Sum of surface pressure and gauge pressure. Standard Atmospheric Pressure:1 a t m = 101 , 300 P a ≈ 10 5 P a = 10 5 N m 2 1 atm = 101,300 Pa \approx 10^5 Pa = 10^5 \frac{N}{m^2} 1 a t m = 101 , 300 P a ≈ 1 0 5 P a = 1 0 5 m 2 N 1 a t m = 760 m m H g = 76 c m H g 1 atm = 760 mmHg = 76 cmHg 1 a t m = 760 mm H g = 76 c m H g Tire Pressure Problem Gauge pressure: 250 kPa. Front tire footprint: 0.012 m² each. Rear tire footprint: 0.01 m² each. Total force: F = P A = 250 , 000 P a ( 0.012 m 2 ∗ 2 + 0.01 m 2 ∗ 2 ) = 11 , 000 N F = P A = 250,000 Pa (0.012 m^2 * 2 + 0.01 m^2 * 2) = 11,000 N F = P A = 250 , 000 P a ( 0.012 m 2 ∗ 2 + 0.01 m 2 ∗ 2 ) = 11 , 000 N Mass of the car: m g = 11 , 000 N ⟹ m = 11 , 000 10 N s 2 m = 1100 k g mg = 11,000 N \implies m = \frac{11,000}{10} \frac{Ns^2}{m} = 1100 kg m g = 11 , 000 N ⟹ m = 10 11 , 000 m N s 2 = 1100 k g Pascal's Principle Explained Changing the pressure at any point in an incompressible fluid causes the same pressure change at all points in the fluid. Otter Example When the otter jumps in, the pressure increases everywhere in the fluid equally. Application of Pascal's Principle Small force F < e m > 1 F<em>1 F < e m > 1 F < / e m > 2 F</em>2 F < / e m > 2 This does not violate conservation of energy. Problem:What force must be applied to a small piston (diameter = 0.1m) to lift a 1500 kg car on a large piston (diameter = 0.3m)? If the small piston is pushed down by 5cm, how far is the car lifted? Hydraulic Brakes Components: brake pedal, wheel drum, brake shoe, pad, rotor. Archimedes' Principle (Buoyancy) Any object partially or wholly immersed in a fluid experiences a buoyant force equal to the weight of the displaced fluid: F < e m > b = W < / e m > d i s p l a c e d fluid F<em>b = W</em>{displaced \text{ fluid}} F < e m > b = W < / e m > d i s pl a ce d fluid Mathematical Proof of Archimedes' Principle ∑ F = F < e m > b o t t o m − F < / e m > t o p = A P < e m > b o t t o m − A P < / e m > t o p = A ( P < e m > a t m + ρ g y < / e m > b o t t o m ) − A ( P < e m > a t m + ρ g y < / e m > t o p ) = A ρ g ( y < e m > b o t t o m − y < / e m > t o p ) = A ρ g h = ρ V g \sum F = F<em>{bottom} - F</em>{top} = A P<em>{bottom} - A P</em>{top} = A (P<em>{atm} + \rho g y</em>{bottom}) - A (P<em>{atm} + \rho g y</em>{top}) = A \rho g (y<em>{bottom} - y</em>{top}) = A \rho g h = \rho V g ∑ F = F < e m > b o tt o m − F < / e m > t o p = A P < e m > b o tt o m − A P < / e m > t o p = A ( P < e m > a t m + ρ g y < / e m > b o tt o m ) − A ( P < e m > a t m + ρ g y < / e m > t o p ) = A ρ g ( y < e m > b o tt o m − y < / e m > t o p ) = A ρ g h = ρ V g The result is an upward force. Ice Cube on the Moon Example On Earth, an ice cube floats with 9/10 of its volume submerged. On the Moon, the ice cube will also float with 9/10 submerged. Earth: F < e m > b = m < / e m > c g < e m > E F<em>b = m</em>c g<em>E F < e m > b = m < / e m > c g < e m > E F < / e m > b = W < e m > f ⟹ m < / e m > c g < e m > E = ρ < / e m > f 9 10 V < e m > c g < / e m > E ⟹ m < e m > c = ρ < / e m > f 9 10 V c F</em>b = W<em>f \implies m</em>c g<em>E = \rho</em>f \frac{9}{10} V<em>c g</em>E \implies m<em>c = \rho</em>f \frac{9}{10} V_c F < / e m > b = W < e m > f ⟹ m < / e m > c g < e m > E = ρ < / e m > f 10 9 V < e m > c g < / e m > E ⟹ m < e m > c = ρ < / e m > f 10 9 V c Moon: F < e m > b = m < / e m > c g < e m > M F<em>b = m</em>c g<em>M F < e m > b = m < / e m > c g < e m > M F < / e m > b = W < e m > f ⟹ m < / e m > c g < e m > M = ρ < / e m > f V ′ < e m > c g < / e m > M ⟹ m < e m > c = ρ < / e m > f V c ′ F</em>b = W<em>f \implies m</em>c g<em>M = \rho</em>f V'<em>c g</em>M \implies m<em>c = \rho</em>f V'_c F < / e m > b = W < e m > f ⟹ m < / e m > c g < e m > M = ρ < / e m > f V ′ < e m > c g < / e m > M ⟹ m < e m > c = ρ < / e m > f V c ′ Thus, V ′ < e m > c = 9 10 V < / e m > c V'<em>c = \frac{9}{10} V</em>c V ′ < e m > c = 10 9 V < / e m > c Density and Pressure Example Compare density and pressure at different points (X, Y, Z) in water. Clarifying Archimedes' Principle The density in F < e m > b = ρ < / e m > f V f g F<em>b = \rho</em>f V_f g F < e m > b = ρ < / e m > f V f g The volume in F < e m > b = ρ < / e m > f V f g F<em>b = \rho</em>f V_f g F < e m > b = ρ < / e m > f V f g Buoyant force does not increase with depth; it only depends on the volume of displaced fluid, density of the fluid, and gravity. Key statement: Every object is buoyed upwards by a force equal to the weight of the fluid the object displaces.Buoyancy and Density Summary Density determines whether an object sinks or floats. Diagrams illustrating masses in a fluid (Fig 12.3):Floating: ρ < e m > o b j < ρ < / e m > f l u i d \rho<em>{obj} < \rho</em>{fluid} ρ < e m > o bj < ρ < / e m > f l u i d m < e m > o b j < m < / e m > Fluid Displaced m<em>{obj} < m</em>{ \text{Fluid Displaced}} m < e m > o bj < m < / e m > Fluid Displaced V < e m > o b j > V < / e m > F l u i d D i s p l a c e d V<em>{obj} > V</em>{Fluid Displaced} V < e m > o bj > V < / e m > Fl u i d D i s pl a ce d Rising to surface: ρ < e m > o b j < ρ < / e m > f l u i d \rho<em>{obj} < \rho</em>{fluid} ρ < e m > o bj < ρ < / e m > f l u i d m < e m > o b j < m < / e m > Fluid Displaced m<em>{obj} < m</em>{ \text{Fluid Displaced}} m < e m > o bj < m < / e m > Fluid Displaced V < e m > o b j = V < / e m > F l u i d D i s p l a c e d V<em>{obj} = V</em>{Fluid Displaced} V < e m > o bj = V < / e m > Fl u i d D i s pl a ce d Neutral buoyancy: ρ < e m > o b j = ρ < / e m > f l u i d \rho<em>{obj} = \rho</em>{fluid} ρ < e m > o bj = ρ < / e m > f l u i d m < e m > o b j = m < / e m > Fluid Displaced m<em>{obj} = m</em>{ \text{Fluid Displaced}} m < e m > o bj = m < / e m > Fluid Displaced V < e m > o b j = V < / e m > F l u i d D i s p l a c e d V<em>{obj} = V</em>{Fluid Displaced} V < e m > o bj = V < / e m > Fl u i d D i s pl a ce d Sinking: ρ < e m > o b j > ρ < / e m > f l u i d \rho<em>{obj} > \rho</em>{fluid} ρ < e m > o bj > ρ < / e m > f l u i d m < e m > o b j > m < / e m > Fluid Displaced m<em>{obj} > m</em>{ \text{Fluid Displaced}} m < e m > o bj > m < / e m > Fluid Displaced V < e m > o b j = V < / e m > F l u i d D i s p l a c e d V<em>{obj} = V</em>{Fluid Displaced} V < e m > o bj = V < / e m > Fl u i d D i s pl a ce d Resting on the bottom: ρ < e m > o b j > ρ < / e m > f l u i d \rho<em>{obj} > \rho</em>{fluid} ρ < e m > o bj > ρ < / e m > f l u i d m < e m > o b j > m < / e m > Fluid Displaced m<em>{obj} > m</em>{ \text{Fluid Displaced}} m < e m > o bj > m < / e m > Fluid Displaced V < e m > o b j = V < / e m > F l u i d D i s p l a c e d V<em>{obj} = V</em>{Fluid Displaced} V < e m > o bj = V < / e m > Fl u i d D i s pl a ce d Variables Affecting Buoyant Force F < e m > b = ρ < / e m > f V f g F<em>b = \rho</em>f V_f g F < e m > b = ρ < / e m > f V f g Fig 1: F < e m > B b > F < / e m > A b , V < e m > B > V < / e m > A F<em>{Bb} > F</em>{Ab}, V<em>B > V</em>A F < e m > B b > F < / e m > A b , V < e m > B > V < / e m > A Fig 2: F < e m > H b > F < / e m > W b , ρ < e m > H > ρ < / e m > W F<em>{Hb} > F</em>{Wb}, \rho<em>H > \rho</em>W F < e m > H b > F < / e m > Wb , ρ < e m > H > ρ < / e m > W Fig 3: F < e m > S b = F < / e m > L b , V < e m > S = V < / e m > L F<em>{Sb} = F</em>{Lb}, V<em>S = V</em>L F < e m > S b = F < / e m > L b , V < e m > S = V < / e m > L Fig 4: F < e m > B b = F < / e m > A b , V < e m > B = V < / e m > A F<em>{Bb} = F</em>{Ab}, V<em>B = V</em>A F < e m > B b = F < / e m > A b , V < e m > B = V < / e m > A Fig 5: F < e m > B b = F < / e m > A b , V < e m > B = V < / e m > A F<em>{Bb} = F</em>{Ab}, V<em>B = V</em>A F < e m > B b = F < / e m > A b , V < e m > B = V < / e m > A Apparent Weight A 7kg rock has an apparent weight of 4kg when submerged in water.Find: F < e m > b F<em>b F < e m > b ρ < / e m > r o c k \rho</em>{rock} ρ < / e m > roc k ρ < e m > o b j = ρ < / e m > f l m < e m > o b j m < / e m > o b j − m a p p \rho<em>{obj} = \rho</em>{fl} \frac{m<em>{obj}}{m</em>{obj} - m_{app}} ρ < e m > o bj = ρ < / e m > f l m < / e m > o bj − m a pp m < e m > o bj Saltwater Density Prediction A student predicts saltwater is denser than drinking water. Measurements needed to verify the prediction: The spring scale reading with the cube in air and submerged in each liquid. Fluids in Motion Equation of Continuity Mass flow rate: m < e m > i n Δ t = m < / e m > o u t Δ t \frac{m<em>{in}}{\Delta t} = \frac{m</em>{out}}{\Delta t} Δ t m < e m > in = Δ t m < / e m > o u t Equation of Continuity: A < e m > 1 U < / e m > 1 = A < e m > 2 U < / e m > 2 A<em>1 U</em>1 = A<em>2 U</em>2 A < e m > 1 U < / e m > 1 = A < e m > 2 U < / e m > 2 Volume Flow Rate: A U = v o l u m e Δ t AU = \frac{volume}{\Delta t} A U = Δ t v o l u m e Examples Shower Head Example Shower head has 80 holes with a diameter of 0.1 cm each. Hose diameter is 1.5 cm. Water flow speed in the hose is 2 m/s. Find the speed of water exiting the shower head. Filling a Container Water exits a 3cm diameter hose at 4m/s. How long to fill a 0.5m x 1.0m x 0.4m container? Bernoulli's Principle As the speed of a fluid increases, its pressure decreases. Bernoulli's Equation W = Δ E W = \Delta E W = Δ E P < e m > 1 + 1 2 ρ v < / e m > 1 2 + ρ g y < e m > 1 = P < / e m > 2 + 1 2 ρ v < e m > 2 2 + ρ g y < / e m > 2 P<em>1 + \frac{1}{2} \rho v</em>1^2 + \rho g y<em>1 = P</em>2 + \frac{1}{2} \rho v<em>2^2 + \rho g y</em>2 P < e m > 1 + 2 1 ρ v < / e m > 1 2 + ρ g y < e m > 1 = P < / e m > 2 + 2 1 ρ v < e m > 2 2 + ρ g y < / e m > 2 Examples Water Tank and Faucet A water tank is positioned 10m above a faucet. What is the speed of the water out of the faucet? Homework Fluids PacketFluids In Motion MC: All Fluids In Motion FR:B2003B6 B2005B5 2007B4 B2007B4 2008B4 B2009B3 pg 216: SUP5 Knowt Play Call Kai