Pre-work week 10 Fluid Dynamics Notes Buoyancy and Archimedes’ Principle If an object's density is less than water, it experiences an upward net force and rises until partially out of the water. For a floating object, the submerged fraction equals the ratio of the object's density to the fluid's density. Example: Hydrometer Calibration A hydrometer measures a liquid's specific gravity by how deep it sinks.
It consists of a weighted glass tube, 25.0 25.0 25.0 2.00 2.00 2.00 2 ^2 2 45.0 45.0 45.0
Calculation:
Average density of hydrometer: ρ = m V = 0.045 25.0 × 2.0 × 10 − 4 = 900 kg/m 3 ρ = \frac{m}{V} = \frac{0.045}{25.0 \times 2.0 \times 10^{-4}} = 900 \text{ kg/m}^3 ρ = V m = 25.0 × 2.0 × 1 0 − 4 0.045 = 900 kg/m 3 Hydrometer floats with 900 1000 = 0.90 \frac{900}{1000} = 0.90 1000 900 = 0.90 0.90 × 25.0 = 22.5 0.90 \times 25.0 = 22.5 0.90 × 25.0 = 22.5 Fluids in Motion; Flow Rate and the Equation of Continuity Fluid flow can be smooth (streamline or laminar) or turbulent.
Turbulent flow features eddies, increasing fluid viscosity.
Streamlines:
Path traced by a tiny fluid element in flow. Local fluid velocity is tangent to the streamline. Streamlines cannot cross, as it would imply two directions at once. Mass flow rate is the mass passing a point per unit time.
The flow rates at any two points are equal if no fluid is added or removed.
Equation of continuity:
ρ < e m > 1 A < / e m > 1 v < e m > 1 = ρ < / e m > 2 A < e m > 2 v < / e m > 2 ρ<em>1 A</em>1 v<em>1 = ρ</em>2 A<em>2 v</em>2 ρ < e m > 1 A < / e m > 1 v < e m > 1 = ρ < / e m > 2 A < e m > 2 v < / e m > 2 Where: A A A v v v ρ ρ ρ If density is constant (typical for liquids):
A < e m > 1 v < / e m > 1 = A < e m > 2 v < / e m > 2 A<em>1v</em>1 = A<em>2v</em>2 A < e m > 1 v < / e m > 1 = A < e m > 2 v < / e m > 2 In wider pipe sections, flow is slower. Example: Blood Flow Blood flows from the heart into the aorta, then to arteries, arterioles, and capillaries, returning via veins.
Aorta radius: 1.2 1.2 1.2 40 40 40
Capillary radius: 4 × 10 − 4 4 \times 10^{-4} 4 × 1 0 − 4 5 × 10 − 4 5 \times 10^{-4} 5 × 1 0 − 4
Calculations:
Aorta area: A < e m > A = π r < / e m > A 2 = π ( 1.2 × 10 − 2 ) 2 A<em>A = π r</em>A^2 = π (1.2 \times 10^{-2})^2 A < e m > A = π r < / e m > A 2 = π ( 1.2 × 1 0 − 2 ) 2 Capillary area: A < e m > C = π r < / e m > C 2 = π ( 4 × 10 − 6 ) 2 A<em>C = π r</em>C^2 = π (4 \times 10^{-6})^2 A < e m > C = π r < / e m > C 2 = π ( 4 × 1 0 − 6 ) 2 N < e m > C = A < / e m > A v < e m > A A < / e m > C v C = π ( 1.2 × 10 − 2 ) 2 ( 0.4 ) π ( 4 × 10 − 6 ) 2 ( 5 × 10 − 4 ) ≈ 7 × 10 9 N<em>C = \frac{A</em>A v<em>A}{A</em>C v_C} = \frac{π (1.2 \times 10^{-2})^2 (0.4)}{π (4 \times 10^{-6})^2 (5 \times 10^{-4})} ≈ 7 \times 10^9 N < e m > C = A < / e m > C v C A < / e m > A v < e m > A = π ( 4 × 1 0 − 6 ) 2 ( 5 × 1 0 − 4 ) π ( 1.2 × 1 0 − 2 ) 2 ( 0.4 ) ≈ 7 × 1 0 9 Example: Heating Duct Air moves at 3.0 3.0 3.0 300 300 300 3 ^3 3 15 15 15
Treat the room as a wider duct
Calculation:
A = V v t = 300 3.0 × 15 × 60 = 0.11 m 2 A = \frac{V}{v t} = \frac{300}{3.0 \times 15 \times 60} = 0.11 \text{ m}^2 A = v t V = 3.0 × 15 × 60 300 = 0.11 m 2 Bernoulli’s Equation Bernoulli’s principle: High fluid velocity corresponds to low pressure, and vice versa.
A force is required to accelerate fluid to higher velocity.
Bernoulli’s equation is derived from conservation of energy, considering work to move a fluid volume and increase its height.
Derivation:
Consider volume ( Δ V ) (\Delta V) ( Δ V ) ( Δ t ) (\Delta t) ( Δ t ) Fluid mass: ( Δ m ) = ρ ( Δ V ) = ρ A < e m > 1 ( Δ x < / e m > 1 ) = ρ A < e m > 2 ( Δ x < / e m > 2 ) (\Delta m) = ρ (\Delta V) = ρ A<em>1 (\Delta x</em>1) = ρ A<em>2 (\Delta x</em>2) ( Δ m ) = ρ ( Δ V ) = ρ A < e m > 1 ( Δ x < / e m > 1 ) = ρ A < e m > 2 ( Δ x < / e m > 2 ) Work done on ( Δ m ) : ( Δ W ) = W < e m > 1 − W < / e m > 2 = F < e m > 1 ( Δ x < / e m > 1 ) − F < e m > 2 ( Δ x < / e m > 2 ) = P < e m > 1 A < / e m > 1 v < e m > 1 ( Δ t ) − P < / e m > 2 A < e m > 2 v < / e m > 2 ( Δ t ) = P < e m > 1 ( Δ V ) − P < / e m > 2 ( Δ V ) = ( P < e m > 1 − P < / e m > 2 ) ( Δ V ) (\Delta m): (\Delta W) = W<em>1 - W</em>2 = F<em>1 (\Delta x</em>1) - F<em>2 (\Delta x</em>2) = P<em>1 A</em>1 v<em>1 (\Delta t) - P</em>2 A<em>2 v</em>2 (\Delta t) = P<em>1 (\Delta V) - P</em>2 (\Delta V) = (P<em>1 - P</em>2) (\Delta V) ( Δ m ) : ( Δ W ) = W < e m > 1 − W < / e m > 2 = F < e m > 1 ( Δ x < / e m > 1 ) − F < e m > 2 ( Δ x < / e m > 2 ) = P < e m > 1 A < / e m > 1 v < e m > 1 ( Δ t ) − P < / e m > 2 A < e m > 2 v < / e m > 2 ( Δ t ) = P < e m > 1 ( Δ V ) − P < / e m > 2 ( Δ V ) = ( P < e m > 1 − P < / e m > 2 ) ( Δ V ) Change in kinetic energy: Δ E < e m > K = 1 2 ( Δ m ) v < / e m > 2 2 − 1 2 ( Δ m ) v < e m > 1 2 = 1 2 ρ ( Δ V ) ( v < / e m > 2 2 − v 1 2 ) \Delta E<em>K = \frac{1}{2} (\Delta m) v</em>2^2 - \frac{1}{2} (\Delta m) v<em>1^2 = \frac{1}{2} ρ (\Delta V) (v</em>2^2 - v_1^2) Δ E < e m > K = 2 1 ( Δ m ) v < / e m > 2 2 − 2 1 ( Δ m ) v < e m > 1 2 = 2 1 ρ ( Δ V ) ( v < / e m > 2 2 − v 1 2 ) Change in potential energy: Δ E < e m > P = ( Δ m ) g y < / e m > 2 − ( Δ m ) g y < e m > 1 = ρ ( Δ V ) g ( y < / e m > 2 − y 1 ) \Delta E<em>P = (\Delta m) g y</em>2 - (\Delta m) g y<em>1 = ρ (\Delta V) g (y</em>2 - y_1) Δ E < e m > P = ( Δ m ) g y < / e m > 2 − ( Δ m ) g y < e m > 1 = ρ ( Δ V ) g ( y < / e m > 2 − y 1 ) Work-energy theorem: Δ W = Δ E < e m > K + Δ E < / e m > P \Delta W = \Delta E<em>K + \Delta E</em>P Δ W = Δ E < e m > K + Δ E < / e m > P ( P < e m > 1 − P < / e m > 2 ) ( Δ V ) = 1 2 ρ ( Δ V ) ( v < e m > 2 2 − v < / e m > 1 2 ) + ρ ( Δ V ) g ( y < e m > 2 − y < / e m > 1 ) (P<em>1 - P</em>2) (\Delta V) = \frac{1}{2} ρ (\Delta V) (v<em>2^2 - v</em>1^2) + ρ (\Delta V) g (y<em>2 - y</em>1) ( P < e m > 1 − P < / e m > 2 ) ( Δ V ) = 2 1 ρ ( Δ V ) ( v < e m > 2 2 − v < / e m > 1 2 ) + ρ ( Δ V ) g ( y < e m > 2 − y < / e m > 1 ) Cancelling ( Δ V ) (\Delta V) ( Δ V ) 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} ρ v</em>1^2 + ρ g y<em>1 = P</em>2 + \frac{1}{2} ρ v<em>2^2 + ρ 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 Or P + 1 2 ρ v 2 + ρ g y = c o n s t a n t P + \frac{1}{2} ρ v^2 + ρ g y = constant P + 2 1 ρ v 2 + ρ g y = co n s t an t If there is no change in height ( y = 0 ) (y = 0) ( y = 0 )
P + 1 2 ρ v 2 = c o n s t a n t P + \frac{1}{2} ρ v^2 = constant P + 2 1 ρ v 2 = co n s t an t Maximum pressure occurs when fluid is stationary ( v = 0 ) (v = 0) ( v = 0 ) Example: Hot-Water Heating System Water is pumped at 0.5 0.5 0.5 4.0 4.0 4.0 3.0 3.0 3.0
Find the flow speed and pressure in a 2.6 2.6 2.6 5.0 5.0 5.0
Solution:
Basement:r < e m > 1 = 2.0 cm r<em>1 = 2.0 \text{ cm} r < e m > 1 = 2.0 cm v < / e m > 1 = 0.5 m/s v</em>1 = 0.5 \text{ m/s} v < / e m > 1 = 0.5 m/s y < e m > 1 = 0 y<em>1 = 0 y < e m > 1 = 0 P < / e m > 1 = 3.0 atm = 3.0 × 10 5 Pa P</em>1 = 3.0 \text{ atm} = 3.0 \times 10^5 \text{ Pa} P < / e m > 1 = 3.0 atm = 3.0 × 1 0 5 Pa ρ = 1000 kg/m 3 ρ = 1000 \text{ kg/m}^3 ρ = 1000 kg/m 3 Second Floor:r < e m > 2 = 1.3 cm r<em>2 = 1.3\text{ cm} r < e m > 2 = 1.3 cm y < / e m > 2 = 5.0 m y</em>2 = 5.0 \text{ m} y < / e m > 2 = 5.0 m Continuity equation:A < e m > 1 v < / e m > 1 = A < e m > 2 v < / e m > 2 A<em>1 v</em>1 = A<em>2 v</em>2 A < e m > 1 v < / e m > 1 = A < e m > 2 v < / e m > 2 v < e m > 2 = v < / e m > 1 ( A < e m > 1 A < / e m > 2 ) = v < e m > 1 ( π r < / e m > 1 2 π r 2 2 ) = 0.5 ( 2.0 1.3 ) 2 = 1.18 m/s v<em>2 = v</em>1 (\frac{A<em>1}{A</em>2}) = v<em>1 (\frac{π r</em>1^2}{π r_2^2}) = 0.5 (\frac{2.0}{1.3})^2 = 1.18 \text{ m/s} v < e m > 2 = v < / e m > 1 ( A < / e m > 2 A < e m > 1 ) = v < e m > 1 ( π r 2 2 π r < / e m > 1 2 ) = 0.5 ( 1.3 2.0 ) 2 = 1.18 m/s Bernoulli's equation: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} ρ v</em>1^2 + ρ g y<em>1 = P</em>2 + \frac{1}{2} ρ v<em>2^2 + ρ 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 P < e m > 2 = P < / e m > 1 + 1 2 ρ ( v < e m > 1 2 − v < / e m > 2 2 ) + ρ g ( y < e m > 1 − y < / e m > 2 ) = 3.0 × 10 5 + 1 2 × 1000 ( 0.5 2 − 1.18 2 ) + 1000 × 9.8 ( 0 − 5.0 ) ≈ 2.5 × 10 5 Pa = 2.5 atm P<em>2 = P</em>1 + \frac{1}{2} ρ (v<em>1^2 - v</em>2^2) + ρ g (y<em>1 - y</em>2) = 3.0 \times 10^5 + \frac{1}{2} \times 1000 (0.5^2 - 1.18^2) + 1000 \times 9.8 (0-5.0) ≈ 2.5 \times 10^5 \text{ Pa} = 2.5 \text{ atm} P < e m > 2 = P < / e m > 1 + 2 1 ρ ( v < e m > 1 2 − v < / e m > 2 2 ) + ρ g ( y < e m > 1 − y < / e m > 2 ) = 3.0 × 1 0 5 + 2 1 × 1000 ( 0. 5 2 − 1.1 8 2 ) + 1000 × 9.8 ( 0 − 5.0 ) ≈ 2.5 × 1 0 5 Pa = 2.5 atm Applications of Bernoulli’s Principle Torricelli’s theorem: Speed of fluid coming from a spigot on an open tank: 2 g h \sqrt{2gh} 2 g h P < e m > 1 = P < / e m > 2 P<em>1 = P</em>2 P < e m > 1 = P < / e m > 2 Airplanes: Lift on a wing is due to different air speeds and pressures on the wing's surfaces. Sailboats: Can move against the wind using pressure differences on the sail and keel. Baseballs: A ball’s path curves due to spin, creating unequal air speeds and a pressure difference. Venturi meter: Measures fluid flow by measuring pressure differences. Bernoulli’s principle applies to spinning balls (cricket, golf, tennis). Further Examples in Fluid Flow Aerodynamic Lift – Airflow over wing A small plane has a mass of 300 300 300 30 30 30 2 ^2 2 Note that the pressure is lower on top of the wing as the air velocity is higher (streamlines are closer together). You do not need to use Bernoulli’s Equation to calculate the pressures, rather you should work out the weight of the aircraft and the pressure difference required to lift that weight.Weight of aircraft:W = m g = 300 × 9.8 = 2940 N W = mg = 300 \times 9.8 = 2940 \text{ N} W = m g = 300 × 9.8 = 2940 N Pressure difference:( Δ P ) = F / A = 2940 / 30 = 98 Pa (\Delta P) = F/A = 2940/30 = 98 \text{ Pa} ( Δ P ) = F / A = 2940/30 = 98 Pa Assume the plane is cruising at 180 180 180 ( 180 × 1000 ) / ( 60 × 60 ) = 50 (180 \times 1000)/(60 \times 60) = 50 ( 180 × 1000 ) / ( 60 × 60 ) = 50 180 180 180 1.29 1.29 1.29 3 ^3 3 98 98 98
Using Bernoulli’s Equation: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} ρ v</em>1^2 + ρ g y<em>1 = P</em>2 + \frac{1}{2} ρ v<em>2^2 + ρ 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 Noting that the height difference is negligible ( y < e m > 1 = y < / e m > 2 ) (y<em>1 = y</em>2) ( y < e m > 1 = y < / e m > 2 ) P < e m > 1 − P < / e m > 2 = 1 2 ρ ( v < e m > 2 2 − v < / e m > 1 2 ) P<em>1 - P</em>2 = \frac{1}{2} ρ (v<em>2^2 - v</em>1^2) P < e m > 1 − P < / e m > 2 = 2 1 ρ ( v < e m > 2 2 − v < / e m > 1 2 ) 98 = 1 2 × 1.29 ( v 2 2 − 50 2 ) 98 = \frac{1}{2} \times 1.29 (v_2^2 - 50^2) 98 = 2 1 × 1.29 ( v 2 2 − 5 0 2 ) v 2 = 2 × 98 1.29 + 50 2 = 51.5 m/s v_2 = \sqrt{\frac{2 \times 98}{1.29} + 50^2} = 51.5 \text{ m/s} v 2 = 1.29 2 × 98 + 5 0 2 = 51.5 m/s 51.5 m/s = 51.5 × 10 − 3 60 × 60 = 185 km/hr 51.5 \text{ m/s} = 51.5 \times \frac{10^{-3}}{60 \times 60} = 185 \text{ km/hr} 51.5 m/s = 51.5 × 60 × 60 1 0 − 3 = 185 km/hr Summary of Chapter 13 Phases of matter: solid, liquid, gas Liquids and gases are called fluids. Density is mass per unit volume. Pressure is force per unit area. Pressure at a depth h h h ρ g h ρgh ρ g h External pressure applied to a confined fluid is transmitted throughout the fluid (Pascal’s Principle). Gauge pressure is the total pressure minus the atmospheric pressure. An object submerged partly or wholly in a fluid is buoyed by a force equal to the weight of the fluid it displaces (Archimedes’ Principle). Fluid flow can be laminar or turbulent. The product of the cross-sectional area and the speed is constant for horizontal flow (Continuity Equation). Where the velocity of a fluid is high, the pressure is low, and vice versa (Bernoulli’s Equation).