F LUID D YNAMICS Created by The North Carolina School of Science and Math.The North Carolina School of Science and Math Copyright North Carolina Department of Public Instruction.North Carolina Department of Public Instruction Created by The North Carolina School of Science and Math.The North Carolina School of Science and Math Copyright North Carolina Department of Public Instruction.North Carolina Department of Public Instruction

Think-Pair-Share: Fluids & Engineering Internal combustion engines Aerospace propulsion systems Waste disposal Pollution dispersal Electric power generation Pipelines Fluid/structure interaction HVAC systems Biomedical Instrumentation Manufacturing

Key Terms Pressure: P= F/A Measured in Pascals (Pa) ; 1 Pa = 1 N/m 2 N= kg m/s 2 Density: ρ= m/v Units: kg/m 3 Liquids are virtually incompressible Density of gases is affected by T & P Viscosity: Viscosity of liquids is much larger than viscosity of gases or steam

Simple Pipe Flow Two types of flow: Laminar Turbulent Type of flow is determined by the fluid’s properties: Laminar- high viscosity, low density, and small velocity Turbulent- low viscosity, high velocity

Laminar & Turbulent Flow Image from:

C ONSERVATION OF M ASS Mass Flow A 1 = Mass Flow A 2 ρ 1 A 1 v 1 = ρ 2 A 2 v 2 For incompressible fluid, ρ 1 = ρ 2, & A 1 v 1 = A 2 v 2 The mass flow rate must be constant across all sections, provided there are no leaks, feeds, bleeds, or storage! Q = AV

E XAMPLE

B ERNOULLI ’ S EQUATION : C ONSERVATION OF E NERGY From the law of conservation of mass: Velocity is greatest in region B and smallest in region A Pressure in regions A & C is larger than pressure in region B

Conservation of Energy Fluid has kinetic energy, potential energy, and force on the fluid due to pressure. Thus work is being done on the fluid KE = ½ mv 2 PE = mgh P = F/A If there are no frictional losses in the system we can apply the law of conservation of energy. P + ρgh + ½ ρv 2 = constant Copyright © 2009 ITL Program, College of Engineering, University of Colorado at Boulder. From (used with permission)

v1v1 v2v2 Ground (h = 0) h1h1 h2h2 P1P1 P2P2 As the water loses elevation from the high end of the pipe to the low end, it gains velocity. To find the exact value of any parameter, we apply the Bernoulli equation to two points anywhere along the same streamline. streamline

h 1 = 250 m P 1 = atmospheric h 2 = 0 m P 2 = atmospheric The water at the top of the reservoir starts at rest, so v 1 is zero, and the first term drops out. Since the final height (h 2 ) is also zero, this term drops out too. Lastly, P 1 = P 2, which is atmospheric pressure, so these terms drop out as well. Plugging in the remaining the known parameters:  water g (250 m) = ½  water v 2 2 Now the  water terms can be cancelled out. Using g = 9.8 m/s 2 we can solve for v 2 : v 2 = sqrt (2*9.8 m/s 2 * 250 m) v 2 = 70 m/s

E XAMPLE Water (density = 1000 kg/m 3 ) flows through a hose with a velocity of 1 m/sec. As it leaves the nozzle the constricted area increases the velocity to 20 m/sec. The pressure on the water as it leaves is atmospheric pressure (1 Atm = 100,000 N/m 2 ). What is the pressure on the water in the hose? Express the answer in N/m 2 Inside: v 1 = 1 m/sec P 1 = ? Outside: v 2 = 20 m/secP 2 = 100,000 N/m 2 Density of water = d = 1000 kg/m 3 Bernoulli’s eqn: P 1 + ½dv 1 2 = P 2 + ½dv 2 2 P 1 = P 2 + ½dv ½dv 1 2 = 100,000 N/m 2 + ½(1000 kg/m 3 )(20 m/sec) 2 - ½(1000 kg/m 3 )(1 m/sec) 2 = 100,000 N/m ,000 N/m N/m 2 = 299,500 N/m 2

Units Force: Newton - N = kg·m/s 2 (pound) Pressure: Pa = N/m 2 = J/m 3 = kg/(m·s 2 ) (psi) Energy, Work, Heat - J = N·m = kg·m 2 /s 2 (BTU) Velocity: v = m/s (ft/s) Density: ρ - kg/m 3 (lbm/ft3 Power: W = J/s = (kg·m 2 )/s 3 = N·m/s (ft-lb or HP)

Bernoulli Demonstration

Bernoulli Interactive

Fluid Dynamics Video Click the image to open the video in YouTube.