Chapter 10 Fluids
Fluids in Motion In steady flow the velocity of the fluid particles at any point is constant as time passes. Unsteady flow exists whenever the velocity of the fluid particles at a point changes as time passes. Turbulent flow is an extreme kind of unsteady flow in which the velocity of the fluid particles at a point change erratically in both magnitude and direction.
Fluids in Motion Fluid flow can be compressible or incompressible. Most liquids are nearly incompressible. Fluid flow can be viscous such that the fluid does not flow readily due to internal frictional forces being present (e.g. honey), or nonviscous such that the fluid flows readily due to no internal frictional forces being present (e.g. water is almost nonviscous). An incompressible, nonviscous fluid is called an ideal fluid.
Fluids in Motion Steady flow is also sometimes called streamline or laminar flow since the neighboring layers of fluid slide by each other smoothly, i.e. each particle of the fluid follows a smooth path and the paths do not cross. Streamlines are often used to represent the trajectories of the fluid particles.
Fluids in Motion Making streamlines with dye in a flowing liquid, and smoke, in a wind tunnel.
The Equation of Continuity The mass of fluid per second that flows through a tube is called the mass flow rate.
The Equation of Continuity Consider the steady (laminar) flow of a fluid through a tube with a varying cross sectional area: Δm = ρ ΔV = ρ A vδt! distance Δm Δt = ρ Av Δm Δt 1 = ρ 1A1 v1 Since mass must be conserved as the fluid flows Δm 1 Δt = Δm Δt
The Equation of Continuity EQUATION OF CONTINUITY The mass flow rate has the same value at every position along a tube that has a single entry and a single exit for fluid flow. ρ = 1A1 v1 ρav SI Unit of Mass Flow Rate: kg/s
The Equation of Continuity Incompressible fluid: A 1v1 = Av Volume flow rate Q: Q Av = A Δl Δt = AΔl Δt = ΔV Δt
The Equation of Continuity Example: A Garden Hose A garden hose has an unobstructed opening with a cross sectional area of.85x10-4 m. It fills a bucket with a volume of 8.00x10-3 m 3 in 30 seconds. Find the speed of the water that leaves the hose through (a) the unobstructed opening and (b) an obstructed opening with half as much area.
The Equation of Continuity (a) Q = Av v = Q A = 3 3 ( 8.00 10 m ) ( 30.0 s).85 10-4 m = 0.936m s (b) A 1v1 = Av v = A 1 v 1 = A 1 A A 1 ( ) v 1 = ( )( 0.936m s) =1.87m s
Bernoulli s Equation Consider the steady flow of an incompressible and nonviscous fluid in two situations: The fluid accelerates from higher to lower pressure regions from Newton s nd law due to the unbalanced non-conservative forces According to the pressure-depth relationship, the pressure is lower at higher levels, provided the area of the pipe does not change. Incorporate both of these situations into a single equation!
Bernoulli s Equation Combined situation: Use the Work-Energy theorem to derive a relationship among P, v, and y at two different points along the tube à look at the fluid elements at 1 and. W nc = E 1 E = 1 mv 1 + mgy 1 ( ) ( 1 mv + mgy ) " $ " $ W nc = ΔW nc = "# ( ΔF)s$ % = (( ΔP)! As) = ( ( ΔP) ) ΔV = ( P P 1 )ΔV 1 1 1# ΔV % # 1 %!# " $# P P 1
Bernoulli s Equation ( P P 1 )ΔV = 1 mv 1 + mgy 1 ( ) ( 1 mv + mgy ) ( ) ( ) ( ) 1 1 P P = ρ v + ρgy ρv + ρ 1 1 1 gy BERNOULLI S EQUATION In steady flow of a nonviscous, incompressible fluid, the pressure, the fluid speed, and the elevation at two points are related by: P 1 1 1 + ρ v1 + ρgy1 = P + ρv + ρgy
Applications of Bernoulli s Equation Example: Efflux Speed The tank is open to the atmosphere at the top. Find and expression for the speed of the liquid leaving the pipe at the bottom.
Applications of Bernoulli s Equation = P P v 0 1 atm P = P 1 1 1 + ρ v1 + ρgy1 = P + ρv + ρgy y y = 1 h 1 ρ v = ρgh 1 For an element of fluid of mass, m 1 mv 1 = mgy v = 1 gh ( 1 gh) = gy y = h
Applications of Bernoulli s Equation When a moving fluid is contained in a horizontal pipe, all parts of it have the same elevation, i.e. y 1 = y, and Bernoulli s equation simplifies to: P 1 + 1 ρv 1 = P + 1 ρv Thus, P + ½ρv remains constant, so that If v increases à P decreases If v decreases à P increases
Applications of Bernoulli s Equation Example: An enlarged horizontal blood vessel à Find P P 1 ρ Blood = 1060 kg/m3 A 1 v 1 = 0.40 m/s v A = 1.7A 1 P 1 + 1 ρv 1 = P + 1 ρv P 1 P P P 1 = 1 ρ v ( 1 v ) Use the continuity equation to find v A 1 v 1 = A v v = A 1 A v 1 * P P 1 = 1 ρ v $ 1 A ' 1, & ) + % A ( v 1 - * / = 1 ρv $ 1 1 A ' 1, & ). + % A ( - / = 1 1060. ( ), 1 ( ) 0.40 *, + = 55 Pa = 0.41 mmhg $ A 1 ' & ) % 1.7A 1 ( - /.
Applications of Bernoulli s Equation Conceptual Example: Tarpaulins and Bernoulli s Equation When the truck is stationary, the tarpaulin lies flat, but it bulges outward when the truck is speeding down the highway. Account for this behavior.
Applications of Bernoulli s Equation
Applications of Bernoulli s Equation