3Shear stress and velocity distribution in pipe for laminar flow
4Typical velocity and shear distributions in turbulent flow near a wall: (a) shear; (b) velocity.
5Solution of Pipe Flow Problems Single PathFind Dp for a given L, D, and QUse energy equation directlyFind L for a given Dp, D, and Q
6Solution of Pipe Flow Problems Single Path (Continued)Find Q for a given Dp, L, and DManually iterate energy equation and friction factor formula to find V (or Q), orDirectly solve, simultaneously, energy equation and friction factor formula using (for example) ExcelFind D for a given Dp, L, and QManually iterate energy equation and friction factor formula to find D, or
7Example 1Water at 10C is flowing at a rate of 0.03 m3/s through a pipe. The pipe has 150-mm diameter, 500 m long, and the surface roughness is estimated at 0.06 mm. Find the head loss and the pressure drop throughout the length of the pipe.Solution:From Table 1.3 (for water): = 1000 kg/m3 and =1.30x10-3 N.s/m2V = Q/A and A=R2A = (0.15/2)2 = m2V = Q/A =0.03/ =1.7 m/sRe = (1000x1.7x0.15)/(1.30x10-3) = 1.96x105 > turbulent flowTo find , use Moody Diagram with Re and relative roughness (k/D).k/D = 0.06x10-3/0.15 = 4x10-4From Moody diagram, 0.018The head loss may be computed using the Darcy-Weisbach equation.The pressure drop along the pipe can be calculated using the relationship:ΔP=ghf = 1000 x 9.81 x 8.84ΔP = 8.67 x 104 Pa
8Example 2Determine the energy loss that will occur as 0.06 m3/s water flows from a 40-mm pipe diameter into a 100-mm pipe diameter through a sudden expansion.Solution:The head loss through a sudden enlargement is given by;Da/Db = 40/100 = 0.4From Table 6.3: K = 0.70Thus, the head loss is
9Example 3Calculate the head added by the pump when the water system shown below carries a discharge of 0.27 m3/s. If the efficiency of the pump is 80%, calculate the power input required by the pump to maintain the flow.
11The velocity can be calculated using the continuity equation: Thus, the head added by the pump: Hp = 39.3 mPin = Watt ≈ 130 kW.
12EGL & HGL for a Pipe System Energy equationAll terms are in dimension of length (head, or energy per unit weight)HGL – Hydraulic Grade LineEGL – Energy Grade LineEGL=HGL when V=0 (reservoir surface, etc.)EGL slopes in the direction of flow
13EGL & HGL for a Pipe System A pump causes an abrupt rise in EGL (and HGL) since energy is introduced here
14EGL & HGL for a Pipe System A turbine causes an abrupt drop in EGL (and HGL) as energy is taken outGradual expansion increases turbine efficiency
15EGL & HGL for a Pipe System When the flow passage changes diameter, the velocity changes so that the distance between the EGL and HGL changesWhen the pressure becomes 0, the HGL coincides with the system
16EGL & HGL for a Pipe System Abrupt expansion into reservoir causes a complete loss of kinetic energy there
17EGL & HGL for a Pipe System When HGL falls below the pipe the pressure is below atmospheric pressure
18FLOW MEASUREMENT Direct Methods Examples: Accumulation in a Container; Positive Displacement FlowmeterRestriction Flow Meters for Internal FlowsExamples: Orifice Plate; Flow Nozzle; Venturi; Laminar Flow Element
30The measured stagnation pressure cannot of itself be used to determine the fluid velocity (airspeed in aviation).However, Bernoulli's equation states:Stagnation pressure = static pressure + dynamic pressureWhich can also be written
31Solving that for velocity we get: Note: The above equation applies only to incompressible fluid.where:V is fluid velocity;pt is stagnation or total pressure;ps is static pressure;and ρ is fluid density.
32The value for the pressure drop p2 – p1 or Δp to Δh, the reading on the manometer: Δp = Δh(ρA-ρ)gWhere:ρA is the density of the fluid in the manometerΔh is the manometer reading
36Main Topics The Boundary-Layer Concept Boundary-Layer Thickness Laminar Flat-Plate Boundary Layer: Exact SolutionMomentum Integral EquationUse of the Momentum Equation for Flow with Zero Pressure GradientPressure Gradients in Boundary-Layer FlowDragLift
48Laminar Flat-Plate Boundary Layer: Exact Solution Equations are Coupled, Nonlinear, Partial Differential EquationsBlassius Solution:Transform to single, higher-order, nonlinear, ordinary differential equation
49Laminar Flat-Plate Boundary Layer: Exact Solution Results of Numerical Analysis
50Momentum Integral Equation Provides Approximate Alternative to Exact (Blassius) Solution
51Momentum Integral Equation Equation is used to estimate the boundary-layer thickness as a function of x:Obtain a first approximation to the freestream velocity distribution, U(x). The pressure in the boundary layer is related to the freestream velocity, U(x), using the Bernoulli equationAssume a reasonable velocity-profile shape inside the boundary layerDerive an expression for tw using the results obtained from item 2
52Use of the Momentum Equation for Flow with Zero Pressure Gradient Simplify Momentum Integral Equation (Item 1)The Momentum Integral Equation becomes
53Use of the Momentum Equation for Flow with Zero Pressure Gradient Laminar FlowExample: Assume a Polynomial Velocity Profile (Item 2)The wall shear stress tw is then (Item 3)
54Use of the Momentum Equation for Flow with Zero Pressure Gradient Laminar Flow Results (Polynomial Velocity Profile)Compare to Exact (Blassius) results!
55Use of the Momentum Equation for Flow with Zero Pressure Gradient Turbulent FlowExample: 1/7-Power Law Profile (Item 2)
56Use of the Momentum Equation for Flow with Zero Pressure Gradient Turbulent Flow Results (1/7-Power Law Profile)
59Drag Pure Friction Drag: Flat Plate Parallel to the Flow Pure Pressure Drag: Flat Plate Perpendicular to the FlowFriction and Pressure Drag: Flow over a Sphere and CylinderStreamlining
60Drag Flow over a Flat Plate Parallel to the Flow: Friction Drag Boundary Layer can be 100% laminar, partly laminar and partly turbulent, or essentially 100% turbulent; hence several different drag coefficients are available
61DragFlow over a Flat Plate Parallel to the Flow: Friction Drag (Continued)Laminar BL:Turbulent BL:… plus others for transitional flow