Presentation is loading. Please wait.

Presentation is loading. Please wait.

Fluid Mechanics Wrap Up CEE 331 June 27, 2015 CEE 331 June 27, 2015 

Similar presentations


Presentation on theme: "Fluid Mechanics Wrap Up CEE 331 June 27, 2015 CEE 331 June 27, 2015 "— Presentation transcript:

1 Fluid Mechanics Wrap Up CEE 331 June 27, 2015 CEE 331 June 27, 2015 

2 Review ä Fluid Properties ä Fluid Statics ä Control Volume Equations ä Navier Stokes ä Dimensional Analysis and Similitude ä Viscous Flow: Pipes ä External Flows ä Open Channel Flow ä Fluid Properties ä Fluid Statics ä Control Volume Equations ä Navier Stokes ä Dimensional Analysis and Similitude ä Viscous Flow: Pipes ä External Flows ä Open Channel Flow

3 Shear Stress change in velocity with respect to distance dimension of Tangential force per unit area Rate of angular deformation rate of shear

4 Pressure Variation When the Specific Weight is Constant Piezometric head

5 Center of Pressure: y p Sum of the moments Transfer equation y = 0 where p = datum pressure

6 Inclined Surface Findings ä The horizontal center of pressure and the horizontal centroid ________ when the surface has either a horizontal or vertical axis of symmetry ä The center of pressure is always _______ the centroid ä The vertical distance between the centroid and the center of pressure _________ as the surface is lowered deeper into the liquid ä What do you do if there isn’t a free surface? ä The horizontal center of pressure and the horizontal centroid ________ when the surface has either a horizontal or vertical axis of symmetry ä The center of pressure is always _______ the centroid ä The vertical distance between the centroid and the center of pressure _________ as the surface is lowered deeper into the liquid ä What do you do if there isn’t a free surface? coincide below decreases

7 Forces on Curved Surfaces: Horizontal Component ä The horizontal component of pressure force on a curved surface is equal to the pressure force exerted on a horizontal ________ of the curved surface ä The horizontal component of pressure force on a closed body is always _____ ä The center of pressure is located on the projected area using the moment of inertia ä The horizontal component of pressure force on a curved surface is equal to the pressure force exerted on a horizontal ________ of the curved surface ä The horizontal component of pressure force on a closed body is always _____ ä The center of pressure is located on the projected area using the moment of inertia projection zero

8 Forces on Curved Surfaces: Vertical Component The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and extending up to the (virtual or real) free surface Streeter, et. al The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and extending up to the (virtual or real) free surface Streeter, et. al

9 C (78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 0 0 0.948 m 1.083 m 89.7kN 78.5kN Cylindrical Surface Force Check ä All pressure forces pass through point C. ä The pressure force applies no moment about point C. ä The resultant must pass through point C. ä All pressure forces pass through point C. ä The pressure force applies no moment about point C. ä The resultant must pass through point C.

10 Uniform Acceleration ä How can we apply our equations to a frame of reference that is accelerating at a constant rate? _______________________________ _______________________ Use total acceleration including acceleration due to gravity. Free surface is always normal to total acceleration

11 Conservation of Mass N = Total amount of ____ in the system h = ____ per unit mass = __ N = Total amount of ____ in the system h = ____ per unit mass = __ mass 1 1 But dm/dt = 0! cv equation mass leaving - mass entering = - rate of increase of mass in cv

12 EGL (or TEL) and HGL ä The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______) ä The decrease in total energy represents the head loss or energy dissipation per unit weight ä EGL and HGL are ____________and lie at the free surface for water at rest (reservoir) ä Whenever the HGL falls below the point in the system for which it is plotted, the local pressures are lower than the __________________ ä The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______) ä The decrease in total energy represents the head loss or energy dissipation per unit weight ä EGL and HGL are ____________and lie at the free surface for water at rest (reservoir) ä Whenever the HGL falls below the point in the system for which it is plotted, the local pressures are lower than the __________________ pump coincident reference pressure

13 Losses and Efficiencies ä Electrical power ä Shaft power ä Impeller power ä Fluid power ä Electrical power ä Shaft power ä Impeller power ä Fluid power IE TwTw TwTw TwTw TwTw g QH p Motor losses bearing losses pump losses

14 Linear Momentum Equation The momentum vectors have the same direction as the velocity vectors Fp1Fp1 Fp1Fp1 Fp2Fp2 Fp2Fp2 W W M1M1 M1M1 M2M2 M2M2 F ss y F ss x

15 Vector Addition cs 1 cs 3 q1q1 q2q2 cs 2 x y q3q3

16 Summary ä Control volumes should be drawn so that the surfaces are either tangent (no flow) or normal (flow) to streamlines. ä In order to solve a problem the flow surfaces need to be at locations where all but 1 or 2 of the energy terms are known ä The control volume can not change shape over time ä When possible choose a frame of reference so the flows are steady ä Control volumes should be drawn so that the surfaces are either tangent (no flow) or normal (flow) to streamlines. ä In order to solve a problem the flow surfaces need to be at locations where all but 1 or 2 of the energy terms are known ä The control volume can not change shape over time ä When possible choose a frame of reference so the flows are steady

17 Summary ä Control volume equation: Required to make the switch from a closed to an open system ä Any conservative property can be evaluated using the control volume equation ä mass, energy, momentum, concentrations of species ä Many problems require the use of several conservation laws to obtain a solution ä Control volume equation: Required to make the switch from a closed to an open system ä Any conservative property can be evaluated using the control volume equation ä mass, energy, momentum, concentrations of species ä Many problems require the use of several conservation laws to obtain a solution

18 Navier-Stokes Equations Navier-Stokes Equation Inertial forces [N/m 3 ] Pressure gradient (not due to change in elevation) Shear stress gradient h is vertical (positive up)

19 Summary ä Navier-Stokes Equations and the Continuity Equation describe complex flow including turbulence, but are difficult to solve ä The Navier-Stokes Equations can be solved analytically for several simple flows ä Navier-Stokes Equations and the Continuity Equation describe complex flow including turbulence, but are difficult to solve ä The Navier-Stokes Equations can be solved analytically for several simple flows

20 Dimensionless parameters ä Reynolds Number ä Froude Number ä Weber Number ä Mach Number ä Pressure Coefficient ä (the dependent variable that we measure experimentally) ä Reynolds Number ä Froude Number ä Weber Number ä Mach Number ä Pressure Coefficient ä (the dependent variable that we measure experimentally)

21 Froude similarity difficult to change g ä Froude number the same in model and prototype ä ________________________ ä define length ratio (usually larger than 1) ä velocity ratio ä time ratio ä discharge ratio ä force ratio ä Froude number the same in model and prototype ä ________________________ ä define length ratio (usually larger than 1) ä velocity ratio ä time ratio ä discharge ratio ä force ratio

22 Laminar Flow through Circular Tubes Velocity Shear True for Laminar or Turbulent flow Shear at the wall Laminar flow

23 Pipe Flow Energy Losses Dimensional Analysis Darcy-Weisbach equation

24 Laminar Flow Friction Factor Slope of ___ on log-log plot Hagen-Poiseuille Darcy-Weisbach

25 Moody Diagram 0.01 0.10 1E+031E+041E+051E+061E+071E+08 R friction factor laminar 0.05 0.04 0.03 0.02 0.015 0.01 0.008 0.006 0.004 0.002 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 smooth 0.02 0.03 0.04 0.05 0.06 0.08

26 l find head loss given (D, type of pipe, Q) l find flow rate given (head, D, L, type of pipe) l find pipe size given (head, type of pipe,L, Q) Solution Techniques

27 Minor Losses ä We previously obtained losses through an expansion using conservation of energy, momentum, and mass ä Most minor losses can not be obtained analytically, so they must be measured ä Minor losses are often expressed as a loss coefficient, K, times the velocity head. ä We previously obtained losses through an expansion using conservation of energy, momentum, and mass ä Most minor losses can not be obtained analytically, so they must be measured ä Minor losses are often expressed as a loss coefficient, K, times the velocity head. High Re

28 Swamee Jain Iterative Technique for D and Q (given h l ) ä Assume all head loss is major head loss. ä Calculate D or Q using Swamee-Jain equations ä Calculate minor losses ä Find new major losses by subtracting minor losses from total head loss ä Assume all head loss is major head loss. ä Calculate D or Q using Swamee-Jain equations ä Calculate minor losses ä Find new major losses by subtracting minor losses from total head loss

29 Darcy Weisbach/Moody Iterative Technique Q (given h l ) ä Assume a value for the friction factor. ä Calculate Q using head loss equations ä Find new friction factor ä Assume a value for the friction factor. ä Calculate Q using head loss equations ä Find new friction factor

30 ä Geometric parameters ä ___________________ ä Write the functional relationship ä Geometric parameters ä ___________________ ä Write the functional relationship Hydraulic radius (R h ) Channel length (l) Roughness ( e ) Open Conduits: Dimensional Analysis

31 Open Channel Flow Formulas Dimensions of n? Is n only a function of roughness? Manning formula (MKS units!) NO! T /L 1/3 Chezy formula

32 Boundary Layer Thickness ä Water flows over a flat plate at 1 m/s. Plot the thickness of the boundary layer. How long is the laminar region? x = 0.5 m

33 Flat Plate: Streamlines U 0 1 2 3 4 PointvC p p 1 2 3 4 0 1 <U >0 >U<0 >p 0 <p 0 Points outside boundary layer!

34 Flat Plate Drag Coefficients 1 x 10 -3 5 x 10 -4 2 x 10 -4 1 x 10 -4 5 x 10 -5 2 x 10 -5 1 x 10 -5 5 x 10 -6 2 x 10 -6 1 x 10 -6

35 Drag Coefficient on a Sphere 0.1 1 1 10 100 1000 0.1 1 1 10 10 2 10 3 10 4 10 5 10 6 10 7 Reynolds Number Drag Coefficient Stokes Law Re=500000 Turbulent Boundary Layer

36 More Fluids? ä Hydraulic Engineering (CEE 332 in 2003) ä Hydrology ä Measurement Techniques ä Model Pipe Networks (computer software) ä Open Channel Flow (computer software) ä Pumps and Turbines ä Design Project ä Pollutant Transport and Transformation (CEE 655) ä Hydraulic Engineering (CEE 332 in 2003) ä Hydrology ä Measurement Techniques ä Model Pipe Networks (computer software) ä Open Channel Flow (computer software) ä Pumps and Turbines ä Design Project ä Pollutant Transport and Transformation (CEE 655)


Download ppt "Fluid Mechanics Wrap Up CEE 331 June 27, 2015 CEE 331 June 27, 2015 "

Similar presentations


Ads by Google