1. Fluid Properties and Units CVEN 311 

2. Continuum ä All materials, solid or fluid, are composed of molecules discretely spread and in continuous motion. ä However, in dealing with fluid-flow relations on a mathematical basis, it is necessary to replace the actual molecular structure by a hypothetical continuous medium, called the continuum. ä All materials, solid or fluid, are composed of molecules discretely spread and in continuous motion. ä However, in dealing with fluid-flow relations on a mathematical basis, it is necessary to replace the actual molecular structure by a hypothetical continuous medium, called the continuum.

3. Continuum ä In a continuum, the physical variable at a point in space is the averaged value of the variable in a small sphere. ä How good is the assumption? ä In a continuum, the physical variable at a point in space is the averaged value of the variable in a small sphere. ä How good is the assumption? 10 -3 cm 3x10 10 molecules of air

4. Dimensions and Units ä The dimensions have to be the same for each term in an equation ä Dimensions of mechanics are ä length ä time ä mass ä force ä temperature ä The dimensions have to be the same for each term in an equation ä Dimensions of mechanics are ä length ä time ä mass ä force ä temperature L T M MLT -2 

5. Dimensions and Units QuantitySymbolDimensions VelocityV LT -1 AccelerationaLT -2 AreaA L 2 Volume  L 3 DischargeQ L 3 T -1 Pressurep ML -1 T -2 Gravityg LT -2 TemperatureT’  Mass concentrationC ML -3 QuantitySymbolDimensions VelocityV LT -1 AccelerationaLT -2 AreaA L 2 Volume  L 3 DischargeQ L 3 T -1 Pressurep ML -1 T -2 Gravityg LT -2 TemperatureT’  Mass concentrationC ML -3

6. Dimensions and Units QuantitySymbolDimensions Density  ML -3 Specific Weight  ML -2 T -2 Dynamic viscosity  ML -1 T -1 Kinematic viscosity L 2 T -1 Surface tension  MT -2 Bulk mod of elasticityE ML -1 T -2 QuantitySymbolDimensions Density  ML -3 Specific Weight  ML -2 T -2 Dynamic viscosity  ML -1 T -1 Kinematic viscosity L 2 T -1 Surface tension  MT -2 Bulk mod of elasticityE ML -1 T -2 These are _______ properties! fluid How many independent properties? _____ 4

7. Definition of a Fluid ä “a fluid, such as water or air, deforms continuously when acted on by shearing stresses of any magnitude.” - Munson, Young, Okiishi Water Oil Air Why isn’t steel a fluid? Water Oil Air Why isn’t steel a fluid?

8. Fluid Deformation between Parallel Plates Side view Force F causes the top plate to have velocity U. What other parameters control how much force is required to get a desired velocity? Distance between plates (b) Area of plates (A) F b U Viscosity!

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

10. Fluid classification by response to shear stress ä Newtonian ä Ideal Fluid ä Ideal plastic ä Newtonian ä Ideal Fluid ä Ideal plastic Newtonian Ideal Fluid Ideal plastic Shear stress  Rate of deformation dy du  1

11. Fluid Viscosity ä Examples of highly viscous fluids ä ______________________ ä Fundamental mechanisms ä Gases - transfer of molecular momentum ä Viscosity __________ as temperature increases. ä Viscosity __________ as pressure increases. ä Liquids - cohesion and momentum transfer ä Viscosity decreases as temperature increases. ä Relatively independent of pressure (incompressible) ä Examples of highly viscous fluids ä ______________________ ä Fundamental mechanisms ä Gases - transfer of molecular momentum ä Viscosity __________ as temperature increases. ä Viscosity __________ as pressure increases. ä Liquids - cohesion and momentum transfer ä Viscosity decreases as temperature increases. ä Relatively independent of pressure (incompressible) molasses, tar, 20w-50 oil increases _______ increases

12. Example: Measure the viscosity of water The inner cylinder is 10 cm in diameter and rotates at 10 rpm. The fluid layer is 2 mm thick and 10 cm high. The power required to turn the inner cylinder is 50x10 -6 watts. What is the dynamic viscosity of the fluid? Outer cylinder Thin layer of water Inner cylinder

13. Solution Scheme ä Restate the goal ä Identify the given parameters and represent the parameters using symbols ä Outline your solution including the equations describing the physical constraints and any simplifying assumptions ä Solve for the unknown symbolically ä Substitute numerical values with units and do the arithmetic ä Check your units! ä Check the reasonableness of your answer ä Restate the goal ä Identify the given parameters and represent the parameters using symbols ä Outline your solution including the equations describing the physical constraints and any simplifying assumptions ä Solve for the unknown symbolically ä Substitute numerical values with units and do the arithmetic ä Check your units! ä Check the reasonableness of your answer Solution

14. Role of Viscosity ä Statics ä Fluids at rest have no relative motion between layers of fluid and thus du/dy = 0 ä Therefore the shear stress is _____ and is independent of the fluid viscosity ä Flows ä Fluid viscosity is very important when the fluid is moving ä Statics ä Fluids at rest have no relative motion between layers of fluid and thus du/dy = 0 ä Therefore the shear stress is _____ and is independent of the fluid viscosity ä Flows ä Fluid viscosity is very important when the fluid is moving zero

15. Dynamic and Kinematic Viscosity ä Kinematic viscosity (__) is a fluid property obtained by dividing the dynamic viscosity (__) by the fluid density [m 2 /s] Connection to Reynolds number!

16. Density and Specific Weight  Density (mass/unit volume)  ä density of water: ä density of air at atmospheric pressure and 15  C:  Specific Weight (weight per unit volume)  ä __________________  Density (mass/unit volume)  ä density of water: ä density of air at atmospheric pressure and 15  C:  Specific Weight (weight per unit volume)  ä __________________ 950 960 970 980 990 1000 050100 Temperature (C) Density (kg/m 3 ) 997 998 999 1000 01020 Temperature (C) Density (kg/m 3 ) 1000 kg/m 3 1.22 kg/m 3   =  g = 9806 N/m 3 Specific mass

17. Perfect Gas Law ä PV = nRT ä R is the universal gas constant ä T is in Kelvin ä PV = nRT ä R is the universal gas constant ä T is in Kelvin Note deviation from the text! Use absolute pressure for P and absolute temperature for T

18. Bulk Modulus of Elasticity ä Relates the change in volume to a change in pressure ä changes in density at high pressure ä pressure waves ä _________ ä ______ __________ ä Relates the change in volume to a change in pressure ä changes in density at high pressure ä pressure waves ä _________ ä ______ __________ 2.00 2.05 2.10 2.15 2.20 2.25 2.30 2.35 020406080100 Temperature (C) Bulk Modulus of elasticity (GPa) sound water hammer Water - speed of sound

19. Vapor Pressure 0 1000 2000 3000 4000 5000 6000 7000 8000 010203040 Temperature (C) Vapor pressure (Pa) liquid What is vapor pressure of water at 100°C? 101 kPa Connection forward to cavitation!

20. Cavitation

21. Cavitation Damage

22.  p  R 2 = 2  R  Surface Tension ä Pressure increase in a spherical droplet pR2pR2 2R2R Surface molecules 0.050 0.055 0.060 0.065 0.070 0.075 0.080 020406080100 Temperature (C) Surface tension (N/m)

23. Example: Surface Tension ä Estimate the difference in pressure (in Pa) between the inside and outside of a bubble of air in 20ºC water. The air bubble is 0.3 mm in diameter. R = 0.15 x 10 -3 m  = 0.073 N/m What is the difference between pressure in a water droplet and in an air bubble? Statics!

24. Outline the solution ä Restate the goal ä Identify the given parameters and represent the parameters using symbols ä Outline your solution including the equations describing the physical constraints and any simplifying assumptions ä Restate the goal ä Identify the given parameters and represent the parameters using symbols ä Outline your solution including the equations describing the physical constraints and any simplifying assumptions

25. Viscosity Measurement: Solution Outer cylinder Thin layer of water Inner cylinder r = 5 cm t = 2 mm h = 10 cm P = 50 x 10 -6 W 10 rpm rr 2  rh FrFr