1. PHAROS UNIVERSITY ME 259 FLUID MECHANICS FOR ELECTRICAL STUDENTS Basic Equations for a Control Volume

2. Main Topics Flow Classification Basic Laws for a System Relation of System Derivatives to the Control Volume Formulation Conservation of Mass Bernoulli Equation

3. Flow Classification Classification of Fluid Dynamics Inviscid µ = 0 Viscous Laminar Turbulent Compressible Incompressible ϱ = constant Internal External

4. Basic Laws for a System Conservation of Mass

5. Relation of System Derivatives to the Control Volume Formulation Extensive and Intensive Properties

6. Relation of System Derivatives to the Control Volume Formulation Reynolds Transport Theorem

7. Continuity Equation Bernoulli’s Equation Momentum Equation Energy Equation FLUID FLOW

8. Basic Laws Conservation of mass: dM/dt=0 for system ∂/∂t ∫ ϱ d +∫ ϱ ⊽. d Ᾱ =0 for control vol. Newton’s 2 nd law Σ F = ma Σ F = ∂/∂t ∫ ⊽ ϱ d + ∫ ⊽ ϱ ⊽. d Ᾱ First Law of Thermo: Q - W = dE/dt Q-w=∂/∂t∫e ϱ d+∫(p/ ϱ +0.5V 2 +gz) ϱ⊽. d Ᾱ

9. Relation of System Derivatives to the Control Volume Formulation Interpreting the Scalar Product

10. Conservation of Mass Basic Law, and Transport Theorem

11. Conservation of Mass

12. Basic Law for a System

13. Conservation of Mass Incompressible Fluids Steady, Compressible Flow

14. Definitions Volume (Volumetric) Flow Rate Q = Cross Sectional Area*Average Velocity of the fluid Q = A*v cms Weight Flow Rate W =  *Q N/s Mass Flow Rate M =  *Q kg/s Volume v Q = Volume/Unit time Q = Area*Distance/Unit Time

15. Flow in non-circular sections Flow rate is determined by: Q = A*v Where, A = Net flow area v = average velocity Example: D large, i = 0.5 m D small, o = 0.25 m D small, i = 0.2 m V small, i = 1 m/sec V large, i = 1 m/sec Find Q large and Q small

16. Continuity Equation Continuity for any fluid (gas or liquid) Mass flow rate In = Mass Flow Rate out M 1 = M 2  1 *A 1 *v 1 =  2 *A 2 *v 2 Continuity for liquids Q 1 = Q 2 A 1 *v 1 = A 2 *v 2 M1M1 M2M2

17. 17/27 Equation of continuity Volume flow rate has units m 3 /s Mass flow rate has units kg/s

18. Units and Conversion Factors Q: m 3 /sec M: kg/sec, Volume Flow Rate: 1 L/min = 0.06 m 3 /h 1 m 3 /sec = 60,000 L/min 1 gal/min = 3.785 L/min

19. Example #1 If d 1 and d 2 are 50 mm and 100 mm, respectively, and water at 70° C is flowing at 8 m/sec in section 1, determine: v 2, Q, W, M. 12d1d1 d2d2 v1v1 v2v2

20. © Pritchard Example # 2 If d 1, d 2 and d 3 are 10 cm, 20 cm, and 50 cm, determine Q and the velocities, v 2 and v 3 if v 1 = 1 m/sec. V 1 = 1 m/sec d 1 = 10 cmd 2 = 20 cm d 3 = 50 cm

21. Example # 3 Determine the required size standard Schedule 40 steel pipe to carry 192 m 3 /hr with a maximum velocity of 6.0 m/sec.

22. Example # 4 The tank is being filled with water by two 1-D inlets. Air is at the top of the tank. The water height is h. (a) Find an expression for the change in water height dh/dt. (b) Compute dh/dt if D 1 = 1 cm, D 2 = 3 cm, V 1 = 3 m/s, V 2 = 2 m/s and A t = 2 m 2. Tank Area A t aa ww h 1 2

23. Example # 5 Consider the entrance region of a circular pipe for laminar flow. What is mean velocity of the fluid.

24. © Pritchard 13-Sep-1524/27 Assume: the flow of fluids is laminar (not turbulent) or steady flow - the fluid has no viscosity (no friction). Ideal Fluids in Motion: Continuity & Bernoulli’s equation A fluid element traces out a streamline as it moves. The velocity vector of the element is tangent to the streamline at every point. The steady flow of a fluid around an air foil, as revealed by a dye tracer that was injected into the fluid upstream of the airfoil

25. Conservation of Energy Bernoulli’s Equation Energy cannot be created or destroyed, just transformed Three forms of energy in fluid system: Potential Kinetic Flow energy

26. Potential Energy Due to the elevation of the fluid element Where, w = weight of fluid element z = elevation with respect to a reference level

27. Kinetic Energy Due to the velocity of the fluid element Where, v = average velocity of the fluid element

28. Flow Energy Flow work or pressure energy Amount of energy necessary to move a fluid element across a certain section against pressure Where, p = pressure on the fluid element

29. Total Energy and Conservation of Energy Principle E = FE + PE + KE Two points along the same pipe: E 1 = E 2 Bernoulli’s Equation:

30. Heads Assumption: No energy is added or lost Assumption: Energy level remains constant

32. Restrictions on Bernoullis’ Equation Valid only for incompressible fluids No energy is added or removed by pumps, brakes, valves, etc. No heat transfer from or to liquid No energy lost due to friction

33. Application of Bernoulli’s Equation Write Bernoulli’s equation in the direction of flow, Label diagram Simplify equation by canceling terms that are zero, or equal on both sides of the equation Solve equation and find desired result(s)

34. Example A hose carries water at a flow rate of 0.01 m 3 /sec. The hose has an internal diameter of 12 mm, and the gauge pressure at faucet is 100 kPa. Determine the pressure at the end of the hose  Z = 10 m

35. Torricelli’s Theorem For a liquid flowing from a tank or reservoir with constant fluid elevation, the velocity through the orifice is given by: where, h is the difference in elevation between the orifice and the top of the tank Example: If h = 3.00 m, compute v 2 h

36. Take Home Experiment A reservoir of water has the surface at 310m above the outlet nozzle of a pipe with diameter 15mm. What is the a) velocity, b) the discharge out of the nozzle and c) mass flow rate. Water Velocity = (2gh) 0.5 = (2x 9.81 x 310) 0.5 = 78 m/s

37. Individual Experiment Pipe Flow: Ideal flow Assumption and Energy Equation The aim is to study Continuity equation and Bernoulli equation as will as pressure losses due to viscous ( frictional) effects in fluid flows through pipes Flow meter Differential Pressure Gauge- measure ΔP L Valve H Reservoir Pipe D Schematic of experimental Apparatus Pipes with different Diameter and Length will be used later for the experiments to study Energy Equation and pressure losses