1. Finite Control Volume Analysis CVEN 311 Application of Reynolds Transport Theorem

2. Moving from a System to a Finite Control Volume ä Mass ä Linear Momentum ä Moment of Momentum ä Energy ä Putting it all together! ä Mass ä Linear Momentum ä Moment of Momentum ä Energy ä Putting it all together!

3. Conservation of Mass B = Total amount of ____ in the system b = ____ per unit mass = __ B = Total amount of ____ in the system b = ____ per unit mass = __ mass 1 1 But DM sys /Dt = 0! cv equation mass leaving - mass entering = - rate of increase of mass in cv Continuity Equation

4. Conservation of Mass 1 2 V1V1V1V1 A1A1A1A1 If mass in cv is constant Unit vector is ______ to surface and pointed ____ of cv normal out We assumed uniform ___ on the control surface  is the spatially averaged velocity normal to the cs [M/T]

5. Continuity Equation for Constant Density and Uniform Velocity Density is constant across cs Density is the same at cs 1 and cs 2 [L 3 /T] Simple version of the continuity equation for conditions of constant density. It is understood that the velocities are either ________ or _______ ________. uniformspatially averaged

6. Example: Conservation of Mass? The flow out of a reservoir is 2 L/s. The reservoir surface is 5 m x 5 m. How fast is the reservoir surface dropping? h Example Constant density Velocity of the reservoir surface

7. Linear Momentum Equation R.T.T. momentum momentum/unit mass Steady state

8. Linear Momentum Equation AssumptionsAssumptions Vectors!!! äUniform density äUniform velocity äV  AäV  A äV  AäV  A äSteady äV fluid velocity relative to cv

9. Steady Control Volume Form of Newton’s Second Law ä What are the forces acting on the fluid in the control volume? äGravity äShear forces at the walls äPressure forces at the walls äPressure forces on the ends Why no shear on control surfaces? _______________________________No velocity gradient normal to surface

10. Resultant Force on the Solid Surfaces ä The shear forces on the walls and the pressure forces on the walls are generally the unknowns ä Often the problem is to calculate the total force exerted by the fluid on the solid surfaces ä The magnitude and direction of the force determines ä size of _____________needed to keep pipe in place ä force on the vane of a pump or turbine... ä The shear forces on the walls and the pressure forces on the walls are generally the unknowns ä Often the problem is to calculate the total force exerted by the fluid on the solid surfaces ä The magnitude and direction of the force determines ä size of _____________needed to keep pipe in place ä force on the vane of a pump or turbine... =force applied by solid surfaces thrust blocks

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

12. Reducing elbow in vertical plane with water flow of 300 L/s. The volume of water in the elbow is 200 L. Energy loss is negligible. Calculate the force of the elbow on the fluid. W = _________ section 1section 2 D50 cm 30 cm A__________________ __________________ V__________________ p 150 kPa_________ M__________________ F p __________________ Reducing elbow in vertical plane with water flow of 300 L/s. The volume of water in the elbow is 200 L. Energy loss is negligible. Calculate the force of the elbow on the fluid. W = _________ section 1section 2 D50 cm 30 cm A__________________ __________________ V__________________ p 150 kPa_________ M__________________ F p __________________ Example: Reducing Elbow 11 2 1 m y x Direction of V vectors 0.196 m 2 0.071 m 2 1.53 m/s ↑ 4.23 m/s → -459 N ↑ 1269 N → 29,400 N ↑ -1961 N ↑ ? ? ? ?

13. Example: What is p 2 ? P 2 = 132 kPa F p2 = -9400 N

14. Example: Reducing Elbow Horizontal Forces 11 22 Pipe wants to move to the _________ left y x Force of pipe on fluid Fp2Fp2 Fp2Fp2 M2M2 M2M2

15. Example: Reducing Elbow Vertical Forces 11 22 Pipe wants to move _________ up y x Fp1Fp1 Fp1Fp1 M1M1 M1M1 28 kN acting downward on fluid W

16. Example: Fire nozzle A small fire nozzle is used to create a powerful jet to reach far into a blaze. Estimate the force that the water exerts on the fire nozzle. The pressure at section 1 is 1000 kPa (gage). Ignore frictional losses in the nozzle. 8 cm 2.5 cm

17. Example: Momentum with Complex Geometry cs 1 cs 3 11 22 cs 2 Find Q 2, Q 3 and force on the wedge. x x y y 33 Unknown: ________________ Q 2, Q 3, V 2, V 3, F x

18. 5 Unknowns: Need 5 Equations cs 1 cs 3 11 cs 2 x x y y 33 Unknowns: Q 2, Q 3, V 2, V 3, F x Continuity Bernoulli (2x) Momentum (in x and y) 22 Identify the 5 equations!

19. Solve for Q 2 and Q 3 x x y y   Component of velocity in y direction Mass conservation Negligible losses – apply Bernoulli atmospheric pressure

20. Solve for Q 2 and Q 3 Why is Q 2 greater than Q 3 ?

21. Solve for F x Force of wedge on fluid

22. Vector solution

23. Vector Addition cs 1 cs 3 11 22 cs 2 x y 33 Where is the line of action of F ss ?

24. Moment of Momentum Equation R.T.T. Moment of momentum Moment of momentum/unit mass Steady state

25. r2r2r2r2 r1r1r1r1 VnVnVnVn VtVtVtVt Application to Turbomachinery cs 1 cs 2 VtVt VnVn

26. Example: Sprinkler Total flow is 1 L/s. Jet diameter is 0.5 cm. Friction exerts a torque of 0.1 N-m-s 2  2. = 30º.  = 30º. Find the speed of rotation. Total flow is 1 L/s. Jet diameter is 0.5 cm. Friction exerts a torque of 0.1 N-m-s 2  2. = 30º.  = 30º. Find the speed of rotation. vtvtvtvt vtvtvtvt  10 cm   V t and V n are defined relative to control surfaces. cs 2

27. = 34 rpm c = -(1000 kg/m 3 )(0.001 m 3 /s) 2 (0.1m)(2sin30)/3.14/(0.005 m) 2 b = (1000 kg/m 3 )(0.001 m 3 /s) (0.1 m) 2 = 0.01 Nms Example: Sprinkler c = -1.27 Nm    = 3.5/s a = 0.1Nms 2

28. Reflections ä What is the name of the equation that we used to move from a system (Lagrangian) view to the control volume (Eulerian) view? ä Explain the analogy to your checking account. ä The velocities in the linear momentum equation are relative to …? ä Why is ma non-zero for a fixed control volume? ä Under what conditions could you generate power from a rotating sprinkler? ä What questions do you have about application of the linear momentum and momentum of momentum equations? ä What is the name of the equation that we used to move from a system (Lagrangian) view to the control volume (Eulerian) view? ä Explain the analogy to your checking account. ä The velocities in the linear momentum equation are relative to …? ä Why is ma non-zero for a fixed control volume? ä Under what conditions could you generate power from a rotating sprinkler? ä What questions do you have about application of the linear momentum and momentum of momentum equations?

29. RTT Energy Equation First law of thermodynamics: The heat Q H added to a system plus the work W done on the system equals the change in total energy E of the system. What is for a system? DE/Dt

30. dE/dt for our System? Heat transfer Shaft work Pressure work

31. General Energy Equation z z RTT 1 st Law of Thermo Potential Kinetic Internal (molecular spacing and forces) Total

32. Steady Simplify the Energy Equation Assume...Assume... But V is often ____________ over control surface! äHydrostatic pressure distribution at cs äŭ is uniform over cs not uniform 0 0

33. If V normal to n kinetic energy correction term V = point velocity V = average velocity over cs Energy Equation: Kinetic Energy Term  = _________________________  =___ for uniform velocity 1 1

34. Energy Equation: steady, one- dimensional, constant density mass flux rate

35. Energy Equation: steady, one- dimensional, constant density hPhP hPhP Lost mechanical energy divide by g mechanical thermal

36. Thermal Components of the Energy Equation Example Water specific heat = 4184 J/(kg*K) For incompressible liquids Change in temperature Heat transferred to fluid

37. Example: Energy Equation (energy loss) datum 2 m 4 m An irrigation pump lifts 50 L/s of water from a reservoir and discharges it into a farmer’s irrigation channel. The pump supplies a total head of 10 m. How much energy is lost? An irrigation pump lifts 50 L/s of water from a reservoir and discharges it into a farmer’s irrigation channel. The pump supplies a total head of 10 m. How much mechanical energy is lost? 2.4 m cs 1 cs 2 h L = 10 m - 4 m

38. We need _______ in the pipe, __, and ____ ____. Example: Energy Equation (pressure at pump outlet) datum 2 m 4 m 50 L/s h P = 10 m The total pipe length is 50 m and is 20 cm in diameter. The pipe length to the pump is 12 m. What is the pressure in the pipe at the pump outlet? You may assume (for now) that the only losses are frictional losses in the pipeline. 2.4 m cs 1 cs 2 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / velocity head loss 

39. ä How do we get the velocity in the pipe? ä How do we get the frictional losses?  What about  ? ä How do we get the velocity in the pipe? ä How do we get the frictional losses?  What about  ? Expect losses to be proportional to length of the pipe h l = (6 m)(12 m)/(50 m) = 1.44 m V = 4(0.05 m 3 /s)/[   0.2 m) 2 ] = 1.6 m/s Example: Energy Equation (pressure at pump outlet) Q = VA A =  d 2 /4 V = 4Q/(  d 2 )

40. Kinetic Energy Correction Term:    is a function of the velocity distribution in the pipe. ä For a uniform velocity distribution ____ ä For laminar flow ______ ä For turbulent flow _____________ ä Often neglected in calculations because it is so close to 1   is a function of the velocity distribution in the pipe. ä For a uniform velocity distribution ____ ä For laminar flow ______ ä For turbulent flow _____________ ä Often neglected in calculations because it is so close to 1  is 1  is 2 1.01 <  < 1.10

41. Example: Energy Equation (pressure at pump outlet) datum 2 m 4 m 50 L/s h P = 10 m V = 1.6 m/s  = 1.05 h L = 1.44 m V = 1.6 m/s  = 1.05 h L = 1.44 m 2.4 m = 59.1 kPa

42. pipe burst Example: Energy Equation (Hydraulic Grade Line - HGL) ä We would like to know if there are any places in the pipeline where the pressure is too high (_________) or too low (water might boil - cavitation). ä Plot the pressure as piezometric head (height water would rise to in a manometer) ä How? ä We would like to know if there are any places in the pipeline where the pressure is too high (_________) or too low (water might boil - cavitation). ä Plot the pressure as piezometric head (height water would rise to in a manometer) ä How?

43. Example: Energy Equation (Hydraulic Grade Line - HGL) datum 2 m 4 m 50 L/s H P = 10 m 2.4 m

44. Pressure head (w.r.t. reference pressure) EGL (or TEL) and HGL velocity head velocity head Elevation head (w.r.t. datum) Piezometric head Energy Grade Line Hydraulic Grade Line

45. pump coincident reference pressure 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 __________________

46. z z Example HGL and EGL z = 0 pump energy grade line hydraulic grade line velocity head pressure head elevation datum

47. Bernoulli vs. Control Volume Conservation of Energy Free jet pipe Find the velocity and flow. How would you solve these two problems?

48. Bernoulli vs. Control Volume Conservation of Energy Point to point along streamline Control surface to control surface No frictional losses Has a term for frictional losses Based on average velocity Requires kinetic energy correction factor Includes shaft work

49. Power 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 T shaft T loss  QH p Motor losses bearing losses pump losses

50. Example: Hydroplant Q = 5 m 3 /s 2100 kW 180 rpm 116 kN·m 50 m Water power = ? total losses = ? efficiency of turbine = ? efficiency of generator = ? solution

51. Energy Equation Review ä Control Volume equation ä Simplifications ä steady ä constant density ä hydrostatic pressure distribution across control surface (cs normal to streamlines) ä Direction of flow matters (in vs. out) ä We don’t know how to predict head loss ä Control Volume equation ä Simplifications ä steady ä constant density ä hydrostatic pressure distribution across control surface (cs normal to streamlines) ä Direction of flow matters (in vs. out) ä We don’t know how to predict head loss

52. Conservation of Energy, Momentum, and Mass ä Most problems in fluids require the use of more than one conservation law to obtain a solution ä Often a simplifying assumption is required to obtain a solution ä neglect energy losses (_______) over a short distance with no flow expansion ä neglect shear forces on the solid surface over a short distance ä Most problems in fluids require the use of more than one conservation law to obtain a solution ä Often a simplifying assumption is required to obtain a solution ä neglect energy losses (_______) over a short distance with no flow expansion ä neglect shear forces on the solid surface over a short distance to heat

53. greater Head Loss: Minor Losses ä Head (or energy) loss due to: outlets, inlets, bends, elbows, valves, pipe size changes ä Losses due to expansions are ________ than losses due to contractions ä Losses can be minimized by gradual transitions ä Losses are expressed in the form where K is the loss coefficient ä Head (or energy) loss due to: outlets, inlets, bends, elbows, valves, pipe size changes ä Losses due to expansions are ________ than losses due to contractions ä Losses can be minimized by gradual transitions ä Losses are expressed in the form where K is the loss coefficient

54. z in = z out What is p in - p out ? Head Loss due to Sudden Expansion: Conservation of Energy 1 2

55. Apply in direction of flow Neglect surface shear Divide by (A out  ) Head Loss due to Sudden Expansion: Conservation of Momentum Pressure is applied over all of section 1. Momentum is transferred over area corresponding to upstream pipe diameter. V in is velocity upstream. Pressure is applied over all of section 1. Momentum is transferred over area corresponding to upstream pipe diameter. V in is velocity upstream. 1 2 A1A1A1A1 A2A2A2A2 x

56. Head Loss due to Sudden Expansion Discharge into a reservoir?_________ Energy Momentum Mass K L =1

57. Example: Losses due to Sudden Expansion in a Pipe (Teams!) ä A flow expansion discharges 2.4 L/s directly into the air. Calculate the pressure immediately upstream from the expansion 1 cm 3 cm We can solve this using either the momentum equation or the energy equation (with the appropriate term for the energy losses)! Solution

58. 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 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 When possible choose a frame of reference so the flows are steady

59. Summary ä Control volume equation: Required to make the switch from Lagrangian to Eulerian 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 Lagrangian to Eulerian 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 end

60. Example: Conservation of Mass (Team Work) ä The flow through the orifice is a function of the depth of water in the reservoir ä Find the time for the reservoir level to drop from 10 cm to 5 cm. The reservoir surface is 15 cm x 15 cm. The orifice is 2 mm in diameter and is 2 cm off the bottom of the reservoir. The orifice coefficient is 0.6. ä CV with constant or changing mass. ä Draw CV, label CS, solve using variables starting with ä The flow through the orifice is a function of the depth of water in the reservoir ä Find the time for the reservoir level to drop from 10 cm to 5 cm. The reservoir surface is 15 cm x 15 cm. The orifice is 2 mm in diameter and is 2 cm off the bottom of the reservoir. The orifice coefficient is 0.6. ä CV with constant or changing mass. ä Draw CV, label CS, solve using variables starting with

61. Example Conservation of Mass Constant Volume h cs 1 cs 2

62. Example Conservation of Mass Changing Volume h cs 1 cs 2

63. Example Conservation of Mass

64. Pump Head hphphphp

65. Example: Venturi

66. Find the flow (Q) given the pressure drop between section 1 and 2 and the diameters of the two sections. Draw an appropriate control volume. You may assume the head loss is negligible. Draw the EGL and the HGL. 1 2 hhhh

67. Example Venturi

68. Fire nozzle: Team Work Identify what you need to know Count your unknowns Determine what equations you will use Identify what you need to know Count your unknowns Determine what equations you will use 8 cm 2.5 cm 1000 kPa

69. Find the Velocities

70. Fire nozzle: Solution 2.5 cm 8 cm 1000 kPa force applied by nozzle on water Is this the force that the firefighters need to brace against? _______ Which direction does the nozzle want to go? ______ NO!

71. Temperature Rise over Taughanock Falls ä Drop of 50 meters ä Find the temperature rise ä Drop of 50 meters ä Find the temperature rise

72. Hydropower

73. Solution: Losses due to Sudden Expansion in a Pipe ä A flow expansion discharges 2.4 L/s directly into the air. Calculate the pressure immediately upstream from the expansion 1 cm 3 cm