CIVE10006 Hydraulic Engineering 4

1. CIVE10006Hydraulic Engineering 4 Dr Martin Crapper Session 2011-2012

2. Details 1 This is a difficult and possibly quite dull course � but it is important! Requires considerable study � lectures can only highlight the important points It is quite mathematical Most problems amenable to spreadsheet solutions Turning up to class is going to help, despite 9.00am start

3. Details 2 Material is mostly same as in previous years, including what is in the exam and how it is asked BUT new material on waves, and (some of) Open Channel/Dam Break stuff removed Open Book exam so no data sheet (but beware) Computer exam as in Fluids 3 No marks for cutting and pasting tutorial solutions � must explain enough to demonstrate understanding

4. Teaching Philosophy http://www.eng.ed.ac.uk/~martc/teaching/teachingphilosophy.html Not my intention that you note down everything on the slide You do need to note verbal commentary as well as things that appear in writing Worked examples must be taken as a whole including verbal commentary Worked examples including verbal commentary are my only source of information on presenting good exam answers

5. Resources WebCT or www.eng.ed.ac.uk/~martc teaching pages. I follow the book, Chadwick Morfett and Borthwick, Hydraulics in Civil and Environmental Engineering 4th Edition, Spon Press, ISBN 0-415-306904-4 closely If you don�t have it already, I advise you to buy the book � it is the best notes. It will be useful in your career as a reference. Tutorials in computer lab to give practice at using spreadsheet solutions Handouts (on webct) for some of the more irritating derivations

6. Section 1: Unsteady Flow in Pipes Revision of fundamentals of flow in pipes Introduction to pressure surge Incompressible unsteady flow Compressible unsteady flow � inelastic and elastic pipes The Method of Characteristics applied to unsteady flow in pipes Finite Difference solution for Characteristic equations

7. Revision Quiz List the three principles of conservation that need to be applied to examine the motion of fluids (or anything else!) Without using equations, define laminar and turbulent flow Write down the formula for Reynolds Number. What ratio does Reynolds Number express? Write down the equation for Darcy�s Law as it is applied to flow in pipes. Give two ways of determining the friction factor ? that appears in Darcy�s law What is the typical order of magnitude for ??

8. The Fundamentals Continuity, or conservation of mass. Put simply, there can be no loss of mass in a system. Conservation of energy. Energy can neither be created or destroyed, although can be transformed , for example from potential to kinetic energy. Conservation of momentum. A body in motion can neither gain or lose momentum unless an external force is applied to it.

9. Laminar and Turbulent Flow http://www.youtube.com/watch?v=nl75BGg9qdA Reynolds No: ? = density � = viscosity U = mean velocity of flow D=characteristic dimension of flow, eg diameter of pipe Laminar flow: Re < 2000, hf a v Transitional flow: 2000 < Re < 4000 Turbulent flow: Re > 4000, hf a v2

10. Flow Resistance Darcy-Weisbach equation Colebrook-White equation for friction factor

11. Derivation of the Friction Factor When the flow is laminar, relative roughness has no influence on the friction factor. Laminar flow can be described by the Hazen-Poiseuille equation:

13. The Colebrook-White Transition Formula Is applicable to the entire turbulent flow region Requires a trial and error solution, and this restricted its widespread use when it was first introduced Other simpler formula eg Hazen-Williams and Manning are still used (with different friction factors)

15. Typical ks Values

17. An Introduction to Pressure Surge

18. Videos

19. Incompressible Unsteady Flow For a pipe line of length L and cross sectional area A, containing water flowing at velocity v: Assuming that the valve at the downstream end of the pipe is suddenly closed, then the liquid will undergo a change of momentum:

20. Force required to change momentum is exerted by the control valve. This results in an immediate build up in pressure upstream of valve: So:

21. Example

23. Example Results(head at valve)

28. Derivation of EquationsCompressible Flow, Inelastic Pipe

30. Example

33. Elastic Pipes

36. Real Behaviour So far we have considered �instantaneous� valve closure and we have neglected friction effects in flow This is not realistic Need to consider Elasticity of pipe Friction losses in fluid Non-instantaneous movement of valves (etc)

39. PDEs and the Method of Characteristics The characteristic equations for unsteady pipe flow Their solution (using Finite Difference approach) Boundary conditions Valves Dead end Reservoir Examples

40. The Method of Characteristics Note that this appears at first glance rather scary. This is because it is very difficult to visualize. Perhaps best thought of as a mathematical device�

41. A PDE is an equation describing the dependence of a quantity (such as velocity) on two or more variables. Very generally: Clearly the different terms don�t all apply in any given physical interpretation of this. For example the wave equation is Describing the propagation of a wave where t is time, x is distance and f (or y) is the surface elevation, has a= 1, c= -c2 (it�s a different c) and all the other coefficients as zero. Solution Behaviour

42. Various different PDEs made up from the general one can be described by reference to different curves elliptic, parabolic or hyperbolic All of these curves are obtainable by cutting a cone in different ways.

43. Hard part about a PDE is that it depends on two independent variables Would be much easier to solve if we could reduce it to dependence on one independent variable Then it becomes an ODE

45. Conservation Equations From previously (book eqns 6.16 slightly re-written & 6.17) we had: Multiplying the latter by c/g gives: These are both functions of x and t � 2 variables.

46. Characteristic Form What follows is some mathematical manipulation � a device. Go with the flow� Add the two equations together (possible since both equal zero) and rearrange a bit: Similarly, subtract the first from the second:

47. But definition of partial differentiation is: With a similar expression for dH/dt instead of du/dt. Now suppose we (arbitrarily) define these possibilities:

48. The preceding equations can be reduced as follows:

49. Finite Difference Form Differential equations are approximated as algebraic equations (no derivatives) by use of finite steps in time and space. Suitable for numerical calculation on a computer (or by hand). Accuracy depends on selection of steps. For example�

50. Forward approximation to value of the 1st derivative of u in space

51. Backward approximation to value of the 1st derivative of u in space

52. Central approximation to value of the 1st derivative of u in space

53. Approximations to values of the 1st derivative of u in space

54. Characteristic Equations in Difference Form Notation and condition for solution:

56. Adding the two finite difference equations gives: Whilst subtracting them gives:

57. These equations can be solved From know conditions at t we can get solution at t + dt Explicit in time, so only conditionally stable � dt small. Simple equations assume velocity and friction at time t apply throughout interval from t to t+dt For more accurate solution should use average value over time interval, requiring iterative solution Using these equations gives a first approximation only but can be done by hand or on a spreadsheet.

58. The Characteristic Equations

59. What does all this mean?

60. In General� Need to find head H and velocity u Need 2 equations to find 2 unknowns Typically 2 characteristic equations (positive and negative) Or one characteristic equation and one boundary condition, such as u=0 at a dead end� First step is always to identify unknowns and available equations

61. Typical Boundary Conditions At a valve, flow and head are related. We introduce the concept of area reduction over time as a valve closes Hence

62. Example This example is nearly the same as the one on pages 519-521 of the book 1km long, 500mm diameter horizontal pipe u/s end is a reservoir of head 100m d/s end is a valve. Head at open valve is 4.75m ? = 0.012 and c=1000m/s Valve is closed over 4s giving a linear retardation Determine pressure rise at valve and at mid point of pipe over 1s A spreadsheet solution is employed Equations can however be made explicit � see book p516-518

63. More Boundaries Valve in mid-pipeline Bifurcations and branches Pumps A complex example

64. Valve in Mid-pipeline

65. Discharge through valve given by the orifice equation: So:

66. Pumps Power input used to determine head across pump:

67. Bifurcations and Branches 3 unknown flows 1 unknown head For solution: 3 characteristic equations + continuity

68. Example