2.  Most of the problems encountered in engineering are of unsteady type.  Two main types of problems under unsteady conditions  The first is where the water level in a reservoir or pressure tank is steadily rising or falling so that rate of flow varies continuously but where change takes place slowly.

3.  The second is where the velocity in a pipeline is changed rapidly by the fast closing or opening of a value  As far as discharge under varying head is concerned, the rate of discharge will continuously vary.  Consider the figure

5.  Let V L represent the volume of liquid contained in the tank at a particular instant of time.  The inflow is Q i and outflow is Q o  The change in volume during a small instant of time dt can be expressed as

6.  If A s is the surface area of volume and dz is the change in level of the surface then  Then (a)

7.  Either Q i or Q o or both may be variable.  The outflow Q o is usually a function of z.  If a liquid is discharged through a pipe of area A under differential head z,

8.  C d is the numerical discharge coefficient and z is variable.  If the liquid flows over a weir or spillway of length L, then

9.  C is the appropriate coefficient and h is the head on the weir or spillway.  Z or h is the variable height of the liquid surface above some appropriate datum.  Q i in (a) is assumed to be either zero or constant as its variation with time makes the case much more complicated.

10.  (a) can also be written for time t, the time for water level to change from z 1 to z 2, by integrating it.  The RHS of above eq. can be integrated if Q i is constant and if A s and Q o can be expressed as function of z.

11.  In the case of natural reservoir the surface area cannot be expressed as a simple mathematical function of z but values of it may be obtained from a topographical map.  In such a case, the above eq. can be solved graphically by plotting values of against simultaneous values of z.  the Area under such a curve to some scale is the numerical value of the integral

14.  It the velocity of a liquid in a pipeline is abruptly decreased by a valve movement, the phenomenon encountered is called water hammer.  This is very important problem in case of hydroelectric plants, where the flow of water must be rapidly varied in proportion to the load changes on the turbine.  Water hammer can burst large penstocks, causing great damage to hydraulic and power generating facilities, in addition to loss of life.

15.  Even though, practically this case is impossible, but still will be used to develop concept of water hammer.  Consider a steady flow occurring in a horizontal pipe provided with a partly open valve at its end.  Consider that the valve at N is closed instantly.

16.  This results in the pressing of the liquid against the valve by the rest of the liquid column.  Walls of the pipe are also stretched due to the pressure being applied.  In a similar way, the next upstream lamina will also be brought to rest and so on.

17.  This stopping of flow and the pressure increase due to it, move upstream like wave with velocity C p.

18.  Ev is the volume modulus of elasticity  E is the modulus of elasticity of material of pipe.  D and t are the diameter and wall thickness.  c is the velocity of pressure wave given by

20.  After a short time interval, it is possible that the liquid might have come to rest in the length BN but it might still be flowing in MB.  When the pressure wave, travelling backward, reaches point M, the entire mass of water will be at rest in the length L.  At the same time, the entire mass will be under tremendous pressure.

21.  During the pressure wave travel from N to M, there will be a transient hydraulic grade line parallel to the original steady flow line XP but at a height p h /γ.  p h is the water hammer pressure.  It is impossible for pressure to exist at M, that is greater than that due to MX.

22.  So when the pressure wave arrives at M, the pressure at M drops instantly to the value it would have for zero flow.  The entire pipe now becomes under excess pressure so the liquid in it is compressed and the pipe walls are stretched.  then some liquid starts flowing back into the reservoir and a wave of pressure unloading travels from M to N.

23.  Assuming that there is no damping, at the instant this unloading wave reaches N, the entire mass of the liquid will be under normal pressure indicated by XP.  But on the other hand, the liquid is still flowing back into the reservoir.  This will result in drop in pressure at N

24.  This pressure drop will be as far below the normal steady flow pressure as the pressure an instant below was above it.  This leads to a travelling of wave back from N to M.  Ideally there would be a series of waves travelling back and forth and alternating between high and low pressures.

25.  The time for a round trip of a pressure wave from N to M and back again is given by  L is the pipe length.  So for instantaneous closure of valve, the excess pressure remains constant for this length of time.

26.  After which it is affected by the return of the unloading pressure wave.  So the pressure deficit remains constant for this length during rarefaction.  At a distance x, say B, the time for the round trip of pressure wave will be

27.  So the time duration at B for excess or deficient pressure will be the same.  At inlet M, where x=0, the excess pressure only occurs for an instant.

28. (p+dp)A

29.  A close up of the scenario created in the vicinity of the valve is shown in figure  On abrupt closure, a pressure wave travels up the pipe with velocity Cp.  In a short time interval, an element of length Cpdt is brought to rest.

30.  From Newton’s second law

31.  For the case of instantaneous valve closure, the velocity is reduced from V to zero.  So  Δp represents the increase in pressure due to valve closure.  So water hammer pressure p h = Δp

32.  The total pressure at the valve immediately after closure is p h +p where p is the pressure in the pie just upstream of the valve prior to closure.

34.  Often it is not possible to determine all the possible problems in fluid flow by theory alone.  Therefore one has to rely on experiments.  The number of tests needed by a system can be efficiently reduced by the knowledge of dimensional analysis and similitude.

35.  This also helps us in applying the test data to other studies as well.  As far as hydraulics is concerned, one can obtain fruitful results at minimum possible costs by the use of small scaled models to analyse actual hydraulic problems.  Similitude helps us to understand and predict the performance of a prototype from the test made on a small scaled model.

36.  Fields of application ◦ Aerodynamics ◦ Naval architecture ◦ Flow machinery (pumps and turbines) ◦ Hydraulic structures ◦ Rivers, estuaries ◦ Sediment transport Data is derived from the experiments performed on the model which is then used as an input to empirical models to analyse to structures.

38. Sediment Transport

39. Construction of model

41.  Similitude consists of construction of a model, then carrying out tests on the model and transfer of the results of the tests to the prototype.

42.  Geometric similarity  Kinematic similarity  Dynamic similarity

43.  Most important feature of similitude analysis.  Means that the model and the prototype have identical shapes but differ only in size.  Important thing to be kept in mind is that the flow patterns must be geometrically similar.  If ‘p’ and ‘m’ denote the prototype and model then

44.  The length scale ratio is given by  Or the ratio of the linear dimension of the prototype to that of the model.  areas vary as and volumes as.

45.  Attaining 100% geometric similarity is perhaps impossible.  For example, we may not be able to reduce the surface roughness of a small model in proportion unless we can make its surface very much smooth than that of the prototype.

46.  The same can be said for sediment transport studies.  It might not be possible to scale down the bed material to such an extent that it becomes fine and cohesive.  A cohesive material cannot exactly simulate the behaviour of sand particles.

47.  Also, for river modelling, if we select a small scale to represent the length of the river because of space limitations.  Then the same scale cannot be used for representing the vertical sides as this will lead to a very shallow flow and the flow becomes laminar.  So a distorted model is used in which the vertical scale is larger than the horizontal scale.

48.  So for this case, if the horizontal scale ratio is The vertical scale ratio becomes And the area ratio as

49.  Kinematics similarity implies that, in addition to geometric similarity, the ratio of the velocities at all corresponding points in the flows are the same.  Given by

50.  Also time scale ratio is represented by  Acceleration scale ratio is given by

51.  Two system have dynamic similarity if, in addition to kinematic similarity, corresponding forces are in the same ratio in both of them.  The force scale ratio is

52.  Forces that may act on a fluid element include ◦ Force due to gravity F G ◦ Pressure F P ◦ Viscosity F V ◦ Elasticity F E ◦ Surface tension F T (in case the fluid is at liquid –gas interface)

53.  If the summation of forces on a fluid element does not add up to zero, then it must accelerate in accordance with Newton’s law.  Such an unbalanced system can be transformed into a balanced system with the addition of inertial force.  So

54. Resultant -Resultant Thus

55.  All these terms can also be explained as  Gravity  Pressure  Viscosity

56.  Elasticity  Surface Tension  Inertia  For most of the cases of flow in Hydraulics, some of these forces might not exist or be insignificant.

57.  If we consider two geometrically similar flow systems, both of which possess kinematic similarity.  The forces acting on any fluid element are and  The dynamic similarity will then be achieved if

58.  where subscripts p and m refer to a prototype and model as before.  With four forces, there are three independent expressions, for three forces, there will be two expressions and so on.

59.  For the case of a fluid flowing full through a pipe, gravity does not effect the flow pattern.  Because of non existence of free liquid surfaces, capillarity is of no practical importance.  Therefore the significant forces are that of inertia and fluid friction due to viscosity.

60.  The ratio of inertial forces to viscous forces is known as Reynolds number (Osborne Reynolds 1882) which was later modified by Lord Rayleigh.  The ratio is given by

61.  R is actually a dimensionless number.  The linear dimension L may be any significant length in the flow pattern.  For a fully filled pipe, it may either be the radius or diameter.  If diameter is considered to be the linear dimension,

62.  If two system, such as a model and its prototype or two pipeline with different fluid, are to be dynamically equivalent so far as inertia and viscous forces are concerned, they must both have the same of R.  Thus, for such cases, we will have dynamic similarity when

63.  So for a same fluid in model and prototype, for the above equation to be true, we should have a model of small dimensions with high velocity.

64.  Question: If the Reynolds number in a model and prototype are the same, find expression for the scale ratios and

66.  The ratio of inertial forces to gravity forces is known as Froude’s number (William Froude).  He experimented with flat plates towed lengthwise through water in order to estimate the resistance of ships due to wave action.

67.  Systems involving gravity and inertia include ◦ the wave action set up by a ship ◦ the flow of water in open channels ◦ force of a stream on a bridge pier, ◦ the flow over a spillway, ◦ the flow of a jet from an orifice, and other cases where gravity is the dominant factor.

68.  Length L is a linear dimension that is significant in the flow pattern.  For a ship, it is taken as the length of the water line.  For an open channel it is taken as the depth of flow.

69.  Dynamic similarity is achieved when

70.  From the equation of Froude number, V varies as and if g is considered to be the same in prototype and model then (for same F and g)

71.  For time ratio, the ratio of time for prototype to model is (for same F and g)

72. Since the velocity varies as and cross sectional area as then the discharge ratio (for same F and g)

73.  For the cases of fluid flow where compressibility is important, the ratio of inertia to elastic forces becomes important.  Mach number is the square root of this ratio. Thus

74.  where c is the sonic velocity (or celerity) in the medium in question.  It can also be said that Mach number is the ratio of fluid velocity (or the velocity of a body through stationary fluid) to that of sound wave in the same medium.  If M is less than 1, the flow is subsonic, 1 for sonic and greater than 1 for supersonic.

75.  For extremely high values of M, the flow is hypersonic.

76.  In few cases, surface tension is also important but normally it is negligible  The ratio of inertial forces to surface tension is the square root of which is known as Weber number.

77.  Ratio of inertial forces to the pressure forces is known as Euler number.

78.  A submerged body is to move horizontally through oil (γ= 52 lbft 3, μ= 0.0006 lb.sec/ft 2 ) at a velocity of 45 fps. To study the characteristics of this motion, an enlarged model of the body is tested in 60 0 F water. The model ratio λ is 8:1. Determine the velocity at which this enlarged model should travel through the water to achieve dynamic similarity. If the drag force on the model is 0.80 lb, predict the drag force on the prototype

79.  The body is submerged; so there is no wave action. Reynolds, criterion must be satisfied Table A.1 for water at 60 0 F: ν m = 12.17 ×10 -6 ft 2 /sec. ft 2 /sec

81. (lb)