HYDRAULICS (HYDRODYNAMICS) SPRING 005 Part. Fluid-Flow Principles. Introduction. Definitions. Notation and fluid properties.3 Hydrostatics.4 Fluid dynamics.5 Control volumes.6 Visualising fluid flow.7 Real and ideal fluids.8 Laminar and turbulent flow. Continuity (mass conservation). Flow rate. The steady continuity equation.3 Unsteady continuity equation 3. The Equation of Motion 3. Forms of the equation of motion 3. Fluid acceleration 3.3 Bernoulli s equation 3.4 Application to flow measurement 3.5 Other applications (flow through an orifice; tank-emptying) 4. The Momentum Principle 4. Steady-flow momentum principle 4. Applications (pipe contractions; pipe bends; jets) 5. Energy 5. Derivation of Bernoulli s equation from an energy principle 5. Fluid head 5.3 Departures from ideal flow (discharge coefficients; loss coefficients; momentum & energy coefficients) Part. Applications (Separate Notes). Hydraulic Jump. Pipe Flow (Dr Lane-Serff) Recommended Reading Hamill, 00, Understanding Hydraulics, nd Edition, Palgrave, ISBN 0-333-77906- Chadwick, Morfett and Borthwick, 004, Hydraulics in Civil and Environmental Engineering, 4 th Edition, Spon Press, ISBN 0-45-30609-4 Massey, 998, Mechanics of Fluids, 7th Edition, (Revised by Ward-Smith, J.), Stanley Thornes, ISBN 0-748-74043-0 White, 003, Fluid Mechanics, 5th Edition, McGraw-Hill, ISBN 0-07-407- Hydraulics David Apsley
. Introduction and Basic Principles. Definitions A fluid is a body of matter that can flow; i.e. continues to deform under a shearing force. Fluids may be liquids (definite volume; free surface) or gases (expand to fill any container). Fluids obey the usual laws of Newtonian mechanics, but as a continuum. Unlike rigid bodies, fluid particles may move relative to each other, interacting via internal forces. These are usually expressed in terms of stresses (stress = force / area), the principal ones being: pressure, p normal stress: pushing or pressing; shear stress, tangential stress: frictional drag; opposing relative motion. An ideal fluid has no viscosity. This is never exactly true, but is often a useful approximation. Fluids may be regarded as incompressible (density or volume not changed by the flow) at speeds much less than that of sound. This is usually the case in hydraulics. Fluid flow may be described as laminar (adjacent layers slide smoothly past each other) or turbulent (irregular, with constant intermingling of adjacent layers).. Notation and Fluid Properties The main flow variables are: p pressure (force/area) u velocity These are field variables because they are functions of position and time t; e.g. u = u(x,t) Vector quantities like position and velocity are often decomposed into components: x ( x, y, z) u ( u, v, w) U (or occasionally V) will be used for the magnitude of velocity. Important fluid properties are: density = mass / volume dynamic viscosity; defined by Newton s viscosity law: du = dy = / is called the kinematic viscosity surface tension = force / length (or surface energy/area) F = l K bulk modulus = pressure change / volumetric strain V p = K( ) V Hydraulics David Apsley
Most gases at normal temperatures and pressures satisfy the ideal gas law, which in fluid mechanics is usually written as p = RT The gas constant R depends on the particular gas (R = R * /m = universal gas constant / molar mass). It has a value of 87 J kg K for dry air..3 Hydrostatics Hydrostatics concerns the balance of forces in a fluid at rest. The principal forces are: pressure; p is the normal force per unit area weight; g is the specific weight (weight per unit volume) surface tension; is the force per unit length In the interior of a stationary fluid, pressure forces balance weight. As a result the change in pressure with depth is given by the hydrostatic equation p = g z () Pressure increases as height decreases and vice versa. As a result there is a net buoyancy force on an immersed body equal to the weight of fluid displaced (Archimedes Principle); i.e. the immersed weight of a body is less than its actual weight. p If density is constant then p + gz = constant () The quantity on the LHS is called the piezometric pressure. Hydrostatic principles are used to measure: Pressure in a flowing liquid, using a piezometer tube. The pressure (relative to atmospheric pressure) is given by p = gh (3) Pressure differences using a U-tube manometer. The difference in pressure between points A and B is proportional to the difference in heights of the manometer fluid in the two arms: p = ( m )gh (4) where m is the density of the manometer fluid and the density of the working fluid.(the latter can be ignored if the working fluid is a gas). ρ h p > A p B h ρ m Note that, in incompressible flow, absolute pressure is unimportant: pressure differences determine the net force. In particular, it is usual to work with the gauge pressure i.e. difference between actual and atmospheric pressures. The pressure at a free surface is then p = 0, whilst a sub-atmospheric (suction or vacuum) pressure has p < 0. Hydraulics 3 David Apsley
Surface tension is vital to survival if you are a water-hopping insect but it can usually be ignored in large-scale hydraulics. The capillary rise in a thin-bore tube is given by h = cos (5) gd where d is the tube diameter and is the contact angle ( 0 for water on glass). θ h.4 Fluid Dynamics Fluid dynamics concerns fluids in motion. Almost all hydraulic problems can be addressed by the application of one or more of three equations/physical principles: continuity (mass conservation); momentum; energy. Continuity is almost invariably required and is usually applied first to reduce the number of unknowns. Continuity (Mass Conservation) Mass is neither created nor destroyed. For a steady flow, mass flow rate in = mass flow rate out or, if the flow is also incompressible, volume flow rate in = volume flow rate out Any mathematical expression of this is called the continuity equation. Q Q Q 3 Q in = Q out Q = Q + Q 3 Hydraulics 4 David Apsley
Momentum Based on Newton s Laws of Motion; specifically: Second Law: force = rate of change of momentum. Third Law (Action and reaction): force exerted by a fluid on its containment is equal and opposite to the force exerted by the containment on the fluid. force on FLUID force on FLUID force on BODY wake force on PIPE Energy Energy principle: change of energy = work done + heat input For incompressible fluids we need only consider the mechanical energy principle: change of kinetic energy = work done For ideal fluids (no losses) this is expressed as Bernoulli s equation p + gz + U = constant (along a streamline) In particular: velocity increases pressure decreases small loss of head LIFT Bernoulli s equation is strictly for ideal fluids (no friction or other fluid energy changes), but it can be amended to account for: energy put in (by pumps) or taken out (by turbines); frictional losses (e.g. pipe flow). Because all forms of fluid energy can be converted to an equivalent gravitational potential energy it is common in hydraulics to measure energy in height units or fluid head. Hydraulics 5 David Apsley
.5 Control Volumes Although it is possible to write the equations of fluid mechanics in terms of differential equations (describing what is happening at every point), engineers are more often concerned with the behaviour of fluid in bulk and the usual method of analysis is to apply mass and momentum principles, etc., to all the fluid in a control volume. For example, consider suitable control volumes for the two pipe fittings shown. F Mass conservation: Net mass flux out of control volume = (mass flow rate) B (mass flow rate) A = 0 Momentum principle: Net force on fluid in control volume = rate of change of momentum = (mass flow rate) change in velocity.6 Visualising Fluid Flow Velocity vectors are arrows of length proportional to magnitude. Streamlines are everywhere tangential to the instantaneous flow direction. The flow is fastest where the streamlines are closest together (why?). Streamlines can never cross (else the velocity would be -valued) and cannot terminate in the interior of the flow. In steady flow one streamline always coincides with a solid boundary or free surface. Hydraulics 6 David Apsley
Path lines or particle paths are the lines followed by individual elements of fluid. (They are what you would see on a long-exposure photograph). Streaklines are swept out by all the fluid elements which have passed through a particular point; e.g. dye injection. In steady (i.e. not time-varying) flow streamlines, streaklines and particle paths are identical. Stream tubes are bundles of streamlines ( virtual pipes ). By definition, there is no flow across the sides and the same flow must enter at one end as leaves at the other..7 Real and Ideal Fluids Ideal fluids have no viscosity there is no internal friction or loss of mechanical energy. No such fluid exists, but many flows can be approximated as ideal if viscous forces are small and do not cause major flow phenomena such as boundary-layer separation. Real fluids have non-zero viscosity. This has two important consequences: () They satisfy the no-slip condition at solid boundaries. i.e. the (relative) velocity at the boundary is zero. ideal real () There are frictional forces between adjacent layers of fluid moving at different speeds and between the fluid and a boundary. shear stress of UPPER fluid on LOWER shear stress of LOWER fluid on UPPER For Newtonian fluids (which includes most fluids of interest), the stress (force per unit area) of the upper fluid on the lower fluid is given by Newton s viscosity law: du = (6) dy An equal and opposite force is exerted by the lower fluid on the upper fluid. Hydraulics 7 David Apsley
.8 Laminar and Turbulent Flow At low flow speeds viscosity ensures that adjacent layers of fluid slide smoothly over one another in a steady fashion without intermingling. This flow regime is called laminar. At higher speeds viscosity is insufficient to smooth out minor perturbations to the flow, which grow rapidly to create unsteadiness and eddying motions. This flow regime is called turbulent. The transition between laminar and turbulent flow can be observed in the rising plume from a cigarette. There is an initial laminar flow which wavers and then breaks up into turbulent eddies as the flow accelerates. turbulent laminar The pioneering experiments determining the occurrence of laminar or turbulent flow in pipes were first performed by Osborne Reynolds (Professor of Engineering at Owens College soon to become the University of Manchester). He demonstrated that which regime occurred depended on a dimensionless parameter later to become known as the Reynolds number: UL UL Re = or (7) where U and L are typical velocity and length scales of the flow, is the density, is the dynamic viscosity and (= / ) the kinematic viscosity. The Reynolds number is probably the single most important parameter in fluid mechanics! laminar turbulent The Reynolds number can be regarded as a measure of the ratio of the total force (= mass acceleration) to the viscous force. For a block of fluid, volume L 3, using order-ofmagnitude estimates: 3 U mass acceleration ( L ) = U L L / U whilst U viscous force A L = UL L Hence, mass acceleration U L = UL = Re viscous force UL When the Reynolds number is large the effects of viscosity are small and conversely. In general, Hydraulics 8 David Apsley
low Re high Re laminar flow turbulent flow Note that low and high is all relative and depends on the particular flow and the choice of velocity and length scales (which should be stated). For pipes it is conventional to take U as average velocity and L as the diameter; then: Re < 000 (laminar flow) Re > 4000 (turbulent flow) (The commonly-accepted critical Reynolds number for transition in a round pipe is 300). These particular numerical values are for pipe flow only, with a particular choice of velocity and length scales. For example, choosing radius rather than diameter as a length scale would immediately halve the Reynolds number, but wouldn t affect the flow regime. The numerical size of viscosity for the common fluids, air and water ( air.5 0 5 m s ; water.0 0 6 m s ), means that most civil-engineering flows are fully turbulent. Even when the flow is turbulent it usually consists of relatively small fluctuations about a mean flow, which is what we are actually interested in. In turbulent flow, however, most of the net transfer of momentum and energy between layers of fluid occurs by the eddying motions mingling fluid elements and so, as far as the mean flow is concerned, the mean stress is no longer given by Newton s viscosity law ( = du/dy) but is substantially increased by the net effect of turbulent eddies. This mixing of fluid streams leads, for example, to a turbulent velocity profile in a pipe being much more uniform than a corresponding laminar velocity profile. Hydraulics 9 David Apsley
. Continuity (Mass Conservation) In fluid mechanics any mathematical expression of the conservation of mass is called the continuity equation.. Flow Rate Consider fluid passing through an area at uniform velocity u. ρ A u In time t the fluid moves a distance u t, and hence the volume of fluid that has passed through that area is A u t u δt A The volumetric flow rate Q (aka volume flux, quantity of flow or discharge ) is given by volume A u t Q = = time t i.e. Q = ua (8) It may be measured in m 3 s or L s ( L = 0 3 m 3 s ). The mass flow rate (or mass flux) through a section, m, is given by mass density volume m = = time time i.e. m = Q = ua (9) It may be measured in kg s. Example. A pipe of internal diameter 00 mm carries water (density 000 kg m 3 ) at average velocity.5 m s. Find the volumetric flow rate and mass flow rate. Note. In general the flow may be at any angle to the area A. In general the U which appears in the formula for Q or m should be the component normal to the area. Hydraulics 0 David Apsley
. The Steady Continuity Equation The continuity equation is invariably applied when either: there is a change of cross section; there is a junction. Change of Cross Section Consider a stream tube with a change of cross section from A to A. ρ ρ u u A A If the flow is steady (i.e. doesn t change with time) then no mass can accumulate between the two cross sections and hence ( mass flux) = ( mass flux) (0) u A = u A In hydraulics the fluid can usually be treated as incompressible (density constant along a streamline) and hence the volume flow rate is constant: Q = u A = u () A Junctions At a junction of more than two pipes, we require that total flow into junction = total flow out of junction e.g. Q Q Q = Q + Q 3 Q 3 Hydraulics David Apsley
Example. A circular pipe carrying water undergoes a smooth contraction from internal diameter 00 mm to one of diameter 80 mm. If the velocity in the larger cross section is 0.6 m s find: (a) the velocity in the smaller-diameter pipe; (b) the volume and mass flow rates. Example. A manifold splits the air supply (unequally) between two components of an engine. The inlet duct has square section with side 50 mm. The outlet ducts have circular cross-section with diameters 80 mm and 0 mm. The air velocity in the intake duct is m s and that in the smaller outlet duct is 4 m s. Find the volumetric flow rate and velocity in the larger outlet duct. 0 mm 50 mm m/s 4 m/s 80 mm Example. Liquid of controlled density can be supplied by injecting saturated brine (s.g..0) into a freshwater stream. If brine is injected at L s into a pipe of internal diameter 00 mm carrying fresh water at 3 L s find, in the well-mixed region downstream: (a) the total quantity of flow; (b) the mass flow rate; (c) the average velocity; (d) the density of fluid. Hydraulics David Apsley
.3 Unsteady Continuity Equation Mass conservation applied to the fluid in a control volume which is either: (a) moving; or (b) changing density; gives d ( mass) = ( mass flux) in ( mass flux) out () dt In many cases the fluid is incompressible and continuity can equally well be applied to volume rather than mass: dv = Q in Q out (3) dt i.e. rate of change of volume = flow rate in flow rate out Example. A cylinder of diameter 0. m contains water and has intake and exit pipes both with cross sections of 4 0 3 m and fitted with non-return valves. A piston oscillates sinusoidally up and down in the cylinder with amplitude 80 mm and frequency Hz. Find the maximum velocity of water in the exit pipe and the volume discharged over a cycle. Hydraulics 3 David Apsley
3. The Equation of Motion 3. Forms of the Equation of Motion The equation of motion is any mathematical expression of Newton s Second Law. As you have seen in the Mechanics module, this can be written down in various ways; e.g. F = ma or, equivalently, as a momentum principle: force = rate of change of momentum or as a mechanical energy principle: work done = change of kinetic energy In fluid mechanics, also, the equation of motion can be expressed in different ways. In Section 3 we examine the first and third of the forms above, leading to an important result known as Bernoulli s equation. In Section 4 we apply the momentum principle using control volumes in order to deduce forces on structures. 3. Fluid Acceleration In order to apply Newton s Second Law (F = ma) to fluid motion one must know the fluid acceleration. Suppose the position of a particle at time t is given by x(t). Its velocity u is given by dx u = dt du Its acceleration is defined as a =, but you will know from Mechanics that this can also be dt written in terms of displacement x as du du dx a = = (by the chain rule) dt dx dt i.e. du a = u dx Consider now the fluid flow through a converging section as shown. u u u 3 a If the flow is steady then the velocity at any particular position does not change with time; i.e. u / t = 0. However the fluid is accelerating because as you move with any particular fluid element along the centreline you are moving with faster velocity. The fluid acceleration i.e. Hydraulics 4 David Apsley
the acceleration of a particular element of fluid is given in this instance by u u / x. In general, the acceleration of a fluid element in a flow u(x,t) is the sum of two parts: u (temporal derivative) because of changes with time at a point; t u u (advective derivative) because of changes as you move with the flow. x The latter usually dominates. When the velocity is also a function of the other space components y and z the fluid acceleration can be written as the total (or material, or substantive ) derivative Du u u u u a + u + v + w (4) Dt t x y z Fortunately, many flows that you will deal with are steady and quasi--dimensional, and this can often be simplified to u a = u (5) x Example. An air supply system consists of ducts with rectangular cross section. The upstream part of the ducting has a width of 0.4 m and height 0.5 m, whilst the downstream part has the same width (0.4 m) but a height of 0. m. These two sections are connected by a simple transition duct where the height varies linearly over a distance of m. 0.4 m 0.5 m 0. m x (a) (b) If the volumetric flow rate along the duct is 0.6 m 3 s, find an expression for the speed of the flow as a function of the distance from the start of the contraction, x. Find an expression for the acceleration of the air as a function of distance along the contraction. Assume that the flow is quasi-d. (Answer: with x in m, u in m s, a in m s, (a) 3.7 u = ; (b) a = ) 3 0.3x ( 0.3x) What force provides this acceleration? Hydraulics 5 David Apsley
3.3 Bernoulli s Equation A (better) derivation of Bernoulli s equation from general energy considerations will be given later. For now we consider the forces on a short fluid element, length s, uniform cross-section A and aligned locally with the flow. δ s θ (p+ δp)a We assume that the flow is inviscid (no friction), steady (no time derivative) and incompressible ( constant along a streamline). Resolving forces in the mg direction of flow: pa ( p + p) A mg sin = ma Writing: du a = U for the acceleration (U is the magnitude of velocity, s is the distance) ds m = ( A s) for the mass sin = z s gives pa ga z = du A su ds d and, on dividing by the volume (A s) and noting that U U = ( U ) : ds ds p z d g = ( U ) s s ds Since the flow is incompressible, is constant along the streamline and can be taken inside the differential. In the limit as s 0: i.e. d ( p + gz + U ) = 0 ds p + gz + U = constant Remember the qualifiers: inviscid (no friction) steady incompressible along a streamline pa (6) Some useful things to remember when applying Bernoulli s theorem are: use continuity first to reduce the number of unknowns; at a free surface the pressure is atmospheric: p = 0; for a free discharge to atmosphere the pressure is atmospheric: p = 0; for a large tank/reservoir velocity at the surface is unaffected by the discharge: U = 0. Hydraulics 6 David Apsley
Example (Exam, May 003 *** modified ***) Air (density. kg m 3 ) is flowing through a smooth contraction from a circular pipe of diameter 0 mm to a tube of diameter 60 mm as shown in the figure. There are tappings at four locations along the contraction, where the diameter is 0 mm, 00 mm, 80 mm and 60 mm (labelled A, B, C and D respectively). (a) (b) (c) If the air flow along the duct is 0. m 3 s, find the speed of the flow at each location A, B, C and D. Assuming no energy losses, calculate the pressure difference between locations A and B, and the pressure difference between locations C and D. A water manometer is connected across tappings A and B, and another manometer is connected across C and D (see the Figure). What are the water-level differences ( z and z ) in the two manometers. A B C D 0 mm 00 mm 80 mm 60 mm z z Hydraulics 7 David Apsley
Example. Water is being siphoned as shown. The internal diameter of the pipe is 0 mm and that of the nozzle is 5 mm. Assuming no losses find: (a) the volumetric flow rate; (b) the pressure at the top..5 m.5 m Example (Exam, June 004) A pipe carrying water with a flow rate of 0.04 m 3 s contracts from a diameter of 00 mm to 60 mm over a distance of 0.5 m, with the diameter changing linearly with distance. (a) (b) (c) What is the flow speed as a function of distance along the pipe? What is the acceleration of the flow as a function of distance along the pipe? If the pressure where the diameter is 60 mm is denoted by p 0, what is the pressure as a function of distance along the pipe, assuming no energy loss? (Answer: With x in m, u in m s, Q in m 3 s, p in Pa, 6 409.6 6 (a) u = ; (b) Q = ; (c) p = p 5 0 + ( ) ( ) ) 4 4 ( 0.8x) ( 0.8x) 0.6 ( 0.8x) Example. A river may be approximated by a rectangular channel of width 0 m. The depth of water is.4 m. The abutments of a bridge reduce the channel width to 7 m and the depth of water to. m. Find the upstream velocity and quantity of flow. Explain, physically, why the depth of water decreases between the abutments. Hydraulics 8 David Apsley
3.4 Application to Flow Measurement Some of the most important applications of Bernoulli s equation are to flow measurement. The main reason is that velocity changes can be related to pressure changes and pressure is usually cheap and easy to measure with a manometer or piezometer tube. Ideal-flow theory is used to compute a theoretical discharge Q. This can then be corrected for friction, turbulence, etc. by applying a discharge coefficient c D (see Section 5.3). 3.4. Pitot and Pitot-Static Tubes A Pitot tube is used for measuring velocity in the flow. It works by converting the kinetic energy into pressure energy. Flow is brought to rest at the stagnation point at the front. By Bernoulli s theorem this raises the pressure to the stagnation or Pitot pressure P 0 = P + U (7) where P and U are the pressure and velocity in the undisturbed flow. If one also measures the static pressure P then the difference P0 P is equal to U, from which the velocity can be deduced. u piezometer p ρ g Pitot tube p0 ρ g In a pitot-static tube both pitot and static pressures are measured with the same instrument. The pressure difference can be measured with a manometer: P P = ( m ) gh. 0 to manometer p 0 p h u p 0 p 0 P P = U = ( m ) gh Example. The velocity of water in a conduit is to be measured using a piezometer and Pitot tube. If the water level in the piezometer rises to 0.3 m and that in the Pitot tube to 0.5 m find the excess pressure and the flow velocity at the measuring point. How do you expect the height in the pitot tube to change as the measuring point is traversed across the conduit? Hydraulics 9 David Apsley
Example. The speed of air is measured with a pitot-static tube connected to a U-tube manometer containing alcohol (s.g. 0.79). (a) Find the air velocity if the difference in fluid levels in the two arms of the manometer is 50 mm. (b) What difference in levels would be expected for a flow speed of 5 m s? How could the accuracy of the reading be enhanced for low flow speeds? 3.4. Venturi Meter A venturi is a smooth contraction in a pipe. The idea is that the contraction causes an increase in velocity and hence a decrease in pressure, which can be measured. Continuity and Bernoulli give two equations for two unknowns (U and U ) and hence either can be found and multiplied by the corresponding area to give the volumetric flow rate Q. p p u u A A Apply Bernoulli between sections and : p + U = p + U U U = ( p p ) / (8) Apply continuity between sections and : U A = U A A U = U (9) A Substitution in (8) yields an expression for U, which, when multiplied by A, gives the volumetric flow rate (exercise: complete the analysis): p / / Q = A = constant p ( A / A ) Notes. This is the ideal flow rate. To account for non-ideal behaviour it can be multiplied by a discharge coefficient c D, so that Q actual = cdqideal However, for a well-defined venturi meter, c D is 0.98 or more, so is often neglected. Hydraulics 0 David Apsley
. The downstream expansion plays no role in the analysis, but should be at a small angle so as not to provoke flow separation and cause energy loss. 3. The same principle is used in an orifice meter. This is a sharp contraction in a pipe. The downstream pressure at the walls is assumed to be the same as at the orifice, due to the recirculating flow region downstream. A A p p An orifice meter is much cheaper to manufacture and install but causes substantial energy loss. If the Bernoulli analysis is used to predict an ideal flow rate then a discharge coefficient with a value of the order 0.6 is needed, mainly due to the continued convergence of streamlines after passing through the orifice. 3.4.3 Sharp-Crested Weir Sharp-crested weirs are slotted plates used to provide accurate measurements of quantity of flow in small open channels for example, in hydraulics laboratories. (Broad-crested weirs are used for larger channels and work by a different principle see open-channel flow in Hydraulics 3). crest, or sill nappe Q H The quantity of flow Q is related to the height over the weir, H. For a given rectangular weir, Bernoulli s theorem predicts 3 / Q H Key to operation is that air at atmospheric pressure reaches the underside of the weir (so that the pressure across the nappe is approximately atmospheric, p = 0. the weir has a sharp crest (so that the flow breaks away on the downstream side); the backwater level does not approach the crest of the weir; the notch does not span the full width of the channel (contracted weir) or, if it is full width, then a vent pipe is used to ensure atmospheric pressure underneath the nappe. air Hydraulics David Apsley
Idealised Analysis Velocity is not uniform over the weir but is a function of height. Let h be the distance from the free surface. Apply Bernoulli between points upstream and over the crest of the weir: p + gz + U = p + gz + U 0 gh + g( H h ) + U = 0 + g( H h) + U = gh + U As a first approximation it is usually assumed that the approach flow U is small; (the notch of the weir is considerably smaller than the cross-section of the channel). Then / U = ( gh), 0 < h < H 0 H 0 U p=0 h h p hydrostatic p=0 H Since the velocity of flow varies with height the total quantity of flow must be obtained by summation over small elements of area A = b h, where b is the width of the weir: Q = U A = (gh) / b h b(g) = 3 / b(g) This is usually multiplied by a discharge coefficient c D ( 0.6) to account for non-ideal flow: / 3 / Q = c b( g) H (0) D 3 Most important, however, is the power-law dependence, Q H 3/ ; the constant of proportionality can be obtained by calibration. / H 0 H h / 3 / dh Notes. For accurate work the upstream velocity U (= Q/A) can be included in the formula for U and the integration leads to an implicit equation for Q which can be solved iteratively; (see Hamill for details and an example).. Another common device you can see one in the second-year centrifugal-pump experiment is the V-notch weir. This has the advantage of large sensitivity for small flow rates. Q H The velocity U is still given by (gh) /, but since the width of flow at any level, b h, integration for the quantity of flow produces a different power-law relationship, 5 / Q H ; (exercise: prove it!). Hydraulics David Apsley
3.5 Other Applications 3.5. Discharge Through an Orifice Consider water discharging from a small hole at depth h below the surface of a large tank. Apply Bernoulli s theorem between free surface () and exit point (): ( p + gz + U ) = ( p + gz + U ) At the free surface, p = 0 (free surface) and U = 0 (large reservoir). At the exit point, p = 0 (free discharge). For convenience, take all heights relative to the orifice. Then + gh + 0 = 0 + 0 + U 0 exit h u exit Hence U exit = gh () Notes. This result is called Torricelli s formula and it is basically a statement of energy conservation (potential energy kinetic energy).. If the orifice is not small then the depth below the free surface and hence the exit velocity will vary. The quantity of flow must then be obtained by integration. 3. The exit velocity can be multiplied by the exit area to compute discharge. However, in practice, this ideal volumetric flow rate must be multiplied by a discharge coefficient (Section 5.3) to account for losses and flow convergence after exit. Example. Water flows from a 3-cm-diameter hole m below the surface of a large tank. (a) Assuming no losses calculate the exit velocity and volumetric flow rate. (b) Using the results of (a) determine the subsequent path of the jet. (c) If a discharge coefficient c D = 0.6 is applied, calculate the discharge. Hydraulics 3 David Apsley
3.5. Tank Emptying By equating the rate of change of fluid volume in a tank to the volumetric flow rate, dv = Q out dt the time to empty the tank can be found. () Using Torricelli s formula for the exit velocity, and allowing for a discharge coefficient c D the RHS of () can be expanded to give dv = cd Aout gh (3) dt The volume of fluid V is a function of the depth of water h, so this becomes a differential equation in h. Note that one assumption of Bernoulli s theorem that of steady-state conditions is strictly not met here! Example. A tank has a rectangular base with sides m by 3 m and initially contains water to a depth of 0.5 m. When the plug is pulled out water exits from the base through a hole of area 0 3 m. Assuming no losses, estimate the time to empty the tank. Hydraulics 4 David Apsley
4. The Momentum Principle Consideration of momentum is necessary whenever one is considering forces. Two mechanical principles are involved. MOMENTUM PRINCIPLE (Newton s nd Law): Force = Rate of change of momentum ACTION/REACTION (Newton s 3 rd Law): The force exerted by a fluid on its containment is equal and opposite to that exerted by its containment on the fluid. These principles are applied to the whole body of fluid in a control volume. Only external forces need be considered (since internal forces cancel in action/reaction pairs). 4. Steady-Flow Momentum Principle Consider a control volume consisting of a (thin) stream tube carrying mass flow rate m. Since every element of fluid has its velocity changed from u to u the net rate of change of momentum in this control volume is m u ). Hence, ( u F = m ( u u) (4) Note:. F is the sum of all external forces on the control volume, including reaction from solid boundaries fluid forces from adjacent fluid (pressure, viscous forces etc.). Force, velocity and momentum are all vector quantities and each direction must be considered. u F u (*) does not generalise easily to non-uniform flow (where velocity varies over a crosssection). Since mass flux is constant along the stream tube (continuity) a better way of writing it is F = ( m u) ( m u) or force = (momentum flux) out (momentum flux) in where momentum flux = mass flux velocity = ( ua)u Can you see how this would generalise to non-uniform inflow or outflow velocity profiles and to time-dependent flow? (These will be addressed in Hydraulics ). Hydraulics 5 David Apsley
4. Applications (i) (ii) (iii) (iv) Pipe contraction Forces on pipe bends Jets and nozzles Hydraulic jump (considered as a special application later). The basic technique is to apply the steady-state momentum principle: F = Q ( u u) to the fluid in a control volume. Technique. Choose a suitable control volume; (usually coincides with boundaries, so that there is only one inflow and outflow; cuts fluid flow where it is simple uniform, if possible).. Mark in all the forces on the fluid (equal and opposite to that on the boundary; usually use gauge pressures so that atmospheric pressure is zero). 3. Mark inflow and outflow velocities (and their directions) and use continuity to relate them if possible. 4. Apply the momentum principle in each relevant direction. 4.. Pipe Contraction Example. For the pipe-reducing section shown, the greater and smaller diameters are 00 mm and 60 mm, respectively and the water discharges to the atmosphere at section. When the volumetric flow rate is 5 L s the differential height in the mercury manometer is h = 700 mm. Calculate the total force resisted by the flange bolts. (Mercury has specific gravity 3.6). water h mercury Hydraulics 6 David Apsley
4.. Forces on Pipe Bends Example (Exam, June 004) Water is flowing at a rate of 0.00 m 3 s through a reducing bend, with the diameter contracting from 0 mm to 5 mm and the flow direction changing by 45 as shown in the figure. The bend is in the horizontal plane and the flow exits to the atmosphere at the narrow end. You may ignore energy losses. (a) (b) What is the pressure (relative to atmospheric pressure) in the 0 mm pipe? What is the force on the pipe bend (expressed in the x-y coordinate system shown)? y x 45 o Example (Exam, May 003 *** slightly modified ***) Water is flowing through a horizontal pipe of diameter 5 mm with a flow rate of 0.003 m 3 s -. At the end of the pipe there is a T-junction (see Figure) with both arms horizontal and each of diameter 0 mm. The water exits the T-junction directly into the atmosphere. (a) (b) (c) Calculate the velocity of the water in the 5 mm pipe (u ) and in the two branches of the T-junction (u ). If the head-loss coefficient for the T-junction (based on the inflow speed u ) is K =., calculate the pressure in the pipe just upstream of the T-junction (ignoring any other energy losses). What is the force exerted by the water on the T-junction? u u u Hydraulics 7 David Apsley
4..3 Jets The most important aspect of these calculations is that the jet is free and the pressure throughout is atmospheric. Example. A jet of diameter 80 mm and carrying a flow of 50 L s impacts normally upon a plate. Calculate the force on the plate. Example. (from White, 003) A liquid jet of velocity V and diameter D strikes a fixed hollow cone and deflects back as a conical sheet at the same speed. Find the cone angle for which 3 the restraining force F = AV, where A is the cross-sectional area of the jet. V V θ Hydraulics 8 David Apsley
5. Energy Principle of Conservation of Energy ( st Law of Thermodynamics): Change of energy = work done + heat input This is an important equation for mechanical engineers. For incompressible fluids, however, we do not need the thermal elements of this equation and can work with the much simpler Mechanical Energy Principle: Change of kinetic + potential energy = work done by non-conservative forces You will remember, from Mechanics, that: work done = force displacement rate of doing work (i.e. power) = force velocity Remember that displacement and velocity both refer to components in the direction of the force. 5. Derivation of Bernoulli s Equation From an Energy Principle Consider steady, frictionless flow through a length of a thin stream tube. Each mass m of fluid has kinetic energy mu and potential energy mgz. The rate at which energy passes section or is, therefore, given by m ( U + gz ), and hence the rate at which (kinetic + potential) energy is created in this section is given by m ( ) U + gz) m ( U + gz where m = Q is the mass flow rate. m is the same at both sections since, by definition, there is no flow through the long sides of the tube. In the absence of viscosity, work is done only by pressure forces acting on the two ends. Rate of working = force velocity = p AU p AU (No work is done on the long sides of the streamtube because the pressure force is perpendicular to velocity there). Hence, applying rate of working = rate of change of (kinetic + potential) energy p A U p A U = m ( U + gz) m ( U + gz ) m ( = U A = U A, Dividing by ) p p = ( U + gz) ( U + gz) u u Hydraulics 9 David Apsley
i.e. p p ( + U + gz) = ( + U + gz ) For incompressible fluids, is also constant along a streamline and hence p + U + gz = constant ( along a streamline ) (5) Notes () Reminder of the qualifiers: inviscid (we shall examine how to include losses shortly) steady (for the unsteady Bernoulli equation see Hydraulics 3) incompressible (but see note below) along a streamline () An advantage of the energy-related derivation is that it is easy to incorporate (i) thermal effects and (ii) frictional losses (or energy input/extraction by pumps/turbines). For thermal flows the total energy is supplemented by the internal energy (per unit mass) e and equation generalised to read h + U + gz = work done + heat input (6) where h = e + p/ is the enthalpy and the RHS represents energy transferred to the fluid (per unit mass) by mechanical means (positive for pumps, negative for frictional forces and turbines) or by heat transfer. The incompressible assumption can be dropped. 5. Fluid Head In fluid mechanics, energy is often measured in length units, referred to as fluid head. If you lift mass m through height H you give it (potential) energy mgh. Thus, the height to which you lift something is a measure of its energy: potential energy H = = energy per unit weight mg The same applies in fluid mechanics. An obvious example is a pumpedstorage power station (e.g. Dinorwig) which pumps water to the top of a hill during the night, storing energy for release to the turbines during the day. Another example is the constant-head tank which pressurises the hotwater supply in your house: it is basically the available head which determines how fast hot water comes out of the taps. h Pressure also represents stored energy, and it is common to measure it in length units; e.g mm Hg (millimeters of mercury) or metres of water. Atmospheric pressure is about 760 mm Hg or 0 m of water. Consider Bernoulli s equation: p + gz + U = constant, p0 (the total pressure ) Each term represents a form of energy per unit volume. If we divide by the weight per unit volume (the specific weight g) we obtain an equivalent form Hydraulics 30 David Apsley
p g U + z + = H, constant (7) g H is called the total head. p/ g is the pressure head and p/ g + z is the piezometric head. U /g is the dynamic head. Power For mass m, energy stored = mgh. energy mass Power = = gh time time Hence, in fluid flow if a mass flow rate Q experiences a change of head H then the power involved is given by power = QgH This formula is applicable to both pumps and turbines. It may be multiplied by an efficiency to account for energy losses. Example. The Hoover dam in Colorado holds back water to a depth of 80 m. If its Francistype hydroelectric turbines can pass 400 cubic metres of water per second with a generating efficiency of 80%, calculate the power available. (Don t quote me on the flow rate or efficiency please!) Key Points. In fluid mechanics pressure and energy are commonly measured in length units, referred to as fluid head.. Pressure and head are connected (as in hydrostatics) by p = gh 3. Bernoulli s equation describes an interchange between different forms of energy: pressure energy gravitational potential energy kinetic energy 4. In p U + z + = H g g H is the total head, and is constant unless there are frictional losses (or energy is input/extracted by pumps/turbines). H can be plotted graphically as the energy line. 5. U / g is the dynamic head. Many energy losses are quantified as multiples of the dynamic head (see Section 5.3). 6. Power = efficiency gqh. Hydraulics 3 David Apsley
5.3 Departures From Ideal Flow Many theoretical results are derived for ideal fluids, in particular assuming no frictional losses and uniform velocity profiles. In practice, compensation is necessary for non-ideal flow. Important examples include: discharge coefficients to correct the quantity of flow deduced by Bernoulli s equation; loss coefficients to quantify energy losses in pipelines; momentum and energy coefficients to account for non-uniform velocity profiles. Such factors are usually determined experimentally and documented in official standards. 5.3. Discharge Coefficients A discharge coefficient c D is the ratio of the actual quantity of flow, Q, to that deduced from Bernoulli s equation on the basis of ideal flow. i.e. Q = c D Q ideal (8) c D can be nearly.0 for a well-designed venturi meter, but as low as 0.6 for a weir or sharpedged orifice. 5.3. Loss Coefficients These are used to quantify energy losses, particularly in pipelines (Dr Lane-Serff will cover this in detail in the section on Pipe Flow ). A loss coefficient K is the ratio of energy loss to the kinetic energy, or head loss h to dynamic head ( V / g) : V h = K (9) g For pipe flow there is a gradual energy loss due to pipe friction over a length L of pipe. This is quantified in terms of a friction factor : L V h =, i.e. D g L = (30) D K Example. A park s water feature is supplied with water from a large tank via a uniform pipe of length 35 m and diameter 50 mm. There is a free discharge at a point 0 m below the water level in the tank. Find how long it takes to deliver 000 L of water: (a) assuming no losses; (b) allowing for an entry loss coefficient K =.0 and pipe friction factor = 0.05. Hydraulics 3 David Apsley
5.3.3 Momentum and Energy Coefficients These may be applied to account for non-uniform velocity profiles in the application of the Momentum Principle or Bernoulli equation. In the control-volume application of the Momentum Principle momentum fluxes are determined as momentum flux = mass flux velocity If the velocity is uniform: momentum flux = ( ua) u = u A If the velocity is not uniform then one has to sum over individual small contributions: momentum flux = u da (For continuous velocity profiles this sum becomes an integral see Hydraulics ). Since it is inconvenient to write down a sum or integral each time it is common to introduce a momentum coefficient as the ratio of the actual momentum flux to that calculated by assuming a uniform velocity equal to the average velocity u av ; i.e. momentum flux = ( u A) (3) av For laminar pipe flow (see later) is 4/3. For turbulent pipe flow, however, the velocity profile is much more uniform and is only slightly more than. Since the kinetic energy flow past a point is given by ( ua ) ( u ) a similar energy coefficient may be defined, which is the ratio of the average value of uniform velocity, u. 3 av 3 u to that assuming a Hydraulics 33 David Apsley