TUTORIAL No. FLUID FLOW THEORY In order to complete tis tutorial you sould already ave completed level or ave a good basic knowledge of fluid mecanics equivalent to te Engineering Council part examination 0. Wen you ave completed tis tutorial, you sould be able to do te following. Explain te meaning of viscosity. Define te units of viscosity. Describe te basic principles of viscometers. Describe non-newtonian flow Explain and solve problems involving laminar flow toug pipes and between parallel surfaces. Explain and solve problems involving drag force on speres. Explain and solve problems involving turbulent flow. Explain and solve problems involving friction coefficient. Trougout tere are worked examples, assignments and typical exam questions. You sould complete eac assignment in order so tat you progress from one level of knowledge to anoter. Let us start by examining te meaning of viscosity and ow it is measured. D.J.DUNN
. BASIC THEORY. VISCOSITY Molecules of fluids exert forces of attraction on eac oter. In liquids tis is strong enoug to keep te mass togeter but not strong enoug to keep it rigid. In gases tese forces are very weak and cannot old te mass togeter. Wen a fluid flows over a surface, te layer next to te surface may become attaced to it (it wets te surface). Te layers of fluid above te surface are moving so tere must be searing taking place between te layers of te fluid. Fig.. Let us suppose tat te fluid is flowing over a flat surface in laminated layers from left to rigt as sown in figure.. y is te distance above te solid surface (no slip surface) L is an arbitrary distance from a point upstream. dy is te tickness of eac layer. dl is te lengt of te layer. dx is te distance moved by eac layer relative to te one below in a corresponding time dt. u is te velocity of any layer. du is te increase in velocity between two adjacent layers. Eac layer moves a distance dx in time dt relative to te layer below it. Te ratio dx/dt must be te cange in velocity between layers so du dx/dt. Wen any material is deformed sideways by a (sear) force acting in te same direction, a sear stress τ is produced between te layers and a corresponding sear strain γ is produced. Sear strain is defined as follows. γ sideways deformation eigt of te layer being deformed dx dy Te rate of sear strain is defined as follows. γ& sear strain time taken γ dt dx dt dy du dy It is found tat fluids suc as water, oil and air, beave in suc a manner tat te sear stress between layers is directly proportional to te rate of sear strain. τ constant x γ& Fluids tat obey tis law are called NEWTONIAN FLUIDS. D.J.DUNN
It is te constant in tis formula tat we know as te dynamic viscosity of te fluid. DYNAMIC VISCOSITY µ sear stress rate of sear τ γ & τ dy du FORCE BALANCE AND VELOCITY DISTRIBUTION A sear stress τ exists between eac layer and tis increases by dτ over eac layer. Te pressure difference between te downstream end and te upstream end is dp. Te pressure cange is needed to overcome te sear stress. Te total force on a layer must be zero so balancing forces on one layer (assumed m wide) we get te following. dp dy + dτ dl 0 dτ dy dp dl It is normally assumed tat te pressure declines uniformly wit distance downstream so te pressure dp gradient is assumed constant. Te minus sign indicates tat te pressure falls wit distance. dl Integrating between te no slip surface (y 0) and any eigt y we get dp dl dp dl du d µ dτ dy dy dy d u µ...(.) dy Integrating twice to solve u we get te following. dp du y µ + A dl dy y dp dl µ u + Ay + B A and B are constants of integration tat sould be solved based on te known conditions (boundary conditions). For te flat surface considered in figure. one boundary condition is tat u 0 wen y 0 (te no slip surface). Substitution reveals te following. 0 0 +0 +B ence B 0 At some eigt δ above te surface, te velocity will reac te mainstream velocity u o. Tis gives us te second boundary condition u u o wen y δ. D.J.DUNN
Substituting we find te following. δ dp µ u o + Aδ dl δ dp µ u o A ence dl δ y dp δ dp µ u µ u + dl dl δ u y δ µ dp dl u o + δ Plotting u against y gives figure.. BOUNDARY LAYER. o y Te velocity grows from zero at te surface to a maximum at eigt δ. In teory, te value of δ is infinity but in practice it is taken as te eigt needed to obtain 99% of te mainstream velocity. Tis layer is called te boundary layer and δ is te boundary layer tickness. It is a very important concept and is discussed more fully in later work. Te inverse gradient of te boundary layer is du/dy and tis is te rate of sear strain γ. Fig.. D.J.DUNN 4
.. UNITS of VISCOSITY.. DYNAMIC VISCOSITY µ Te units of dynamic viscosity µ are N s/m. It is normal in te international system (SI) to give a name to a compound unit. Te old metric unit was a dyne.s/cm and tis was called a POISE after Poiseuille. Te SI unit is related to te Poise as follows. 0 Poise Ns/m wic is not an acceptable multiple. Since, owever, Centi Poise (cp) is 0.00 N s/m ten te cp is te accepted SI unit. cp 0.00 N s/m. Te symbol η is also commonly used for dynamic viscosity. Tere are oter ways of expressing viscosity and tis is covered next... KINEMATIC VISCOSITY ν Tis is defined as : ν dynamic viscosity /density ν µ/ρ Te basic units are m/s. Te old metric unit was te cm/s and tis was called te STOKE after te Britis scientist. Te SI unit is related to te Stoke as follows. Stoke (St) 0.000 m/s and is not an acceptable SI multiple. Te centi Stoke (cst),owever, is 0.00000 m/s and tis is an acceptable multiple... OTHER UNITS cst 0.00000 m/s mm/s Oter units of viscosity ave come about because of te way viscosity is measured. For example REDWOOD SECONDS comes from te name of te Redwood viscometer. Oter units are Engler Degrees, SAE numbers and so on. Conversion carts and formulae are available to convert tem into useable engineering or SI units...4 VISCOMETERS Te measurement of viscosity is a large and complicated subject. Te principles rely on te resistance to flow or te resistance to motion troug a fluid. Many of tese are covered in Britis Standards 88. Te following is a brief description of some types. D.J.DUNN 5
U TUBE VISCOMETER Te fluid is drawn up into a reservoir and allowed to run troug a capillary tube to anoter reservoir in te oter limb of te U tube. Te time taken for te level to fall between te marks is converted into cst by multiplying te time by te viscometer constant. ν ct Te constant c sould be accurately obtained by calibrating te viscometer against a master viscometer from a standards laboratory. Fig.. REDWOOD VISCOMETER Tis works on te principle of allowing te fluid to run troug an orifice of very accurate size in an agate block. 50 ml of fluid are allowed to fall from te level indicator into a measuring flask. Te time taken is te viscosity in Redwood seconds. Tere are two sizes giving Redwood No. or No. seconds. Tese units are converted into engineering units wit tables. Fig..4 D.J.DUNN 6
FALLING SPHERE VISCOMETER Tis viscometer is covered in BS88 and is based on measuring te time for a small spere to fall in a viscous fluid from one level to anoter. Te buoyant weigt of te spere is balanced by te fluid resistance and te spere falls wit a constant velocity. Te teory is based on Stokes Law and is only valid for very slow velocities. Te teory is covered later in te section on laminar flow were it is sown tat te terminal velocity (u) of te spere is related to te dynamic viscosity (µ) and te density of te fluid and spere (ρ f and ρ s ) by te formula Fig..5 µ F gd (ρ s -ρ f )/8u F is a correction factor called te Faxen correction factor, wic takes into account a reduction in te velocity due to te effect of te fluid being constrained to flow between te wall of te tube and te spere. ROTATIONAL TYPES Tere are many types of viscometers, wic use te principle tat it requires a torque to rotate or oscillate a disc or cylinder in a fluid. Te torque is related to te viscosity. Modern instruments consist of a small electric motor, wic spins a disc or cylinder in te fluid. Te torsion of te connecting saft is measured and processed into a digital readout of te viscosity in engineering units. You sould now find out more details about viscometers by reading BS88, suitable textbooks or literature from oil companies. ASSIGNMENT No.. Describe te principle of operation of te following types of viscometers. a. Redwood Viscometers. b. Britis Standard 88 glass U tube viscometer. c. Britis Standard 88 Falling Spere Viscometer. d. Any form of Rotational Viscometer Note tat tis covers te E.C. exam question 6a from te 987 paper. D.J.DUNN 7
. LAMINAR FLOW THEORY Te following work only applies to Newtonian fluids.. LAMINAR FLOW A stream line is an imaginary line wit no flow normal to it, only along it. Wen te flow is laminar, te streamlines are parallel and for flow between two parallel surfaces we may consider te flow as made up of parallel laminar layers. In a pipe tese laminar layers are cylindrical and may be called stream tubes. In laminar flow, no mixing occurs between adjacent layers and it occurs at low average velocities.. TURBULENT FLOW Te searing process causes energy loss and eating of te fluid. Tis increases wit mean velocity. Wen a certain critical velocity is exceeded, te streamlines break up and mixing of te fluid occurs. Te diagram illustrates Reynolds coloured ribbon experiment. Coloured dye is injected into a orizontal flow. Wen te flow is laminar te dye passes along witout mixing wit te water. Wen te speed of te flow is increased turbulence sets in and te dye mixes wit te surrounding water. One explanation of tis transition is tat it is necessary to cange te pressure loss into oter forms of energy suc as angular kinetic energy as indicated by small eddies in te flow. Fig... LAMINAR AND TURBULENT BOUNDARY LAYERS In capter it was explained tat a boundary layer is te layer in wic te velocity grows from zero at te wall (no slip surface) to 99% of te maximum and te tickness of te layer is denoted δ. Wen te flow witin te boundary layer becomes turbulent, te sape of te boundary layers waivers and wen diagrams are drawn of turbulent boundary layers, te mean sape is usually sown. Comparing a laminar and turbulent boundary layer reveals tat te turbulent layer is tinner tan te laminar layer. D.J.DUNN Fig.. 8
.4 CRITICAL VELOCITY - REYNOLDS NUMBER Wen a fluid flows in a pipe at a volumetric flow rate Q m/s te average velocity is defined Q u m A is te cross sectional area. A ρu md u md Te Reynolds number is defined as R e µ ν If you ceck te units of Re you will see tat tere are none and tat it is a dimensionless number. You will learn more about suc numbers in a later section. Reynolds discovered tat it was possible to predict te velocity or flow rate at wic te transition from laminar to turbulent flow occurred for any Newtonian fluid in any pipe. He also discovered tat te critical velocity at wic it canged back again was different. He found tat wen te flow was gradually increased, te cange from laminar to turbulent always occurred at a Reynolds number of 500 and wen te flow was gradually reduced it canged back again at a Reynolds number of 000. Normally, 000 is taken as te critical value. WORKED EXAMPLE. Oil of density 860 kg/m as a kinematic viscosity of 40 cst. Calculate te critical velocity wen it flows in a pipe 50 mm bore diameter. SOLUTION R u e m u md ν R eν D 000x40x0 0.05 6.6 m/s D.J.DUNN 9
.5 DERIVATION OF POISEUILLE'S EQUATION for LAMINAR FLOW Poiseuille did te original derivation sown below wic relates pressure loss in a pipe to te velocity and viscosity for LAMINAR FLOW. His equation is te basis for measurement of viscosity ence is name as been used for te unit of viscosity. Consider a pipe wit laminar flow in it. Consider a stream tube of lengt L at radius r and tickness dr. y is te distance from te pipe wall. y R r Fig.. dy dr du dy du dr Te sear stress on te outside of te stream tube is τ. Te force (F s ) acting from rigt to left is due to te sear stress and is found by multiplying τ by te surface area. Fs τ x πr L du For a Newtonian fluid, τ µ dy µ du dr. Substituting for τ we get te following. du F s - πr Lµ dr Te pressure difference between te left end and te rigt end of te section is p. Te force due to tis (F p ) is p x circular area of radius r. F p p x πr du Equating forces we ave - πrµ L pπr dr p du rdr µ L In order to obtain te velocity of te streamline at any radius r we must integrate between te limits u 0 wen r R and u u wen r r. u r p du - rdr µ L 0 p u 4µ L p u 4µ L R ( r R ) ( R r ) D.J.DUNN 0
Tis is te equation of a Parabola so if te equation is plotted to sow te boundary layer, it is seen to extend from zero at te edge to a maximum at te middle. Fig..4 pr For maximum velocity put r 0 and we get u 4µ L Te average eigt of a parabola is alf te maximum value so te average velocity is pr u m 8µ L Often we wis to calculate te pressure drop in terms of diameter D. Substitute RD/ and rearrange. µ Lum p D Te volume flow rate is average velocity x cross sectional area. πr pr Q 8µ L Tis is often canged to give te pressure drop as a friction ead. Te friction ead for a lengt L is found from f p/ρg µ Lum f ρgd Tis is Poiseuille's equation tat applies only to laminar flow. 4 πr p 8µ L 4 πd p 8µ L D.J.DUNN
WORKED EXAMPLE. A capillary tube is 0 mm long and mm bore. Te ead required to produce a flow rate of 8 mm/s is 0 mm. Te fluid density is 800 kg/m. Calculate te dynamic and kinematic viscosity of te oil. SOLUTION Rearranging Poiseuille's equation we get f ρgd µ Lu m πd π x A 0.785 mm 4 4 Q 8 u m 0.8 mm/s A 0.785 0.0 x 800 x 9.8 x 0.00 µ 0.04 N s/m or 4.cP x 0.0 x 0.008 µ 0.04-6 ν 0. x 0 m / s or 0.cSt ρ 800 WORKED EXAMPLE No.. Oil flows in a pipe 00 mm bore wit a Reynolds number of 50. Te dynamic viscosity is 0.08 Ns/m. Te density is 900 kg/m. Determine te pressure drop per metre lengt, te average velocity and te radius at wic it occurs. SOLUTION Reρu m D/µ. Hence u m Re µ/ ρd u m (50 x 0.08)/(900 x 0.) 0.05 m/s p µl u m /D p x 0.08 x x 0.05/0. p.88 Pascals. u { p/4lµ}(r - r) wic is made equal to te average velocity 0.05 m/s 0.05 (.88/4 x x 0.08)(0.05 - r) r 0.05 m or 5. mm. D.J.DUNN
.6. FLOW BETWEEN FLAT PLATES Consider a small element of fluid moving at velocity u wit a lengt dx and eigt dy at distance y above a flat surface. Te sear stress acting on te element increases by dτ in te y direction and te pressure decreases by dp in te x dp d u direction. It was sown earlier tat µ dx dy It is assumed tat dp/dx does not vary wit y so it may be regarded as a fixed value in te following work. Fig..5 dp du Integrating once - y µ + A dx dy y dp Integrating again - µ u + Ay + B...(.6A) dx A and B are constants of integration. Te solution of te equation now depends upon te boundary conditions tat will yield A and B. WORKED EXAMPLE No..4 Derive te equation linking velocity u and eigt y at a given point in te x direction wen te flow is laminar between two stationary flat parallel plates distance apart. Go on to derive te volume flow rate and mean velocity. SOLUTION Wen a fluid touces a surface, it sticks to it and moves wit it. Te velocity at te flat plates is te same as te plates and in tis case is zero. Te boundary conditions are ence u 0 wen y 0 Substituting into equation.6a yields tat B 0 u0 wen y Substituting into equation.6a yields tat A (dp/dx)/ Putting tis into equation.6a yields u (dp/dx)(/µ){y - y} (Te student sould do te algebra for tis). Te result is a parabolic distribution similar tat given by Poiseuille's equation earlier only tis time it is between two flat parallel surfaces. D.J.DUNN
FLOW RATE To find te flow rate we consider flow troug a small rectangular slit of widt B and eigt dy at eigt y. Fig..6 Te flow troug te slit is dq u Bdy (dp/dx)(/µ){y - y} Bdy Integrating between y 0 and y to find Q yields Te mean velocity is ence Q -B(dp/dx)(/µ) u m Q/Area Q/B u m -(dp/dx)(/µ) (Te student sould do te algebra).7 CONCENTRIC CYLINDERS Tis could be a saft rotating in a bus filled wit oil or a rotational viscometer. Consider a saft rotating in a cylinder wit te gap between filled wit a Newtonian liquid. Tere is no overall flow rate so equation.a does not apply. Fig.7 Due to te stickiness of te fluid, te liquid sticks to bot surfaces and as a velocity u ωr i at te inner layer and zero at te outer layer. D.J.DUNN 4
If te gap is small, it may be assumed tat te cange in te velocity across te gap canges from u to zero linearly wit radius r. τ µ du/dy But since te cange is linear du/dy u/(r o -R i ) ω R i /(R o -R i ) τ µ ω R i /(R o -R i ) Sear force on cylinder F sear stress x surface area πri µω F πri τ R R Torque F x R i o i πri µω T Fr Ro Ri In te case of a rotational viscometer we rearrange so tat T ( Ro R) µ πri ω In reality, it is unlikely tat te velocity varies linearly wit radius and te bottom of te cylinder would ave an affect on te torque..8 FALLING SPHERES Tis teory may be applied to particle separation in tanks and to a falling spere viscometer. Wen a spere falls, it initially accelerates under te action of gravity. Te resistance to motion is due to te searing of te liquid passing around it. At some point, te resistance balances te force of gravity and te spere falls at a constant velocity. Tis is te terminal velocity. For a body immersed in a liquid, te buoyant weigt is W and tis is equal to te viscous resistance R wen te terminal velocity is reaced. R W volume x density difference x gravity πd g( ρ s ρ f ) R W 6 ρ s density of te spere material ρ f density of fluid d spere diameter Te viscous resistance is muc arder to derive from first principles and tis will not be attempted ere. In general, we use te concept of DRAG and define te DRAG COEFFICIENT as C D Resistance force Dynamic pressure x projected Area D.J.DUNN 5
ρu Te dynamic pressure of a flow stream is πd Te projected area of a spere is 4 8R C D ρu πd Researc sows te following relationsip between C D and R e for a spere. Fig..8 For R e <0. te flow is called Stokes flow and Stokes sowed tat R πdµu ence C D 4µ/ρ f ud 4/R e For 0. < R e < 500 te flow is called Allen flow and C D 8.5R e -0.6 For 500 < R e < 0 5 C D is constant C D 0.44 An empirical formula tat covers te range 0. < R e < 0 5 is as follows. C D 4 6 + R + R e e + 0.4 For a falling spere viscometer, Stokes flow applies. Equating te drag force and te buoyant weigt we get πdµu (πd /6)(ρ s - ρ f ) g µ gd (ρ s - ρ f )/8u for a falling spere vicometer Te terminal velocity for Stokes flow is u d g(ρ s - ρ f )8µ Tis formula assumes a fluid of infinite widt but in a falling spere viscometer, te liquid is squeezed between te spere and te tube walls and additional viscous resistance is produced. Te Faxen correction factor F is used to correct te result. D.J.DUNN 6
.9 THRUST BEARINGS Consider a round flat disc of radius R rotating at angular velocity ω rad/s on top of a flat surface and separated from it by an oil film of tickness t. Fig..9 Assume te velocity gradient is linear in wic case du/dy u/t ωr/t at any radius r. du ωr Te sear stress on te ring isτ µ µ dy t ω Te sear force is df πr drµ t ω Te torque is dt rdf πr drµ t Te total torque is found by integrating wit respect to r. T R 0 πr ω 4 ω drµ πr µ t t In terms of diameter D tis is T µπωd t Tere are many variations on tis teme tat you sould be prepared to andle. 4 D.J.DUNN 7
.0 MORE ON FLOW THROUGH PIPES Consider an elementary tin cylindrical layer tat makes an element of flow witin a pipe. Te lengt is δx, te inside radius is r and te radial tickness is dr. Te pressure difference between te ends is δp and te sear stress on te surface increases by dτ from te inner to te outer surface. Te velocity at any point is u and te dynamic viscosity is µ. Fig..0 Te pressure force acting in te direction of flow is {π(r+dr) -πr }δp Te sear force opposing is {(τ+δτ)(π)(r+dr) - τπr}δx Equating, simplifying and ignoring te product of two small quantities we ave te following result. δp τ dτ du + τ µ for Newtonian fluids. δx r dr dy If y is measured from te inside of δp µ du d u µ δx r dr dr du d u δp + r dr dr µ δx du dr Using partial differentiation to differentiate d r ence dr Integrating we get du dr rd u + dr r δp µ δx du dr r δp + µ δx r δp µ δx A r du r δp r + A dr µ δx...(a) were A is a constant of integration. du te pipe ten r -y and dy - dr soτ µ dr du dr d r dr yields te result du rd u + dr dr D.J.DUNN 8
integration. were B is anoter constant of )...( ln 4 Integrating again we get B B r A x p r u + + δ δ µ Equations (A) and (B) may be used to derive Poiseuille's equation or it may be used to solve flow troug an annular passage..0. PIPE At te middle r0 so from equation (A) it follows tat A0 At te wall, u0 and rr. Putting tis into equation B yields { } again. s equation and tis ispoiseuille' 4 4 4 4 0 were A ln 4 0 r R x p x p R x p r u x p R B B R A x p R + + + δ δ µ δ δ µ δ δ µ δ δ µ δ δ µ.0. ANNULUS Fig.. { } { } { } + + + + + + + + + i o i i o i o i i o o R R A R R x p R R A R R x p B R A x p R C B R A x p R B r A x p r u ln 4 0 ln ln 4 0 subtract D from C...(D) ln 4 0 )...( ln 4 0. R and r R r at 0 Te boundary conditions are u ln 4 0 o i δ δ µ δ δ µ δ δ µ δ δ µ δ δ µ D.J.DUNN 9
{ } { } { } { } { } { } { } { } + + + + + + + ln ln 4 ln ln ln ln 4 u ln ln 4 ln ln 4 4 r - put into equation B Tis is ln ln 4 ln ln 4 4 0 be obtained from C. may be substituted back into equation D. Te same result will Tis ln 4 A r R R r R R R R x p u R R R R R R r R R R R r x p R R R R R R x p r R R R R x p x p u R R R R R R x p B B R R R R R x p x p R R R R R x p i i i o i o i i o i o i i o i o i i o i o i i o i o i i o i o i i i o i o i i o i o δ δ µ δ δ µ δ δ µ δ δ µ δ δ µ δ δ µ δ δ µ δ δ µ δ δ µ For given values te velocity distribution is similar to tis. Fig.. D.J.DUNN 0
ASSIGNMENT. Oil flows in a pipe 80 mm bore diameter wit a mean velocity of 0.4 m/s. Te density is 890 kg/m and te viscosity is 0.075 Ns/m. Sow tat te flow is laminar and ence deduce te pressure loss per metre lengt. (50 Pa per metre).. Oil flows in a pipe 00 mm bore diameter wit a Reynolds Number of 500. Te density is 800 kg/m. Calculate te velocity of a streamline at a radius of 40 mm. Te viscosity µ 0.08 Ns/m. (0.6 m/s). A liquid of dynamic viscosity 5 x 0- Ns/m flows troug a capillary of diameter.0 mm under a pressure gradient of 800 N/m. Evaluate te volumetric flow rate, te mean velocity, te centre line velocity and te radial position at wic te velocity is equal to te mean velocity. (u av 0.0 m/s, u max 0.0 m/s r.06 mm) 4. Similar to Q6 998 a. Explain te term Stokes flow and terminal velocity. b. Sow tat a sperical particle wit Stokes flow as a terminal velocity given by u d g(ρ s - ρ f )/8µ Go on to sow tat C D 4/R e c. For sperical particles, a useful empirical formula relating te drag coefficient and te Reynold s number is 4 6 C D + + 0.4 R + R e Given ρ f 000 kg/m, µ cp and ρ s 60 kg/m determine te maximum size of sperical particles tat will be lifted upwards by a vertical stream of water moving at m/s. d. If te water velocity is reduced to 0.5 m/s, sow tat particles wit a diameter of less tan 5.95 mm will fall downwards. e D.J.DUNN
5. Similar to Q5 998 A simple fluid coupling consists of two parallel round discs of radius R separated by a a gap. One disc is connected to te input saft and rotates at ω rad/s. Te oter disc is connected to te output saft and rotates at ω rad/s. Te discs are separated by oil of dynamic viscosity µ and it may be assumed tat te velocity gradient is linear at all radii. ( ω ω ) 4 πd µ Sow tat te Torque at te input saft is given by T Te input saft rotates at 900 rev/min and transmits 500W of power. Calculate te output speed, torque and power. (747 rev/min, 5. Nm and 44 W) Sow by application of max/min teory tat te output speed is alf te input speed wen maximum output power is obtained. 6. Sow tat for fully developed laminar flow of a fluid of viscosity µ between orizontal parallel plates a distance apart, te mean velocity u m is related to te pressure gradient dp/dx by u m - (/µ)(dp/dx) Fig.. sows a flanged pipe joint of internal diameter d i containing viscous fluid of viscosity µ at gauge pressure p. Te flange as an outer diameter d o and is imperfectly tigtened so tat tere is a narrow gap of tickness. Obtain an expression for te leakage rate of te fluid troug te flange. Fig.. Note tat tis is a radial flow problem and B in te notes becomes πr and dp/dx becomes -dp/dr. An integration between inner and outer radii will be required to give flow rate Q in terms of pressure drop p. Te answer is Q (πp/µ)/{ln(d o /d i )} D.J.DUNN
. TURBULENT FLOW. FRICTION COEFFICIENT Te friction coefficient is a convenient idea tat can be used to calculate te pressure drop in a pipe. It is defined as follows. Wall Sear Stress C f Dynamic Pressure.. DYNAMIC PRESSURE Consider a fluid flowing wit mean velocity u m. If te kinetic energy of te fluid is converted into flow or fluid energy, te pressure would increase. Te pressure rise due to tis conversion is called te dynamic pressure. KE ½ mu m Flow Energy p Q Q is te volume flow rate and ρ m/q Equating ½ mu m p Q p mu /Q ½ ρ u m.. WALL SHEAR STRESS τo Te wall sear stress is te sear stress in te layer of fluid next to te wall of te pipe. Fig.. du Te sear stress in te layer next to te wall is τ o µ dy Te sear force resisting flow is F τ πld Te resulting pressure drop produces a force of Equating forces gives τ o D p 4L s o F p pπd 4 wall D.J.DUNN
.. FRICTION COEFFICIENT for LAMINAR FLOW C f Wall Sear Stress Dynamic Pressure D p 4Lρu µ Lu m From Poiseuille s equation p Hence D..4 DARCY FORMULA m D µ Lu 6µ C f 4L u ρ m D ρu md 6 R e Tis formula is mainly used for calculating te pressure loss in a pipe due to turbulent flow but it can be used for laminar flow also. Turbulent flow in pipes occurs wen te Reynolds Number exceeds 500 but tis is not a clear point so 000 is used to be sure. In order to calculate te frictional losses we use te concept of friction coefficient symbol Cf. Tis was defined as follows. C f Wall Sear Stress Dynamic Pressure D p 4Lρu Rearranging equation to make p te subject 4Cf Lρu m p D Tis is often expressed as a friction ead f p 4Cf Lu m f ρg gd Tis is te Darcy formula. In te case of laminar flow, Darcy's and Poiseuille's equations must give te same result so equating tem gives 4Cf Lu m µ Lu m gd ρgd C f 6µ ρu D m 6 R Tis is te same result as before for laminar flow. e Turbulent flow may be safely assumed in pipes wen te Reynolds Number exceeds 000. In order to calculate te frictional losses we use te concept of friction coefficient symbol Cf. Note tat in older textbooks C f was written as f but now te symbol f represents 4C f...5 FLUID RESISTANCE Fluid resistance is an alternative approac to solving problems involving losses. Te above equations may be expressed in terms of flow rate Q by substituting u Q/A 4C Lu 4C LQ gd gda f m f f Substituting A πd /4 we get te following. m C LQ gπ D f f RQ R is te fluid resistance or restriction. 5 R C f L gπ D 5 D.J.DUNN 4
If we want pressure loss instead of ead loss te equations are as follows. ρc LQ p π D f f ρg f RQ R is te fluid resistance or restriction. 5 ρc f L R 5 π D It sould be noted tat R contains te friction coefficient and tis is a variable wit velocity and surface rougness so R sould be used wit care.. MOODY DIAGRAM AND RELATIVE SURFACE ROUGHNESS In general te friction ead is some function of u m suc tat f φu m n. Clearly for laminar flow, n but for turbulent flow n is between and and its precise value depends upon te rougness of te pipe surface. Surface rougness promotes turbulence and te effect is sown in te following work. Relative surface rougness is defined as ε k/d were k is te mean surface rougness and D te bore diameter. An American Engineer called Moody conducted exaustive experiments and came up wit te Moody Cart. Te cart is a plot of C f vertically against R e orizontally for various values of ε. In order to use tis cart you must know two of te tree co-ordinates in order to pick out te point on te cart and ence pick out te unknown tird co-ordinate. For smoot pipes, (te bottom curve on te diagram), various formulae ave been derived suc as tose by Blasius and Lee. BLASIUS C f 0.079 R e 0.5 LEE C f 0.008 + 0.5 R e 0.5. Te Moody diagram sows tat te friction coefficient reduces wit Reynolds number but at a certain point, it becomes constant. Wen tis point is reaced, te flow is said to be fully developed turbulent flow. Tis point occurs at lower Reynolds numbers for roug pipes. A formula tat gives an approximate answer for any surface rougness is tat given by Haaland. 6.9 ε.6 log0 + Cf R e. 7. D.J.DUNN 5
D.J.DUNN Fig.. CHART 6
WORKED EXAMPLE. Determine te friction coefficient for a pipe 00 mm bore wit a mean surface rougness of 0.06 mm wen a fluid flows troug it wit a Reynolds number of 0 000. SOLUTION Te mean surface rougness ε k/d 0.06/00 0.0006 Locate te line for ε k/d 0.0006. Trace te line until it meets te vertical line at Re 0 000. Read of te value of C f orizontally on te left. Answer C f 0.0067.Ceck using te formula from Haaland. C C C C C f f f f f.6 log.6 log.6 log.06 0.0067 0 0 0 6.9 ε + R e.7 6.9 0000 6.9 0000. 0.0006 +.7 0.0006 +.7.. WORKED EXAMPLE. Oil flows in a pipe 80 mm bore wit a mean velocity of 4 m/s. Te mean surface rougness is 0.0 mm and te lengt is 60 m. Te dynamic viscosity is 0.005 N s/m and te density is 900 kg/m. Determine te pressure loss. SOLUTION Re ρud/µ (900 x 4 x 0.08)/0.005 57600 ε k/d 0.0/80 0.0005 From te cart C f 0.005 f 4C f Lu/dg (4 x 0.005 x 60 x 4 )/( x 9.8 x 0.08).7 m p ρg f 900 x 9.8 x.7. kpa. D.J.DUNN 7
ASSIGNMENT. A pipe is 5 km long and 80 mm bore diameter. Te mean surface rougness is 0.0 mm. It carries oil of density 85 kg/m at a rate of 0 kg/s. Te dynamic viscosity is 0.05 N s/m. Determine te friction coefficient using te Moody Cart and calculate te friction ead. (Ans. 075 m.). Water flows in a pipe at 0.05 m/s. Te pipe is 50 mm bore diameter. Te pressure drop is 40 Pa per metre lengt. Te density is 000 kg/m and te dynamic viscosity is 0.00 N s/m. Determine i. te wall sear stress (67.75 Pa) ii. te dynamic pressure (980 Pa). iii. te friction coefficient (0.00575) iv. te mean surface rougness (0.0875 mm). Explain briefly wat is meant by fully developed laminar flow. Te velocity u at any radius r in fully developed laminar flow troug a straigt orizontal pipe of internal radius ro is given by u (/4µ)(ro - r)dp/dx dp/dx is te pressure gradient in te direction of flow and µ is te dynamic viscosity. Sow tat te pressure drop over a lengt L is given by te following formula. p µlu m /D Te wall skin friction coefficient is defined as C f τ o /( ρum ). Sow tat C f 6/Re were Re ρumd/µ and ρ is te density, um is te mean velocity and τ o is te wall sear stress. 4. Oil wit viscosity x 0- Ns/m and density 850 kg/m is pumped along a straigt orizontal pipe wit a flow rate of 5 dm/s. Te static pressure difference between two tapping points 0 m apart is 80 N/m. Assuming laminar flow determine te following. i. Te pipe diameter. ii. Te Reynolds number. Comment on te validity of te assumption tat te flow is laminar 4. NON-NEWTONIAN FLUIDS D.J.DUNN 8
A Newtonian fluid as discussed so far in tis tutorial is a fluid tat obeys te law A Non Newtonian fluid is generally described by te non-linear law n τ τ + kγ& y du τ µ µ & γ dy τ y is known as te yield sear stress and γ& is te rate of sear strain. Figure 4. sows te principle forms of tis equation. Grap A sows an ideal fluid tat as no viscosity and ence as no sear stress at any point. Tis is often used in teoretical models of fluid flow. Grap B sows a Newtonian Fluid. Tis is te type of fluid wit wic tis book is mostly concerned, fluids suc as water and oil. Te grap is ence a straigt line and te gradient is te viscosity µ. Tere is a range of oter liquid or semi-liquid materials tat do not obey tis law and produce strange flow caracteristics. Suc materials include various foodstuffs, paints, cements and so on. Many of tese are in fact solid particles suspended in a liquid wit various concentrations. Grap C sows te relationsip for a Dilatent fluid. Te gradient and ence viscosity increases wit γ& and suc fluids are also called sear-tickening. Tis penomenon occurs wit some solutions of sugar and starces. Grap D sows te relationsip for a Pseudo-plastic. Te gradient and ence viscosity reduces wit γ& and tey are called sear-tinning. Most foodstuffs are like tis as well as clay and liquid cement.. Oter fluids beave like a plastic and require a minimum stress before it sears τ y. Tis is plastic beaviour but unlike plastics, tere may be no elasticity prior to searing. Grap E sows te relationsip for a Bingam plastic. Tis is te special case were te beaviour is te same as a Newtonian fluid except for te existence of te yield stress. Foodstuffs containing ig level of fats approximate to tis model (butter, margarine, cocolate and Mayonnaise). Grap F sows te relationsip for a plastic fluid tat exibits sear tickening caracteristics. Grap G sows te relationsip for a Casson fluid. Tis is a plastic fluid tat exibits seartinning caracteristics. Tis model was developed for fluids containing rod like solids and is often applied to molten cocolate and blood. Fig.4. D.J.DUNN 9
MATHEMATICAL MODELS Te graps tat relate sear stress τ and rate of sear strain γ are based on models or equations. Most are matematical equations created to represent empirical data. Hirscel and Bulkeley developed te power law for non-newtonian equations. Tis is as follows. n τ τ + Kγ& K is called te consistency coefficient and n is a power. y In te case of a Newtonian fluid n and τ y 0 and K µ (te dynamic viscosity) τ µ γ& For a Bingam plastic, n and K is also called te plastic viscosity µ p. Te relationsip reduces to τ τ y + µ γ& For a dilatent fluid, τ y 0 and n> For a pseudo-plastic, τ y 0 and n< p Te model for bot is τ Kγ& n Te Hercel-Bulkeley model is as follows. τ τ y + Kγ& n Tis may be developed as follows. n τ τ + K & γ τ τ n K & γ dividing by & γ n τ τ y & γ n K K & γ & γ & γ & γ τ τ y n + K & γ Te ratio is called te apparent viscosity µ & γ & γ µ app y y τ τ y + K & γ & γ & γ n sometimes written asτ τ µ & γ n For a Bingam plastic n so app τ y + K & γ For a Fluid wit no yield sear value τ 0 so µ µ y y p app were µ is called K & γ p app n te plastic viscosity. Te Casson fluid model is quite different in form from te oters and is as follows. y τ τ + Kγ& D.J.DUNN 0
THE FLOW OF A PLASTIC FLUID Note tat fluids wit a sear yield stress will flow in a pipe as a plug. Witin a certain radius, te sear stress will be insufficient to produce searing so inside tat radius te fluid flows as a solid plug. Fig. 4. sows a typical situation for a Bingam Plastic. Fig.4. MINIMUM PRESSURE Te sear stress acting on te surface of te plug is te yield value. Let te plug be diameter d. Te pressure force acting on te plug is p x πd /4 Te sear force acting on te surface of te plug is τ y x π d L Equating we find p x πd /4 τ y x π d L d τ y x 4 L/ p or p τ y x 4 L/d Te minimum pressure required to produce flow must occur wen d is largest and equal to te bore of te pipe. p (minimum) τ y x 4 L/D Te diameter of te plug at any greater pressure must be given by d τ y x 4 L/ p For a Bingam Plastic, te boundary layer between te plug and te wall must be laminar and te velocity must be related to radius by te formula derived earlier. p p u ( R r ) ( D d ) 4µ L 6µ L FLOW RATE Te flow rate sould be calculated in two stages. Te plug moves at a constant velocity so te flow rate for te plug is simply Q p u x cross sectional area u x πd /4 Te flow witin te boundary layer is found in te usual way as follows. Consider an elementary ring radius r and widt dr. p dq u x πr dr ( R r ) x πr dr 4µ L pπ Q µ L Q pπ r µ L r ( rr r ) R R dr 4 r 4 R r pπ R µ L R 4 r R 4 r 4 4 4 pπ R r R r Q + µ L 4 4 Te mean velocity as always is defined as u m Q/Cross sectional area. 4 4 D.J.DUNN
WORKED EXAMPLE 4. n Te Hercel-Bulkeley model for a non-newtonian fluid is as follows. τ τ + Kγ&. Derive an equation for te minimum pressure required drop per metre lengt in a straigt orizontal pipe tat will produce flow. Given tat te pressure drop per metre lengt in te pipe is 60 Pa/m and te yield sear stress is 0. Pa, calculate te radius of te slug sliding troug te middle. SOLUTION y Fig.. Te pressure difference p acting on te cross sectional area must produce sufficient force to overcome te sear stress τ acting on te surface area of te cylindrical slug. For te slug to move, te sear stress must be at least equal to te yield value τy. Balancing te forces gives te following. p x πr τ y x πrl p/l τ y /r 60 x 0./r r 0.4/60 0.0066 m or 6.6 mm D.J.DUNN
WORKED EXAMPLE 4. A Bingam plastic flows in a pipe and it is observed tat te central plug is 0 mm diameter wen te pressure drop is 00 Pa/m. Calculate te yield sear stress. Given tat at a larger radius te rate of sear strain is 0 s - and te consistency coefficient is 0.6 Pa s, calculate te sear stress. SOLUTION For a Bingam plastic, te same teory as in te last example applies. p/l τ y /r 00 τ y /0.05 τ y 00 x 0.05/ 0.75 Pa A matematical model for a Bingam plastic is τ τ + Kγ& y 0.75 + 0.6 x 0.75 Pa D.J.DUNN
ASSIGNMENT 4. Researc as sown tat tomato ketcup as te following viscous properties at 5 o C. Consistency coefficient K 8.7 Pa s n Power n 0.7 Sear yield stress Pa Calculate te apparent viscosity wen te rate of sear is, 0, 00 and 000 s - and conclude on te effect of te sear rate on te apparent viscosity. Answers γ µ app 50.7 γ 0 µ app 6.68 γ 00 µ app 0.968 γ 000 µ app 0.5. A Bingam plastic fluid as a viscosity of 0.05 N s/m and yield stress of 0.6 N/m. It flows in a tube 5 mm bore diameter and m long. (i) Evaluate te minimum pressure drop required to produce flow. (480 N/m ) Te actual pressure drop is twice te minimum value. Sketc te velocity profile and calculate te following. (ii) Te radius of te solid core. (.75 mm) (iii) Te velocity of te core. (67.5 mm/s) (iv) Te volumetric flow rate. (7.46 cm /s). A non-newtonian fluid is modelled by te equation du τ K dr n were n 0.8 and K 0.05 N s 0.8 /m. It flows troug a tube 6 mm bore diameter under te influence of a pressure drop of 6400 N/m per metre lengt. Obtain an expression for te velocity profile and evaluate te following. (i) Te centre line velocity. (0.95 m/s) (ii) Te mean velocity. (0.5 m/s) D.J.DUNN 4
FLUID MECHANICS 0 TUTORIAL No. APPLICATIONS OF BERNOULLI On completion of tis tutorial you sould be able to derive Bernoulli's equation for liquids. find te pressure losses in piped systems due to fluid friction. find te minor frictional losses in piped systems. matc pumps of known caracteristics to a given system. derive te basic relationsip between pressure, velocity and force.. solve problems involving flow troug orifices. solve problems involving flow troug Venturi meters. understand orifice meters. understand nozzle meters. understand te principles of jet pumps solve problems from past papers. Let's start by revising basics. Te flow of a fluid in a pipe depends upon two fundamental laws, te conservation of mass and energy. D.J.DUNN www.freestudy.co.uk
. PIPE FLOW Te solution of pipe flow problems requires te applications of two principles, te law of conservation of mass (continuity equation) and te law of conservation of energy (Bernoulli s equation). CONSERVATION OF MASS Wen a fluid flows at a constant rate in a pipe or duct, te mass flow rate must be te same at all points along te lengt. Consider a liquid being pumped into a tank as sown (fig.). Te mass flow rate at any section is m ρau m ρ density (kg/m) u m mean velocity (m/s) A Cross Sectional Area (m) Fig.. For te system sown te mass flow rate at (), () and () must be te same so ρ A u ρ A u ρ A u In te case of liquids te density is equal and cancels so A u A u A u Q D.J.DUNN www.freestudy.co.uk
. CONSERVATION OF ENERGY ENERGY FORMS FLOW ENERGY Tis is te energy a fluid possesses by virtue of its pressure. Te formula is F.E. pq Joules p is te pressure (Pascals) Q is volume rate (m) POTENTIAL OR GRAVITATIONAL ENERGY Tis is te energy a fluid possesses by virtue of its altitude relative to a datum level. Te formula is P.E. mgz Joules m is mass (kg) z is altitude (m) KINETIC ENERGY Tis is te energy a fluid possesses by virtue of its velocity. Te formula is K.E. ½ mu m Joules u m is mean velocity (m/s) INTERNAL ENERGY Tis is te energy a fluid possesses by virtue of its temperature. It is usually expressed relative to 0 o C. Te formula is U mcθ c is te specific eat capacity (J/kg o C) θ is te temperature in o C In te following work, internal energy is not considered in te energy balance. SPECIFIC ENERGY Specific energy is te energy per kg so te tree energy forms as specific energy are as follows. F.E./m pq/m p/ρ Joules/kg P.E/m. gz Joules/kg K.E./m ½ u Joules/kg ENERGY HEAD If te energy terms are divided by te weigt mg, te result is energy per Newton. Examining te units closely we ave J/N N m/n metres. It is normal to refer to te energy in tis form as te energy ead. Te tree energy terms expressed tis way are as follows. F.E./mg p/ρg P.E./mg z K.E./mg u /g Te flow energy term is called te pressure ead and tis follows since earlier it was sown tat p/ρg. Tis is te eigt tat te liquid would rise to in a vertical pipe connected to te system. Te potential energy term is te actual altitude relative to a datum. Te term u /g is called te kinetic ead and tis is te pressure ead tat would result if te velocity is converted into pressure. D.J.DUNN www.freestudy.co.uk
. BERNOULLI S EQUATION Bernoulli s equation is based on te conservation of energy. If no energy is added to te system as work or eat ten te total energy of te fluid is conserved. Remember tat internal (termal energy) as not been included. Te total energy E T at () and () on te diagram (fig..) must be equal so : u u E T pq + mgz + m p Q + mgz + m Dividing by mass gives te specific energy form E T p u p u + gz + + gz + m ρ ρ Dividing by g gives te energy terms per unit weigt E T p u p u + z + + z + mg gρ g gρ g Since p/ρg pressure ead ten te total ead is given by te following. u u T + z + + z + g g Tis is te ead form of te equation in wic eac term is an energy ead in metres. z is te potential or gravitational ead and u /g is te kinetic or velocity ead. For liquids te density is te same at bot points so multiplying by ρg gives te pressure form. Te total pressure is as follows. ρu ρu p T p + ρgz + p + ρgz + In real systems tere is friction in te pipe and elsewere. Tis produces eat tat is absorbed by te liquid causing a rise in te internal energy and ence te temperature. In fact te temperature rise will be very small except in extreme cases because it takes a lot of energy to raise te temperature. If te pipe is long, te energy migt be lost as eat transfer to te surroundings. Since te equations did not include internal energy, te balance is lost and we need to add an extra term to te rigt side of te equation to maintain te balance. Tis term is eiter te ead lost to friction L or te pressure loss p L. u u + z + + z + + L g g Te pressure form of te equation is as follows. ρu ρu p + ρgz + p + ρgz + + p L Te total energy of te fluid (excluding internal energy) is no longer constant. Note tat if a point is a free surface te pressure is normally atmosperic but if gauge pressures are used, te pressure and pressure ead becomes zero. Also, if te surface area is large (say a large tank), te velocity of te surface is small and wen squared becomes negligible so te kinetic energy term is neglected (made zero). D.J.DUNN www.freestudy.co.uk 4
WORKED EXAMPLE No. Te diagram sows a pump delivering water troug as pipe 0 mm bore to a tank. Find te pressure at point () wen te flow rate is.4 dm /s. Te density of water is 000 kg/m. Te loss of pressure due to friction is 50 kpa. SOLUTION Fig.. Area of bore A π x 0.0 /4 706.8 x 0-6 m. Flow rate Q.4 dm /s 0.004 m /s Mean velocity in pipe Q/A.98 m/s Apply Bernoulli between point () and te surface of te tank. ρu ρu p + ρ gz + p + ρgz + + p L Make te low level te datum level and z 0 and z 5. Te pressure on te surface is zero gauge pressure. P L 50 000 Pa Te velocity at () is.98 m/s and at te surface it is zero. 000x.98 p + 0 + 0 + 000x9.95 + 0 + 50000 p 9.9kPa gauge pressure D.J.DUNN www.freestudy.co.uk 5
WORKED EXAMPLE Te diagram sows a tank tat is drained by a orizontal pipe. Calculate te pressure ead at point () wen te valve is partly closed so tat te flow rate is reduced to 0 dm /s. Te pressure loss is equal to m ead. SOLUTION Fig.. Since point () is a free surface, 0 and u is assumed negligible. Te datum level is point () so z 5 and z 0. Q 0.0 m/s A πd /4 π x (0.05 )/4.96 x 0 - m. u Q/A 0.0/.96 x 0-0.8 m/s Bernoulli s equation in ead form is as follows. u u + z + + z + + L g g 0 + 5 + 0 7.7m 0.8 + 0 + + x 9.8 D.J.DUNN www.freestudy.co.uk 6
WORKED EXAMPLE Te diagram sows a orizontal nozzle discarging into te atmospere. Te inlet as a bore area of 600 mm and te exit as a bore area of 00 mm. Calculate te flow rate wen te inlet pressure is 400 Pa. Assume tere is no energy loss. SOLUTION Fig..4 Apply Bernoulli between () and () ρu ρu p + ρgz + p + ρgz + + pl Using gauge pressure, p 0 and being orizontal te potential terms cancel. Te loss term is zero so te equation simplifies to te following. ρu ρu p + From te continuity equation we ave Q Q u 666.7Q -6 A 600 x 0 Q Q u 5 000 Q -6 A 00 x 0 Putting tis into Bernoulli s equation we ave te following. ( 666.7Q) ( 5000Q) 400 + 000 x 000 x 400 +.89 x0 400. x0 Q 400. x0 Q 89.7 x 0 9-6 9 9 Q Q.5 x 0 6 x0 9 Q m /s or 89.7 cm /s 9 D.J.DUNN www.freestudy.co.uk 7
.4 HYDRAULIC GRADIENT Consider a tank draining into anoter tank at a lower level as sown. Tere are small vertical tubes at points along te lengt to indicate te pressure ead (). Relative to a datum, te total energy ead is T + z + u /g and tis is sown as line A. Te ydraulic grade line is te line joining te free surfaces in te tubes and represents te sum of and z only. Tis is sown as line B and it is always below te line of T by te velocity ead u /g. Note tat at exit from te pipe, te velocity ead is not recovered but lost as friction as te emerging jet collides wit te static liquid. Te free surface of te tank does not rise. Te only reason wy te ydraulic grade line is not orizontal is because tere is a frictional loss f. Te actual gradient of te line at any point is te rate of cange wit lengt i δ f /δl Fig..5 D.J.DUNN www.freestudy.co.uk 8
SELF ASSESSMENT EXERCISE. A pipe 00 mm bore diameter carries oil of density 900 kg/m at a rate of 4 kg/s. Te pipe reduces to 60 mm bore diameter and rises 0 m in altitude. Te pressure at tis point is atmosperic (zero gauge). Assuming no frictional losses, determine: i. Te volume/s (4.44 dm/s) ii. Te velocity at eac section (0.566 m/s and.57 m/s) iii. Te pressure at te lower end. (.06 MPa). A pipe 0 mm bore diameter carries water wit a ead of m. Te pipe descends m in altitude and reduces to 80 mm bore diameter. Te pressure ead at tis point is m. Te density is 000 kg/m. Assuming no losses, determine i. Te velocity in te small pipe (7 m/s) ii. Te volume flow rate. (5 dm/s). A orizontal nozzle reduces from 00 mm bore diameter at inlet to 50 mm at exit. It carries liquid of density 000 kg/m at a rate of 0.05 m/s. Te pressure at te wide end is 500 kpa (gauge). Calculate te pressure at te narrow end neglecting friction. (96 kpa) 4. A pipe carries oil of density 800 kg/m. At a given point () te pipe as a bore area of 0.005 m and te oil flows wit a mean velocity of 4 m/s wit a gauge pressure of 800 kpa. Point () is furter along te pipe and tere te bore area is 0.00 m and te level is 50 m above point (). Calculate te pressure at tis point (). Neglect friction. (74 kpa) 5. A orizontal nozzle as an inlet velocity u and an outlet velocity u and discarges into te atmospere. Sow tat te velocity at exit is given by te following formulae. u { p/ρ + u } ½ and u {g + u } ½ D.J.DUNN www.freestudy.co.uk 9
PRESSURE LOSSES IN PIPE SYSTEMS. REVIEW OF EARLIER WORK FRICTION COEFFICIENT Te friction coefficient is a convenient idea tat can be used to calculate te pressure drop in a pipe. It is defined as follows. Wall Sear Stress C f Dynamic Pressure p ½ ρ u m From Poiseuille s equation DARCY FORMULA D p τ o 4L Wall Sear Stress D p Cf Dynamic Pressure 4Lρu m µ Lu D µ Lu 6µ C f D 4L u ρ m D ρu md m p Hence 6 R e 4Cf Lρu m p D Tis is often expressed as a friction ead f p 4Cf Lu m f ρg gd Tis is te Darcy formula. In te case of laminar flow, Darcy's and Poiseuille's equations must give te same result so equating tem gives 4Cf Lu m µ Lu m gd ρgd 6µ 6 Cf ρu D R Tis is te same result as before for laminar flow. m e A formula tat gives an approximate answer for any surface rougness is tat given by Haaland. 6.9 ε.6 log0 + Cf R e. 7 Tis gives a very close model of te Moody cart covered earlier.. D.J.DUNN www.freestudy.co.uk 0
WORKED EXAMPLE 4 Determine te friction coefficient for a pipe 00 mm bore wit a mean surface rougness of 0.06 mm wen a fluid flows troug it wit a Reynolds number of 0 000. SOLUTION Te mean surface rougness ε k/d 0.06/00 0.0006 Locate te line for ε k/d 0.0006. Trace te line until it meets te vertical line at Re 0 000. Read of te value of C f orizontally on te left. Answer C f 0.0067 Ceck using te formula from Haaland. C C C C C f f f f f.6 log.6 log.6 log.06 0.0067 0 0 0. 6.9 ε + R e.7 6.9 0.0006 + 0000.7 6.9 0.0006 + 0000.7.. D.J.DUNN www.freestudy.co.uk
WORKED EXAMPLE 5 Oil flows in a pipe 80 mm bore wit a mean velocity of 4 m/s. Te mean surface rougness is 0.0 mm and te lengt is 60 m. Te dynamic viscosity is 0.005 N s/m and te density is 900 kg/m. Determine te pressure loss. SOLUTION Re ρud/µ (900 x 4 x 0.08)/0.005 57600 ε k/d 0.0/80 0.0005 From te cart C f 0.005 f 4C f Lu/dg (4 x 0.005 x 60 x 4 )/( x 9.8 x 0.08).7 m p ρg f 900 x 9.8 x.7. kpa. D.J.DUNN www.freestudy.co.uk
. MINOR LOSSES Minor losses occur in te following circumstances. i. Exit from a pipe into a tank. ii. Entry to a pipe from a tank. iii. Sudden enlargement in a pipe. iv. Sudden contraction in a pipe. v. Bends in a pipe. vi. Any oter source of restriction suc as pipe fittings and valves. Fig.. In general, minor losses are neglected wen te pipe friction is large in comparison but for sort pipe systems wit bends, fittings and canges in section, te minor losses are te dominant factor. In general, te minor losses are expressed as a fraction of te kinetic ead or dynamic pressure in te smaller pipe. Minor ead loss k u /g Minor pressure loss ½ kρu Values of k can be derived for standard cases but for items like elbows and valves in a pipeline, it is determined by experimental metods. Minor losses can also be expressed in terms of fluid resistance R as follows. u Q 8Q A π D L k k k RQ Hence R 4 4 π D 8ρgQ p 8kρg π L k RQ ence R 4 4 π D D Before you go on to look at te derivations, you must first learn about te coefficients of contraction and velocity. 8k D.J.DUNN www.freestudy.co.uk