CE10048 - TRANSPORT PHENOMENA I Dr. SEAN P. RIGBY TOPICS: Revision and background: Newtonian Mechanics Types of fluids and fluid properties Continuity, Bernoulli and Momentum Equations Application of the above equations Flow measurement Pressure drop in pipes and fittings Laminar and turbulent flow Dimensional analysis Why are we studying Transport Phenomena ? In order to design a chemical plant we need to be able to construct a mathematical model of our new idea to see whether it will make the required amount of product from the given raw materials. Plantsgenerally involve chemicals, in the form of fluids (liquids, gases), moving around in pipes, vessels etc. Hence, in order to know howour design idea will perform, we need to be able to predict how fluids will behave in our plant. In order to monitor and control our plant oncewe have built it, we also need to be able to measure fluid flow rates. BOOKS (i) Chemical Engineering series Volume 1, Coulson and Richardson(ii)Transport Phenomena, Bird, Stewart and Lightfoot (iii) Introduction to Fluid Mechanics & Transfer Processes, Kay & Nedderman 1. Fluids 1.1 Properties Density Liquid: slight variation with temperature Gases: marked variation with temperature Units: mass density = kg/m3;molar density = mol/m3 - differ by molecular mass Specific gravity: ratio of density of substance to that of water Viscosity, = Resistance to shear e.g. 2 planes, one fixed, one moving, both of area A Velocity gradient = du/dy = u/h (if uniform gradient) Shear stress = t = FORCE/AREA = .du/dy Dynamic viscosity, = ( F.h )/( A.u )Units = N.s/m2 = kg/m.s Fixed h Force, FVelocity, u Fluid filled gap Kinematic viscosity, v = /Units = m2/s Pressure p = Force per unit area Measured using a barometer and other devices/methods In a stagnant fluid, pressure variation with depth: Force on bottomA.p Force on topA.( p + op ) WeightA.oh..g A.p = A. ( p + op ) +A.oh..g Hence dp/dh = -.g p p + op Area, A h oh U-tube manometer Pressure at A= p = pressure at A/ Pressure at B = p - .g.h= atmospheric pressure, pa Thus: p - pa = .g.h - measures pressure differences = density of manometer fluid Absolute pressure- genuine pressure, force/unit area, N/m2 More often in bar = 105 N/m2 Gauge pressure- pressure above pressure in the surroundings(usually atmospheric) Absolute pressure - bar absolute, bara Gauge pressure- bar gauge, barg Gasfilledvessel p A B A/ h 1.2. Fluid manometer Pressure at A = p1 Pressure at C = p1 + w.g.h1 Pressure at D = p1 + w.g.h1 + Hg.g.h Pressure at B = p1 + w.g.h1 + Hg.g.h- w.g.(h+h1) = p2 Thus:p2 - p1 = (Hg - w).g.h = A.g.h so if A is small, h is large - hence high accuracy Pressures often expressed asm of fluid, e.g. mH2O, mHg, mHg/H2O Pressures also often expressed as head:H = p/(.g) where is the density of the fluid A p1 B p2 Pump h1 C h D Hg Water 1.3 Laminar and Turbulent flowReynolds 1883 FlowDye Low velocities Laminar Flow High velocities Turbulent Flow Laminar flow- when viscous forces are dominating the situation - viscous flow Transition is sudden Transition point U = mean velocity of fluid in pipe (= volumetric flow/area) d = pipe diameter Units: 2200 is dimensionless, and known as the Reynolds number 2200 = . d . U=kgs . m.mkg.m.sm3U t U t U t Laminar Flow Small scale turbulence Re = 2000 to 105 Small fluctuations about a well- defined mean Gross Turbulence Variation in time of instantaneous velocity at a point 2. IDEAL FLUIDS Assume no viscosity, thus velocity is uniform across a cross-sectionof flow - fits turbulent flow better than laminar flow (see later) 2.1. Continuity Equation Flow 1 2 A1, v1 A2, v2 Control surface - boundary of space under consideration Apply Conservation of mass between 1 and 2 Mass flow rate, W = A1.v1.1 = A2.v2.2

Often 1 = 2 , e.g. for liquids If not ideal, v not constant (across cross-section): }=1AdA v W . . 2.2. Bernoullis Equation In steady flow a streamline is a line along which the fluid flows. Takea control surface with the sides along streamlines (i.e. no flow across sides) 1 2 h1 h2 Small area Stream tube Consider a small mass of fluid m entering at section 1 Volume = mV1where V1 is the specific volume = 1/1 = 1/ Work done by fluid entering tube: = p1.m.V1 = m.p1/(pressure x volume or F.d ) If incompressible Fluid leaving tube does work = p2.m.V2 = m.p2 / Net work done by surroundings on fluid in tube =m.(p1 - p2)/ By Conservation of energy, this appears as: 1) gain of kinetic energy =m(v22 - v12)/2;and 2) gain in potential energy =m(h2 - h1)g( )( )1 22122 2 12 2h h mgv vmp p m +||.|

\| =

222 2121 12 2ghv pghv p+ + = + + Bernoullis equation Note units of each term are J/kg - i.e. energy per unit mass of fluid Equation allows you to keep track of the conversion of energy In this crude derivation we have assumed: 1) Constant density 2) Steady state (no allowance for accumulation of energy) 3) No interchange of mechanical and thermal energy 4) No forces on the side of the stream tube (e.g. friction from apipe wall) More correct form: L ghv pghv p+ + + = + +222 2121 12 2 Where L= lost energy (per unit mass) due to any of the above factors. L is negative if there is a pump in the stream tube (an energy gain) Bernoulli is best over short distances in smoother turbulent flows. Never exact - but often good enough A y h v H pa pa Consider an element of fluid that starts at A and comes out of orifice At A: v1 ~ 0 p1 = pa + .g.y h1 = -y In jet: v2 = v;p2 = pa;h2 = -H Apply Bernoullis equation: gHv pgygy pa a + = ++2 202 2 pa cancels, also y, thus gH v 2 =However, this is the prediction of an oversimplified theory... Arbitrary datum level Actual velocity = Cv - velocity coefficient - ratio of actual to predicted velocity - usually Cv ~ 0.99 Theory works because velocity gradients are small except in a very small region near orifice. Observe Jet Minimum area = Vena Contracta Contraction due to momentum of the fluidapproaching from sides. For a sharp edgedcircular orifice, Cc ~ 0.65 (Cc = contraction coefficient = ratio of areas: vena contracta:orifice) Mass flow rate: gH C A C Av Wv o c2 = =gH A Co D2 =Ao = orifice area (x Cc gives vena contracta area) Ao Discharge coefficient gH Cv2CD - ratio of actual mass flow rate to that predicted assuming no energy losses and no contraction. CD = CcCv ~ 0.64- due almost entirely to contraction Terminology gLhgvgphgvgp+ + + = + +222 2121 12 2 Pressurehead Velocity head Potential head TOTAL HEAD Lost head (all terms have units of length) 2.3. Measurement of VelocityPitot Tube At A: pressure, pvelocity, v At B:p = p0v = 0 AB horizontal Apply Bernoullis equation from A to B: FLOW A B Velocity zero No flow in tube To somepressuremeasuring device 2 22002vp pp v p + = = +p0 is the stagnation pressure (pressure in fluid when brought to rest) v2/2 is the dynamic pressure (the extra pressure associated with K.E.) p is the static pressure (the pressure in the moving liquid) B A C p0 = p + v2/2 Cannot measure pA but can measure pC But pA = pC, otherwise there would be a transverse force on thefluid leading to an acceleration perpendicular to the surface pBpA No pressure gradient normal to straight parallel streamlines. Connect tapping to manometer - record height h pB - pC = (m - ).g.h = .v2/2Hence find v If flow not in duct or if distance AC is large, need better way ofmeasuring pA. Pitot-Static Tube Stagnation pressure Pitot tube Static pressure Coaxial tube with a ring of small holes in side h As before: (m ).g.h = .v2/2 Problems If front too blunt, streamlines curved near static holes - static pressure measurement is then unreliable Example: a Pitot-static tube in air-water manometer - records 25 cm water Ap = h.(w - air ).g = (0.25).(1000).(9.81) (ignore air as small) = 245 N/m2 = .v2/2 = (0.5).1.29.v2Hence v = 19.5 m/s Pressure difference leads to density difference - but we have assumedconstant density. Provided velocity is less than one third of the velocity of sound, we can ignore density changes. 2.4 Measurement of Flow Rate 2.4.1. Venturi Meter Gradual contraction, followed by gradual expansion (1) (2) (3) p1p2 Continuity Equation (assume constant density) A1v1 = A2v2 = A3v3Usually A3 = A1, hence v3 = v1 Apply Bernoullis equation: 2 222 221 1v p v p+ = + ( )(((

||.|

\|= =12 212212121222 1AA vv vp p( ) ( )2122 122212 11 1 11212|.|

\|=|.|

\|= =AAp pAAAp pA v A W Because of losses, extra contraction, non-uniform velocity, insert adischarge coefficient, CD ( )122212 11|.|

\|=AAp pA C WDFor a well designed Venturi: CD~0.96-0.98 Usually A3 = A1, v3 = v1apply Bernoulli 2 2 223 322 221 1v p v p v p+ = + = + Ideally p3 = p1 In fact large losses in the divergent portion often means p3 < p1 Define Diffuser Efficiency as q =Bernoulli by predicted rise essure Prrise pressure Actual( )2322 2 321v v p p = qIf angle too large get stagnant fluid here Jet Diffuser angle must be small >A2, v3 Compare with:p1 - p2 = .( v22 - v12)/2from Bernoulli above Ratio: ( )( )( )( )2 1121221 2 12 12 322v vvloss essure Prery cov re essure Prv vv v vp pp p+= ==( ) ( )2 1 2 3 1 2p p p p v v < >i.e. pressure recovery ALWAYS less than original pressure loss Summary Apply Bernoulli from point (1) to (2) (assuming no energy losses) p1 - p2 = .( v22 - v12)/2 (1) Apply Momentum equation from point (2) to (3) (assuming no force on the wall) (2) Subtract:(3) ( )1 2 1 2 3v v v p p = ( )22 1 3 12v v p p = Apply Momentum equation from point (1) to (2) R v v A p A v A p A + + = +2 1 1 2 121 1 1 1 W Force on orifice plate ( ) ( )2 1 1 1 2 1 1v v v A p p A R + = eliminate (p1 - p2 ) using eq. (1)... ( )22 112v vAR =Force on orifice plate [Overall Momentum balance from point (1) to (3): A1p1 + A1v12 = A1p3 + A1v12 + R thus R = A1(p1 - p3 )(as before)] Apply Bernoulli from (2) to (3): By Continuity Eq. v1 = v3 Lv p v p+ + = +2 223 322 2 Substitute for (p3 - p2) using eq. (2) from above: L = (v1 - v2 )2/2 Apply Bernoullis eq. From pt. (1) to (3)( )2 2 222 1 3 123 321 1v v p pL Lv p v p == + + = + v3 = v1 From eq. (3) What happens to the lost energy ? - goes to thermal energy L = lost energy per unit mass = cAT, where c is the specific heat capacity ( )cv vT222 1= AExample:v1 = 5 m/s(very fast flow) v2 = 10 m/s (area of vena contracta = A1/2) Assume flowing fluid is water: ( )K . T 003 04187 210 52== A2.5.2. Sudden Pipe Expansion A1 v1 A2 v2 (1)(2) As the fluid goesfrom point (1) to point (2) there is a loss of momentum: W.(v1 - v2 ) = A2.v2..(v1 - v2 ) = applied force (on region (1) to (2) ) = p2A2 - p1A2 ( )2 1 2 1 2v v . v . p p = Because p1 acts overwhole cross-section Loss ofenergy = ( )2 2 2 222212 1 222 221 1v vv v vv p v p + =||.|

\|+ + ( )(((

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\| ==2212222 112 2 AA v v vSudden Pipe Contraction (1)(2)(3) Exactly the same as orifice plate calculation except that A1 is not equal to A3 2.5.3. Effect of External Pressure - (use of absolute and gauge pres.) (1)(2) pa pa pa pa Bernoullis equation 2 222 221 1v p v p+ = + p1, p2 - absolute pressures Gauge pressure p/, hence p1 = p1/ + pa p2 = p2/ + pawhere pa is atmospheric pressure2 222 221 1v p p v p pa/a/+ + = + + Exactly the same equation - Bernoullis eq. can use either absolute or gauge pressures,OR measure pressure above any arbitrary datum. Momentum equation: (p1A1 + A1v12) - (p2A2 + A2v22) = F(Force on wall) If we substitute p1 = p1/ + pa, terms A1pa and A2pa do not cancel. MUST use ABSOLUTE pressures when using the Momentum Eq. F is force on inside surface of the wall. There is also a force on theoutside surface of the wall. Put cork in each end. Net force due topa must be zero. Net force is due to force on curved surface and force on cork (1) (A1pa) and force on cork (2) (A2pa ). Hence: Force on outer surface of pipe = pa ( A1 - A2 ) to the left Total right-ward force on tube: FT = (p1A1 + A1v12) - (p2A2 + A2v22) - pa ( A1 - A2 )On inside surfaceForce on outside surface FT = (p1/A1 + A1v12) - (p2/ A2 + A2v22) If we want force on INSIDE surface we must use ABSOLUTE pressures. If we want TOTAL force on pipe we must use GAUGE pressures. 2.5.4. Force on a Pipe Bend Arithmetical Example: 60o D1 = 40 mm (1) (2) 331.4 N 155.9 N Water discharges to atmosphere at (2) at a flow rate of 0.007 m3/sD2 = 20 mm s / m ...AQv 57 504 04007 0211== =ts / m ...AQv 28 2202 04007 0222== =tp2/ = 0 since fluid discharges to atmosphere at point (2) Bernoullis eqn. :-p1/ = p2/ + .( v22 - v12 )/2 = 0+1000(22.282 - 5.572)/2 = 2.327x105 Nm-2 Impulse function at (1): F1= A1p1/+ A1..v12 F1 = (t/4)x0.042{2.327x105 + 1000x5.572} = 331.4 N F2 = (t/4)x0.022{0 + 1000x22.282} = 155.9 N To get overall force 60o 331.4 155.9 X YR o X = 331.4 - ( 155.9.cos60o ) = 253.4 N Y = 155.9 sin60o = 135.0 N R = ( X2 + Y2 )1/2 = 287 N tan o = Y/X, thus o = 28o 2.6. Unsteady Flows Bernoulli can only be used in steady flow - we ignore any accumulation of energy. Problems of moving geometry can often be converted into steady flow problems by giving the whole system some velocity, e.g. A B v vvelocity = 0 Measure velocity of an aircraft by Pitot tube Coordinate system with plane moving: Apply Bernoulli from point Ato point B (top): - wrong, since not steady state Now give whole system velocity -v (bottom) to bring aircraft to rest with respect to coordinate system. Now have steady state. 22v p pB A+ = B Ap v p= +22- Correct Have a force on the aircraft. If aircraft moves, then work will be done and this must be incorporated into the energy balance. However, if the aircraft is stationary, no work is done and Bernoulli is CORRECT. 3. Laminar Flow Situations in which viscous forces dominate. The Reynolds number (Re)is less than the critical value for the Laminar/Turbulent transition. For acylindrical pipe the transition occurs at Re ~2000-2300. For other geometries the value of Re at the transition takes other values. A Newtonian Fluid is one for which the expression for the shear stress takes the form: dyduareaforce t = =where is the viscosity, u is the fluid velocity in the x-direction. This formula is an approximation (i.e. a special case) of the fullexpression: which you will meetin later years. ||.|

\|cc+cc=xvyu ty x u v For simple flow patterns we can put the x-axis in the direction of flow, hence v = 0, and dydu t =Because of the sign convention for shear stressesAA y u1 u2 t t Velocity curve Fluid above plane A pulls the fluid below to the right. Fluid below pulls fluid above to the left. (1) u2>u1 Experimental Observation Gases - kinetic theory of gases predicts agreement with Eq. (1) above which agrees well with experiment. Liquids - Experiments show that Eq. (1) is valid formost lowmolecular weight materials. Minor differences are observed for high molecular weight materials. Polymers show elastic as well as viscous effects. Aside An example of a Non-Newtonian fluid is a Power Law fluid for which there is a different relationship between t and du/dy:- ndydum||.|

\|= twhere m and n are constants characteristic of a given fluid. 3.1. Viscous Liquid Flow down a Vertical Plane y u velocity profile Velocity at the wall = zero Known as The No-Slip Boundary Condition Generally valid, except for liquid polymers andrarefied gases Velocity increases with distance from the wall so du/dy is a positive quantity. y h L Take shaded element, with volume L x ( h - y ) x unit distance into plane ofpaper Consider unit distance normal to paper: Volume of element = L.( h - y ). 1 Weight = L..g.( h - y) = t x L x 1 Hence: t = .g.( h - y ) y 1 the flow pattern is very complicated - so very little theory exists to predict the relationship between CD and Re. Forces on smooth spheres - Experimental results Re < 1000 1000 < Re < 105 CD = 0.44 Re > 105 CD = 0.1 Critical Re of 105 depends on surface roughness. 4.2 Example - Determining Drag Forces when Designing Ships Ships travel over the sea where there are waves. Waves involvevertical displacements. Therefore gravity is involved. Assume drag force, F = f(d, v, , , g) where d is a characteristic dimension of the ship, v is the velocity ofthe ship, and and are the properties of the liquid on which it floats. ( )687 015 0 124.DRe .ReC + =We have: six parameters, 3 dimensions - thus 3 groups Choose standard groups: vdRe=|||.|

\|=2 2 2 2214v . . . d .Forv . . dFCDtThird group must contain g: d . gvdv~TL2 22is dimensionlessand is called the Froude number All three groups are independent, since: only Re contains only CD contains F only Fr contains g In order to make a model ship to enable the drag force on the real ship to be determined, need to operate at the same Reynolds and Froudenumbers as the full size ship by e.g. changing the speed or fluid. 5. Turbulent Flow Through Pipes L L Consider length L - measure pressure drop AP. Next section of pipe - same pressure drop - unless there is a markedchange in density. Unless the velocity is large ~ 100 m/s OR thepipe is very long, density changes will be small. Overall pressure drop o Length of tube Consider length L of diameter D: P Flow t L P - AP t = wall shear stress ( ) DLDP PDP tttAt+ =4 42 2LD . P4At = Constant along tube In turbulent flow, need to consider any 2 of the 3 properties t, D and AP/L. It is convenient to work in terms of t and D. t = f( v, D, , ) where v = Q/A i.e. the mean velocity Have 5 parameters and 3 dimensions thus there are 2 dimensionless groups, of which the Re is the obvious and convenient one. 2nd group includes t which is a force/unit area =2 2 2LTML TML=Eliminate mass by either (or ) 2222 32vv ~TLML.LTMtt = = is dimensionless 5.1. Fannings Friction Factor is defined as:221vCft=Note: There are other definitions of friction factor Darcy Weissbach Friction Factor = 4Cf (used mainly by civil engineers) Coulson & Richardson define friction factor as For a smooth walled pipe Cf = f(Re) only In laminar flow:2v t48aLQPtA =where v = Q/ta2andD = 2a 2 232 48DLVPDLV PA A = = DvLD . P At84= =From force balance: Re D . v .vCf1616212= = =tValid for Re