Modelling of dispersed, multicomponent, multiphase flows in resource industries Section 3: Examples of analyses conducted for Newtonian fluids Globex Julmester 017 Lecture # 04 July 017 Agenda Lecture # Student projects students top 3 selections Section 3: Examples of analyses conducted for Newtonian fluids 3.1 Motivation & context 3. Energy loss (friction) in pipe flow 3.3 Constitutive relationships for laminar Newtonian flow 3.3.1 Pipe flow 3.3. Concentric cylinder viscometer 3.4 Some basics re: turbulent flow 3.5 Terminal settling velocity 1
Review of Lecture #1: Applications of the fundamental laws Fluid flows obey conservation laws for mass, momentum and energy The laws can be stated in differential form or in integral form The integral and differential forms can be derived from each other! Identify important process variables Apply relevant physical laws Physical problem Make reasonable assumptions and approximations Apply appropriate boundary and initial conditions A set of differential equations Apply applicable solution technique Solution of the problem Section 3: Examples of analyses conducted for Newtonian fluids 3.1. Motivation and context The primary goal of this course is to demonstrate the differences between the flow behaviour of simple Newtonian fluids and more complex industrial multiphase flows In order to do this we need to have a good foundation in the behaviour of Newtonian fluids Additionally, this section allows us to use some of the different control volume analyses we introduced in Lecture #1
3. Energy loss (friction) due to flow in pipes: the Mechanical Energy Balance 1 4 Assumptions: Incompressible flow Steady state 3 h (3.1a) (3.1b) P P V V g g g 4 3 4 3 h4 h3 hp ht hl P P V V 4 3 4 3 gh 4 h3 Wp WT w HEAD form (units = meters of fluid) Energy form (units = J/kg fluid) The relationship between lost work (lw) maj and wall shear stress ( w ) Consider incompressible, steady-state, pipe flow: We consider a fluid element of length L : Q R r z L Just to make things easier, we look at horizontal flow; but the relationship between lost work and shear stress is valid for all pipe orientations! 3
Z-direction force balance on fluid element F s w SA P 1 A P A PA PA SA 1 w 0 (P P ) R ( RL) 1 w 0 P L R w 4 w P dp L dz D (3.N) Horizontal, steady-state, incompressible pipe flow Consider same pipe flow, this time using the Mechanical Energy Balance: 1 Q r z MEB, 1 P P 1 w or P L L w or w 4w Thus: 4 w L w (3.3) L D D 4
The shear stress decay law Again, let s consider that same FLUID ELEMENT in incompressible, steady-state, horizontal pipe flow: Q r z Here, we found that: P L D 4 w Now, let s consider a new FLUID ELEMENT with r < R: Q r z dz A z-direction force balance on this fluid element: PA (P dp)a ( r dz) 0 rz dp dz rz r dp w dz R r rz rz w r R (3.4) 5
Shear stress decay law rz w r R (3.4) Applies to laminar and turbulent flow Applies to Newtonian and non- Newtonian fluids Must be steady state Must be a constant-density system (e.g. no variation in density from top to bottom of pipe) Impact: r = R (@ pipe wall) Q r z rz w r R 6
Example 3.1 Basic wall shear stress calculations The wall shear stress in a portion of a 0.5 m (diameter) pipe that carries water is 11.7 Pa. The flow is fully developed. Determine the pressure gradient dp/dx (where x is in the flow direction), if the pipe is a) horizontal b) Vertical with flow up c) Vertical with flow down Also, what is the shear stress at r = 0.8R for each of the three pipe orientations described above? Calculating friction losses (i.e. w ) for single-phase flows Steady state or transient? Incompressible or compressible? Newtonian or non-newtonian? Laminar or turbulent? 7
Wall shear stress ( w ) for Newtonian fluids w fv This equation defines the Fanning friction factor, f LAMINAR FLOW Derived relationship between w and V exists Can directly show that: f 16 Re TURBULENT FLOW Experimental relationship between w and V found: f fn Re,k D Re DV f (3.5) f Can be obtained from dimensional analysis of the Navier-Stokes equation Represents the ratio of carried (inertial) momentum to molecular (viscous) momentum transfer Re < 100: Laminar flow (typically) Re > 4000: Turbulent flow (typically) 8
Determining friction losses for Newtonian fluids h L w f Calculating f Moody diagram Correlations e.g. Churchill, Swamee-Jain 9
Moody diagram Correlations: the Churchill Equation (1977) 1 8 1 f Re A B 3 1 1 (3.6a) 0.9 7 A.457 ln 0.7 Re D 16 (3.6b) 16 37530 B Re (3.6c) NOTE: the hydrodynamic roughness is sometimes referred to as and sometimes k! 10
Correlations: the Swamee-Jain equation (1976) f log 0.065 D 5.74 3.7 Re 10 0.9 (3.7) Example 3.: Water pipeline calculation 1 Fluid properties = 1 mpa.s = 1000 kg/m 3 D = 100 mm; L = 1. km Q = 746 m 3 /day; P 1 =P k = 0.045 mm Find: (a) Wall shear stress (in Pa) and dp/dz (in Pa/m); (b) Pump head required (in meters of water) 11
Course project updates Assignment of projects Comments on Oral presentation #1 Motivation what is the industrial relevance of your project, why is the topic important Objectives what are your goals for this project? Example: To determine the 3 main sources of air pollution in Beijing Key resources (maximum 3) How will you decide which ones are best? What criteria could you use to make this determination? Agenda Lecture # Student projects students top 3 selections Section 3: Examples of analyses conducted for Newtonian fluids 3.1 Motivation & context 3. Energy loss (friction) in pipe flow 3.3 Constitutive relationships for laminar Newtonian flow 3.3.1 Pipe flow 3.3. Concentric cylinder viscometer 3.4 Some basics re: turbulent flow 3.5 Terminal settling velocity 1
3.3. Constitutive relationships for laminar Newtonian flow 3.3.1 Pipe flow 3.3. Concentric cylinder viscometry 3.3.1 Laminar, Newtonian pipe flow Part 1: Review of the microscopic conservation laws Part : Write and simplify the equations for steady, incompressible, 1D, laminar, Newtonian pipe flow 13
The microscopic equations: continuity, momentum Assumptions: Cartesian (x, y, z) co-ordinates Velocity components: vx v x(x,y,z,t); vy v y(x,y,z,t); vz v z(x,y,z,t) Continuity Equation: vx vy vz0 t x y z Momentum Equations: P x-direction: v v v v v v v g t x y z x y z x P y-direction: v v v v v v v g t x y z x y z y z-direction: x x x y x z x xx yx zx x y x y y y z y xy yy zy y P v v v v v v v g t x y z x y z z z x z y z z z xz yz zz z The microscopic equations: continuity, momentum Assumptions: Cartesian (x, y, z) co-ordinates Velocity components: vx v x(x,y,z,t); vy v y(x,y,z,t); vz v z(x,y,z,t) Continuity Equation: vx vy vz0 t x y z 14
The microscopic equations: continuity, momentum Assumptions: Cartesian (x, y, z) co-ordinates Velocity components: vx v x(x,y,z,t); vy v y(x,y,z,t); vz v z(x,y,z,t) Momentum Equations: (only x-direction shown here) P v v v v v v v g t x y z x y z x x x x y x z x xx yx zx x The microscopic equations for steady, incompressible, 1D, laminar, Newtonian pipe flow What we know: Cylindrical coordinates Note: In Section 3, u and v are used interchangeably to mean local fluid velocity 15
The microscopic equations for steady, incompressible, 1D, laminar, Newtonian pipe flow (continued ) Continuity 1 1 rur u u z0 t r r r z Conservation of momentum: z-direction 1 u u u u u u u 1 r 1 P g t r r z r r r z z z r z z z z rz z zz z so: duz dr Pz r P u r C 4 z z Since u z = 0 at r = R, we can find C and write the Poiseuille equation: PR r 4 R z uz 1 (3.8) Q u da + Q VA A z PR z PD V 8 3 z Since the flow is horizontal: 4w Pz D and V 8 w D (3.9) 16
Steady, laminar pipe flow of a Newtonian fluid (summary) Microscopic: Macroscopic: du dr z rz + Newton s Law of Viscosity rz w r R Shear stress decay law w 16 f V Re VD Re wr r uz 1 R Poiseuille s Equation (3.8) w 8V D Q u da + Q VA A z w 8V D (3.9) Example 3.3 A new heat exchange fluid (SG 1.90) has been developed by our company s R&D department. They have asked us to determine its flow properties. A 38 mm horizontal pipe is used for this purpose. The data shown in the table below are collected at a constant operating temperature of 130ºC. a) Is the fluid Newtonian? Use a graph to explain your answer. b) Calculate the fluid s best-fit rheological parameters. c) Prove that none of the data points listed here should be rejected from our analysis. V (m/s) 0.75 1.05 1.45 1.9.7 3.5 -dp/dz (kpa/m) 1.08.0.43 3.96 5.83 6.55 17
70 60 50 40 w (Pa) 30 0 10 0 0 0.5 1 1.5.5 3 3.5 4 V (m/s) 70 60 50 40 y = 18.581x R² = 0.9683 w (Pa) 30 0 10 0 0 0.5 1 1.5.5 3 3.5 4 V (m/s) 18
Solution Newtonian, based on shape of curve (straight line, through origin) m = slope of line = 18.58 = 8/D Therefore: = 18.58 D / 8 = 0.088 Pa.s Check for laminar flow: We need to ensure that Re 100 for all data points V D max 100 or V max 100 3.8m/s D We see therefore that all the data points collected here were collected in laminar flow and all can be used to determine. 3.3. Laminar Newtonian flow in a concentric cylinder viscometer Spindle Cup 19
Concentric cylinder viscometer parameters T = Applied Torque = Angular Velocity L R 1 = Spindle Radius R = Cup Radius L = Spindle Length 3.4. A (very) brief introduction to turbulence Turbulence is ubiquitous in natural and industrial processes How would you describe turbulence? 40 0
*Possible descriptions of fluid turbulence Not laminar (!) Three-dimensional, unsteady, chaotic motion - characterized by random and rapid fluctuations of swirling regions of fluid The fluctuations have a wide range of different length and time scales The fluctuations provide an additional mechanism for momentum and energy transfer (over laminar flow) Turbulence is a continuum phenomenon. The smallest scale of turbulence is MUCH larger than the molecular scale of the fluid It is a characteristic of the flow and NOT the fluid Associated with high Reynolds numbers *G. Ahmadi, ME 637 course notes, Clarkson University 41 Fluctuating and time-smoothed velocity components: u z (t) t u(t) u u z z 4 1
Evaluation of the time-smoothed velocity components in the direction of flow V = Q/A 1 y TURBULENT y/d 0 v There are three regions in a turbulent pipe flow: Inertial region Distance from pipe wall Buffer zone Viscous sublayer Velocity
Shear velocity Shear or friction velocity: u* w f (3.11a) or u 1 fv / V f / * f f (3.11b) Fundamental (and useful) parameter in describing turbulent flow The velocity fluctuations in a turbulent flow have the same order of magnitude as the shear velocity The time-averaged velocity distribution can be described based on the shear velocity! Turbulence scaling 5 0 u u u * 15 10 5 Viscous sublayer (inner layer) Inertial region (outer layer) 0 0.1 1 10 100 1000 yu y * Viscous sublayer (inner layer) u y (0 y 5) (3.1a) Inertial region (outer layer) u.5 ln y 5 (30 y 500) (3.1b) 3
3.5. Terminal settling velocity (v ) Critically important parameter in assessing / predicting the performance of many industrial systems, including: Fluidized beds Separation equipment (gravity separation, hydrocyclones, mechanical flotation) Slurry pipelines Sphere in Newtonian fluid Consider a single spherical particle immersed in a Newtonian liquid. f d s Net downward gravity force d = Upward drag force d 3 s fg 0.5CDf v 6 4 v volume Drag coefficient projected area so v 4gd 3C 1/ s D f f (3.13) 4
d: Particle diameter (m) s : Particle density (kg/m 3 ); often must be measured C D : Must calculate, based on Re p v fd (3.14) f Stokes Law Intermediate Newton s Law Boundary Layer Separation C D Re p 5
Drag Coefficient Correlations (3.15a) Stokes Law C D = 4/Re p Re p 0.3 (3.15b) Intermediate C D = Re 4 0. (1 0.15 Re 687 p p ) 0.3< Re p 1000 (3.15c) Newton s Law C D ~ 0.445 1000<Re p <10 5 (3.15d) Boundary Layer Separation C D ~ 0.05 Important equation to know: Stokes settling v 4gd 3C 1/ s D f f v fd (3.13) Re p (3.14) f When Stokes Law applies (Re p 0.3): C D Therefore: 4 Re p v dg( s f) 18 f (3.16) Stokes Law 6
Solution methods **Spherical particles, Newtonian fluid** Iterative Direct Wilson method Wilson et al.* direct method Spherical particles, Newtonian fluid *Wilson et al., Direct prediction of fall velocities in non-newtonian materials, Int. J. Miner. Process. 71(003) 17-30. Pipe flow Particle settling Shear velocity u* w f v* f (3.17) Characteristic shear stress Shear Reynolds number w = wall shear stress Re * f ud * f = particle surficial stress Re * f vd * f s f gd 6 (3.18) (3.19) Velocity ratio uu * v v* 7
Wilson et al. direct method **Spherical particles, Newtonian fluid** v* f Combining Equations (3.17) and (3.18) gives: 6 s f gd (3.17) (3.18) (3.0) v gd 6 * s f f v v * Wilson et al. direct method **Spherical particles, Newtonian fluid** v gd 6 * s f f (3.0) Re * vd f * (3.19) f v Re*.80 v* 1. 4 3. 31 0.08Re* 13x10 Re* For Re* 10 (Includes Stokes regime) 1.7 log v v 0.069 0.500x 0.158x * where x logre 10 * For 10 < Re* < 60 v v* 4.4 For Re* 60 8
Example 3.4 In the 1970 s and 80 s, the idea of transporting coal-in-oil slurries was proposed, as the feed could be sent directly to a thermal power plant without drying the coal. Use the Wilson et al. direct method to determine the terminal settling velocity of a coal particle (d = 150 m, s = 1500 kg/m 3 ) in a Newtonian oil ( f = 876 kg/m 3 ; f = 4.0 mpa.s) Homework for tomorrow Assign #1 (see Globex site) 9