Entropy Generation and Exergy Destruction in Flow of Multiphase Dispersions of Droplets and Particles in a Polymeric Liquid

fluids

Article Entropy Generation and Exergy Destruction in Flow of Multiphase Dispersions of Droplets and Particles in a Polymeric Liquid

Rajinder Pal

Department of Chemical Engineering, University of Waterloo, Waterloo, ON N2L 3G1, Canada; [email protected]; Tel.: +1-519-888-4567 (ext. 32985)

Received: 31 December 2017; Accepted: 26 February 2018; Published: 2 March 2018

Abstract: The theoretical background for entropy generation and exergy destruction in the flow of fluids is reviewed briefly. New experimental results are presented on the quantification of exergy destruction rates in flows of emulsions (oil droplets dispersed in a polymeric liquid), suspensions (solid particles dispersed in a polymeric liquid), and blends of emulsions and suspensions (dispersions of oil droplets and solid particles in a polymeric liquid). A new model is proposed to estimate the exergy destruction rate, and hence power loss, in the flow of multi-phase dispersions of oil droplets, solid particles, and polymeric matrix.

Keywords: exergy; entropy; irreversibility; multiphase; dispersion; emulsion; suspension; rheology; non-Newtonian; thermodynamics

1. Introduction Two-phase dispersions (such as oil/water emulsions and solid particles/liquid suspensions) and three-phase dispersions (such as blends of oil droplets, solid particles, and aqueous phase) are encountered in a number of industrial applications [1–10]. A major portion of all the commercial products produced in a modern industrial society are multi-phase dispersions. For example, many food, cosmetic, household, pharmaceutical, and other products of industrial significance, are manufactured and sold in the form of multi-phase mixtures. The manufacturing of these multi-phase products involves mixing, pumping, and flow of multi-phase dispersions in pipelines and other process equipment. Each of these operations are irreversible in nature, and consequently, exergy destruction and power loss occurs in these operations. In order to minimize the power loss and to improve the thermodynamic efficiency of the process, it is important to quantify the irreversibilities in terms of exergy destruction [11,12]. It should be noted that the irreversibility in flow of multi-phase dispersions is due to frictional effects, that is, viscous dissipation of mechanical energy. The viscous dissipation effect in the flow of dispersions is responsible for entropy generation and exergy destruction. In this paper, new experimental results are presented on rheology and exergy destruction rates in flows of two-phase and three-phase dispersions. The systems investigated are emulsions of oil droplets dispersed in a polymeric liquid, suspensions of solid particles (glass beads) in a polymeric liquid, and three-phase blends of oil droplets, solid particles, and polymeric liquid. The theoretical background necessary for the analysis of exergy destruction in flow of fluids is also reviewed. This work is in continuation of our earlier studies on the production of entropy and destruction of exergy in flow of dispersions [13–16]. Our earlier studies were restricted to only two-phase oil and water emulsions, where the matrix phase was a Newtonian fluid. This is the first study on the second law analysis (entropy production and exergy destruction) of multi-phase dispersions of droplets and particles suspended in a non-Newtonian polymeric matrix.

Fluids 2018, 3, 19; doi:10.3390/ﬂuids3010019 www.mdpi.com/journal/ﬂuids Fluids 2018, 3, 19 2 of 18

2. Theoretical Background

2.1. Entropy Balance in Fluid Flow Entropy is an equilibrium property. It can be expressed as a function of other state variables (such as temperature, volume, pressure, internal energy, etc.), provided that the system is in thermodynamic equilibrium. The state variables do not vary with space and time when the system is in thermodynamic equilibrium. There are no gradients or driving forces present in the system that cause the changes. The fluid under flow condition, however, is not in thermodynamic equilibrium. In general, there are present velocity gradients, temperature gradients, and concentration gradients (if two or more chemical components are present) in fluid, causing transports of momentum, heat, and mass, respectively from one region to another. Although the bulk fluid under flow conditions is not in thermodynamic equilibrium, it is often necessary to assume that local equilibrium exists everywhere in the fluid. In other words, it is assumed that all the variables defined in equilibrium are defined locally under non-equilibrium condition as well. This is the key hypothesis of classical irreversible thermodynamics (CIT). The CIT assumes that there exists local equilibrium everywhere in the system, and that the state variables vary with space and time. One way to imagine local equilibrium in a system is to subdivide the system into a large number of tiny cells. The size of the cells is selected such that it is large enough for microscopic fluctuations to be negligible, but small enough for local equilibrium to be realized within the individual cells [17]. Although equilibrium exists locally within each cell, the state of equilibrium varies from one cell to another, so that mass, momentum, and energy exchanges are permitted between neighboring cells. Furthermore, the state of equilibrium within a given cell is allowed to vary with time. Consider an arbitrary volume V, fixed in space and bounded by surface A. The amount of entropy in that portion of the fluid instantaneously occupying the volume V is given by the integral Z S = ρsdV (1) V where S is the total entropy of the system (fluid) instantaneously occupying the volume V, ρ is the local density of the fluid which may vary from one position to another, and s is the local specific entropy of the fluid. The rate of increase of entropy of the system is the sum of two terms: one term that represents the rate at which entropy enters the system through its bounding surface A due to presence of some gradient, and the other term that represents the rate of production of entropy within the system. Thus, entropy balance over the fluid instantaneously occupying the volume V could be expressed as

DS D Z Z → Z . = ρsdV = − nˆ · J dA + σ dV (2) Dt Dt s s V A V

→ where nˆ is a unit vector normal to the bounding surface of the system pointing outwardly, J s is the → entropy ﬂux at the bounding surface of the system ( J s is the rate at which entropy enters the system . through the bounding surface per unit area), σs is the rate of entropy production within the system per unit volume, and D/Dt represents material or substantial time derivative. Using the Reynolds transport theorem, it can be readily shown that

D Z Z ∂ → ρsdV = (ρs)dV + nˆ · ρ v sdA (3) Dt ∂t { V V A

→ where v is the ﬂuid velocity vector. Converting the second integral on the right-hand side of Equation (3) to a volume integral with the help of the Gauss divergence theorem, one can re-write Equation (3) as Fluids 2018, 3, 19 3 of 18

D Z Z ∂ → ρsdV = (ρs) + ∇ · ρ v s dV (4) Dt ∂t V V Combining Equations (2) and (4) gives

Z ∂ → Z → Z . (ρs) + ∇ · ρ v s dV = − nˆ · J dA + σ dV (5) ∂t s s V A V

Converting the ﬁrst integral on the right-hand side of Equation (5) to a volume integral with the help of the Gauss divergence theorem and rearranging gives

Z ∂ → → . (ρs) + ∇ · ρ ν s + ∇ · J − σ dV = 0 (6) ∂t s s V

Because this equation holds for an arbitrary volume V, the integrand must be zero. Thus,

∂ → → . (ρs) + ∇ · ρ v s = −∇ · J + σ (7) ∂t s s Using the following continuity (mass balance) equation along with the deﬁnition of the substantial derivative given below, ∂ρ → + ∇ · ρ v = 0 (8) ∂t D ∂ → = + v · ∇ (9) Dt ∂t it can be readily shown that ∂ → Ds (ρs) + ∇ · ρ v s = ρ (10) ∂t Dt Combining Equations (7) and (10), it follows that

Ds → . ρ = −∇ · J + σ (11) Dt s s . According to the second law of thermodynamics, σs ≥ 0, and therefore,

. Ds → σ = ρ + ∇ · J ≥ 0 (12) s Dt s The “greater than” sign in the above relation is valid for real (irreversible) processes and the “equal to” sign is valid for reversible processes.

2.2. Momentum and Energy Balance Equations for Fluid Flow According to Newton’s second law of motion, the time rate of change of linear momentum of a fluid instantaneously occupying an arbitrary volume, V, fixed in space, is equal to the sum of forces acting upon the fluid occupying V. This principle leads to the following equation of motion for fluids [18]: → D v → ρ = ∇ · π + ρ g (13) Dt → where g is acceleration due to gravity and π is total stress tensor that can be expressed as

π = −pδ + τ (14) Fluids 2018, 3, 19 4 of 18 where p is the local thermodynamic pressure, δ is the unit tensor, and τ is the extra or deviatoric stress tensor (often referred to as viscous stress tensor). Upon substitution of the expression for π from Equation (14) into Equation (13), the equation of motion becomes:

→ D v → ρ = −∇p + ∇ · τ + ρ g (15) Dt Applying the first law of thermodynamics (energy balance) to fluid flow, the following total energy equation for fluids can be obtained:

De →00 → →00 → → ρ = −∇ · q + ∇ · π · v = −∇ · q − ∇ · p v + ∇ · τ · v (16) Dt

→00 where e is the specific total energy of the fluid and q is the heat flux. The mechanical energy equation can be obtained from the equation of motion (Equation (15)) by taking the scalar (dot) product between Equation (15) and the fluid velocity vector as

D → → → → ρ v2/2 = − v · ∇p + v · ∇ · τ + ρ v · g (17) Dt → where v is the magnitude of the velocity vector v . Upon subtraction of the mechanical energy equation, Equation (17), from the total energy equation, Equation (16), the following internal energy equation (also referred to as thermal energy equation) is obtained:

Du →00 → → ρ = −∇ · q − p∇ · v + τ : ∇ v (18) Dt where u is the speciﬁc internal energy of the ﬂuid.

2.3. Application of the Gibbs Equation to Fluid Flow For a single-component ﬂuid (pure substance), the relationship between entropy and other state variables is expressed by the Gibbs equation given below:

ds = T−1du + pT−1dυ (19) where T is the absolute temperature, and υ is the speciﬁc volume of ﬂuid. Strictly speaking, this relationship is valid for a system in thermodynamic equilibrium. It can be used to calculate entropy changes when the system traverses from one equilibrium state to another. However, in classical irreversible thermodynamics (CIT), it is assumed that the local state variables are related to each other by the same equations of state as in equilibrium. Thus, the Gibbs relation (Equation (19)) can be applied to systems locally, even though the systems are in non-equilibrium state. The Gibbs relation, Equation (19), could be re-written in terms of rates as follows:

Ds Du Dυ = T−1 + pT−1 (20) Dt Dt Dt At any given point in the flow field, this equation describes the time rate of change of specific entropy of a fluid element instantaneously located at that point. Combining Equations (18) and (20), the following relation is obtained:

Ds →00 → → Dυ ρ = T−1[−∇ · q − p∇ · v + τ : ∇ v ] + ρpT−1 (21) Dt Dt Using the second law of thermodynamics (Equation (12)) and Equation (21), it follows that Fluids 2018, 3, 19 5 of 18

. →00 → → Dυ → σ = T−1[−∇ · q − p∇ · v + τ : ∇ v ] + ρpT−1 + ∇ · J ≥ 0 (22) s Dt s Using the continuity equation (Equation (8)), Equation (22) further simpliﬁes to

→ . −1 →00 → σs = T [−∇ · q + τ : ∇ v ] + ∇ · J s ≥ 0 (23)

→ In a single-component fluid (pure substance), the entropy flux J s is due to temperature gradient → →00 in the fluid. The entropy flux J s is related to the heat flux q as follows

→ q00 J = (24) s T Using Equations (23) and (24), the rate of entropy production per unit volume of fluid can be expressed as . →00 −1 −1 → σs = q · ∇(T ) + T (τ : ∇ v ) ≥ 0 (25) Thus, the rate of entropy production in flow of single-component fluids (or multi-component fluids of uniform composition, without any concentration gradients) is due to two mechanisms: irreversible heat transfer caused by temperature gradient, and viscous dissipation of mechanical energy caused by velocity gradient. In the absence of temperature gradients in the fluid, Equation (25) reduces to . −1 → σs = T (τ : ∇ v ) ≥ 0 (26)

2.4. Exergy Destruction in Fluid Flow The exergy of a ﬂuid per unit mass (ψ) is given as

ν2 ψ = (h − h ) − T (s − s ) + + gz (27) o o o 2 where h is the specific enthalpy of fluid, ho and so are specific enthalpy and specific entropy of fluid in the dead state, respectively, To is the absolute temperature of the environment (surroundings), ν is the magnitude of velocity vector of fluid, g is the magnitude of acceleration due to gravity, and z is the elevation of the fluid stream with respect to the dead state. In this expression of exergy, it is assumed that the chemical composition of the fluid is constant. Using the Gouy–Stodola theorem of thermodynamics [12], one can calculate exergy destruction in flow of fluids as . . −1 → ψD = Toσs = ToT (τ : ∇ v ) ≥ 0 (28) . where ψD is the rate of exergy destruction per unit volume during the flow of fluid.

3. Experimental Work

3.1. Experimental Set Up In this work, the entropy generation, and hence exergy destruction, in torsional flow of multiphase dispersions is investigated experimentally using a parallel plate geometry. The multiphase dispersions consisted of oil droplets and solid particles (glass beads) dispersed in a continuum of polymeric matrix fluid. Figure1 shows a schematic diagram of the parallel plate geometry used in the experiments. The fluid (multiphase dispersion) is placed in the gap between the parallel plates. The upper plate is rotated at an angular velocity Ω, and the bottom plate is kept stationary. The torque required to rotate the upper plate at Ω is measured. FluidsFluids2018 2018, ,3 3,, 19 x FOR PEER REVIEW 66 of of 18 18

Fluid H

Stationary Plate R

FigureFigure 1. 1.Schematic Schematic diagram diagram of of a a parallel parallel plate plate geometry. geometry.

InIn order order to to calculate calculate the the rate rate of of exergy exergy destruction destruction or or the the rate rate of of entropy entropy production, production, the the term term → → τ: :∇v is is needed. needed. InIn cylindricalcylindrical coordinates,coordinates,τ : :∇v is is given given as as [ 19[19]]

→   1        1    : ∂νrr  1 ∂νrr νθ  ∂νrr  ∂νθ   1 ∂νθ  νr r  τ : ∇ν = τrrrr ∂r  +τrrθ r ∂θ − r +τrz  ∂z +τθrr ∂r +τθθ r ∂θ + r  r   r  r   z   r   r  r  (29) ∂νθ ∂νz 1 ∂νz ∂νz (29) +τθz ∂z + τz r ∂r + τzθ r ∂θ + τzz ∂z       z   1  z    z  z    z r    z    zz    z   r   r    z  The velocity distribution in the gap between the parallel plates of Figure1 is given as [19] The velocity distribution in the gap between the parallel plates of Figure 1 is given as [19] rz ν = Ω (30) θ Hrz    (30)  H where r is the radial position coordinate from the origin fixed at the center of the bottom plate, and z is thewhere vertical r is the position radial coordinate position coordina from thete origin.from the Using origin the fixed above at velocitythe center distribution, of the bottom Equation plate, and (29) z reducesis the vertical to position coordinate from the origin. Using the above velocity distribution, Equation → Ωr (29) reduces to τ : ∇ ν = τ (31) θ z H .   r  The shear rate (γ) at any radial location is:  given   asz  [19,20 ] (31)  H  . r The shear rate (  ) at any radial locationγ is= givenΩ as [19,20] (32) H r Thus,    (32) → H . τ : ∇ ν = τzθγ (33) Thus, Thus, the rate of energy dissipation in frictional heating of the fluid per unit volume of fluid, →  that is, the term τ : ∇ v , at any location in the :  gap  between z the parallel plates can be determined(33) . → from the knowledge of shear stress τ and shear rate γ at that location. Note that the term τ : ∇ v is Thus, the rate of energy dissipationzθ in frictional heating of the fluid per unit volume of fluid, always positive, and that the stress tensor τ is symmetric. In the present work, Bohlin controlled-stress that is, the term  :v , at any location in the gap between the parallel plates can be determined from rheometer (Bohlin CS-50) was used with a parallel plate geometry. The parallel plate geometry consistedthe knowledge of a stainless of shear steel stress upper  z plate and (diametershear rate 25  mm) at that and location. an aluminum Note that lower the plate term (diameter  :v is 40always mm). positive, The gap widthand that between the stress the tensor plates was is kept symmetric. at 1 mm. In The the rheologicalpresent work, measurements Bohlin controlled- were carried out at 25 ◦C. From the measurement of the torque required to rotate the upper plate at Ω, stress rheometer (Bohlin CS-50) was used with a parallel plate geometry. The parallel plate geometry the shear stress τzθ is determined at the edge of the upper plate, where the shear rate is given as . consisted of a stainless steel upper plate (diameter 25 mm) and an aluminum lower plate (diameter γ = ΩR/H [19,20]. 40 mm). The gap width between the plates was kept at 1 mm. The rheological measurements were carried out at 25 °C. From the measurement of the torque required to rotate the upper plate at Ω, the Fluids 2018, 3, x FOR PEER REVIEW 7 of 18 shearFluids 2018stress, 3, 19 z is determined at the edge of the upper plate, where the shear rate is given7 of as 18   R / H [19,20]. 3.2. Preparation of Multiphase Dispersions 3.2. Preparation of Multiphase Dispersions The multiphase dispersions of droplets and particles in a polymeric liquid were prepared The multiphase dispersions of droplets and particles in a polymeric liquid were prepared by by blending emulsions and suspensions gently in different proportions, while keeping the total blending emulsions and suspensions gently in different proportions, while keeping the total volume volume fraction of the dispersed phase (φ) the same. The emulsions and suspensions were prepared fraction of the dispersed phase (  ) the same. The emulsions and suspensions were prepared individually with the same dispersion medium (aqueous polymer–surfactant solution) using a individually with the same dispersion medium (aqueous polymer–surfactant solution) using a Gifford–Wood homogenizer. The dispersion medium of emulsions and suspensions and their blends Gifford–Wood homogenizer. The dispersion medium of emulsions and suspensions and their blends consisted of 1.02 wt % polymer and 2 wt % surfactant dispersed in deionized water. The polymer consisted of 1.02 wt % polymer and 2 wt % surfactant dispersed in deionized water. The polymer used was Hercules cellulose gum (type 7H4F PM), which is purified sodium carboxymethyl cellulose. used was Hercules cellulose gum (type 7H4F PM), which is purified sodium carboxymethyl cellulose. The molecular weight of the polymer was 7 × 105 approximately. The cellulose gum is an anionic The molecular weight of the polymer was 7 × 105 approximately. The cellulose gum is an anionic water-soluble polymer used in a number of applications in various industries. The surfactant used was water-soluble polymer used in a number of applications in various industries. The surfactant used Triton X-100, which is a non-ionic water-soluble surfactant supplied by Union Carbide. The oil used was Triton X-100, which is a non-ionic water-soluble surfactant supplied by Union Carbide. The oil in the preparation of emulsions was odorless kerosene supplied by Fisher Scientific. The viscosity of used in the preparation of emulsions was odorless kerosene supplied by Fisher Scientific. The the oil was 1.5 mPa·s at 25 ◦C. The emulsions prepared were of oil-in-water (O/W) type consisting viscosity of the oil was 1.5 mPa·s at 25 °C. The emulsions prepared were of oil-in-water (O/W) type of oil droplets dispersed in the aqueous polymer–surfactant dispersion medium. The glass beads consisting of oil droplets dispersed in the aqueous polymer–surfactant dispersion medium. The glass used in the preparation of suspensions were supplied by Flex-O-Lite Inc. (Crestwood, MO, USA). beads used in the preparation of suspensions were supplied by Flex-O-Lite Inc. (Crestwood, MO, The Sauter mean diameters of glass beads and oil droplets were 92 µm and 3.45 µm, respectively. USA). The Sauter mean diameters of glass beads and oil droplets were 92 µm and 3.45 µm, The typical photomicrographs of glass beads and oil droplets are shown in Figure2. The emulsions respectively. The typical photomicrographs of glass beads and oil droplets are shown in Figure 2. The and suspensions, and their blends, were prepared with four different volume fractions of the dispersed emulsions and suspensions, and their blends, were prepared with four different volume fractions of phase: φ = 0.176, φ = 0.289, φ = 0.377, and φ = 0.449. the dispersed phase:   0.176 ,   0.289 ,   0.377 , and   0.449 .

Figure 2. SampleSample photomicrographs photomicrographs of glass beads (suspension) and oil droplets (emulsion).

4. Results and Discussion

4.1. Two-Phase Dispersions

4.1.1. Rheology of Two-Phase Dispersions Figure 33 showsshows the shear stress versus versus shear shear rate rate plots plots of of suspensions suspensions of of spherical spherical glass glass beads beads in polymericin polymeric matrix. matrix. The The experimental experimental data data for for susp suspensionsensions with with different different concentrations concentrations of of solid particles follow a power-law behavior described by thethe followingfollowing equation:equation: . n τ = Kγ (34) Fluids 2018, 3, x FOR PEER REVIEW 8 of 18

n Fluids 2018, 3, 19   K 8(34) of 18

where K is the consistency index and n is the flow behavior index. According to the power-law model, whereEquationK is (34), the consistency the shear stress index versus and n isshear the ﬂowrate plot behaviors are index.linear on According a log–log to scale, the power-law with slopes model, of n. EquationTable 1 summarizes (34), the shear the stress statistical versus information shear rate plotsregarding are linear the power-law on a log–log fit scale,of shear with stress slopes versus of n. Tableshear1 ratesummarizes data for thesuspensions statistical of information glass beads regarding in a polymeric the power-law matrix. ﬁt of shear stress versus shear rate data for suspensions of glass beads in a polymeric matrix.

Figure 3. Shear stress versus shear rate plots of suspensions of glass beads in polymeric solution. Figure 3. Shear stress versus shear rate plots of suspensions of glass beads in polymeric solution.

Table 1. Goodness of fit of power-law model and confidence intervals for the power-law parameters Interestingly, the suspension plots with different volume fractions of the dispersed phase (glass related to suspensions of glass beads in a polymeric matrix. beads) are nearly parallel to each other, indicating that the flow behavior index n does not change appreciably with the introduction of particles in the polymeric matrix. The average flow behavior Volume Regression Consistency 95% 95% Flow Behavior indexPercent is 0.52, of indicatingCorrelation that the suspensions and the matrixIndex, phaseK areConfidence pseudoplastic non-NewtonianConfidence Index, n fluids.Glass ParticlesHowever,Coefficient, the suspensionsR2 become more(Units viscous of Pa.s nwith) Intervalthe increase of n inInterval the particle of K concentration,0 as reflecte 0.9863d in the increase 0.540 in the K value (upward 6.05 shift [0.507, in the 0.573] shear stress [5.45, versus 6.73] shear rate plots).17.6 0.9884 0.557 9.76 [0.527, 0.587] [8.94, 10.58] 28.9 0.9866 0.518 16.91 [0.479, 0.558] [15.10, 18.94] Table37.7 1. Goodness 0.9908of fit of power-law 0.524 model and confid 26.85ence intervals for [0.488, the power-law 0.560] parameters [24.61, 29.26] 44.9 0.9913 0.468 55.09 [0.452, 0.484] [53.36, 56.88] related to suspensions of glass beads in a polymeric matrix.

Volume Regression Consistency 95% 95% Interestingly, the suspension plotsFlow with Behavior different volume fractions of the dispersed phase (glass Percent of Correlation Index, K (Units Confidence Confidence Index, n beads)Glass are Particles nearly parallelCoefficient, to each R2 other, indicating thatof the Pa.s flown) behaviorInterval index of nn doesInterval not change of K appreciably0 with the introduction 0.9863 of particles 0.540 in the polymeric 6.05 matrix. [0.507, The 0.573] average flow [5.45, behavior 6.73] index is17.6 0.52, indicating0.9884 that the suspensions0.557 and the matrix 9.76 phase are pseudoplastic[0.527, 0.587] non-Newtonian [8.94, 10.58] fluids. However,28.9 the suspensions0.9866 become0.518 more viscous with16.91 the increase[0. in479, the 0.558] particle concentration, [15.10, 18.94] as reflected37.7 in the increase0.9908 in the K value0.524 (upward shift in26.85 the shear stress[0.488, versus 0.560] shear rate[24.61, plots). 29.26] Figure44.9 4 shows the0.9913 shear stress versus0.468 shear rate data55.09 for O/W emulsions[0.452, 0.484] with different [53.36, volume 56.88] concentrations of oil droplets. The same data are also shown as viscosity versus shear rate in Figure5. EmulsionsFigure of 4 oil shows droplets the shear behave stress very versus differently shear from rate the data suspensions for O/W emulsions of rigid particles. with different The emulsion volume dataconcentrations for different of oiloil concentrationsdroplets. The same closely data matches are also the shown matrix as viscosity behavior. versus Thus, shear the addition rate in Figure of oil droplets5. Emulsions to the of polymeric oil droplets matrix beha doesve notvery cause differently any appreciable from the changesuspensions in the of consistency rigid particles. (K value) The andemulsion flow behavior data for (differentn value). oil This concentrations is not surprising closely as emulsion matches the droplets matrix are behavior. deformable Thus, in the nature, addition and undergo internal circulation when subjected to shear flow. The increase in flow resistance due to the FluidsFluids 20182018,, 33,, xx FORFOR PEERPEER REVIEWREVIEW 99 ofof 1818 ofof oiloil dropletsdroplets toto thethe polymericpolymeric matrixmatrix doesdoes notnot causecause anyany appreciableappreciable changechange inin thethe consistencyconsistency ((KK Fluidsvalue)2018 and, 3 ,flow 19 behavior (n value). This is not surprising as emulsion droplets are deformable9 of in 18 value) and flow behavior (n value). This is not surprising as emulsion droplets are deformable in nature,nature, andand undergoundergo internalinternal circulationcirculation whenwhen subjectesubjectedd toto shearshear flow.flow. TheThe increaseincrease inin flowflow resistanceresistance due to the introduction of interfaces in emulsions is balanced by the low viscosity of dispersed phase introductiondue to the introduction of interfaces of interfaces in emulsions in emulsions is balanced is balanced by the low by viscosity the low viscosity of dispersed of dispersed phase (oil) phase as (oil) as compared with the matrix fluid. All the emulsion data could be described by the power-law compared(oil) as compared with the with matrix the fluid.matrix All fluid. the emulsionAll the emulsion data could data be could described be described by the power-lawby the power-law model n (Equationmodelmodel (Equation(Equation (34)) with (34))(34))K =withwith6.84 K.K.Pa.s684Pa.s684Pa.sn and n =n 0.52.andand Thenn == 0.52.0.52. 95% The confidenceThe 95%95% confidenceconfidence intervals of intervalsintervalsK and n ofofare KK as andand follows nn areare as[6.53,as followsfollows 7.16] [6.53,[6.53, and [0.506, 7.16]7.16] andand 0.533], [0.506,[0.506, respectively. 0.533],0.533], respectively.respectively.

FigureFigure 4. 4. ShearShear stress stress versus versus shear shear rate rate data data of of emul emulsionsemulsionssions of of oil oil dropletsdroplets inin polymericpolymeric solution.solution.

Figure 5. Apparent viscosity versus shear rate data of emulsions of oil droplets in polymeric solution. FigureFigure 5.5. Apparent viscosity versus shearshear raterate data ofof emulsions ofof oil droplets inin polymeric solution.solution. Fluids 2018, 3, 19 10 of 18 Fluids 2018, 3, x FOR PEER REVIEW 10 of 18

Figure 66 comparescompares thethe power-lawpower-law parametersparameters ofof suspensionssuspensions andand emulsions.emulsions. TheThe figurefigure revealsreveals the following points: (a) the the flow flow behavior behavior index n of emulsions emulsions and suspensions does not differ from that of the polymeric matrix; (b) the consistency index K of emulsions is nearly constant with respect to oil concentration variation; (c) the consistency index of suspensions rises sharply with the increase in the particle concentration; and (d) the suspensionssuspensions are much more viscous than the emulsions as reflectedreflected in the values of the consistencyconsistency index.index.

Figure 6. Comparison of power-law paramete parametersrs of suspensions and emulsions.

4.1.2. Exergy Destruction in Flow of Two-Phase DispersionsDispersions From EquationsEquations (28)(28) andand (33), (33), it it follows follows that that the the rate rate of of exergy exergy destruction destruction per per unit unit volume volume of the of theﬂuid fluid is given is given as as . . ψ = τγ (35) DD  (35) Note that in our experiments T = T = 298.15 K. Using the shear stress versus shear rate data Note that in our experiments TTo 298 . 15K . Using the shear stress versus shear rate data obtained from the rheometer, the rate of exergyo destruction per unit volume of the ﬂuid was calculated obtained from the rheometer, the rate of exergy destruction per unit volume of the fluid was as a function of shear rate from Equation (35). calculated as a function of shear rate. from Equation. (35). Figure7 shows the plots of ψ versus γ for suspensions of glass particles in polymeric matrix. Figure 7 shows the plots of D versus  for suspensions of glass particles in polymeric The plots of exergy destruction rate forD suspensions with different volume fractions of the dispersed phasematrix. (glass The beads)plots of are exergy nearly destruction parallel to eachrate other,for suspensions as expected with from different the shear volume stress versus fractions shear of ratethe plotsdispersed of Figure phase3. Note(glass that beads) Equation are nearly (35) in parallel conjunction to each with other, the power as expected law model from yields the shear stress versus shear rate plots of Figure 3. Note that Equation (35) in conjunction with the power law model . yields . n+1 ψD = Kγ (36) n 1   K  (36) Thus, the slope of the exergy destructionD rate plot is simply (n + 1). The average value of the slopesThus, of different the slope lines of shownthe exergy in Figure destruction7 is 1.52 (averagerate plot valueis simply of n (wasn + 1). 0.52, The as notedaverage earlier). value Atof anythe given shear rate, the exergy destruction rate increases with the increase in the particle concentration. slopes of different. lines. shown in Figure 7 is 1.52 (average value of n was 0.52, as noted earlier). At any Thegivenψ Dshearversus rate,γ data the forexergy emulsions destruction of oil droplets rate increases in polymeric with matrixthe increase are shown in the in Figureparticle8 concentration.for different concentrations of oil droplets. The data for the emulsions and matrix overlap with each other indicating that for a given shear rate, the exergy destruction rate is independent of the oil droplet concentration. This is because the consistency of emulsions, and hence viscous dissipation of Fluids 2018, 3, x FOR PEER REVIEW 11 of 18

Fluids 2018, 3, 19 11 of 18 mechanical energy in emulsions, does not vary to any appreciable extent with the increase in the oil concentrationFluids 2018, 3, x FOR at anyPEER given REVIEW shear rate. 11 of 18

Figure 7. Plots of exergy destruction rates in suspensions of glass beads in polymeric matrix.

The  D versus  data for emulsions of oil droplets in polymeric matrix are shown in Figure 8 for different concentrations of oil droplets. The data for the emulsions and matrix overlap with each other indicating that for a given shear rate, the exergy destruction rate is independent of the oil droplet concentration. This is because the consistency of emulsions, and hence viscous dissipation of mechanical energy in emulsions, does not vary to any appreciable extent with the increase in the oil Figure 7. Plots of exergy destruction rates in suspensionssuspensions of glass beads in polymeric matrix. concentration at any given shear rate.

Figure 8. ExergyExergy destruction destruction rate data for emulsions of oil droplets in polymeric matrix.

4.2. Multi-Phase Dispersions of Oil and Water in a Polymeric Matrix

4.2.1. Exergy Destruction in Multi-Phase Dispersions As noted earlier, the multiphase dispersions of droplets and particles in a polymeric liquid were prepared by blending emulsions and suspensions gently in different proportions, while keeping Figure 8. Exergy destruction rate data for emulsions of oil droplets in polymeric matrix. the total volume fraction of the dispersed phase (φ) the same. Figures9 and 10 show the typical

Fluids 2018, 3, x FOR PEER REVIEW 12 of 18

4.2. Multi-Phase Dispersions of Oil and Water in a Polymeric Matrix

4.2.1. Exergy Destruction in Multi-Phase Dispersions As noted earlier, the multiphase dispersions of droplets and particles in a polymeric liquid were Fluids 2018, 3, 19 12 of 18 prepared by blending emulsions and suspensions gently in different proportions, while keeping the total volume fraction of the dispersed phase ( ) the same. Figures 9 and 10 show the typical exergy destructionexergy destruction behavior behavior of blends of blends of emulsions of emulsions and suspensions and suspensions at given at given total total volume volume fractions fractions of dispersedof dispersed phase. phase. The The total total volume volume percent percent of the of di thespersed dispersed phase phase (oil droplets (oil droplets + glass + beads) glass beads) is 37.7% is in37.7% Figure in Figure9, and9 ,44.9% and 44.9% in Figure in Figure 10. Upon 10. Upondilution dilution of the of suspension the suspension with emulsion with emulsion of the of same the dispersed-phasesame dispersed-phase concentration, concentration, the exergy the exergy destruct destructionion rate rate decreases. decreases. The The data data for for blends, blends, with different proportionsproportions ofof emulsion emulsion in in the the blends, blends, fall fa inll betweenin between the the lines lines for purefor pure suspension suspension and pureand pureemulsion. emulsion. However, However, the variation the variation in the exergyin the destructionexergy destruction rate with rate the increasewith the in increase the proportion in the proportionof emulsion, of keepingemulsion, total keeping volume total fraction volume ﬁxed, fraction is not fixed, linear. is not This linear. can beThis seen can more be seen clearly more in clearlyFigures in11 Figures and 12, 11 where and the12, exergywhere destructionthe exergy ratedestru is plottedction rate as ais functionplotted as of emulsiona function content of emulsion of the contentblend of of emulsion the blend and of emulsion suspension, and keeping suspension, the total keeping volume the total fraction volume of the fr dispersedaction of the phase dispersed (glass phaseparticles (glass and particles oil droplets) and andoil droplets) shear rate and ﬁxed. shear The rate experimental fixed. The data experimental show negative data deviationshow negative from deviationthe linear behavior.from the linear behavior.

Figure 9. Exergy destruction rate in blends of emulsion and suspension at a constant total volume Figurefraction 9. of Exergy dispersed destruction phase of rate 0.377. in blends of emulsion and suspension at a constant total volume fraction of dispersed phase of 0.377. The exergy destruction rate as a function of the total volume fraction of the dispersed phase (glass beads + oil droplets) is plotted in Figure 13 for different contents of emulsion in the blend of emulsion and suspension, at a ﬁxed shear rate of 10 s−1. As the suspension is diluted with an emulsion having the same volume fraction of the dispersed phase, the exergy destruction rate decreases. In other words, the substitution of solid particles with oil droplets, keeping the total volume fraction of dispersed phase constant, results in a decrease in the exergy destruction rate. This is not unexpected, as the emulsion is relatively much less viscous than the suspension at the same value of the dispersed phase concentration. Also note that for a given content of emulsion in the blend, the exergy destruction rate of the blend increases with the increase in the total volume fraction of the dispersed phase. Fluids 2018, 3, x FOR PEER REVIEW 13 of 18

Fluids 2018, 3, 19 13 of 18 Fluids 2018, 3, x FOR PEER REVIEW 13 of 18

Figure 10. Exergy destruction rate in blends of emulsion and suspension at a constant total volume fraction ofFigureFigure dispersed 10.10. ExergyExergy phase destructiondestruction of 0.449. raterate inin blendsblends ofof emulsionemulsion andand suspensionsuspension atat aa constantconstant totaltotal volumevolume fractionfraction ofof disperseddispersed phasephase of of 0.449. 0.449.

Figure 11. Exergy destruction in blends of emulsion and suspension as a function of emulsion content (vol %) at a constant total volume fraction of dispersed phase of 0.377 and shear rate of 10 s−1. Figure 11. Exergy destruction in blends of emulsion and suspension as a function of emulsion content −1 Figure 11.(vol Exergy %) at a destruction constant total in volume blends fraction of emulsion of dispersed and phase suspension of 0.377 and as sheara function rate of 10of semulsion. content (vol %) at a constant total volume fraction of dispersed phase of 0.377 and shear rate of 10 s−1. Fluids 2018, 3, x FOR PEER REVIEW 14 of 18

Fluids 2018, 3, 19 14 of 18 Fluids 2018, 3, x FOR PEER REVIEW 14 of 18

Figure 12. Exergy destruction in blends of emulsion and suspension as a function of emulsion content (vol %) at a constant total volume fraction of dispersed phase of 0.449 and shear rate of 10 s−1.

The exergy destruction rate as a function of the total volume fraction of the dispersed phase (glass beads + oil droplets) is plotted in Figure 13 for different contents of emulsion in the blend of emulsion and suspension, at a fixed shear rate of 10 s−1. As the suspension is diluted with an emulsion having the same volume fraction of the dispersed phase, the exergy destruction rate decreases. In other words, the substitution of solid particles with oil droplets, keeping the total volume fraction of dispersed phase constant, results in a decrease in the exergy destruction rate. This is not unexpected, as the emulsion is relatively much less viscous than the suspension at the same value of the dispersed phase concentration. Also note that for a given content of emulsion in the blend, the exergy destruction rate of the blend increases with the increase in the total volume fraction of the dispersed FigureFigure 12. 12. ExergyExergy destruction destruction in in blends blends of of emulsion emulsion and and suspension suspension as as a afunction function of of emulsion emulsion content content phase. −1 (vol(vol %) %) at at a a constant constant total total volume volume fraction fraction of of dispersed dispersed phase phase of of 0.449 0.449 and and shear shear rate rate of of 10 10 s s−1. .

Figure 13. Exergy destruction in blends of emulsion and suspension as a function of total volume Figure 13. Exergy destruction in blends of emulsion and suspension as a function of total volume fraction of dispersed phase for different contents of emulsion at a shear rate of 10 s−1. fraction of dispersed phase for different contents of emulsion at a shear rate of 10 s−1.

4.2.2. Model for Calculation of Exergy Destruction in Multi-Phase Dispersions The droplets of emulsion are much smaller in size as compared with the solid particles (glass beads) of suspension. The Sauter mean diameter of solid particles is approximately 27 times larger than that of emulsion droplets. Furthermore, the rheology of emulsions is not signiﬁcantly different than that of the polymer matrix alone. Thus, it is reasonable to treat oil droplets and polymeric matrix together as an effective matrix for the solid particles. This is shown schematically in Figure 14. The three-phase system of oil droplets/glass beads/polymeric matrix is equivalent to a two-phase system consisting of solid particles dispersed in an effective matrix of polymeric liquid and oil droplets. Based on this reasoning, it is expected that the relative exergy destruction in three-phase blends is a unique function Figure 13. Exergy destruction in blends of emulsion and suspension as a function of total volume fraction of dispersed phase for different contents of emulsion at a shear rate of 10 s−1. Fluids 2018, 3, x FOR PEER REVIEW 15 of 18

4.2.2. Model for Calculation of Exergy Destruction in Multi-Phase Dispersions The droplets of emulsion are much smaller in size as compared with the solid particles (glass beads) of suspension. The Sauter mean diameter of solid particles is approximately 27 times larger than that of emulsion droplets. Furthermore, the rheology of emulsions is not significantly different than that of the polymer matrix alone. Thus, it is reasonable to treat oil droplets and polymeric matrix together as an effective matrix for the solid particles. This is shown schematically in Figure 14. The three-phase system of oil droplets/glass beads/polymeric matrix is equivalent to a two-phase system consisting of solid particles dispersed in an effective matrix of polymeric liquid and oil droplets. Fluids 2018, 3, 19 15 of 18 Based on this reasoning, it is expected that the relative exergy destruction in three-phase blends is a unique function of the concentration (volume fraction) of solid particles alone, where the relative exergyof the concentrationdestruction is (volume defined fraction)as a ratio of of solid the exer particlesgy destruction alone, where rate the in relativethree-phase exergy blend destruction to exergy is destructiondeﬁned as arate ratio in of the the effective exergy destruction matrix composed rate in three-phaseof oil droplets blend and to polymeric exergy destruction liquid (that rate is, in the the emulsioneffective matrixphase). composed of oil droplets and polymeric liquid (that is, the emulsion phase).

FigureFigure 14. 14. Three-phaseThree-phase blend blend of of oil oil droplets, droplets, solid solid particles, particles, and and polymeric polymeric liquid liquid can can be be treated treated as as equivalentequivalent toto aa two-phase two-phase system system consisting consisting of solid of solid particles particles dispersed dispersed in an effective in an matrixeffective composed matrix composedof oil droplets of oil and droplets polymeric and polymeric liquid, that liquid, is, the that emulsion is, thephase. emulsion phase.

TheThe experimental experimental data data of of Figure Figure 1313 areare re-plotted re-plotted in in Figure Figure 1515 asas relative relative exergy exergy destruction destruction rate rate versusversus volume volume fraction fraction of of solid solid (glass (glass beads) beads) partic particlesles in in the the blend. blend. Clearly Clearly al alll the the data data of of three-phase three-phase blendsblends with with different different emulsion emulsion cont contentsents fall fall on on the the same same curve. curve. Also Also note note that that the the pure pure suspension data,data, with with polymeric polymeric liquid liquid as as the the matrix, matrix, also also fall fall on on the the same same curve. curve. The The relative relative exergy exergy destruction destruction datadata can can be be described described by by the the following following model, model, shown shown as solid curve in Figure 15:15:

. 2.5m    −2.5φm . ψD,blendD,blend SφS  D=,r  = 1−  (37) ψD,r .   1  (37)  D,emulsion  mφm ψD,emulsion where S is the volume fraction of solid particles in the blend, and m is the maximum packing where φS is the volume fraction of solid particles in the blend, and φm is the maximum packing volume volumefraction fraction of solid of particles solid particles where jammingwhere jamming of suspension of suspension takes place. takes place. In the In model the model prediction prediction curve curveshown shown in Figure in Figure 15, φ 15,m is  takenm is taken to be to 0.64. be 0.64. At φ Atm, them , suspension the suspension viscosity, viscosity, and and consequently, consequently, the therelative relative exergy exergy destruction destruction rate, rate, is expected is expected to diverge. to diverge. The value The ofvalueφm whereof m jamming where jamming of particles, of particles,and hence, and divergence hence, divergence of suspension of suspension viscosity viscos takes place,ity takes is stillplace, an is open stillquestion an open inquestion the literature. in the literature.However, aHowever, large number a large of experimentalnumber of experiment studies [21al– 26studies] available [21–26] on theavailable viscosity on ofthe suspensions viscosity of of suspensionsspherical particles of spherical have usedparticlesφm ofhave 0.64 used to successfully m of 0.64 correlateto successfully the relative correlate viscosity the relative versus viscosity particle concentration data. It should be noted that φm of 0.64 corresponds to random close packing of uniform versus particle concentration data. It should be noted that m of 0.64 corresponds to random close spheres. As discussed in Section 4.1.2, the exergy destruction rate in emulsions can be expressed as

. . n+1 ψD,emulsion = Keγ (38) where Ke is the consistency index of emulsion and n is the ﬂow behavior index of emulsion. From Equations (37) and (38), it follows that

−2.5φm . . n+1 φS ψD,blend = Keγ 1 − (39) φm Fluids 2018, 3, x FOR PEER REVIEW 16 of 18

packing of uniform spheres. As discussed in Section 4.1.2, the exergy destruction rate in emulsions can be expressed as

n1  D,emulsion  Ke (38)

where Ke is the consistency index of emulsion and n is the flow behavior index of emulsion. From Equations (37) and (38), it follows that

2.5 Fluids 2018, 3, 19   m 16 of 18  n1 S   D,blend  Ke 1  (39)  m  This equation can be used to calculate the exergy destruction rate in three-phase blends of this This equation can be used to calculate the exergy destruction rate in three-phase blends of this study at any given shear rate, and volume fraction of solid particles in the blend. Note that for the study at any given shear rate, and volumen fraction of solid particles in the blend. Note that for the system under consideration, Ke = 6.84 Pa.s ; n = 0.52 ; and φm = 0.64. n 0.64 system under consideration, K.e  684Pa.s ; n = 0.52 ; and m  .

Figure 15. Relative exergy destruction rate in three-phasethree-phase blend of oil droplets, solid particles, and polymeric liquid.

5. Practical Implications of This Work The rate of exergy exergy destruction destruction is is a a measure measure of of level level of of irreversibility irreversibility in in the the process. process. According According to tothe the Gouy–Stodola Gouy–Stodola theorem, theorem, the the power power loss loss in a in process a process due due to toirreversibilities irreversibilities is equal is equal to tothe the rate rate of ofexergy exergy destruction destruction in inthe the process. process. Thus, Thus, the the mechan mechanicalical energy energy dissipation dissipation rate rate (power (power loss) loss) due due to tofrictional frictional effects effects in influid fluid flow flow can can be be quantified quantified in in terms terms of of the the exergy exergy destruction destruction rate. rate. For For a finitefinite control volume, the global rate of exergy destruction can be calculated once the local rate of exergy destruction is known, using the followingfollowing integration:integration:  . D,CV  Z . D,localdV =  (40) ΨD,CV CV ψD,localdV (40) CV where  is the global or overall rate of exergy destruction in the finite control volume, and . D,CV . where Ψ is the global or overall rate of exergy destruction in the finite control volume, and ψ  D,local isD, CVthe local rate of exergy destruction per unit volume of the fluid. In the present work,D, localthe is the local rate of exergy destruction per unit volume of the fluid. In the present work, the local rate of local rate of exergy destruction in three-phase blends is given by Equation (39). Thus, exergy destruction in three-phase blends is given by Equation (39). Thus,

. . n+1 −2.5φm Ψ = R K γ 1 − φS dV D,CV e φm CV (41) −2.5φm . n+1 = K 1 − φS R γ dV e φm CV

The constant terms are taken outside the integral. As the shear rate may vary with position in the control volume, it is kept inside the integral. From the knowledge of the velocity ﬁeld, the local shear rate and its variation inside the control volume can be calculated. Hence, the global or overall rate of exergy destruction can be determined for a given ﬁnite control volume using Equation (41). Fluids 2018, 3, 19 17 of 18

6. Conclusions Based on the experimental work and analysis, the following conclusions can be made:

(a) The addition of oil droplets to a polymeric matrix does not alter the consistency index and flow behavior index of the system to any appreciable extent. (b) The addition of solid particles to a polymeric matrix does not alter the flow behavior index significantly, but the consistency index increases sharply with the increase in the particle concentration. (c) The exergy destruction rate in flow of emulsions of oil droplets and polymeric matrix does not change to any appreciable extent with the increase in the concentration of oil droplets. (d) The exergy destruction rate in flow of suspensions of solid particles and polymeric matrix increases with the increase in particle concentration. (e) Three-phase blends of oil droplets, solid particles, and polymeric matrix, can be modelled as two-phase blends of solid particles dispersed in an effective matrix composed of oil droplets and polymeric matrix. (f) A model is proposed to estimate the exergy destruction rates in flow of three-phase dispersions of oil droplets, solid particles, and polymeric matrix.

Conﬂicts of Interest: The author declares no conﬂict of interest.

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