Consider Again the Fully developed Laminar Blood Flow in the Annulus a Blood Vessel

Direct Numerical Simulation of Rotating Turbulent Flows Through Concentric Annuli

Thou. Okamoto , North. Shima , in Engineering Turbulence Modelling and Experiments six, 2005

INTRODUCTION

Turbulent flow through concentric annuli is important in various engineering science applications and is investigated by many researchers. Rehme (1974) studied the concentric annular menstruum experimentally. From the viewpoints of numerical investigation, Satake & Kawamura (1993) performed the large eddy simulation of the flow and Chung, Rhee & Sung (2001) and the authors (2002) simulated the flow by the directly numerical simulation (DNS). In this flow, due to the curvature effect the distributions of hateful quantities are disproportionate different symmetric ones of plane channel flow. In the nowadays newspaper, we written report the concentric annular menstruation with the inner-wall and outer-wall rotations. The wall-rotation furnishings induce the azimuthal move and have similarities with three-dimensional boundary layers. In the laminar menstruum, the outer-wall rotation produces an approximately forced vortex and the inner-wall one produces an approximately gratuitous vortex. However, the effects of both wall-rotations in the turbulent flow are complicated. In this work, we systematically perform DNS of the rotating turbulent flows through concentric annuli with various radius ratios and rotation numbers and investigate the mean quantities including the high-guild statistics and instantaneous fields.

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EVALUATION OF TRANSFER COEFFICIENTS: Engineering CORRELATIONS

Ismail Tosun , in Modeling in Transport Phenomena (Second Edition), 2007

4.6.1 Friction Factor Correlations

The friction cistron for packed beds, f _lead , is defined by

(4.6-i) $f_{pb}=\frac{{\epsilon }^3}{1-\epsilon }\frac{D_P\left|\Delta P\right|}{\rho {\upsilon }_o^{2}L}$

where ε is the porosity (or void volume fraction), D _P is the particle diameter, and υ _o is the superficial velocity. The superficial velocity is obtained by dividing the volumetric menses rate by the total cantankerous-sectional area of the bed. Notation that the bodily flow area is a fraction of the total cross-sectional area.

Example 4.19

Water flows through a concentric annulus at a volumetric flow rate of 5 thou ³/min. The diameters of the inner and the outer pipes are thirty cm and 50 cm, respectively. Calculate the superficial velocity.

Solution

If the inner and outer pipe diameters are designated by D _i and D _o , respectively, the superficial velocity, υ _o, is defined past

${\upsilon }_o=\frac{\mathcal{Q}}{\pi D_o^2/4}=\frac{5}{\pi {\left(\mathrm{0.v}\right)}^{\mathrm{two}}/\mathrm{iv}}=\mathrm{25.five}\text{yard}/\min$

The actual boilerplate velocity, 〈υ〉 _act , in the annulus is

${〈\upsilon 〉}_{act}=\frac{\mathcal{Q}}{\pi \left(D_o^{\mathrm{ii}}-D_i^2\right)/\mathrm{four}}=\frac{5}{\pi \left[{\left(0.5\right)}^2-{\left(\mathrm{0.iii}\right)}^{\mathrm{two}}\right]/4}=40\text{m}/\min$

Comment: The superficial velocity is always lower than the bodily average velocity past a factor of porosity, which is equal to [1 – (D _i /D _o )²] in this example.

For packed beds, the Reynolds number is defined by

(4.6-2) ${\mathrm{Re}}_{pb}=\frac{D_P{\upsilon }_o\rho }{\mu }\frac{\mathrm{i}}{\mathrm{i}-\epsilon }$

For laminar flow, the relationship betwixt the friction cistron and the Reynolds number is given by

(4.vi-three) $\begin{array}{cc}{f}_{pb}=\frac{150}{{\mathrm{Re}}_{pb}}& {\mathrm{Re}}_{pb}\end{array}<10$

which is known as the Kozeny-Carman equation.

In the case of turbulent flow, i.e., Re _pb > 1000, the human relationship between Re_pb and f _pb is given by the Burke-Plummer equation in the course

(4.6-4) $\begin{array}{cc}{f}_{pb}=1.75& {\mathrm{Re}}_{pb}\end{array}>1000$

The so-called Ergun equation (1952) is simply the summation of the Kozeny-Carman and the Burke-Plummer equations

(4.6-5) $f_{pb}=\frac{150}{{\mathrm{Re}}_{pb}}+1.75$

Example 4.20

A column of 0.viii m² cross-section and 30 thou top is packed with spherical particles of bore half-dozen mm. A fluid with ρ = 1.2 kg/thouⁱⁱⁱ and μ = i.eight × 10^–five kg/m·s flows through the bed at a mass catamenia rate of 0.65 kg/s. If the pressure drop is measured every bit 3200 Pa, calculate the porosity of the bed:

a)

Analytically,

b)

Numerically.

Solution

Assumption

1.

The organization is isothermal.

Analysis

The superficial velocity through the packed bed is

${\upsilon }_o=\frac{0.65}{\left(\mathrm{one.2}\right)\left(0.8\right)}=0.677\text{m}/\text{s}$

Substitution of the values into Eqs. (4.half-dozen-i) and (4.6-ii) gives the friction cistron and the Reynolds number as a function of porosity in the form

(1) $f_{pb}=\frac{{\epsilon }^3}{\mathrm{i}-\epsilon }\frac{D_P\left|\Delta P\right|}{\rho {\upsilon }_o^{2}L}=\frac{{\epsilon }^3}{1-\epsilon }\left[\frac{\left(\mathrm{vi}\times {\mathrm{x}}^{-3}\right)\left(3200\right)}{\left(\mathrm{1.two}\right){\left(0.677\right)}^2\left(\mathrm{thirty}\right)}\right]=1.164\left(\frac{{\epsilon }^{\mathrm{three}}}{1-\epsilon }\right)$

(2) ${\mathrm{Re}}_{pb}=\frac{D_P{\upsilon }_o\rho }{\mu }\frac{1}{1-\epsilon }=\left[\frac{\left(6\times {10}^{-3}\right)\left(0.677\right)\left(\mathrm{1.ii}\right)}{1.8\times {10}^{-\mathrm{five}}}\right]\frac{\mathrm{one}}{1-\epsilon }=\mathrm{270.viii}\left(\frac{1}{1-\epsilon }\right)$

Substitution of Eqs. (1) and (ii) into Eq. (iv.6-v) gives

(3) ${\epsilon }^{\mathrm{iii}}-0.476{\epsilon }^2+2.455\epsilon -1.979=0$

a)

Equation (3) can be solved analytically by using the procedure described in Section A.7.ane.2 in Appendix A. In guild to calculate the discriminant, the terms M and N must be calculated from Eqs. (A.7-5) and (A.7-6), respectively:

$\begin{array}{ll}\mathrm{Yard}\hfill & =\frac{\left(3\right)\left(2.455\right)-{\left(0.476\right)}^2}{\mathrm{ix}}=0.793\hfill \\ N\hfill & =\frac{-\left(9\right)\left(0.476\right)\left(2.455\right)+\left(27\right)\left(\mathrm{ane.979}\right)+\left(2\right){\left(0.476\right)}^3}{54}=0.799\hfill \end{array}$

Therefore, the discriminant is

$\Delta ={\mathrm{One\; thousand}}^3+{\mathrm{North}}^2={\left(0.793\right)}^3+{\left(0.799\right)}^2=1.137$

Since Δ > 0, Eq. (3) has just one real root as given by Eq. (A.vii-vii). The terms South and T in this equation are calculated equally

$\begin{array}{c}S={\left(\mathrm{Northward}+\sqrt{\Delta }\right)}^{1/\mathrm{three}}={\left(0.799+\sqrt{1.137}\right)}^{1/\mathrm{three}}=1.231\\ T={\left(N-\sqrt{\Delta }\right)}^{1/\mathrm{iii}}={\left(0.799-\sqrt{1.137}\right)}^{\mathrm{i}/\mathrm{iii}}=-0.644\end{array}$

Hence the average porosity of the bed is

$\epsilon =1.231-0.644+\frac{0.476}{\mathrm{iii}}=0.746$

b)

Equation (3) is rearranged as

(four) $F(\epsilon )={\epsilon }^{\mathrm{three}}-0.476{\epsilon }^2+\mathrm{two.455}\epsilon -\mathrm{ane.979}=0$

From Eq. (A.7-25) the iteration scheme is

(5) ${\epsilon }_k={\epsilon }_{k-1}-\frac{0.02{\epsilon }_{k-\mathrm{i}}F\left({\epsilon }_{\mathrm{1000}-\mathrm{i}}\right)}{F\left(1.01{\epsilon }_{\mathrm{grand}-1}\right)-F\left(0.99{\epsilon }_{k-\mathrm{i}}\right)}$

Assuming a starting value of ε _o = 0.7, the iterations are given in the table below:

thou	ε k
0	0.7
one	0.746
2	0.745
3	0.745

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Casing equipment

Alfredo SanchezP.E., in Applied Well Cementing Applied science, 2021

3.2.two.7 Equivalent circulating density

Frictional pressure losses of non-Newtonian fluids in concentric annuli can be significantly college than in eccentric annuli. Furthermore, casing centralizers volition reduce the annular cross-sectional area, and may besides strength fluid to change direction. Almost currently available cement simulators do not account for the latter two. When working in formations with tight pressure windows, peculiarly in horizontal and extended reach wells, consider the effects of boosted pressure losses at the centralizers on equivalent circulating densities. Annotation that the effect of centralizer geometry, specifically blade angle, may be significant, as documented in Yao and Samuel (2008).

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Period in Pipes and in Conduits of Non-circular Cross-sections

R.P. Chhabra , J.F. Richardson , in Non-Newtonian Flow and Practical Rheology, 2008

Instance 3.xiii

A molten chocolate (density=1500   kg/g³ ) flows through a concentric annulus of inner and outer radii 10  mm and xx   mm, respectively, at 30°C at the abiding flow rate of 0.03   m³/min. The steady-shear behaviour of the chocolate can be approximated by a Bingham plastic model with ${\tau }_0^B$ =35   Pa and μ_β =1   Pa·southward.

(a)

Estimate the required pressure gradient to maintain the flow, and decide the velocity and the size of the plug.

(b)

Owing to a pump malfunction, the available pressure level slope drops by 25% of the value calculated in (a), what volition be the new flow charge per unit?

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DIRECT NUMERICAL AND LARGE Eddy SIMULATIONS OF TURBULENT FLOWS THROUGH CONCENTRIC ANNULI

M. Okamoto , N. Shima , in Applied science Turbulence Modelling and Experiments 5, 2002

CONCLUSION

In this written report, we performed the DNS and LES for fully developed catamenia through concentric annuli and obtained mean quantities and budgets of the Reynolds stress. Every bit the radius ratio decreases, the mean velocity profiles shift downward from the log-law, the inner normal stresses decrease and the position of the zero shear stress moves to the inner wall. We found the differences in the upkeep distribution between the inner and outer regions. In particular, the pressure strain for the shear stress at pocket-size radius ratio destructs the shear stress near the outer wall and produces that near the inner wall. There are good agreements between the LES with a one-equation blazon SGS model and DNS in the inner region. It is considered that the present information are effective to test Reynolds-averaged turbulence models.

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Flow in pipes and in conduits of not-circular cross-sections

R.P. Chhabra , J.F. Richardson , in Non-Newtonian Menstruation in the Procedure Industries, 1999

three.6.2 Bingham plastic fluids

The laminar axial flow of Bingham plastic fluids through a concentric annulus has generated fifty-fifty more interest than that for the power-police force fluids, east.g. come across refs. [ Laird, 1957; Fredrickson and Bird, 1958; Bird et al., 1983; Fordham et al., 1991]. The master feature which distinguishes the flow of a Bingham plastic fluid from that of a ability-law fluid is the existence of a plug region in which the shear stress is less than the yield stress. Effigy 3.18 shows qualitatively the salient features of the velocity distribution in an annulus; the corresponding contour for a fluid without the yield stress (e.g. power-constabulary fluid) is besides shown for the sake of comparison.

Figure 3.eighteen. Schematics of velocity profiles for Bingham plastic and power-law fluids in an concentric annulus

In principle, the velocity distribution and the mean velocity of a Bingham plastic fluid flowing through an annulus can exist deduced by substituting for the shear stress in equation (3.76) in terms of the Bingham plastic model, equation (3.10). However, the signs of the shear stress (considered positive in the same sense equally the catamenia) and the velocity gradients in the two menstruum regions need to be treated with special care. With reference to the sketch shown in Figure three.18, the shearing force on the fluid is positive (σR ≤ r ≤ λ_– R) where the velocity gradient is also positive. Thus, in this region:

(3.83) ${\tau }_{\mathrm{r}\mathit{z}}\mathrm{=}{\tau }_{\mathrm{0}}^{\mathit{B}}\mathrm{+}{\mu }_{\mathit{B}}\left(\frac{\text{d}{\mathit{V}}_{\mathit{z}}}{\text{d}\mathit{r}}\right)$

On the otherhand, in the region λ₊ R ≤ r ≤ R, the velocity gradient is negative and the shearing force is also in the negative r-direction and hence

(three.84) $\mathrm{-}{\tau }_{\mathit{r}\mathrm{z}}\mathrm{=}{\tau }_{\mathrm{0}}^{\mathit{B}}\mathrm{+}{\mu }_{\mathit{B}}(\mathrm{-}\frac{\text{d}{\mathit{Five}}_{\mathit{z}}}{\mathit{dx}}\mathrm{1}$

Equations (3.83) and (3.84) tin can at present be substituted in equation (3.76) and integrated to deduce the velocity distributions. The constants of integration can be evaluated by using the no-sideslip boundary condition at both r = σR and r = R. However, the boundaries of the plug existing in the middle of the annulus are not yet known; nor is the plug velocity known. These unknowns are evaluated by applying the following three conditions, namely, the continuity of velocity at r = λ_– R and r = λ₊ R, the velocity slope is as well zero at these boundaries and finally, the force balance on the plug of fluid:

(3.85) $\mathrm{ii}\pi \mathit{R}\left({\lambda }_{+}\mathrm{+}{\lambda }_{\mathrm{-}}\right){\tau }_{\mathrm{0}}^{\mathit{B}}\mathrm{=}\left(\frac{\mathrm{-}\Delta \mathit{p}}{\mathit{L}}\right)\mathrm{\cdot }\pi ({\left({\lambda }_{\mathrm{+}}\mathit{R}\right)}^{\mathrm{2}}\mathrm{-}{\left({\lambda }_{\mathrm{-}}\mathit{R}\right)}^{\mathrm{2}})$

Unfortunately, the algebraic steps required to carry out the necessary integrations and the evaluation of the constants are quite involved and tedious. Thus, these are not presented hither and readers are referred to the original papers [Laird, 1957] or to the volume by Skelland [1967] for detailed derivations. Instead consideration is given here to the practical trouble of estimating the necessary pressure slope to maintain a stock-still flow rate of a Bingham plastic fluid or vice versa. Fredrickson and Bird [1958] organised their numerical solutions of the equations presented above in terms of the following dimensionless parameters:

$\begin{array}{l}\text{dimensionless}\text{velocity}:V_z^{\ast }=\frac{2{\mu }_B{V}_z}{R^2\left(\frac{-\Delta p}{L}\right)}\\ \text{dimensionless}\text{yield}\text{stress}:{\varphi }_0=\frac{\mathrm{two}{\tau }_0^B}{R\left(\frac{-\Delta p}{\mathrm{Fifty}}\right)}\\ \text{dimensionless}\text{flowrate}:\Omega =\frac{Q}{Q_N}\end{array}$

where Q_N is the flow rate of a Newtonian liquid of viscosity, μ _B . Thus,

$Q_N=\frac{\pi R^{\mathrm{iv}}}{8{\mu }_B}\left(\frac{-\Delta p}{L}\right)$

Fredrickson and Bird [1958] presented 3 charts (Figures 3.nineteen–three.21) showing relationships between $V_z^{*}$ , ϕ₀, Ω, σ and λ₊.

Figure 3.nineteen. Dimensionless plug velocity and plug size for laminar Bingham plastic menses in an annulus

Effigy three.20. Dimensionless flowrate for Bingham plastic fluids in laminar flow through an annulus

Effigy three.21. Chart for the interpretation of force per unit area gradient for laminar period of Bingham plastic fluids in an annulus

For given values of the Theological constants (μ _B , ${\tau }_0^B$ ), pressure gradient (−Δp/L) and the dimensions of the annulus (σ, R), the values of λ₊ and the plug velocity $V_z^{*}$ tin be read from Figure 3.19 and the value of Ω from Effigy three.20 from which the volumetric rate of flow, Q, can be estimated. For the reverse calculation, the group (Ω/ϕ₀) is independent of the pressure slope and ane must use Figure 3.21 to obtain the value of ϕ₀ and thus evaluate the required pressure level gradient (−Δp/L).

Instance 3.11

A molten chocolate (density = 1500 kg/thou^three) flows through a concentric annulus of inner and outer radii 10 mm and 20 mm, respectively, at thirty°C at the constant flow rate of 0.03 m³/min. The steady-shear behaviour of the chocolate can be approximated by a Bingham plastic model with ${\tau }_0^B$ = 35 Pa and μ _B = one Pa·;s.

(a)

Estimate the required pressure slope to maintain the menstruation, and determine the velocity and the size of the plug.

(b)

Owing to a pump malfunction, the available pressure slope drops by 25% of the value calculated in (a), what will be the new menstruum rate?

Solution Part (a):

In this instance,

$\begin{array}{l}{\tau }_0^B=35\text{P}a,{\mu }_B=\mathrm{i}\text{P}a.\mathrm{south}\\ Q=0.03{\text{m}}^3/\min =\frac{0.03}{60}{\text{m}}^{\mathrm{iii}}/s\\ \sigma =\frac{\mathrm{x}}{\mathrm{xx}}=0.5;R=\mathrm{twenty}\times {10}^{-3}\text{m}\end{array}$

Since the pressure level gradient (−Δp/L) is unknown, i must use Figure 3.21.

$\therefore \frac{\Omega }{{\varphi }_0}=\frac{4{\mu }_{B}Q}{\pi R^3{\tau }_0^B}=\frac{4\times 1\times \left(0.03/\mathrm{threescore}\right)}{3.14\times {\left(20\times {10}^{-3}\right)}^{\mathrm{three}}\times 35}=2.28$

For $\frac{\Omega }{{\varphi }_0}$ = two.28 and σ = 0.v, Figure three.21 gives:

$\begin{array}{l}{\varphi }_0=\sim 0.048\\ \therefore \left(\frac{-\Delta p}{L}\right)=\frac{\mathrm{two}{\tau }_0^B}{R{\varphi }_0}=\frac{\mathrm{two}\times 35}{20\times {10}^{-\mathrm{iii}}\times 0.048}\\ =73000\text{P}a/\text{m}\\ =73\text{kP}a/\text{g}\end{array}$

Now from Figure 3.nineteen, ϕ₀ = 0.048 and σ = 0.5,

$\begin{array}{l}{V}_{z,p}^{\ast }=\sim 0.05\\ {\lambda }_{+}=0.76\end{array}$

From the definition of $V_z^{*}$ , we have

$\begin{array}{l}{V}_{zp}^{\ast }=\frac{2{\mu }_B{\mathrm{Five}}_{zp}}{R^2\left(\frac{-\Delta p}{\mathrm{50}}\right)}\\ 0.05=\frac{\mathrm{two}\times 1\times V_{zp}}{{\left(20\times {10}^{-3}\right)}^2\left(73000\right)}\\ \therefore V_{zp}=0.73\text{m}/s\end{array}$

i.e. the plug in the central region has a velocity of 0.73 chiliad/s (compared with the mean velocity of Q/πR ²(one − σ²), i.e. 0.53 chiliad/s).

From equation (3.85):

$\left({\lambda }_{+}-{\lambda }_{-}\right)\frac{R}{2}\left(\frac{-\Delta p}{L}\right)={\tau }_0^B$

Substitution of values gives λ_– = 0.71. Thus the plug region extends from λ_– R to λ₊ R, i.eastward. from 14.2 to 15.2 mm. These calculations assume the flow to be laminar. As a outset approximation, one can define the respective Reynolds number based on the hydraulic diameter, D_h .

$\begin{array}{l}{D}_h=\frac{4\times \text{Flow}\text{area}}{\text{wetted}\text{perimeter}}=\frac{4\pi R^{\mathrm{two}}\left(1-{\sigma }^2\right)}{2\pi R\left(\mathrm{ane}+\sigma \right)}=2R\left(1-\sigma \right)\\ =2\times 20\times {\mathrm{ten}}^{-3}\left(1-\mathrm{0.five}\right)=0.02\text{m}\end{array}$

Reynolds number,

$\text{Re=}\frac{\rho 5D_h}{{\mu }_B}=\frac{1500\times 0.53\times 0.02}{\mathrm{ane}}=\mathrm{sixteen}$

The menstruation is thus likely to be streamline.

Part (b): In this case, the available pressure gradient is only 75% of the value calculated above,

$\left(\frac{-\Delta p}{\mathrm{50}}\right)=73\times 0.75=54.75\text{kPa}/\text{thou}$

We tin can now evaluate ϕ₀

${\varphi }_0=\frac{2{\tau }_0^B}{R\left(\frac{-\Delta p}{L}\right)}=\frac{2\times 35}{\mathrm{xx}\times {\mathrm{ten}}^{-3}\times 54.75\times 1000}=0.064$

From Figure 3.20, for ϕ₀ = 0.064 and σ = 0.5,

$\begin{array}{l}\Omega =\frac{Q}{Q_N}=\sim 0.1\\ \therefore Q=\mathrm{0.ane}\times Q=0.1\times \frac{\pi R^4}{8{\mu }_B}\left(\frac{-\Delta p}{\mathrm{50}}\right)\\ =\frac{\mathrm{0.ane}\times \mathrm{3.fourteen}\times {\left(20\times {\mathrm{x}}^{-3}\right)}^{\mathrm{iv}}}{8\times 1}\times 54.75\times 1000{\text{k}}^{\mathrm{iii}}/\text{s}\\ =0.000344{\text{m}}^{\mathrm{three}}/\text{s}\text{or}0.0206{\text{grand}}^3/\min \end{array}$

Ii observations tin be made here. The 25% reduction in the bachelor pressure gradient has lowered the menstruation rate past 31%. Secondly, in this case the flow rate is but 1 tenth of that of a Newtonian fluid of the aforementioned viscosity as the plastic viscosity of the molten chocolate!□

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Fluid Mechanics Challenges and Engineering science Overview

Wilson C. Chin Ph.D. , in Managed Pressure Drilling, 2012

Annular flow solutions

The merely known exact, airtight-course belittling solution is a classic one describing Newtonian flow in a concentric annulus. Let R be the outer radius and κR exist the inner radius, so that 0<κ<i. Then

(1.7a) $\begin{array}{l}\mathrm{u}(\mathrm{r})=[{\mathrm{R}}^2\mathrm{\Delta }\mathrm{p}\mathrm{/}(4\mathrm{\mu }\mathrm{50})\\ \times [1-{(\mathrm{r}\mathrm{/}\mathrm{R})}^2+(1-{\kappa }^2){\log }_{\mathrm{east}}(\mathrm{r}\mathrm{/}\mathrm{R}){\mathrm{/}\log }_{\mathrm{e}}(\mathrm{ane}\mathrm{/}\kappa )]\end{array}$

(1.7b) $\mathrm{Q}=[\mathrm{\pi }{\mathrm{R}}^4\mathrm{\Delta }\mathrm{p}\mathrm{/}(\mathrm{viii}\mathrm{\mu }\mathrm{Fifty})][1-{\kappa }^4-{(1-{\kappa }^{\mathrm{ii}})}^{\mathrm{ii}}{\mathrm{/}\log }_{\mathrm{eastward}}(\mathrm{one}\mathrm{/}\kappa )]$

noting that this solution assumes stationary walls. Here, the slope formed past Q versus Δp/L is adamant in one case and for all by the geometry and the value of viscosity. In fact, Q is inversely proportional to μ, varies directly with Δp/L, and depends only on the lumped quantity ane/μ Δp/L. The internet proportionality abiding but given can be adamant by experiment if desired.

Note that for not-Newtonian flows, even for concentric geometries, numerical procedures are required—encounter Fredrickson and Bird (1958), Skelland (1967), or Bird, Stewart, and Lightfoot (2002). The express number of verbal nonrotating solutions unfortunately summarizes the state of the art, and for this reason recourse must exist fabricated to computational rheology for the smashing majority of practical problems. We will, however, derive an exact belittling solution for Herschel-Bulkley yield stress fluids in concentric annuli without pipe motility in Chapter 5.

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Transient, Three-Dimensional, Multiphase Pipe and Annular Flow

Wilson C. Mentum Ph.D. , in Managed Pressure Drilling, 2012

Discussion 9.five Transient, Three-Dimensional Eccentric Multiphase Period Analysis for Nonrotating Newtonian Fluids

Here nosotros innovate multiphase catamenia computations for a special limit of the general problem, one assuming Newtonian mixtures in concentric or eccentric annuli (with possible cross-sectional changes in the axial direction)—however, without pipe or casing rotation. Subsequently, we will remove our Newtonian, nonrotating flow restrictions and consider general non-Newtonian fluids in eccentric annuli with steady pipe rotation. Software for the nowadays limit was developed because the solution process could be automated and Newtonian applications do be. Just our purposes are twofold: commencement to illustrate bones flow concepts and 2nd to demonstrate that our formulation, solution, and software foundation for subsequent evolution are sound and correct.

Give-and-take ix.5, Example 9.1

We first show that our exact, steady concentric Newtonian menses solution and the transient numerical model under consideration are consistent in the concentric single-phase flow limit. This is intended to validate the software architecture, which is complicated and forms the basis for other models. The simulator for our exact solution is launched from the before "Steady 2D" menu in Figure 9.18, leading to the applications plan in Figure 9.nineteen. Notation how the causeless parameters yield a flow rate of 947.ane   gpm.

Figure 9.eighteen. General "Steady 2nd" menu.

Figure 9.19. Exact two-dimensional Newtonian catamenia solution.

Next nosotros launch the "Transient 3D, Multiphase, Newtonian, Non-Rotating" menses simulator in Effigy ix.20. For multiphase problems, it is non meaningful to specify force per unit area gradients every bit in single-stage calculations; these gradients vary with infinite and time equally local fluids mix, and it is impossible to state clearly what they are. One is therefore forced to specify total menstruation rate, at least approximately, and this specification must be used when dealing with multiphase applications. Our simulator operates in a "Specify flow rate" mode.

Figure 9.twenty. Consequent transient simulation parameters.

To be completely consistent with Figure 9.19, we presume a 1-cp viscosity for both "left" and "right" fluids, zip pipage speed, and identical geometries. We also assume identical small specific gravities; low mechanical inertias allow larger time steps and reduce integration times needed for convergence. Internal to the software, C=0 ways left backdrop, (i.e., μ_left and ρ_left), while C=1 means right; since left and right backdrop are identical, the choice C_left=C_correct=ane ensures that C=i continuously throughout and the fluid is homogeneous.

Note that, in Figure 9.20, nosotros have entered 947.1 as the target menstruation rate. Once numerical integrations begin, the imposed motion must overcome "nonuniformities" associated with the uniform (unsheared) flow used to initialize the adding and, of course, the furnishings of inertia. After some time, the calculations converge. For instance, Figure 9.21 gives a flow rate of 949.1   gpm for an error of 0.two percent. For the 10,000 time steps shown, the computing time is nearly five minutes for this iii-dimensional run. We have used the transient, 3-dimensional, two-phase flow solver to reproduce an verbal, steady, two-dimensional, single-phase flow outcome. In general, single-phase flows can be calculated this fashion, although this is obviously suboptimal. Withal, the example was designed to show that the numerical model is basically correct.

Effigy 9.21. Example ane, smoothly convergent flow rate history.

Transient Flow Subtleties

Again nosotros remind the reader of certain difficulties encountered in transient menstruum modeling. In steady flow analysis, whether concentric or eccentric, computations for menstruum rate (when pressure gradients are given) are very rapid and vice versa. For linear Newtonian flows, these are particularly fast. If (∂P/∂z)₁ corresponding to Q₁ is known from just one eccentric or concentric calculation or experiment, then the identity (∂P/∂z)₂/Q₂=(∂P/∂z)₁/Q₁ allows u.s. to immediately obtain (∂P/∂z)₂ when Q_ii is given or Q₂ when (∂P/∂z)₂ is given. For non-Newtonian flows, the nonlinearity of the pressure gradient and flow rate relationship disallows this simple rescaling. However, the "Specify volumetric flow rate" option in Effigy ix.xviii does use a speedily convergent half-step method to estimate the pressure gradient respective to a target flow rate to within i percent accuracy.

In transient calculations, one can in principle specify total volumetric menstruum charge per unit at each instant in time. However, to reach the required solution, numerous attempts using different pressure gradients will have to exist made at each time step. When this is repeated for the entire range of time integration, the computations needed are voluminous and require hours or overnight runs. This is particularly unacceptable if, perhaps during the calculations, instabilities are encountered; then all of the numerical effort expended will be wasted. Thus, we ask if at that place is an acceptable compromise: "Is there an judge pressure slope we can use in a constant menstruum rate process?"

For Newtonian nonrotating flows, the respond is yes. Nosotros recall from our theoretical discussion of steady single-phase flow that volumetric flow rate is directly proportional to the pressure gradient ∂P/∂z and inversely related to the viscosity μ. If "one" and "ii" now announce two positions along the 3-dimensional channel (at a fixed instant in time) without area changes, then constancy of flow charge per unit implies that (∂P/∂z)₁/μ₁=(∂P/∂z)₂/μ₂. Suppose that the volumetric menstruation rate at the (left) inlet and the starting viscosity are specified. Then the pressure gradient required for the eccentric Newtonian flow tin can be obtained from the "Steady 2D" solver in Figure 9.18. As the fluid at the inlet flows downstream, it mixes with the "right" fluid and local concentrations will modify. The underlying viscosity volition consequently change, in a fashion consequent with an assumed mixing relationship (taken once more as the Todd-Longstaff law). If the local viscosity is at present μ₂, then the corresponding pressure gradient is (∂P/∂z)_ii=(∂P/∂z)_ane μ₂/μ_i, showing correctly, for instance, that an increase in viscosity will require an increment in pressure gradient.

This procedure has been programmed into the solver of Figure nine.twenty; there is no demand to operate the simulator in Figure 9.18 considering the procedure has been completely automated. Over again, starting pressure level gradients are obtained from inlet conditions and local values are obtained past concentration-dependent rescaling. This automation is but convenient for Newtonian mixtures where at that place is no pipage rotation. The "(∂P/∂z)_one/μ₁=(∂P/∂z)₂/μ_ii" law does not apply to eccentric issues with rotation, although information technology remains valid for concentric rotating flow considering axial and azimuthal modes decouple. For more than complicated issues, a more complete arroyo applies, with different degrees of complexity, depending on the nature of the underlying flow. The general trouble volition be considered in a split up discussion.

Give-and-take 9.5, Examples ix.2 and 9.iii

For our 2d adding, we repeat the in a higher place simulation except that nosotros double the inlet-outlet viscosity and density ratios, equally shown in Effigy 9.22. Note that, in club to rail two different phases, the concentrations at the inlet and outlet are set to 0 and 1, respectively. The calculation yields virtually identical menstruum rates and flow rate history curves. Why? This occurs because, in Newtonian mixtures, the ratio of density to viscosity controls the dynamics and not either parameter alone; in that location is, however, an effect associated with the ratio of density to diffusion coefficient, which need not ever be small-scale. Thus, the effects of the doubling almost cancel. In our 3rd simulation, we set our inlet-outlet viscosity and density ratios to five and ii, respectively. Figure ix.23 shows that the volumetric flow rate history changes somewhat, with the predominant effect being the time required to accomplish equilibrium.

Effigy ix.22. Instance ii adding.

Figure ix.23. Example 3 calculation.

A detailed clarification of the simulator appears in Discussion 9.6. The reader should study it, since many of its software features are shared past the more general solver introduced in Discussion ix.7, which deals with real 2-phase flows in which the mixing of non-Newtonian fluids, in the presence of rotation, is addressed. Mixing is controlled by numerous factors: convection, diffusion, annular geometry, rheology, period rate, and initial conditions. This complication means that general conclusions are hard to formulate and that each flow solution must be interpreted on a case-by-example basis. Predictions should be substantiated by laboratory experiment and field information whenever possible.

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STEADY MICROSCOPIC BALANCES WITH GENERATION

Ismail Tosun , in Modeling in Send Phenomena (Second Edition), 2007

9.1.iv.two Investigation of the limiting cases

■ Instance (i) κ → 1

When the ratio of the radius of the inner pipe to that of the outer pipe is close to unity, i.east., κ → 1, a concentric annulus may be considered a thin-aeroplane slit and its curvature tin be neglected. Approximation of a concentric annulus as a parallel plate requires the width, Westward, and the length, L, of the plate to be defined as

(nine.ane-109) $\mathcal{W}=\pi R\left(1+\kappa \right)$

(9.one-110) $B=R\left(\mathrm{ane}-\kappa \right)$

Therefore, the product WB ³ is equal to

(9.i-111) $\begin{array}{ccc}\mathcal{W}{B}^3=\pi R^4\left(\mathrm{one}-{\kappa }^2\right){\left(1-\kappa \right)}^2& \Rightarrow & \pi \end{array}{R}^{\mathrm{iv}}=\frac{\mathcal{W}{B}^3}{\left(1-{\kappa }^2\right){\left(1-\kappa \right)}^2}$

and so that Eq. (9.1-99) becomes

(9.1-112) $\mathcal{Q}=\frac{\left({\mathcal{P}}_o-{\mathcal{P}}_{\mathrm{50}}\right)\mathcal{Due\; west}{B}^3}{8\mu L}\underset{\kappa \to 1}{\text{lim}}\left[\frac{\mathrm{i}+{\kappa }^2}{{\left(\mathrm{ane}-\kappa \right)}^{\mathrm{ii}}}+\frac{1+\kappa }{\left(1-\kappa \right)\text{ln}\kappa }\right]$

Substitution of ψ = one – κ into Eq. (nine.ane-112) gives

(9.i-113) $\mathcal{Q}=\frac{\left({\mathcal{P}}_o-{\mathcal{P}}_{\mathrm{50}}\right)\mathcal{West}{B}^3}{8\mu L}\underset{\psi \to 0}{\text{lim}}\left[\frac{{\psi }^2-2\psi +2}{{\psi }^2}+\frac{2-\psi }{\psi \text{ln}\left(1-\psi \right)}\right]$

The Taylor series expansion of the term ln(1 – ψ) is

(9.1-114) $\text{ln}\left(1-\psi \right)=-\psi -\frac{1}{2}{\psi }^2-\frac{1}{3}{\psi }^{\mathrm{iii}}-\cdots$

Using Eq. (9.1-114) in Eq. (9.1-113) and conveying out the divisions yield

(9.i-115) $\mathcal{Q}=\frac{\left({\mathcal{P}}_o-{\mathcal{P}}_L\right)\mathcal{W}{B}^{\mathrm{iii}}}{8\mu \mathrm{50}}\underset{\psi \to 0}{\text{lim}}\left[\mathrm{ane}-\frac{2}{\psi }+\frac{\mathrm{ii}}{{\psi }^2}+\left(-\frac{2}{{\psi }^2}+\frac{2}{\psi }-\frac{1}{\mathrm{three}}-\frac{\psi }{2}+\cdots \right)\right]$

or,

(ix.i-116) $\mathcal{Q}=\frac{\left({\mathcal{P}}_o-{\mathcal{P}}_{\mathrm{Fifty}}\right)\mathcal{W}{B}^3}{8\mu L}\underset{\psi \to 0}{\text{lim}}\left(\frac{2}{3}-\frac{\psi }{\mathrm{two}}+\cdots \right)=\frac{\left({\mathcal{P}}_o-{\mathcal{P}}_L\right)\mathcal{Due\; west}{B}^3}{12\mu \mathrm{Fifty}}$

which is equivalent to Eq. (9.1-26).

■ Instance (ii) κ → 0

When the ratio of the radius of the inner piping to that of the outer pipage is close to zero, i.due east., κ → 0, a concentric annulus may be considered a round pipe of radius R. In this case, Eq. (9.1-99) becomes

(ix.1-117) $\mathcal{Q}=\frac{\pi \left({\mathcal{P}}_o-{\mathcal{P}}_L\right)R^4}{\mathrm{eight}\mu L}\underset{\kappa \to 0}{\text{lim}}\left[1-{\kappa }^4+\frac{{\left(1-{\kappa }^2\right)}^2}{\text{ln}\kappa }\right]$

Since ln 0 = −∞, Eq. (9.i-117) reduces to

(9.1-118) $\mathcal{Q}=\frac{\pi \left({\mathcal{P}}_o-{\mathcal{P}}_L\right)R^4}{\mathrm{eight}\mu \mathrm{Fifty}}$

which is identical to Eq. (9.ane-83).

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Bones Biomechanics of the Lumbar Spine

Hans-Joachim Wilke , David Volkheimer , in Biomechanics of the Spine, 2018

Intervertebral Disc

Neighboring vertebral bodies are connected by flexible fibrocartilaginous joints, termed intervertebral discs. The intervertebral discs consist of the central nucleus pulposus, which is surrounded past the concentric annulus fibrosus. At that place is no articulate stardom between the amorphous nucleus pulposus and the highly oriented structure of the annulus fibrosus. Nowadays, this zone is referred to as "transition zone." The nucleus pulposus and the inner annulus fibrosus are separated from the bony vertebral endplate by a thin layer of hyaline cartilage. Mainly consisting of proteoglycans and collagen fibers, the fibrocartilaginous endplates' chemic composition is similar to the other components of the intervertebral disc, meaning that, in a good for you state, they do not constitute a further barrier for improvidence of nutrients through the endplates, which is the key pathway for nutrient support to the intervertebral discs. The supply of nutrients via improvidence, and to a minor extent by majority transport, is of crucial importance for the intervertebral disc, because a healthy developed disc has no or simply a very limited number of claret vessels, which are restricted to the outer annulus.

The central nucleus pulposus mainly consists of proteoglycans (about l% of the dry out weight of the nucleus of a child), which correspond long chains of glycosaminoglycans (GAGs) linked to proteins. Proteoglycans are highly negatively charged, which allows them to attract positively charged cations from the interstitial fluid. Therefore they accept the capacity to attract and imbibe water (Urban and Mcmullin, 1988), which amounts to about lxxx% of the total weight of the nucleus pulposus in young individuals. With slightly less than twenty% of the dry weight of the nucleus pulposus, collagen type Two is the cardinal collagen of the nucleus pulposus and serves to hold the proteoglycans together. Furthermore, the nucleus pulposus contains small amounts of collagen type I, elastin fibers, and other non-collagenous proteins.

The nucleus pulposus is surrounded by the highly ordered circular annulus fibrosus, which consists of up to 20 layers of circular collagen type I-rich sheets, called lamellae. The lamellae occasionally are interconnected past radial collagen bundles (Schollum et al., 2008), just most of the space is filled with proteoglycans. Within a lamella, the collagen fibers are bundled at an approximate angle of xxx caste with respect to the vertebral endplates. This angle is reversed betwixt adjacent lamellae. The annulus fibrosus is greater and thicker anteriorly than posteriorly in the lumbar spine, which is why the nucleus pulposus lies posteriorly to the geometric center of the intervertebral disc. The fibers in the outer zone of the annulus fibrosus attach directly to the bony apophyseal ring via Sharpey's fibers, whereas the inner annulus fibers join the cartilaginous vertebral endplate. Compared to the nucleus pulposus, the annulus fibrosus of a healthy disc contains significantly fewer proteoglycans and consequently approximately x% less water, but more than double the amount of collagens. In contrast with the nucleus pulposus, where collagen type II is predominant, the main collagen of the annulus fibrosus is blazon I, which is more suitable to transmit and sustain tensile forces.

The low permeability of the endplates and the surrounding annulus fibrosus, in conjunction with the highly water-attracting nucleus pulposus, causes the nucleus pulposus of healthy specimens to exhibit an intrinsic pressure ranging from 0.v to 1.5   MPa in a neutral position (i.e., no flexion or extension). Thus the annulus fibrosus becomes pre-tensed, which allows the intervertebral disc to sustain loftier compressive loads. The intradiscal pressure generally increases with deflection of the motion segments. Loading of the intervertebral disc in movement planes other than by flexion/extension and lateral bending, such as shear movements or axial rotations, is primarily resisted by the unique lamellar structure of the annulus fibrosus, where resistance is primarily provided by tensioning the collagen fibers, like to the cobweb reinforcement of a tire.

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