When blood flows through the branching area in the aorta and large arteries it changes direction. When a fluid runs through tubes a change of the fluid direction also occurs, and this change is evaluated in terms of the centrifugal force that acts on particles toward the external rounding off. This action results in a secondary flow, and redistribution of velocities takes place. On the external wall, the velocity of flow and the shear stress increase, but they decrease on the internal surface of the wall 2829. The results of experimental studies showed that the increase in the shear stress and the influence of the curvature of the vessel on resistance are considerably higher during laminar flow than turbulent flow 2829.

Assuming that blood flow in the cardiovascular system is predominantly laminar 29, let me consider the shear stress within the laminar flow on the external wall of the branching part of the vessel. It is known 29 that branching regions of arteries have various angles. In general, for their description, the coefficient of branching out is applied (i.e. the relationship among sum of areas, angles of branching, divisor of flow, profiles, velocities, Reynolds number, radius of curvature of internal wall at branching, diameters of diverging vessels) 29. Various approaches to evaluate the degree of branching of the aorta make the classification of branching difficult.

As in previous research 28, let me take the Dean number as a parameter to determine the influence of curvature on the resistance to blood flow. I am interested in considering the following Reynolds number range that most closely corresponds to human (physiological) range:

101.6 < Re ⁡ ( ( R r ) 1 / 2 ) < 10 3 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqaIXaqmcqaIWaamcqaIXaqmcqGGUaGlcqaI2aGncqGH8aapcyGGsbGucqGGLbqzcqGGOaakcqGGOaakdaWcaaqaaiabdkfasbqaaiabdkhaYbaacqGGPaqkdaahaaWcbeqaaiabigdaXiabc+caViabikdaYaaakiabcMcaPiabgYda8iabigdaXiabicdaWmaaCaaaleqabaGaeG4mamdaaaaa@41EC@

where R = radius of the tube (vessel), r = radius of the curvature that can be applied to various branching areas of the entire cardiovascular system 29. Let me show how the Dean number describes conditions arising from various anatomical variations of the canine aorta. For this I shall take three average curvature values that correspond closely to the curvature of the aorta in general: R = 7 mm, r = 25 mm; the maximal value of curvature, R = 7 mm, r = 15 mm; the minimal value of curvature of the ascending aorta, R = 7 mm, r = 40 mm (Figure 1).

If flow is laminar, then the Dean number is determined as:

D = 0.5   Re ⁡ ( ( R r ) 1 / 2 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqqGebarcqGH9aqpcqaIWaamcqGGUaGlcqaI1aqncqqGGaaicyGGsbGucqGGLbqzcqGGOaakcqGGOaakdaWcaaqaaiabdkfasbqaaiabdkhaYbaacqGGPaqkdaahaaWcbeqaaiabigdaXiabc+caViabikdaYaaakiabcMcaPaaa@3DDF@

where Re = Re⁡=(dρU∞)μ MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacyGGsbGucqGGLbqzcqGH9aqpdaWcaaqaamaabmGabaGaemizaqgcciGae8xWdiNaemyvau1aaSbaaSqaaiabg6HiLcqabaaakiaawIcacaGLPaaaaeaacqWF8oqBaaaaaa@3970@

and d = diameter of the vessel, μ = fluid viscosity, ρ = density and U_∞ = blood flow velocity.

The value of the coefficient of resistance (λ) is calculated by the following formula 28:

( λ λ 0 ) = 0.37  D 0.36 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqGGOaakdaWcaaqaaGGaciab=T7aSbqaaiab=T7aSnaaBaaaleaacqaIWaamaeqaaaaakiabcMcaPiabg2da9iabicdaWiabc6caUiabiodaZiabiEda3iabbccaGiabbseaenaaCaaaleqabaGaeGimaaJaeiOla4IaeG4mamJaeGOnaydaaaaa@3D87@

where λ₀ = the coefficient of resistance of a straight tube by the formula 28:

( 1 λ 0 ) = 2.0  lg ( u d λ 0 v ) − 0.8       ( 1 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqGGOaakdaWcaaqaaiabigdaXaqaamaakaaabaacciGae83UdW2aaSbaaSqaaiabicdaWaqabaaabeaaaaGccqGGPaqkcqGH9aqpcqaIYaGmcqGGUaGlcqaIWaamcqqGGaaicqqGSbaBcqqGNbWzcqGGOaakdaWcaaqaaiabdwha1jabdsgaKnaakaaabaGae83UdW2aaSbaaSqaaiabicdaWaqabaaabeaaaOqaaGqaciab+zha2baacqGGPaqkcqGHsislcqaIWaamcqGGUaGlcqaI4aaocaWLjaGaaCzcamaabmGabaGaeGymaedacaGLOaGaayzkaaaaaa@49F2@

where v = kinematic viscosity 28.

The shear stress is determined by the formula 29:

τ = λ ρ u 2 8       ( 2 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFepaDcqGH9aqpdaWcaaqaaiab=T7aSjab=f8aYjabdwha1naaCaaaleqabaGaeGOmaidaaaGcbaGaeGioaGdaaiaaxMaacaWLjaWaaeWaceaacqaIYaGmaiaawIcacaGLPaaaaaa@3A53@

Where u is the average flow velocity.

Calculations for the three modes of curvature are given in Tables 1, 2 and 3.

According to these tables, the degree of curvature (the Dean number) is highly informative with regard to the shear stress on the external surface of the vessel. For instance, for the ascending aorta with an average blood velocity of 0.2 m/s, the shear stress is 1.5–2 times greater than that calculated for a straight (plane) tube (τ = 0.43 N/m²) 29. Also, according to tables 1, 2, 3, the curvature may play a significant role when the vessel is compressed as the result, for example, of injury. The compressed part of the vessel can increase the shear stress 27; however, when the curvature is significant, the shear stress may become even greater owing to the influence of both curvature and compression. Thus, the Dean number is an important factor to consider when determining the shear stress acting on the external wall of the vessel. In addition, according to tables 1, 2, 3, the shear stress may exceed the physiological threshold calculated for the endothelium (i.e. 40 N/m²) at the extreme value of blood velocity (1.2 m/s) 2729.

According to the above and to our previous research 2127, three internal factors have been identified (the Dean number, compression and blood flow velocity) that can play a dominant role with respect to endothelium damages and resulting vessel rupture. Let me now investigate the conditions that appear as the result of fluid (plasma) and particle (erythrocyte) movement, i.e. the multiphase character of the medium. It will allow me to look at the role of erythrocyte concentration, blood viscosity and rouleaux formation on the external surface of the vessel.

Let me now determine the association between yield velocity and shear stress 2729. Calculations are given in Tables 4, 5 and 6.

According to tables 4, 5, 6, shear stress exceeds the maximal physiological value (40 N/m2) when blood viscosity is around 3 mNcm^-2 and the concentration of erythrocytes 28.7 %. Alternatively, at high yield velocity (10 – 50 m^-1), only a high concentration of erythrocytes (48% and higher) can result in abnormal values of blood viscosity 29.

When the blood flow velocity is moderate (i.e. 0.1 – 0.4 m/s) 29 and the shear stress calculated by the formulae (1, 2) equals 0.15 mNm^-2 (i.e. physiological value), then according to table 6 it is possible that the viscosity will increase up to 9–11 mNm^-2 if the erythrocyte concentration equals 48% or more and the yield velocity is within the range 10 to 50 c^-1 (calculations are made at blood flow rate = 0.1 m/s) Therefore, it can be concluded that only quite specific conditions such as an increase in the concentration of erythrocytes (48% or more) and relatively slow motion of blood (equal to or less than 0.1 mc^-1) 21 may lead to the formation of rouleaux and increase the shear stress on the external vessel wall.

In addition, according to tables 4, 5, 6, it can be seen that the shear stress values are different at equal yield velocities. This indicates that two phases participate in creating the shear stress, since if the flow of blood were homogeneous the stress would be the same.

For two-phase flow, the total shear stress is the sum of the shear stresses of the two phases considered separately:

τ = μ 1 f 1 ( ∂ u 1 ∂ y 1 ) + μ 2 f 2 ( ∂ u 2 ∂ y 2 )       ( 3 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFepaDcqGH9aqpcqWF8oqBdaWgaaWcbaGaeGymaedabeaakiabdAgaMnaaBaaaleaacqaIXaqmaeqaaOGaeiikaGYaaSaaaeaacqWFciITcqWG1bqDdaWgaaWcbaGaeGymaedabeaaaOqaaiab=jGi2kabdMha5naaBaaaleaacqaIXaqmaeqaaaaakiabcMcaPiabgUcaRiab=X7aTnaaBaaaleaacqaIYaGmaeqaaOGaemOzay2aaSbaaSqaaiabikdaYaqabaGccqGGOaakdaWcaaqaaiab=jGi2kabdwha1naaBaaaleaacqaIYaGmaeqaaaGcbaGae8NaIyRaemyEaK3aaSbaaSqaaiabikdaYaqabaaaaOGaeiykaKIaaCzcaiaaxMaadaqadiqaaiabiodaZaGaayjkaiaawMcaaaaa@5243@

where phase 1 (f₁) is liquid plasma and phase 2 (f₂) is a relatively solid phase (erythrocytes and rouleaux). From this point of view, let me consider the theory of a multi-phase medium 212730, and in particular the relationship between the longitudinal and transverse velocities of the phases in the boundary layer during flow around a flat surface 27.

Let us assume that the carrying flow, in a direction parallel to the flat surface of the vessel, is viscous, and the flow itself (solid particles = erythrocytes) is ideal. Also, we will assume that axis x is along the direction of flow while axis y is perpendicular to the surface. Then the two equations for plasma (1^st phase) will take the following form:

ρ 1 u 1 ∂ u 1 ∂ x + ρ 1 v 1 ∂ u 1 ∂ y = μ ∂ 2 u 1 ∂ y 2 + k ( u 2 − u 1 ) ,       ( 4 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFbpGCdaWgaaWcbaGaeGymaedabeaakiabdwha1naaBaaaleaacqaIXaqmaeqaaOWaaSaaaeaacqWFciITcqWG1bqDdaWgaaWcbaGaeGymaedabeaaaOqaaiab=jGi2kabdIha4baacqGHRaWkcqWFbpGCdaWgaaWcbaGaeGymaedabeaakiabdAha2naaBaaaleaacqaIXaqmaeqaaOWaaSaaaeaacqWFciITcqWG1bqDdaWgaaWcbaGaeGymaedabeaaaOqaaiab=jGi2kabdMha5baacqGH9aqpcqWF8oqBdaWcaaqaaiab=jGi2oaaCaaaleqabaGaeGOmaidaaOGaemyDau3aaSbaaSqaaiabigdaXaqabaaakeaacqWFciITcqWG5bqEdaahaaWcbeqaaiabikdaYaaaaaGccqGHRaWkcqWGRbWAcqGGOaakcqWG1bqDdaWgaaWcbaGaeGOmaidabeaakiabgkHiTiabdwha1naaBaaaleaacqaIXaqmaeqaaOGaeiykaKIaeiilaWIaaCzcaiaaxMaadaqadiqaaiabisda0aGaayjkaiaawMcaaaaa@60F6@

∂ ( ρ 1 u 1 ) ∂ x + ∂ ( ρ 1 v 1 ) ∂ y = 0.       ( 5 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=jGi2kabcIcaOiab=f8aYnaaBaaaleaacqaIXaqmaeqaaOGaemyDau3aaSbaaSqaaiabigdaXaqabaGccqGGPaqkaeaacqWFciITcqWG4baEaaGaey4kaSYaaSaaaeaacqWFciITcqGGOaakcqWFbpGCdaWgaaWcbaGaeGymaedabeaakiabdAha2naaBaaaleaacqaIXaqmaeqaaOGaeiykaKcabaGae8NaIyRaemyEaKhaaiabg2da9iabicdaWiabc6caUiaaxMaacaWLjaWaaeWaceaacqaI1aqnaiaawIcacaGLPaaaaaa@4B23@

where κ is a coefficient of phase interaction 31. The three equations for erythrocytes (2^nd phase) will take the following form:

ρ 2 u 2 ∂ u 2 ∂ x + ρ 2 v 2 ∂ u 2 ∂ y = k ( u 2 − u 1 ) ,       ( 6 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFbpGCdaWgaaWcbaGaeGOmaidabeaakiabdwha1naaBaaaleaacqaIYaGmaeqaaOWaaSaaaeaacqWFciITcqWG1bqDdaWgaaWcbaGaeGOmaidabeaaaOqaaiab=jGi2kabdIha4baacqGHRaWkcqWFbpGCdaWgaaWcbaGaeGOmaidabeaakiabdAha2naaBaaaleaacqaIYaGmaeqaaOWaaSaaaeaacqWFciITcqWG1bqDdaWgaaWcbaGaeGOmaidabeaaaOqaaiab=jGi2kabdMha5baacqGH9aqpcqWGRbWAcqGGOaakcqWG1bqDdaWgaaWcbaGaeGOmaidabeaakiabgkHiTiabdwha1naaBaaaleaacqaIXaqmaeqaaOGaeiykaKIaeiilaWIaaCzcaiaaxMaadaqadiqaaiabiAda2aGaayjkaiaawMcaaaaa@553F@

ρ 2 u 2 ∂ v 2 ∂ x + ρ 2 v 2 ∂ v 2 ∂ y = k ( v 2 − v 1 ) ,       ( 7 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFbpGCdaWgaaWcbaGaeGOmaidabeaakiabdwha1naaBaaaleaacqaIYaGmaeqaaOWaaSaaaeaacqWFciITcqWG2bGDdaWgaaWcbaGaeGOmaidabeaaaOqaaiab=jGi2kabdIha4baacqGHRaWkcqWFbpGCdaWgaaWcbaGaeGOmaidabeaaieGakiab+zha2naaBaaaleaacqaIYaGmaeqaaOWaaSaaaeaacqWFciITcqWG2bGDdaWgaaWcbaGaeGOmaidabeaaaOqaaiab=jGi2kabdMha5baacqGH9aqpcqWGRbWAcqGGOaakcqWG2bGDdaWgaaWcbaGaeGOmaidabeaakiabgkHiTiabdAha2naaBaaaleaacqaIXaqmaeqaaOGaeiykaKIaeiilaWIaaCzcaiaaxMaadaqadiqaaiabiEda3aGaayjkaiaawMcaaaaa@554F@

∂ ( ρ 2 u 2 ) ∂ x + ∂ ( ρ 2 v 2 ) ∂ y = 0.       ( 8 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=jGi2kabcIcaOiab=f8aYnaaBaaaleaacqaIYaGmaeqaaOGaemyDau3aaSbaaSqaaiabikdaYaqabaGccqGGPaqkaeaacqWFciITcqWG4baEaaGaey4kaSYaaSaaaeaacqWFciITcqGGOaakcqWFbpGCdaWgaaWcbaGaeGOmaidabeaakiabdAha2naaBaaaleaacqaIYaGmaeqaaOGaeiykaKcabaGae8NaIyRaemyEaKhaaiabg2da9iabicdaWiabc6caUiaaxMaacaWLjaWaaeWaceaacqaI4aaoaiaawIcacaGLPaaaaaa@4B31@

An equation for the 1^st and 2^nd phases will take the following form:

The boundary conditions for the system of partial differential equations (4)-(9) depend on x and y. Thus, for x = x₀ and any y, I have:

ρ 2 ρ 20 + ρ 1 ρ 10 = 1.       ( 9 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=f8aYnaaBaaaleaacqaIYaGmaeqaaaGcbaGae8xWdi3aaSbaaSqaaiabikdaYiabicdaWaqabaaaaOGaey4kaSYaaSaaaeaacqWFbpGCdaWgaaWcbaGaeGymaedabeaaaOqaaiab=f8aYnaaBaaaleaacqaIXaqmcqaIWaamaeqaaaaakiabg2da9iabigdaXiabc6caUiaaxMaacaWLjaWaaeWaceaacqaI5aqoaiaawIcacaGLPaaaaaa@41C7@

u₁ = u₂ = f₁, v₁ = v₂ = f₂, ρ₂ = ρ2∗ MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFbpGCdaqhaaWcbaGaeGOmaidabaGaey4fIOcaaaaa@3081@, ρ₁ = ρ1∗ MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFbpGCdaqhaaWcbaGaeGymaedabaGaey4fIOcaaaaa@307F@;     (10)

for x > x₀ and y = 0, I have:

u₁ = u₂ = 0, v₁ = 0, v₂ = f₃(τ₁), ρ₁ = 0;     (11)

and for x > x₀, y = δ

u₁ = u₂ = u_∞.     (12)

In the above, u₁ and u₂ are the longitudinal components of velocity; v₁ and v₂ are the transverse components of velocity; f₁ and f₂ = initial distribution of the velocities in the boundary layer; ρ2∗ MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFbpGCdaqhaaWcbaGaeGOmaidabaGaey4fIOcaaaaa@3081@ and ρ1∗ MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFbpGCdaqhaaWcbaGaeGymaedabaGaey4fIOcaaaaa@307F@ = initial distribution of the densities. The beginning of the 2^nd phase (due to separation) was determined from the calculated shear stress and the corresponding separation determined from the experimental data.

The system of equations (4)-(9) with boundary conditions (10)-(12) was solved numerically by applying the finite-difference method with order of approximation equal to 0.0001. For two-phase flow, the shear stress is calculated as a sum of the shear stresses of the two phases. The numerical solution of the system for the equation of the boundary layer for transverse velocities is shown in Figures 2 and 3, where the longitudinal and transverse velocities of the two phases differ negligibly from each other.

This allows me to determine the approximate value of the viscosity of f₂ and the shear stress corresponding to f₁ and f₂. The calculations are given in Tables 7, 8 and 9

Let me determine the volume fraction of rouleaux in f₂. It is known that in the absence of blood motion, when the yield velocity equals zero, erythrocytes form rouleaux 29. First, let me consider the process of particle precipitation in a fluid. The change in dimensionless velocity (β) of the precipitation of particles depends on the volume fraction of f₂ (where β is the relationship of the velocity of group precipitation to the velocity of a single precipitation). The calculations have been made by other authors, who have proved that the relative velocity can be determined by the formula due to A. D. Maude 32:

β = (1-f₂)^α/m

Where α/m depends on the Reynolds number; if the Reynolds number equals zero then α/m equals 5; if the Reynolds number ranges from 10 to 100 then α/m ranges from 4 to 3.5, or:

μ_m = (1+cf₂)(1+2.5 f₂ + 10.05 f₂)

or

β = − c f 2 + [ c 2 ( 1 − f 1 2 ) + f 1 3 ] 1 2 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFYoGycqGH9aqpcqGHsislcqWGJbWycqWGMbGzdaWgaaWcbaGaeGOmaidabeaakiabgUcaRmaadmGabaGaem4yam2aaWbaaSqabeaacqaIYaGmaaGcdaqadiqaaiabigdaXiabgkHiTiabdAgaMnaaDaaaleaacqaIXaqmaeaacqaIYaGmaaaakiaawIcacaGLPaaacqGHRaWkcqWGMbGzdaqhaaWcbaGaeGymaedabaGaeG4mamdaaaGccaGLBbGaayzxaaWaaWbaaSqabeaadaWccaqaaiabigdaXaqaaiabikdaYaaaaaaaaa@46AA@

If one considers the sedimentation of a particle in a suspension with viscosity μ_m and density ρ_m, then the equilibrium equation can be expressed as 31:

f 2 ρ 2 i g − f 2 ρ m g + 9 2 f 2 μ m a − 2 ( V 1 − V 2 ) = 10       ( 13 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqWGMbGzdaWgaaWcbaGaeGOmaidabeaaiiGakiab=f8aYnaaBaaaleaacqaIYaGmcqWGPbqAaeqaaOGaem4zaCMaeyOeI0IaemOzay2aaSbaaSqaaiabikdaYaqabaGccqWFbpGCdaWgaaWcbaGaemyBa0gabeaakiabdEgaNjabgUcaRmaalaaabaGaeGyoaKdabaGaeGOmaidaaiabdAgaMnaaBaaaleaacqaIYaGmaeqaaOGae8hVd02aaSbaaSqaaiabd2gaTbqabaGccqWGHbqydaahaaWcbeqaaiabgkHiTiabikdaYaaakmaabmGabaGaemOvay1aaSbaaSqaaiabigdaXaqabaGccqGHsislcqWGwbGvdaWgaaWcbaGaeGOmaidabeaaaOGaayjkaiaawMcaaiabg2da9iabigdaXiabicdaWiaaxMaacaWLjaWaaeWaceaacqaIXaqmcqaIZaWmaiaawIcacaGLPaaaaaa@57B2@

V c = 2 9 ( ρ 2 − ρ 1 ) g μ 1 a 2       ( 14 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqWGwbGvdaWgaaWcbaGaem4yamgabeaakiabg2da9maalaaabaGaeGOmaidabaGaeGyoaKdaamaalaaabaWaaeWaceaaiiGacqWFbpGCdaWgaaWcbaGaeGOmaidabeaakiabgkHiTiab=f8aYnaaBaaaleaacqaIXaqmaeqaaaGccaGLOaGaayzkaaGaem4zaCgabaGae8hVd02aaSbaaSqaaiabigdaXaqabaaaaOGaemyyae2aaSbaaSqaaiabikdaYaqabaGccaWLjaGaaCzcamaabmGabaGaeGymaeJaeGinaqdacaGLOaGaayzkaaaaaa@461C@

ρ_m = f₁ρ_1i + f₂ρ_2i     (15)

Using equations (13), (14) and (15) and the condition V₁ = 0 (velocity of the f₁ phase) it follows that:

μ m μ 1 = f 1 V c V 2 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=X7aTnaaBaaaleaacqWGTbqBaeqaaaGcbaGae8hVd02aaSbaaSqaaiabigdaXaqabaaaaOGaeyypa0ZaaSaaaeaacqWGMbGzdaWgaaWcbaGaeGymaedabeaakiabdAfawnaaBaaaleaacqWGJbWyaeqaaaGcbaGaemOvay1aaSbaaSqaaiabikdaYaqabaaaaaaa@3B87@

where α is the diameter, μ is viscosity and ρ is the density of the mixture

Taking into account β, I deduce the following:

μ m μ 1 = f 1 / ( ( c 2 ( 1 − f 1 ) 2 + f 1 3 ) − c f 2 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=X7aTnaaBaaaleaacqWGTbqBaeqaaaGcbaGae8hVd02aaSbaaSqaaiabigdaXaqabaaaaOGaeyypa0JaemOzay2aaSbaaSqaaiabigdaXaqabaGccqGGVaWlcqGGOaakdaGcaaqaaiabcIcaOiabdogaJnaaCaaaleqabaGaeGOmaidaaOGaeiikaGIaeGymaeJaeyOeI0IaemOzay2aaSbaaSqaaiabigdaXaqabaGccqGGPaqkdaahaaWcbeqaaiabikdaYaaakiabgUcaRiabdAgaMnaaDaaaleaacqaIXaqmaeaacqaIZaWmaaGccqGGPaqkaSqabaGccqGHsislcqWGJbWycqWGMbGzdaWgaaWcbaGaeGOmaidabeaakiabcMcaPaaa@4D8F@

Taking into account that rouleaux sediment in a medium that contains plasma, erythrocytes and small number of rouleaux (i.e. rouleaux are almost destroyed when the yield velocity exceeds 500 c^-1 21), μ₁ can be calculated according to Einstein formula:

μ m μ 1 = 1 + c f 2 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=X7aTnaaBaaaleaacqWGTbqBaeqaaaGcbaGae8hVd02aaSbaaSqaaiabigdaXaqabaaaaOGaeyypa0JaeGymaeJaey4kaSIaem4yamMaemOzay2aaSbaaSqaaiabikdaYaqabaaaaa@3983@

My previous work 21 shows the relationship to blood viscosity when the yield velocity is zero. This allows me now to calculate the concentrations of rouleaux inside the branching part of the vessel on both the external and internal parts of the vessel wall.

Tables 7, 8 and 9 show that if the yield velocity slightly increases then the shear stress of f₂ increases sharply, and then it will result in the destruction of rouleaux (the yield velocity is 5). The viscosity of f₂ considerably exceeds the viscosity of f₁; thus, an increase of the viscosity of a mixture results in decreased yield velocity. Furthermore, the increase in yield velocity results in a decrease of viscosity of f₂, although the shear stress of f₂ increases owing to the high concentration of erythrocytes. From tables 7, 8 and 9 one can observe that the major stress created by blood flow can be expressed as the shear stress of f₂.

According to tables 7, 8, 9, the shear stress of f₂ is the major factor. The shear stress of this second phase depends on both erythrocytes and rouleaux 21. The concentration of rouleaux can be calculated taking into account the shear stress and viscosity of the mixture. Tables 10, 11 and 12 show the dependence of rouleau concentration on erythrocyte concentration.

Table 10, 11, 12 show that the volume fraction of rouleaux depends to a greater extent on the concentration of erythrocytes (which is expressed as the viscosity of the mixture) than on the shear stress. This might be explained by the increase in rouleaux formation when concentration of erythrocytes is high.

Therefore, very importantly, for a specific part of the external wall of the vessel, a blood viscosity of 9 mNcm^-2 and a volume fraction of rouleaux = 0.044 may appear only when the concentration of erythrocytes is relatively high (48% or more) and when the yield velocity equals 50 m^-1. Thus, only very specific conditions result in the possible formation of rouleaux on the external wall of the vessel.

Let me consider the shear stress on the internal wall, where such stress is usually considered to be low 33. I take account of the fact that, with the division of flow, redistribution of velocity occurs, which may result in a further 2-3-fold decrease of velocity on the internal wall 29. The shear stress acting on the internal wall can be calculated according to the formula:

τ = λ ρ u 2 8 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFepaDcqGH9aqpdaWcaaqaaiab=T7aSjab=f8aYjabdwha1naaCaaaleqabaGaeGOmaidaaaGcbaGaeGioaGdaaaaa@3692@

where λ = 0.06 28

Tables 13, 14, 15, 16, 17 show that rouleaux are more likely to form on the internal wall of the vessel than on the external one. Blunt trauma can lead to conditions of short-term boundary layer separation on the internal wall of the vessel where rouleaux can become attached 212729. In addition, as it can be seen in Tables 13, 14, 15, 16, 17 that the higher the initial concentration of erythrocytes, the more rouleaux can be formed on the internal wall of the vessel. Therefore, the smaller the yield velocity, the more rouleaux may form, and these can stick to the internal wall of the vessel. Caro and colleagues 29 conducted an experiment using actual erythrocytes and microspheres made from polyvinyl latex: it was shown that the coefficient of difference changed from 3 10^-8 cm²c^-1 to 1.5 10^-7cm²c^-1, which significantly exceeded the coefficient of difference for Brownian movement of particles (approximately 4 10^-10 cm²c^-1).