3-1

Chapter 3

The Physical and Flow Properties of Blood 3.1 Introduction Blood is a viscous fluid mixture consisting of plasma and cells. The composition of the plasma is shown in Table 3.1-1 with protein accounts for 7-8 wt% of the plasma. Plasma is a pale yellow fluid that consists of about 91% water and 9% other substances, such as proteins, ions, nutrients, gases and waste products.

Table 3.1-1 Composition of the plasma Solute Plasma MW g/ml Na+ K+ Ca++

Mg++

Cl− HCO3

− HPO4

−−, H2PO4−−

SO4−−

Phosphocreatine Carnosine Amino acids Creatine Lactate Adenosine triphosphate Hexose monophosphate Glucose Protein albumin globulins fibrinogen Urea Others Total (mOsmole/L)

(mOsm/L) 143 4.2 1.3 0.8 108 24 2

0.5 2

0.2 1.2

5.6 1.2

---------- ---------- ----------

4 4.8

302.8

69,000 35,000-1,000,000

400,000

4.5 2.5 0.3

The major proteins found in the plasma are albumin, fibrinogen, and the globulins that consist of the alpha, beta, and gamma types. Albumin is involved in the regulation of pH and the colloid osmotic pressure. Fibrinogen is involved in the process of blood clotting through its conversion to long strands of fibrin. The alpha and beta globulins are involved in solute transport, whereas the gamma globulins are the antibodies that fight infection and form the basis of the humoral immunity. When plasma from an immune animal is injected into the

3-2

blood of a nonimmune animal, the nonimmune animal will become immune. Because this process involved body fluids (humors), it is called humoral immunity. Homeostasis1 is the existence and maintenance of a relative constant environment within the body. Blood helps to maintain homeostasis due to its circulation through the blood vessels extended throughout the body. Oxygen enters blood in the lungs and is carried to cells. Carbon dioxide, produced by cells, is carried in the blood to the lungs, from which it is expelled. Ingested nutrients, ions, and water are transported by the blood from the digestive tract to cells. Waste products of cells are transported by the blood to the kidneys for elimination. Many of the hormones and enzymes that regulate body processes are carried from one part of the body to another by the blood. The precursor to vitamin D is produced in the skin and transported by the blood to the liver and then to the kidneys for processing into active vitamin D, which in turn is transported in the blood to the small intestines, where it promotes the uptake of calcium. Lactic acid, produced by skeletal muscles during anaerobic respiration, is carried by the blood to the liver, where it is converted into glucose. The water, proteins, and other substances in the blood, such as ions, nutrients, waste products, gases, and regulatory substances, are maintained within narrow limits. Normally, water intake through the digestive tract closely matches water loss through the kidneys, lungs, digestive tract, and skin. Therefore, plasma volume remains relatively constant. Suspended or dissolved substances in the blood come from the liver, kidneys, intestines, endocrine glands, and immune tissues like the spleen. The cellular component of blood consists of three main cell types: red blood cells (RBCs) or erythrocytes, platelets, and white blood cells (WBCs) or leukocytes. The RBCs are the most abundant cells comprising about 95% of the cellular component of blood. The platelets and white blood cells only comprise about 5% of the cellular component of blood, therefore their effect on the macroscopic flow characteristics of blood is negligible. The primary functions of RBCs are to transport oxygen from the lungs to the various tissues of the body and to transport carbon dioxide from the tissues to the lungs. Approximately 98.5% of the oxygen transported in the blood is by hemoglobin within the red blood cells, and the remaining 1.5% is dissolved in the water part of the plasma. If red blood cells rupture, the hemoglobin leaks out into the plasma and becomes nonfunctional because the shape of the molecule changes as a result of denaturation. Carbon dioxide is transported in the blood in three major ways: approximately 7% is transported as carbon dioxide dissolved in the plasma, approximately 23% is transported in combination with blood proteins (mostly hemoglobin), and 70% is transported in the form of bicarbonate ions. The bicarbonate ions (HCO3

−) are produced when carbon dioxide and water combine to form carbonic acid (H2CO3), which dissociates to form hydrogen and bicarbonate ions. Carbon dioxide is combined with water with the help of an enzyme called carbonic anhydrase, which is located primarily within red blood cells. Carbon monoxide, which is produced by the incomplete combustion of hydrocarbon, binds to the iron of hemoglobin to form the relatively stable compound carboxyhemoglobin. Due to the stable binding of carbon monoxide, hemoglobin cannot transport oxygen, and death may occur. Carbon monoxide is found in cigarette smoke, and the blood of smokers can contains 5%-15% carboxyhemoglobin.

1 Seeley R.R, Stephens T.D., Tate P., Anatomy & Physiology, McGraw Hill, 2003, p. 640

3-3

3.2 Rheology Rheologh is the study of the deformation and flow behavior of fluids. For a Newtonian fluid, we have a linear relationship between shear stress (τ) and the shear rate (γ& ) or rate of shear strain. τ = µγ& (3.2-1) In this equation, the proportional constant µ is called the viscosity of the fluid. The viscosity is the property of a fluid to resist the rate at which deformation takes place when the fluid is acted upon by a shear forces. As a property of the fluid, the viscosity depends upon the temperature, pressure, and composition of the fluid, but is independent of the shear rate. Most simple homogeneous liquids and gases are Newtonian fluid.

δ

∆y

∆x

(v | - v | ) tx y+ y x y∆ ∆

Element at time t

Element attime t+ t∆

y

x

Figure 3.2-1 Deformation of a fluid element.

The rate of deformation of a fluid element for a simple one-dimensional flow is illustrated in Figure 3.2-1. The flow parallel to the x-axis will deform the element if the velocity at the top of the element is different than the velocity at the bottom. The shear rate at a point is defined as

γ& = dtdδ =

0,,lim

→∆∆∆ tyx tttt

∆

−∆+

δδ

dtdδ =

0,,lim

→∆∆∆ tyx t

ytvvyxyyx

∆

−∆∆−−∆+

2/]/)arctan[(2/ ππ

dtdδ = −

0,,lim

→∆∆∆ tyx t

ytvvyxyyx

∆

∆∆−∆+

]/)arctan[(

For small angle θ, arctan(θ) = θ, therefore

dtdδ = −

0,,lim

→∆∆∆ tyx t

ytvvyxyyx

∆

∆∆−∆+

/)( = −

0,,lim

→∆∆∆ tyx y

vvyxyyx

∆

−∆+

)(

3-4

γ& = dtdδ = −

dydvx

The shear stress for this simple flow is also the molecular momentum flux in the y-direction and is given as

τyx = − µdydvx (3.2-2)

The subscript yx on τyx denotes the viscous flux of x momentum in the y direction. We may use equation (3.2-2) to obtain an expression for shear stress as a function of the fluid velocity and the system dimension. Consider the situation shown in Figure 3.2-2 where a fluid is contained between two large parallel plates both of area A. The plates are separated by a distance h. The system is initially at rest then a force F is suddenly applied to the lower plate to set the plate into motion in the x direction at a constant velocity V. Momentum is transferred from a region of higher velocity to a region of lower velocity. As time proceeds, momentum is transferred in the y direction to successive layers of fluid from the plate that is in motion in the x direction.

x

y

t < 0rest

t = 0lower plate moves

t > 0velocitydevelops

t >> 0steady velocityprofile

h

V V V F

Figure 3.2-2 Velocity profile development for a flow between two parallel plates.

The velocity profile of the fluid between the parallel plates may be obtained by applying the momentum balance, which states that

Time rate of change of linear momentum =

within the CV

rate of linear momentum enters −

the CV

rate of linear momentum exits +

the CV

sum of external forces acting on

the CV Since the velocity in the x direction vx is dependent on the y direction, we choose the control volume CV to be A∆y as shown in Figure 3.3-3.

3-5

y

y+ y∆τyx y+ y| ∆

τyx y|

Figure 3.3-3 x-Momentum entering and leaving the CV = A∆y

Applying the x-momentum balance on the CV yields

t∂∂ ( A∆yρ vx) =

yyxτ A − yyyx ∆+

τ A

Dividing the equation by A∆y and letting ∆y → 0, we obtain for constant physical properties

ρt

vx

∂∂ = −

0lim→∆y y

yyxyyyx

∆

−∆+

ττ = −

yyx

∂∂τ

(3.2-3)

Substituting τyx = − µyvx

∂∂ into equation (3.2-3) yields a second order partial differential

equation (PDE)

ρt

vx

∂∂ = µ 2

2

yvx

∂∂ ⇒

tvx

∂∂ = ν 2

2

yvx

∂∂ (3.2-4)

where ν = µ/ρ is the kinematic viscosity of the fluid. Equation (3.2-4) can be solved with the following initial and boundary conditions: Initial condition: t = 0, vx = 0 (3.2-4a) Boundary conditions: y = 0, vx = V and y = h, vx = 0 (3.2-4b) Equation (3.2-4) with the auxiliary conditions (3.4-2a,b) can be solved by the separation of variables method with the following result

vx = V(1 − hy ) −

πV2 ∑

∞

=1

1n n

exp

− 2

22

htn νπ sin

hynπ

The solution can also be expressed in dimensionless form with vx* =

Vvx , y* =

hy , t* = 2h

tν

3-6

vx* = 1 − y* −

π2 ∑

∞

=1

1n n

exp ( )*22 tn π− sin(nπ y*) (3.2-5)

Table 3.2-1 lists the Matlab program to plot dimensionless velocity profiles at various dimensionless times. The results are shown in Figure 3.2-4.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

vx/V

y/h

t*=.01

t*=.05

t*=.1

t*=1

t*=.2

Figure 3.2-4 Dimensionless velocity profiles for flow between two parallel plates.

____ Table 3.2-1 Matlab program to plot vx* = 1 − y* −

π2 ∑

∞

=1

1n n

exp ( )*22 tn π− sin(nπ y*)___

yoh=0.05:.05:.95; np=length(yoh);u=yoh; thetav=[.01 .05 .1 .2 1]; nt=length(thetav); n=1:20; ns=n.*n; hold on for k=1:nt theta=thetav(k); for i=1:np y=yoh(i); sum1=(1.0./n).*exp(-ns*pi*pi*theta)*sin(n*pi*y)'; u(i)=(1-y)-2*sum1/pi; end yp=[0 yoh 1];up=[1 u 0]; plot(up,yp) end xlabel('v_x/V');ylabel('y/h'); grid on

3-7

As time approaches infinity, the system reaches steady state and the summation terms in equation (3.2-5) become zero. The steady state velocity profile is then vx

* = 1 − y* (3.2-6) The steady state solution can also be obtained directly from equation (3.2-4) by setting the temporal derivative equal to zeo.

2

2

dyvd x = 0 (3.2-7)

Integrating equation (3.2-7) twice, we obtain vx = Ay + B The two constants of integration are evaluated from the boundary conditions: y = 0, vx = V and y = h, vx = 0 Therefore B = V and A = − V/h Hence vx = V(1 − y/h) ⇒ vx

* = 1 − y*

The shear rate at any position y in the fluid is given as γ& = − dydvx =

hV

The force to pull the lower plate at velocity V can be evaluated: F = A0=yyxτ = Aµ

hV

Fluids are classified as Newtonian or non-Newtonian, depending upon the relation between shear stress and shear rate. In Newtonian fluids the relation is linear while in non-Newtonian fluids, the shear stress is not a linear function of shear rate as shown in Figure 3.2-5.

Yieldstress

Shear rate

τ

Ideal plastic

Real plastic

Pseudo plastic

Newtonian fluid

Dilatant

Figure 3.2-5 Behaviors of Newtonian and non-Newtonian fluids.

3-8

The slope of the Newtonian fluid line is the viscosity. For the non-Newtonian fluids, the slope is not constant therefore its value at a given shear rate is called the apparent viscosity. The apparent viscosity of a dilatant fluid increases with shear rate while the apparent viscosity of a pseudo platic decreases with shear rate. The ideal or Bingham plastic has a linear shear stress-shear rate relation for stresses greater than the yield stress. Real plastic or Carson fluid also flows with stresses greater than the yield stress. The apparent viscosity however decreases with shear rate and at some point the Carson fluid behaves as a Newtonian fluid. Heterogeneous fluids that contain a particulate phase that forms aggregates at low rates of shear require a yield stress. Blood is a heterogeneous fluid with the particulates consisting primarily of the red blood cells. Therefore blood follows the curve shown for real plastic. At low shear rates, red blood cells clump together to form aggregates. This behavior results in high value of apparent viscosity. However, at shear rate higher than 100/s, red blood cells do not clump together, therefore blood behaves as a Newtonian fluid with an apparent viscosity of about 3 cP. The properties of blood change rapidly if removed from the system and so it is extremely difficult to perform experiments on it under laboratory conditions. The heart delivers blood in short busts during contraction into the aorta, therefore the flow in the artery is pulsatile and not uniform. The arteries are elastic and a typical cross section may change significantly with time due to the pulsating nature of the flow of blood. Thus it may be unreasonable to treat the arteries as rigid tubes. As blood arrives at the arterioles and capillaries, the variation in blood pressure becomes negligible as shown in Figure 1.1-3. The flow in the arterioles and capillaries can be treated as flow in a rigid pipe. It should be noted that Poiseuille who developed the theory of pipe flow was in fact a physician, and his interest was the study of blood flow.

Figure 1.1-3 Pressure variation in the systemic circulation.

3-9

3.3 Fully Developed Laminar Flow in Capillary We want to develop a relationship for shear stress-shear rate given volume flow rate Q and pressure drop ∆P across the horizontal capillary as shown in Figure 3.3-1. We use cylindrical coordinates with the following assumptions: the length of the tube (L) is much larger than the tube radius (R) (i.e. L/R > 100) to eliminate entrance effect; steady incompressible and isothermal flow; one-dimensional flow in the z direction only, therefore vz = vz(r); and no-slip boundary condition at the wall.

∆z

z

Rr

Pz Pz+ z∆

τrz

L

PoPL

Figure 3.3-1 Forces acting on a cylindrical fluid element within a capillary.

Consider the control volume πr2∆z shown in Figure 3.5-1. For steady flow, the summation of the viscous and pressure forces acting on the control volume must be equal to zero. P|zπr2 − τrz2πr∆z − P|z+∆zπr2 = 0 Dividing the equation by the control volume yields

− z

PPzzz

∆

−∆+ =

r2 τrz

In the limit when ∆z → 0, we obtain the differential equation for the shear stress distribution

− dzdP =

r2 τrz (3.3-1)

Since τrz = − µdrdvz , the right hand side of equation (3.3-1) is a function of r only and the left

hand side of equation (3.3-1) is a function of z only. They both must be equal to a constant

− dzdP =

r2 τrz =

LPP Lo −

The equation is rearranged to

τrz = − 2r

dzdP = ( )

LrPP Lo

2− (3.3-2)

The shear stress vanishes at the centerline of the capillary and achieves its highest value, τw, at the wall.

3-10

τw = τrz|r=R = − 2R

dzdP = ( )

LRPP Lo

2− (3.3-3)

Equations (3.3-2) and (3.3-3) are valid for both Newtonian and non-Newtonian fluids since we has not specified a relationship between the shear stress and shear rate. Solving equations

(3.3-2) and (3.3-3) for dzdP yields

− dzdP =

r2 τrz =

R2 τw

Therefore τrz = Rr τw = ( )

LrPP Lo

2− (3.3-4)

If γτ&rz = µ = constant, the fluid is a Newtonian fluid and µ is called the viscosity. If

γτ&rz = η ≠

constant, the fluid is a non-Newtonian fluid and η is called the apparent viscosity. We will follow different procedures to determine a relationship between shear stress and shear rate depending on whether or notγ& and τrz are known directly. A)γ& and τrz are not known directly We want to find a general relationship between the shear rate and some function of the shear stress in terms of the measurable quantities Qmeas., Po − PL, L, and R. That is:

γ& = − drdvz = γ& (τrz) (3.3-5)

We can follow the following procedures to obtain a relationship between the shear rate γ& and shear stress τrz. 1) We calculate the volumetric flow rate from the axial velocity profile as follows.

Qcal = 2π ∫R

z rv0

)( rdr

2) We express Qcal in terms of shear rate using integration by part. d(uv) = vdu + udv ⇒ ∫ v du = ∫ )(uvd − ∫ u dv

Let v = vz(r) ⇒ dv = dvz(r)

du = rdr ⇒ u = 21 r2

3-11

Therefore Qcal = 2π21 r2 R

zv0

− 2π ∫R

r0

2

21

drdvz dr = − π ∫

Rr

0

2

drdvz dr

3) Next, we change the integration variable from r to τrz using equation (3.3-4)

τrz = Rr τw ⇒ r =

w

Rτ

τrz ⇒ dr = w

Rτ

dτrz

Qcal = π ∫rz R

τ

0

22

w

rz

ττ γ&

w

Rτ

dτrz , (Note: γ& = −

drdvz )

Qcal = 3

3

w

Rτπ

∫rzτγ

0& 2

rzτ dτrz (3.3-6)

4) We then assume a relationship between γ& and τrz (for example γ& = µτ 2/1

rz ).

Equation (3.3-6) is then integrated to obtain Qcal. We will accept the assumed expression between γ& and τrz if Qcal ≈ Qmeas. Otherwise step (4) is repeated. B)γ& and τrz are known directly The shear stress and shear rate can be determined using a cup-and-bob or Couette viscometer. As the name implies, the Couette viscometer consists of two concentric cylinders as shown in Figure 3.3-2. The fluid is in the annular gap between the outer cylinder (cup) and the inner cylinder (bob).

T

ΩL

RiRo

Figure 3.3-2 Couette viscometer.

The outer cylinder is rotated at a fixed angular velocity (Ω). The shearing force is transmitted to the fluid, causing it to deform or flow. The inner cylinder is kept stationary by a torque (T)

3-12

that can be measured by a torsion spring. The shear stress at any position r within the gap (Ri ≤ r ≤ Ro) is determined by a balance of moments on a cylindrical surface 2πrL T = τrθ(2πrL)r Solving for the shear stress, we have

τrθ = Lr22π

T (3.3-7)

Setting r = Ri gives the stress on the bob surface (τi), and setting r = Ro gives the stress on the cub surface (τo). If the gap is small [i.e., (Ro − Ri)/Ro ≤ 0.02], the flow in the annular gap can be approximated by the flow between two parallel plates. In this case, an average shear stress should be used

τrθ = 2

oi ττ +≈

LR 22πT where R = (Ri + Ro)/2 (3.3-8)

The average shear rate is given by

γ& = drdvθ ≈

io

io

RRVV

−− =

io

o

RRR−Ω =

oi RR /1−Ω (3.3-9)

Equations (3.3-8, 9) provide the experimental values for the shear stress and the shear rate that can be fitted by a non-Newtonian fluid model. Example 3.3-1.0 ---------------------------------------------------------------------------------- The viscosity of a fluid sample is measured in a cup-and-bob viscometer. The bob is 15 cm long with a diameter of 9.8, and the cup has a diameter of 10 cm. The cup rotates, and the torque is measured on the bob. The following data were obtained: Ω(rpm) 2 4 10 20 40 T (dyn⋅cm) 3.6×105 3.8×105 4.4×105 5.4×105 7.4×105 (a) Determine the viscosity of the sample. (b) Fit the data with the following model equations τ = τo + µ∞γ& (Bingham Plastic Model) and τ = mγ& n (Power Law Model) (c) Determine the viscosity of this sample at a cup speed of 100 rpm in the viscometer using the above models.

0 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 74

3-13

Solution ------------------------------------------------------------------------------------------ Since (Ro − Ri)/Ro = (10 − 9.8)/10 = 0.02, we can use the equation (3.3-8, 9) to determine the shear stress and the shear rate

τrθ = LR 22π

T , and γ& = oi RR /1−

Ω

The apparent viscosity η is determined from

η = γτ θ

&r

Table 3.3-1 lists the results from the calculation. Table 3.3-2 lists the Matlab program to fit data with the Bingham plastic and power law model.

Table 3.3-1 Fluid apparent viscosity at different shear rates Ω(rpm) T (dyn⋅cm) γ& (1/s) τrθ(dyne/cm2) η(Poise = g/cm⋅s) 2 4 10 20 40 100

360000 380000 440000 540000 740000

10.5 20.9 52.5 105 209 524

156 165 191 234 320

14.89 7.86 3.64 2.23 1.53

For the Bingham plastic model, we obtain τ (dyne/cm2) = τo + µ∞γ& = 147 + 0.827γ& (1/s) For the power law model, we obtain τ(dyne/cm2) = mγ& n = 83.2γ& 0.234 At 100 rpm or γ& = 524 s-1, for the Bingham plastic model τ = 147 + 0.827×524 = 580 dyne/cm2 η = 580/524 = 1.11 g/cm⋅s For the power law model

η = γτ θ

&r = 83.2γ& 0.234−1 = 0.69 g/cm⋅s

3-14

______ Table 3.3-2 Matlab program to fit shear stress and shear rate data ______

% Example 3.3-1 % rpm=[2 4 10 20 40]; Torque=[36 38 44 54 74]*1e4; ndata=length(rpm); Ri=9.8/2;Ro=10/2;L=15; omega=rpm*2*pi/60; shear_rate=omega/(1-Ri/Ro) Rave=(Ri+Ro)/2; stress=Torque/(2*pi*Rave^2*L) vis=stress./shear_rate co=polyfit(shear_rate,stress,1); tao=co(2) vis_inf=co(1) x=log(shear_rate);y=log(stress); co=polyfit(x,y,1); n=co(1) m=exp(co(2)) drate=(shear_rate(ndata)-shear_rate(1))/25; s_rate=shear_rate(1):drate:shear_rate(ndata); tao1=tao+vis_inf*s_rate;vis1=tao1./s_rate; vis2=m*s_rate.^(n-1); loglog(shear_rate,vis,'d',s_rate,vis1,s_rate,vis2,':') xlabel('Shear rate (1/s)');ylabel('Viscosity (Poise)') legend('Data','Bingham plastic','Power law') grid on % Evaluate the correlation coefficient vis_ave=mean(vis); St=(vis-vis_ave)*(vis-vis_ave)'; tao1=tao+vis_inf*shear_rate;vis1=tao1./shear_rate; vis2=m*shear_rate.^(n-1); S1=(vis-vis1)*(vis-vis1)';r1=sqrt(1-S1/St); S2=(vis-vis2)*(vis-vis2)';r2=sqrt(1-S2/St); fprintf('Correlation coefficient for Bingham plastic = %8.4f\n',r1) fprintf('Correlation coefficient for Power law = %8.4f\n',r2) >> e3d1 shear_rate = 10.4720 20.9440 52.3599 104.7198 209.4395 stress = 155.8910 164.5516 190.5334 233.8365 320.4426 vis = 14.8865 7.8568 3.6389 2.2330 1.5300 tao = 147.2304 vis_inf = 0.8270 n = 0.2337 m = 83.1932 Correlation coefficient for Bingham plastic = 1.0000 Correlation coefficient for Power law = 0.9938

3-15

A crude measure of the how well the data is fitted by an expression is given by the correlation coefficient r, which is defined as

r = tS

S−1

In this expression St = ∑=

−N

ii YY

1

2)( is the spread of the data around the mean Y of the

dependent variable and S = ∑=

−N

iii yY

1

2)( is the sum of the square of the difference between

the data (Yi) and the calculated value (yi). Figure 3.3-3 shows a plot of viscosity versus flow rate for the Bingham plastic and the Power law models. The Bingham plastic model fits the data better as evident by its higher correlation coefficient (1.0) in comparison with that (0.9938) of the Power law model.

101

102

103

100

101

102

Shear rate (1/s)

Visc

osity

(Poi

se)

DataBingham plasticPower law

Figure 3.3-3 Behavior of non-Newtonian fluid.

---------------------------------------------------------------------------------------------------

3-16

3.4 The Hagan-Poiseuille Equation We now consider the case of a Newtonian fluid flowing through a capillary. The shear rate-

shear stress relation γ& (τrz) = µτ rz is substituted into equation (3.3-6) to obtain

Q = 3

3

w

Rµτπ

∫rz

rz

ττ

0

3 dτrz = 3

3

w

Rµτπ w

rzττ

04 = 3

3

w

Rµτπ

4

4wτ

Q = µ

π4

3R τw = µ

π4

3R ( )L

RPP Lo

2− = ( )

LPPR Lo

µπ

8

4 −

The velocity profile inside the capillary can also be obtained by integrating equation (3.3-2)

τrz = − µdrdvz = ( )

LrPP Lo

2− (3.3-2)

∫zv

zdv0

= − ( )LPP Lo

2−

∫r

Rrdr

vz = ( )L

RPP Lo

µ4

2−

−

2

1Rr

3.5 The Velocity Profile for a Casson Fluid in Tube Flow Above a shear rate of 100/sec, blood is a Newtonian fluid with a constant viscosity of about 3 cP. Over the whole range of shear rate, blood may be described by the following empirical equation known as the Casson equation. τ1/2 = 2/1

yτ + sγ& 1/2 (3.5-1) In this equation, τy is the yield stress and s is a constant, both of which must be determined from experimental data. The force applied to stagnant blood must provide a stress greater than the yield stress before the blood will flow. The reduced average velocity U is defined as the ratio of the average velocity and the tube diameter

U = D

Vave = 34DQ

π (3.5-2)

3-17

The reduced average velocity U has the same dimension as the shear rate (sec-1) and a plot of τw versus U will show the similar behavior as a plot of shear stress versus shear rate. We can

substitute γ& = 2

22/12/1 )(s

yrz ττ −into equation (3.3-6)

Q = 3

3

w

Rτπ

∫wτ γ

0& 2

rzτ dτrz (3.3-6)

to obtain

U = 34DQ

π = 32

1

wτ∫

w

yrz

τ

ττ 2

2

22/12/1 )(s

yrz ττ −dτrz

The lower limit is changed to the yield stress τy since the fluid will not move until τ > τy.

U = 3221

ws τ ∫w

yrz

τ

ττ 2 )2( 2/12/1

yyrzrz ττττ +− dτrz

U = 3221

ws τ

+−−

+− y

yyy

yy

wyw

w ττ

τττ

τττττ37

4437

44

32/12/7

432/12/7

4

U = 221s

+−−

3841

74

4 3

42/12/1 y

w

yyw

w τττ

τττ (3.3-7)

The measured wall stress τw is obtained from τw = − 2R

dzdP = ( )

LRPP Lo

2− and the measured

U is obtained from U = 34DQ

π. Equation (3.3-7) is then fitted to the data providing values

for the two parameters: τy = 0.0289 dynes/cm2 and s = 0.229 (dynes-sec/cm2)1/2. The physiological values of U are in the range of 5 to 1,000 s-1. The flow in a small vessel is often observed to be intermittent. It can stop completely and then start again1 The axial velocity profile for a Casson fluid is shown in Figure 3.5-1. For this fluid to flow, the wall shear stress, τw, must be greater than the yield stress, τy. Since the value of the shear stress reduces with r, there will be a critical radius, rcrtitical, where the local shear stress will equal τy. From the tube centerline to this critical radius, the core fluid will have a flat velocity profile with value vz = vcore. From the critical radius to the pipe wall the velocity will decrease with r.

1 Merrill, E. W., A. M. Benis, E. R. Gillialand, T. K. Sherwood, and E. W. Salzman. 1965. Pressure flow relations of human blood in hollow fibers at low flow rates. J. Appl. Physiol. 20:954-967.

3-18

Vcore

rcritical

At r = r = (r)

critical

y rzτ τ

R

Figure 3.5-1 Velocity profile of a Casson fluid.

We can determine the velocity profile by integrating the expression for the shear stress from r = rcritical to r = R. τrz

1/2 = 2/1yτ + sγ& 1/2 ⇒ γ& = (τrz

1/2 − 2/1yτ )2/s2 (3.5-1)

From equation (3.3-2)

τrz = − 2r

dzdP = ( )

LrPP Lo

2− = ar (3.3-2)

where a = ( )LPP Lo

2− =

Rwτ

Therefore γ& = − drdvz = 2

1s

(a1/2 r1/2 − 2/1yτ )2 = 2

1s

(ar − 2 a1/2r1/2 2/1yτ + τy)

∫zv

zdv0

= − 21s ∫

r

R(ar − 2 a1/2r1/2 2/1

yτ + τy)dr

Integrating the equation and putting in the limits we obtain the velocity profile for the interval rcritical ≤ r ≤ R

vz(r) = 22sR wτ

−

2

1Rr −

38

2/1

w

y

ττ

−

2/3

1Rr + 2

w

y

ττ

−

Rr1 (3.5-3)

At r = rcritical, τrz = τy = − 2

criticalrdzdP

Since τw = − 2R

dzdP , we have

Rrcritical =

w

y

ττ

3-19

The core velocity is obtained by substituting Rr =

Rrcritical =

w

y

ττ

into equation (3.5-3)

vcore = 22sR wτ

−

2

1w

y

ττ

− 38

2/1

w

y

ττ

−

2/3

1w

y

ττ

+ 2

w

y

ττ

−

w

y

ττ

1 (3.5-4)

Example 3.5-1. ---------------------------------------------------------------------------------- Consider blood flows in an arteriole with a diameter of 0.01 cm and an average velocity of 0.05 cm/s. Blood is a Casson fluid (τrz

1/2 = 2/1yτ + sγ& 1/2) with yield stress τy = 0.0289

dynes/cm2 and s = 0.229 (dynes-sec/cm2)1/2. Determine the wall shear stress τw and plot the velocity profile. Solution ------------------------------------------------------------------------------------------ We can solve for the wall shear stress from equation (3.3-7)

U = 221s

+−−

3841

74

4 3

42/12/1 y

w

yyw

w τττ

τττ (3.3-7)

where U = D

Uave = 01.005.0 = 5 sec-1

The unknown wall shear stress τw can be solved by Newton’s method using the Matlab program listed in Table 3.5-1.

______ Table 3.5-1 Matlab program to solve for the wall shear stress and plot vz ______ % f='(taow/4-4*(taow*taoy)^.5/7-taoy^4/(84*taow^3)+taoy/3)/(2*s^2)'; taoy=0.0289;% dynes/cm2 s=0.229; % dynes-sec/cm2 taow=.4; % dynes/cm2 dtao = .01; U=.05; % cm/s D=0.01; % cm Ubar=U/D; for i=1:20 ft=Ubar-eval(f); taow=taow+dtao; ft2=Ubar-eval(f);df=(ft2-ft)/dtao; etao=ft/df; taow=taow-dtao-etao; if abs(etao)<1e-4, break, end end fprintf('taow(dynes/cm2) = %9.5f\n',taow) rc=taoy/taow; dr=(1-rc)/20;

3-20

r=rc:dr:1; R=D/2; vz=R*taow*((1-r.^2)-8*(rc.^.5)*(1-r.^1.5)/3+2*rc*(1-r))/(2*s*s); rp=[0 r];vzp=[vz(1) vz]; plot(vzp,rp) ylabel('r/R (R = 0.005 cm)');xlabel('vz(cm/s)'); grid on >> tubeflow taow(dynes/cm2) = 2.69727 The velocity profile is plotted in Figure 3.5-2 where the critical radius is given by

R

rcritical = w

y

ττ

= 69727.20289.0 = 0.0107

The velocity profile is evaluated from equation (3.5-3)

vz(r) = 22sR wτ

−

2

1Rr −

38

2/1

w

y

ττ

−

2/3

1Rr + 2

w

y

ττ

−

Rr1 (3.5-3)

It should be noted that the velocity profile is plotted against the dimensionless distance Rr

.

Since the wall stress is much larger than the yield stress, the core velocity (vcore = 0.0958 cm/s) is limited to a small region near the centerline.

3-21

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

r/R

(R =

0.0

05 c

m)

vz(cm/s) Figure 3.5-2 Velocity profile of the Casson fluid in problem 3.5-1.

3.6 Generalized Mechanical Energy Balance Equation For laminar flow of a fluid in a cylindrical tube of radius R and length L, the Hagan-Poiseuille equation provides a relationship between volumetric flow rate and pressure drop across the tube as follows.

Q = µ

π4

3R τw = µ

π4

3R ( )L

RPP Lo

2− = ( )

LPPR Lo

µπ

8

4 −

3-22

Z1 Z2

PV

1

1

PV

2

2

Figure 3.6-1 A general piping system.

For a general piping system shown in Figure 3.6-1, we need the generalized relationship, equation (3.6-1), that can account for the effect of pressure drop on incompressible fluid flow, changes in elevation, tube cross section, changes in fluid velocity, sudden contractions or expansions, and friction loss through pipe and fittings such as valves and flow meters.

ρ

1P + gZ1 + 2

211Vα + ηWp =

ρ2P + gZ2 +

2

222Vα + hfriction (3.6-1)

Each term in this equation has units of energy per unit fluid mass flow rate or (length/time)2. P = pressure ρ = fluid density g = acceleration of gravity Z = elevation relative to a reference surface V = average fluid velocity α = kinetic energy correction factor α = 2 for laminar flow α = 1 for turbulent flow Wp = work done per unit mass flow rate η = pump efficiency (η < 1) hfriction = friction loss due to piping and fitting The kinetic correction factor is due to the fact that the velocity profile is not uniform over the cross-sectional area of flow. For uniform flow, the rate of kinetic energy entering a C.V. is given as

kE& =

2

21 Vρ VA

The kinetic energy per unit mass flow rate is then

3-23

VAEk

ρ

& =

21 V2

For turbulent flow, the velocity profile is almost flat, therefore

turbulent

k

VAEρ

&≈

21 V2 ⇒ α = 1 for turbulent flow

Laminarvelocity profile

Uniformvelocity profile

V

vz

Figure 3.6-2 Laminar velocity profile in a pipe.

The velocity for laminar flow in a pipe is given as

vz = 2V

−

2

1Rr = 2V (1 − δ2), where δ =

Rr

The rate of kinetic energy entering a C.V. is

kE& = ∫

R

zv0

2

21 ρ 2πrdr

Therefore kE& = πρ ∫R

zv0

3 rdr = πρR2 ∫R

zv0

3

Rr d

Rr

kE& = πρR2 ∫1

08V3(1 − δ2)3δdδ = 8πρR2V3 ∫ −

1

0

32 )1( δ δdδ

kE& = 8πρR2V3

81

= πρR2V3 = ρA V3

Therefore arla

k

VAE

minρ

&=

22 V2, and α = 2 for laminar flow

3-24

The friction loss is given by the following equation

hfriction = 4∑i

ifi

ii

DVL

2

2

+ ∑j

jV2

2

Kfitting,j (3.6-2)

where

fi = 2

21 V

w

ρ

τ = friction factor in tube segment i with length Li and diameter Di.

Vi = average velocity within tube segment i. Kfitting = friction loss factor or loss coefficient for pipe fittings, some typical values are

given in Table 3.6-1. The velocity Vj in the summation is for the fluid just downstream of the contraction, expansion, or fitting.

Table 3.6-1 Friction loss factor for various pipe fittings.

Fitting Kfitting Globe valve, wide open Angle valve, wide open Gate valve, wide open Gate valve, half open Standard 90o elbow Standard 45o elbow Tee, through side outlet Tee, straight through Sudden contraction (turbulent flow) Sudden expansion (turbulent flow)

7.5 3.8 0.15 4.4 0.7 0.35 1.5 0.4

0.4

−

1

21AA

2

1

2 1

−

AA

A1 A2

A1 A2

Sudden contraction

Sudden expansion

The friction factor for laminar flow (Re = µ

ρVD < 2000) is given by

f = Re16 (3.6-3)

The friction factor for turbulent flow (Re > 4000) can be estimated by

f = − 1.737 ln[0.269Dε −

Re185.2 ln (0.269

Dε +

Re14 )]-2 (3.6-4)

3-25

In this equation ε is the surface pipe roughness and D is the inside pipe diameter. Representative values for surface roughness are given in Table 3.6-2.

Table 3.6-2 Surface roughness Surface ε (ft) ε (mm) Concrete Cast iron Galvanized iron Commercial steel Drawn tubing

0.001-0.01 0.00085 0.0005 0.00015 0.000005

0.3-3.0 0.25 0.15 0.046 0.0015

If the fluid flows through a noncircular duct, then the equivalent diameter, Deq, can be used in equations (3.6-2, 3, 4). The equivalent diameter is defined as

Deq = 4rH = 4wet

cross

PA

where rH = hydraulic radius Across = cross sectional area of the flow Pwet = wetted perimeter of the duct

Do Di

Figure 3.6-3 Flow through an annular tube.

For the flow through an annular tube, the equivalent diameter is given as

Deq = 4)(

4/)( 22

io

io

DDDD+

−π

π = Do − Di

Example 3.6-1. ---------------------------------------------------------------------------------- Water is pumped from the upper reservoir to the lower reservoir through the piping system shown. Determine the power required for the pump if the water flow rate is 60 kg/s. The fittings from pipe D1 to pipe D2 and from pipe D2 to pipe D3 can be considered to be standard 90o elbows. Data: h1 = 10 m, h2 = 3 m, L1 = 50 m, L2 = 300 m, L3 = 2 m, D1 = 0.2 m, D2 = 0.5 m, D3 = 0.03 m, water viscosity = 1 cP = 10-3 kg/m⋅s, ρ = 1000 kg/m3. The pipe roughness is 0.05 mm. The pump efficiency is 75%.

3-26

(1)

(2)

h1

D , L1 1

D , L2 2D , L3 3

h2

Globe valve

Solution ------------------------------------------------------------------------------------------ Applying the mechanical energy balance between (1) and (2) we have

ρ

1P + gZ1 + 2

211Vα + ηWp =

ρ2P + gZ2 +

2

222Vα + hfriction

Let the reference level be at (2), the end of pipe 3, the energy equation becomes

ρatmP + g(h1 + L1 − L3) + 0 + ηWp =

ρρ 2ghPatm + + 0 +

2

233Vα + hfriction

g(h1 + L1 − L3) + ηWp = gh2 + 2

233Vα + hfriction

D(m) A(m2) V(m/s) Re ε/D f .2 .5 .03

3.14×10-2 1.96×10-1 7.07×10-4

1.91 0.306 84.9

3.82×105 1.53×105 2.55×106

2.50×10-4

1.00×10-4

0.0017

0.00406 0.00431 0.00600

hfriction = 4∑i

ifi

ii

DVL

2

2

+ ∑j

jV2

2

Kfitting,j

4∑i

ifi

ii

DVL

2

2

= 2× 10-3[4.06×2.091.150 2× + 4.31×

2.0306.0300 2× + 6×

2.09.842 2× ]

= 5.77× 103 m2/s2

∑j

jV2

2

Kfitting,j = 0.5×1.912×0.4 sudden contraction, Kfitting = 0.4

+ 0.5×0.3062×0.7 standard 90o elbow, Kfitting = 0.7

3-27

+ 0.5×0.3062×7.5 open globe valve, Kfitting = 7.5 + 0.5×84.92×0.7 standard 90o elbow, Kfitting = 0.7

∑j

jV2

2

Kfitting,j = 2.52× 103 m2/s2

Therefore hfriction = 5.77×103 + 2.52×103 = 8.29×103 m2/s2

g(h1 + L1 − L3) + ηWp = gh2 + 2

233Vα + hfriction

9.81(10 + 50 − 2) + 0.75 Wp = 9.81×3 + 29.84 2

+ 8.29×103

Wp = 1.51×104 m2/s2 The power required for the pump is pW& = m& Wp = 60×1.51×104 = 9.08×105 W = 1220 hp Note: 1 hp = 746 W

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