Question

4. BONUS PROBLEM TO CHANGE ECTION OF THE COURSE, At a central Texas unhversty's rew investigating therapies for modeling blood flow In brand flow for a varkety of micre-vascular condrions. Your team h dameters and lengths of the blood vessels are displayed in the table folowing ONE MINIMUM QUIZ GRADE (0% or 30%) TO 100% FOR THE FIRST mmedical school, reseat dhers are N (# of vessels) inner Diameter (um) |Generation Length (μm) 1800 16 1200 2200 Biood is a complex fluld often modeled as a power law or Bingham plastic luid, at times with vico elastic behavior. Because of the presence of blood cells in the plasma, such simple models of shear rate and stress become especially inadequate in smaller vessels. In the blood of your study a common behavior is observed in that the apparent viscosity is a function of the vessel Gameter. The following data is available. Inner Diameter (um) Viscosity (aP) 15 18 25 30 45 1.25 1-40 1.45 1.65 1.70 1.85 2.05 2.30 2.40 75 90 A simple inear correlation for viscosity as a function of the Log of the inner dameter is adequate for the present model. The curve fit yields (with R'-0.9557) viscosity in cP 1.1334 lo10(D) +0.0918 where D is the diameter in um. Using this correlation, what is the volumetric flow rate of blood in uml's through the above network if the pressure drop between A and B is 40 mm Hg? Ignore entrance effects and use any other needed common assumption for laminar flow calculations. Assume the density of the blood is constant at 1,080 kg/m [Working with the flow network of problem 3 should make this problem easier to set up and solve. Note įkg = i0° ㎍ and im-1 m.)

Answer 1

SOLUTION

Basic relations

$Q=\frac{\pi}{4}D^2$ (1)

$Re=\frac{\rho V D}{\mu}$ (2)

In a viscous flow the change in pressure head is equal to loss of friction head, so

$\frac{\Delta p}{\rho g}=h_{L1}+h_{L2}+h_{L3}+h_{L4}+h_{L5}$

where, Friction loss is given by darcy equation $h_L=\frac{fLV^2}{2gD}=\frac{0.81fLQ^2}{gD^5}$ (Using relation 1)

Thus, $\frac{\Delta p}{\rho g}=\frac{0.81f_1L_1Q_1^2}{gD_1^5}+\frac{0.81f_2L_2Q_2^2}{gD_2^5}+\frac{0.81f_3L_3Q_3^2}{gD_3^5}+\frac{0.81f_4L_4Q_4^2}{gD_4^5}+\frac{0.81f_5L_5Q_5^2}{gD_5^5}$

$\frac{1.234\Delta p}{\rho }=\frac{f_1L_1Q_1^2}{D_1^5}+\frac{f_2L_2Q_2^2}{D_2^5}+\frac{f_3L_3Q_3^2}{D_3^5}+\frac{f_4L_4Q_4^2}{D_4^5}+\frac{f_5L_5Q_5^2}{D_5^5}$

From Continuity equation,

$Q_1=Q_2+Q_2\Rightarrow Q_2=0.5Q_1$

$Q_2=Q_3+Q_3\Rightarrow Q_3=0.5Q_2=0.25Q_1$

$Q_3+Q_3=Q_4\Rightarrow Q_4=2Q_3=0.5Q_1$

$Q_4+Q_4=Q_5\Rightarrow Q_5=2Q_4=Q_1$

Let Q₁=Q and so Q₂=0.5Q, Q₃=0.25Q ,Q₄=0.5Q and Q₅=Q

Substituting these in the above equation

$\frac{1.234\Delta p}{\rho }=\frac{f_1L_1Q^2}{D_1^5}+\frac{f_2L_2Q^2}{4D_2^5}+\frac{f_3L_3Q^2}{16D_3^5}+\frac{f_4L_4Q^2}{4D_4^5}+\frac{f_5L_5Q^2}{D_5^5}$

Since all flow is laminar, so coefficient of friction $f=\frac{64}{Re}=\frac{50.265 \mu D}{\rho Q}$ (Using relation 1 and 2)

Substituing again f in the pressure head equation,

$\frac{1.234\Delta p}{\rho }=\frac{50.265 \mu_1L_1Q}{\rho D_1^4}+\frac{50.265 \mu_2 L_2Q}{2\rho D_2^4}+\frac{50.265 \mu_3 L_3Q}{4\rho D_3^4}+\frac{50.265 \mu_4 L_4Q}{2\rho D_4^4}+\frac{50.265 \mu_5 L_5Q}{\rho D_5^4}$

0.024-p = μǐDiQ +-+-+ 4D3

$Q=\frac{0.024\Delta p}{\frac{\mu_1L_1}{ D_1^4}+\frac{ \mu_2 L_2}{2 D_2^4}+\frac{ \mu_3 L_3}{4 D_3^4}+\frac{ \mu_4 L_4}{2 D_4^4}+\frac{ \mu_5 L_5}{ D_5^4}}$

Calculation of various parameters

We have

$\mu = 1.1334 \times \log_{10}D+0.0918$

$\mu_1 = 1.1334 \times \log_{10}D_1+0.0918= 1.1334 \times \log_{10}65+0.0918=2.1465$ cP

$\mu_2 = 1.1334 \times \log_{10}D_2+0.0918= 1.1334 \times \log_{10}33+0.0918=1.813$ cP

$\mu_3 = 1.1334 \times \log_{10}D_3+0.0918= 1.1334 \times \log_{10}16+0.0918=1.456$ cP

$\mu_4 = 1.1334 \times \log_{10}D_4+0.0918= 1.1334 \times \log_{10}39+0.0918=1.895$ cP

$\mu_5 = 1.1334 \times \log_{10}D_5+0.0918= 1.1334 \times \log_{10}88+0.0918=2.296$ cP

Also Given

L1 = 1800e-6 L2 = 1000e-6 L3 = 400e-6 L4 = 1200e-6 L5 = 2200e-6

$\frac{\mu_1 L_1}{D_1^4}=\frac{2.14e^{-2} \times 1800e^{-6}}{\left (65e^{-6} \right )^4}=2.16e^{12}$

$\frac{\mu_2 L_2}{2D_2^4}=\frac{1.81e^{-2} \times 1000e^{-6}}{2\left (33e^{-6} \right )^4}=7.64e^{12}$

$\frac{\mu_3 L_3}{4D_3^4}=\frac{1.456e^{-2} \times 400e^{-6}}{4\left (16e^{-6} \right )^4}=22.2e^{12}$

$\frac{\mu_4 L_4}{2D_4^4}=\frac{1.895e^{-2} \times 1200e^{-6}}{2\left (39e^{-6} \right )^4}=4.91e^{12}$

$\frac{\mu_5 L_5}{D_5^4}=\frac{2.296e^{-2} \times 2200e^{-6}}{\left (88e^{-6} \right )^4}=0.84e^{12}$

$\Delta p=40$ mm of Hg (Since 1 mm of Hg = 133.32 Pa)

Substituting all values in the relation for Q we get, Q= 3.39e^-12 m³/s = 3.39e⁶ micro meter cube per second