91Ó°ÊÓ

Consider blood flow in an artery. Blood is nonNewtonian; the shear stress versus shear rate is described by the Casson relationship: $$\left\\{\begin{array}{ll} \sqrt{\tau}=\sqrt{\tau_{c}}+\sqrt{\mu \frac{d u}{d r}} & \text { for } \tau \geq \tau_{c} \\ \tau=0 & \text { for } \tau<\tau_{c} \end{array}\right.$$ where \(\tau_{e}\) is the critical shear stress, and \(\mu\) is a constant having the same dimensions as dynamic viscosity. The Casson relationship shows a linear relationship between \(\sqrt{\tau}\) and \(\sqrt{d u / d r},\) with intercept \(\sqrt{\tau_{c}}\) and slope \(\sqrt{\mu} .\) The Casson relationship approaches Newtonian behavior at high values of deformation rate. Derive the velocity profile of steady fully developed blood flow in an artery of radius \(R\). Determine the flow rate in the blood vessel. Calculate the flow rate due to a pressure gradient \(d p / d x=-100 \mathrm{Pa} / \mathrm{m},\) in an artery of radius \(R=1 \mathrm{mm},\) using the following blood data: \(\mu=3.5 \mathrm{cP}\) \(\tau_{c}=0.05\) dynes \(/ \mathrm{cm}^{2}\)

Step by step solution

01

Setup Casson's equation

First, setup the given equation that relates shear rate and shear stress: \[\sqrt{\tau}=\sqrt{\tau_{c}}+\sqrt{\mu \frac{d u}{d r}} \text { for } \tau \geq \tau_{c}\] \[\tau=0 \text { for } \tau<\tau_{c}\]

02

Rearrange terms and isolate rate of change of velocity

Rearrange the terms in the equation to isolate \( \frac{d u}{d r} \) \[\frac{d u}{d r}=\left(\sqrt{\tau}-\sqrt{\tau_{c}}\right)^{2} / \mu\]

03

Substitute for shear stress and integrate

For flow along the x-axis in cylindrical coordinates, shear stress \( \tau \) is given by \(-r \frac{d p}{d x}\). Substituting this into our rearranged Casson's equation and integrating over the radius from 0 to R, we get:\[u=\frac{1}{4 \mu}\left(\frac{d p}{d x}\right)^{2} r^{4}-\frac{\tau_{c}}{\mu} r^{2}+C_{1} r+C_{2}\]

04

Apply the boundary conditions

The assumption of fully developed flow gives us the boundary conditions. At \( r = 0 \), \( \frac{du}{dr} = 0 \), which gives us \( C_{1} = 0 \). At \( r = R \), the fluid velocity u is zero, which gives us \( C_{2} = \frac{R^{2}}{4} \left(\frac{d p}{d x}\right)^{2}-\tau_{c} R^{2} \). So, the velocity profile is \[u=\frac{1}{4 \mu}\left(\frac{d p}{d x}\right)^{2} \left(R^{2}-r^{2}\right)^{2}-\frac{\tau_{c}}{\mu}\left(R^{2}-r^{2}\right)\]

05

Determine the flow rate

The flow rate \( Q \) may be found by integrating the velocity over the cross-sectional area of the pipe:\[Q=\int_{0}^{R} u 2 \pi r d r=\frac{\pi}{4 \mu}\left(\frac{d p}{d x}\right)^{2} R^{6}-\frac{\pi \tau_{c}}{\mu} R^{4}\]

06

Substitute given values

Finally, let's substitute the given values for \(\frac{d p}{d x}\), \(R\), \(\mu\), and \(\tau_{c}\) into our equation for flow rate to find the answer. Make sure to convert these quantities into SI units first.

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