Chapter 2

For a particle with a velocity distribution of \(v_{x}=5 x, v_{y}=-5 y, v_{z}=0\), determine the particle's acceleration vector. Also, determine whether this velocity profile has a local and/or convective acceleration.

Consider a velocity vector \(v=\left(x t^{2}-y\right) \vec{i}+\left(x t-y^{2}\right) \vec{j}\). (i) Determine whether this flow is steady (hint: no changes with time). (ii) Determine whether this is an incompressible flow (hint: check if \(\nabla \cdot=0\) ).

Given the velocity vector \(\vec{v}=(2 x-y) \vec{i}+(x-2 y) \vec{j}\), determine whether it is irrotational.

The velocity vector for a steady incompressible flow in the \(x y\) plane is given by \(\vec{v}=\) \(\frac{4}{x} \vec{i}+\frac{4 y}{x^{2}} \vec{j}\), where the coordinates are measured in centimeters. Determine the time it takes for a particle to move from \(x=1 \mathrm{~cm}\) to \(x=4 \mathrm{~cm}\) for a particle that passes through the point \((x, y)=(1,4)\).

Flow in a two-dimensional channel of width \(W\) has a velocity profile defined by $$ v_{x}(y)=k\left[\left(\frac{W}{2}\right)^{2}-y^{2}\right] $$ where \(y=0\) is located at the center of the channel. Sketch the velocity distribution (with \(k=1\) and \(W=10\) ) and find the shear stress/unit width of the channel at the wall.

A velocity field is given by \(v(x, y, z)=20 x y \vec{i}-10 y^{2} \vec{j}\). Calculate the acceleration, the angular velocity, and the vorticity vector at the point \((-1,1,1)\), where the units of the velocity equation are millimeters per second.

Considering one-dimensional fluid (density, \(\rho\); viscosity, \(\mu\) ) flow in a tube with an inlet pressure of \(p_{i}\) and outlet pressure of \(p_{o}\) and tube radius of \(r\) and length \(l\), the density of the fluid can be represented as \(\rho\). Express the wall shear stress as a function of these variables (hint: balance Newton's second law of motion).

Consider a red blood cell that originates from the origin of our \(x y\)-coordinate system. The velocity of the fluid is unsteady and is described by $$ v=\left\\{\begin{array}{cll} v_{x}=\frac{1 \mathrm{~cm}}{\mathrm{~s}} & v_{y}=\frac{0.5 \mathrm{~cm}}{\mathrm{~s}} & 0 \mathrm{~s} \leq t \leq 2 \mathrm{~s} \\ v_{x}=\frac{0.25 \mathrm{~cm}}{\mathrm{~s}} & v_{y}=\frac{1 \mathrm{~cm}}{\mathrm{~s}} & 2 \mathrm{~s}

The velocity distribution for laminar flow between two parallel plates can be represented as $$ \frac{u}{u_{\max }}=1-\left(\frac{2 y}{h}\right)^{2} $$ where \(h\) is the separation distance between the two flat plates and the origin is located half-way between the plates. Consider the flow of blood at \(37^{\circ} \mathrm{C}(\mu=3.5 \mathrm{cP})\) with maximum velocity of \(25 \mathrm{~cm} / \mathrm{s}\) and a separation distance of \(10 \mathrm{~mm}\). Calculate the force on a \(0.25 \mathrm{~m}^{2}\) section of the lower plate.

A biofluid is flowing down an inclined plane. The velocity profile of this fluid can be described by $$ u=\frac{\rho g}{\mu}\left(h y-\frac{y^{2}}{2}\right) \sin \theta $$ if the coordinate axis is aligned with the inclined plane. Determine a function for the shear stress along this fluid. Plot the velocity profile and shear stress profile, if the fluid density is \(900 \mathrm{~kg} / \mathrm{m}^{3}\) and viscosity is \(2.8 \mathrm{cP}\). The fluid thickness \(h\) is equal to \(10 \mathrm{~mm}\) and the plane is inclined at an angle of 40 degrees.

What type of fluid can be classified by the following shear stress-strain rate data? Plot the data and classify. Determine the viscosity of the fluid (Table 2.4). $$ \begin{aligned} &\text { TABLE 2.4 Shear Stress versus shear rate data for Homework Problem 2.17. }\\\ &\begin{array}{llllll} \hline \text { Shear stress }\left(\mathrm{N} / \mathrm{m}^{2}\right) & 0.4 & 0.82 & 2.50 & 5.44 & 8.80 \\ \text { Shear rate }\left(\mathrm{s}^{-1}\right) & 0 & 10 & 50 & 120 & 200 \\ \hline \end{array} \end{aligned} $$

Two data points on the rheological diagram of a biofluid are provided. Determine the consistency index and the flow behavior index and the strain rate if the shear stress is increased to 3 dynes \(/ \mathrm{cm}^{2}\). Assume that this is a two-dimensional flow. $$ \begin{aligned} &\frac{d V}{d y}=15 \frac{\mathrm{rad}}{\mathrm{s}}, \tau=0.868 \text { dynes } / \mathrm{cm}^{2} \\ &\frac{d V}{d y}=30 \frac{\mathrm{rad}}{\mathrm{s}}, \tau=0.355 \text { dynes } / \mathrm{cm}^{2} \end{aligned} $$

Given that the volumetric flow rate for a fluid within a circular cross- section tube can be represented by $$ Q=\frac{\pi r^{4} \Delta P}{8 \mu L} $$ where \(r\) is the tube radius, \(\mu\) is the fluid viscosity, \(P\) is the pressure drop across the tube, and \(L\) is the tube length, calculate the pressure drop across a tube of length \(1 \mathrm{~m}\) and diameter of \(23 \mathrm{~mm}\). The fluid is blood \((\mu=3.5 \mathrm{cP})\) and has a volumetric flow rate of \(4.5\) \(\mathrm{L} / \mathrm{min}\). Assuming the same conditions, what would the required pressure drop be for water \((\mu=1 c P)\) and chocolate syrup \((\mu=15,000 \mathrm{cP})\).

At a stenosis, the energy in the flowing fluid is skewed from potential energy to kinetic energy. Why? What effect may this have on locations downstream of the stenosis.