Question

Find the maximum average velocity of blood flow through an artery of radius $2 \times {10^{ - 3}}{\text{m}}$ for the flow to remain laminar, the corresponding flow rate is
(Take viscosity of blood as $\eta = 2.084 \times {10^{ - 3}}{\text{Pa s}}$).

Answer 1

Hint: The flow of the fluid is said to be laminar (smooth) or turbulent (chaotic) based on the Reynolds number. Above $R = 2000$ the flow becomes transitional (i.e., a mixture of laminar and turbulent).

Formulas used:
Average velocity of flow is given by, $v = \dfrac{{R\eta }}{{\rho d}}$ ,where $R$ is the Reynolds number, $\eta $ is viscosity of the liquid, $\rho $ is the density of the liquid and $d$ is the diameter of the tube.
Flow rate is given by ${\text{area}} \times {\text{average velocity}}$ .

Complete step by step solution:
Step 1: List the information given in the question.
Radius of the artery is $r = 2 \times {10^{ - 3}}{\text{m}}$ .
Then the diameter of the artery ( $d = 2r$ ) will be $4 \times {10^{ - 3}}{\text{m}}$ .
Given, viscosity of blood is $\eta = 2.084 \times {10^{ - 3}}{\text{Pa s}}$ .
We know, the density of blood is $\rho = 1.06 \times {10^3}{\text{kg }}{{\text{m}}^{ - 3}}$ .
Step 2: Expressing the relation for average velocity.
The average velocity of a liquid of density $\rho $ and viscosity $\eta $ flowing through a tube of diameter $d$ is given by $v = \dfrac{{R\eta }}{{\rho d}}$ .
Here, $R$ is the Reynolds number, it determines whether the flow is laminar or turbulent.
 The maximum value of the Reynolds number for the flow to be laminar is $R = 2000$.
Step 3: Substituting the corresponding values of $\rho $, $\eta $, $d$ and $R$ to find maximum average velocity.
We have, the average velocity $v = \dfrac{{R\eta }}{{\rho d}}$.
Substituting $d = 4 \times {10^{ - 3}}{\text{m}}$ , $\eta = 2.084 \times {10^{ - 3}}{\text{Pa s}}$ , $\rho = 1.06 \times {10^3}{\text{kg }}{{\text{m}}^{ - 3}}$ in the above equation we get, $v = \dfrac{{R \times 2.084 \times {{10}^{ - 3}}}}{{1.06 \times {{10}^3} \times 4 \times {{10}^{ - 3}}}}$.
Now to obtain the maximum average velocity substitute $R = 2000$ in the above equation.
i.e., $v = \dfrac{{2000 \times 2.084 \times {{10}^{ - 3}}}}{{1.06 \times {{10}^3} \times 4 \times {{10}^{ - 3}}}} = 0.98{\text{m/s}}$.
Therefore, the largest average velocity is $0.98{\text{m/s}}$.
Step 4: Expressing the relation for flow rate.
The flow rate is described as the volume of liquid flowing per second.
It is given by, ${\text{area}} \times {\text{average velocity}}$ .
Area of the artery of radius $r = 2 \times {10^{ - 3}}{\text{m}}$ is $A = \pi {r^2}$ .
So, we get $A = 3.14 \times {\left( {2 \times {{10}^3}} \right)^2} = 1.256 \times {10^{ - 5}}{{\text{m}}^2}$
Step 5: Calculating the flow rate.
We have, ${\text{Flow rate = area}} \times {\text{average velocity}}$ .
Substituting for area and average velocity,
flow rate $ = 1.256 \times {10^{ - 5}} \times 0.98 = 1.23 \times {10^{ - 5}}{{\text{m}}^3}{\text{/s}}$ .

Therefore, the flow rate corresponding to the blood flow’s maximum average velocity is $1.23 \times {10^{ - 5}}{{\text{m}}^3}{\text{/s}}$.

Note:
Laminar flow is the smooth flow of fluids. Here, the viscous forces are dominant and thereby all the fluid particles follow smooth paths in layers. Each layer will drag behind its above layer.