Chapter 10, Problem 55P

To determine

The pressure drop per centimeter along aorta is to be calculated on the basis of given information.

Answer to Problem 55P

The pressure drop per centimeter along the aorta is $0.89Pa/cm$ .

Explanation of Solution

Given info:

The radius, $r=1.2\times {10}^{-2}m$

Viscosity coefficient, $\eta =4\times {10}^{-3}Pas$

Velocity, $v=0.4m/\sec$

Formula used:

• According to the Poiseuille's equation,

  $Q=\frac{\pi r^4\left(P_2-P_1\right)}{8\eta l}$

Where, $Q$ is the volume discharge flow

  $\left(P_2-P_1\right)$ , Pressure difference between any two ends of object

  $\eta$ Viscosity of fluid

  $r$ Is the inner radius

  $l$ Is the length of the pipe

• The continuity equation,

  $Q=Av$

Calculation:

By using the continuity equation,

  $\begin{array}{l}Q=Av\\ =\pi r^2\left(0.4m/s\right)\end{array}$

Substitute the values in the given equation, and then the result will be,

  $\begin{array}{l}Q=\frac{\pi r^4\left(P_2-P_1\right)}{8\eta l}\Rightarrow \frac{\Delta P}{l}=\frac{Q8\eta }{\pi r^4}=\frac{\pi r^2\times \left(0.4m/s\right)\times 8\eta }{\pi r^4}\\ =\frac{\left(0.4m/s\right)\times 8\times 4\times {10}^{-3}}{r^2}=\frac{\left(0.4m/s\right)\times 8\times 4\times {10}^{-3}}{{\left(1.2\times {10}^{-2}\right)}^2}\\ =\frac{12.800\times {10}^{-3}}{1.44\times {10}^{-4}}=88.8Pa/m\end{array}$

Or,

  $\frac{\Delta P}{l}=0.89Pa/cm$

Conclusion:

The pressure drop per centimeter along the aorta is $0.89Pa/cm$ .