Question

(II) Calculate the pressure drop per cm along the aorta using the data of Example 10–12 and Table 10–3.

Short answer

The pressure drop per cm along the aorta is \(0.889\;{\rm{Pa/cm}}\).

Step-by-step solution

Concept

When a fluid flows through a tube with uniform fluid speed, then the pressure drop along the length of the tube relies upon the radius of the tube, velocity of the fluid, and viscosity. Poiseuille’s law shows the relation between the pressure drop along the tube length and the fluid’s velocity.

Given data

Using the data of Example 10-12:

The radius of aorta is, \(R = 1.2\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 0.012\;{\rm{m}}\).

The speed of blood flowing through aorta is, \(v = 40\;{\rm{cm/s}} \times \frac{{1\;{\rm{m/s}}}}{{100\;{\rm{cm/s}}}} = 0.4\;{\rm{m/s}}\).

Calculation

Using the Table 10-3, the coefficient of viscosity of the blood at temperature \(37^\circ {\rm{C}}\) is given by,

\(\eta = 4 \times {10^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}}\)

Applying thePoiseuille’s law, the expression for the pressure drop per unit length along the aorta is given by,

\(\begin{array}{c}Q = \frac{{\pi {R^4}\Delta P}}{{8\eta L}}\\\pi {R^2} \times v = \frac{{\pi {R^4}\Delta P}}{{8\eta L}}\\\frac{{\Delta P}}{L} = \frac{{8\eta v}}{{{R^2}}}\end{array}\)

Here, \(Q\) is the volume blood flow rate, \(\Delta P\) is the pressure drop, \(L\) is the length of aorta, \(A\) is the cross-sectional area of aorta and \(v\) is the speed of blood flowing.

Substitute the values in the above expression,

\(\begin{array}{l}\frac{{\Delta P}}{L} = \frac{{8 \times 4 \times {{10}^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}} \times 0.4\;{\rm{m/s}}}}{{{{\left( {0.012\;{\rm{m}}} \right)}^2}}}\\\frac{{\Delta P}}{L} = 88.89\;\frac{{{\rm{Pa}}}}{{\rm{m}}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}\\\frac{{\Delta P}}{L} = 0.889\;\frac{{{\rm{Pa}}}}{{{\rm{cm}}}}\end{array}\)

Therefore, the pressure drop per cm along the aorta is \(0.889\;{\rm{Pa/cm}}\).