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Q87PE

Expert-verifiedFound in: Page 399

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**The alveoli in emphysema victims are damaged and effectively form larger sacs. Construct a problem in which you calculate the loss of pressure due to surface tension in the alveoli because of their larger average diameters. (Part of the lung’s ability to expel air results from pressure created by surface tension in the alveoli.) Among the things to consider are the normal surface tension of the fluid lining the alveoli, the average alveolar radius in normal individuals and its average in emphysema sufferers.**

The pressure loss due to surface tension is obtained as:\(2000 N/{m^2}\).

**The force acting per unit length on a surface film is called surface tension. It is a scalar quantity measures in newton per meter.**

The diameter of alveoli of an emphysema victim is increased twice of its initial value. Calculate the loss of pressure due to surface tension in the alveoli because of their larger average diameters, if the average diameter of alveoli is \(100 \mu m\) and the normal surface tension of the fluid lining the alveoli is \(0.05 N/m\).

The initial radius of the alveoli is:

\({r_i} = \frac{{{d_i}}}{2}\)

Here, \({d_i} = 100 \mu m\) is the initial diameter of the alveoli.

\(\begin{array}{c}{r_i} = \frac{{100 \mu m}}{2}\\ = 50 \mu m\end{array}\)

The initial pressure inside the alveoli is:

\({P_i} = \frac{{4\gamma }}{{{r_i}}}\)

Here, \(\gamma = 0.05 N/m\) is the surface tension of the fluid lining the alveoli, and \({r_i} = 50 \mu m\) is the initial radius of the alveoli.

\(\begin{array}{c}{P_i} = \frac{{4 \times 0.05 N/m}}{{50 \mu m}}\\ = \frac{{4 \times 0.05 N/m}}{{\left( {50 \mu m} \right) \times \left( {\frac{{1{0^{ - 6}} m}}{{1 \mu m}}} \right)}}\\ = 4000 N/{m^2}\end{array}\)

The final radius of the alveoli is,

\({r_f} = \frac{{{d_f}}}{2}\)

Here, \({d_f}\) is the initial diameter of the alveoli.

Since, final diameter is twice of the initial diameter i.e., \({d_f} = 2{d_i}\). Therefore,

\(\begin{array}{c}{r_f} = \frac{{2{d_i}}}{2}\\ = {d_i}\end{array}\)

Substitute \(100 \mu m\)for \({d_i}\):

\({r_f} = 100 \mu m\)

The final pressure inside the alveoli is:

\({P_f} = \frac{{4\gamma }}{{{r_f}}}\)

Substitute \(0.05 N/m\)for \(\gamma \), and \(100 \mu m\) for \({r_f}\),

\(\begin{array}{c}{P_f} = \frac{{4 \times 0.05 N/m}}{{100 \mu m}}\\ = \frac{{4 \times 0.05 N/m}}{{\left( {100 \mu m} \right) \times \left( {\frac{{1{0^{ - 6}} m}}{{1 \mu m}}} \right)}}\\ = 2000 N/{m^2}\end{array}\)

The pressure loss is:

\(\Delta P = {P_i} - {P_f}\)

Substitute \({P_i} = 4000 N/{m^2}\)and \({P_f} = 2000 N/{m^2}\)

\(\begin{array}{c}\Delta P = \left( {4000 N/{m^2}} \right) - \left( {2000 N/{m^2}} \right)\\ = 2000 N/{m^2}\end{array}\)

Hence, the pressure loss due to surface tension is \(2000 N/{m^2}\).

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