Question

Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 smaller parallel pipes of the same total cross-sectional area, 4.0 mm2. Total volume flow is 1000 mm3/is. The pressure drop for fluid passing through the single pipe is lower than that through the 100 vessel array by a factor of:_____.A. 10.B. 100.
C. 1000.

Answer 1

Answer:

A. 10

Explanation:

For a single straight vessel; we can express the equation as;

$H_{f_1} = \dfrac{8 \ fl \ Q_1^2}{\pi ^2 gd_1^5} \ \ \ \ \ ... (1)$

Given that:

The total volume Q₁ = 1000 m/s²

Then the Q₂ = 1000/100 = 10 mm/s₂

However, the question proceeds by stating that 100 pipes of the same cross-section is being used.

Therefore, the formula for the area can be written as:

$\dfrac{\pi}{4}d_1^2 = 100 \bigg ( \dfrac{\pi}{4} d_2^2\bigg)$

Divide both sides by $\dfrac{\pi}{4}$

$d_1^2 = 100 \ d_2^2$

Making $d_1$ the subject of the formula;

$d_1 = 10d_2$

However, considering a pipe in parallel

$H_{f_2} = (H_f_2)_1 = (H_f_2)_2=...= (H_f_2)_{10}= \dfrac{8 \ fl Q_2^2}{\pi^2 \ gd _2^5} \ \ \ \ \ \ \ ...(2)$

Relating equation (1) by (2); then solving; we have;

$\dfrac{H_{f_1}}{H_{f_2}} = \dfrac{\dfrac{8flQ_1^2}{\pi^2 \ gd _1^5} }{\dfrac{8\ fl Q_2^2 }{\pi^2 gd_2^5} }$

$\dfrac{H_{f_1}}{H_{f_2}} =\dfrac{Q_1^2}{Q_2^2} \times \dfrac{d_2^5}{d_1^5}$

$\dfrac{H_{f_1}}{H_{f_2}} =\dfrac{(1000)^2}{(10)^2} \times \dfrac{d_2^5}{(10 \ d_2)^5}$

$\dfrac{H_{f_1}}{H_{f_2}} =\dfrac{1}{10}$

$H_{f_2} =10H_{f_1}$