Questions: An artery has a diameter of 1.2 cm and blood flows through it with a velocity of 3.2 cm / s. If at some point there is plaque buildup, such that the artery has its diameter narrowed to 30% of its initial diameter. What velocity is the blood in the region with plaque buildup?

Solution Steps

The initial diameter of the artery is $1.2 \, \text{cm}$. The cross-sectional area $A_1$ of the artery can be calculated using the formula for the area of a circle:

$$A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \pi \left(\frac{1.2 \, \text{cm}}{2}\right)^2 = \pi (0.6 \, \text{cm})^2 = \pi \times 0.36 \, \text{cm}^2 = 1.1304 \, \text{cm}^2$$

The diameter of the artery is narrowed to $30\%$ of its initial diameter. Therefore, the new diameter $d_2$ is:

$$d_2 = 0.3 \times 1.2 \, \text{cm} = 0.36 \, \text{cm}$$

The new cross-sectional area $A_2$ is:

$$A_2 = \pi \left(\frac{d_2}{2}\right)^2 = \pi \left(\frac{0.36 \, \text{cm}}{2}\right)^2 = \pi (0.18 \, \text{cm})^2 = \pi \times 0.0324 \, \text{cm}^2 = 0.1018 \, \text{cm}^2$$

The continuity equation for incompressible fluid flow states that the product of the cross-sectional area and the velocity remains constant:

$$A_1 v_1 = A_2 v_2$$

Given:

• $A_1 = 1.1304 \, \text{cm}^2$
• $v_1 = 3.2 \, \text{cm/s}$
• $A_2 = 0.1018 \, \text{cm}^2$

We need to find $v_2$:

$$v_2 = \frac{A_1 v_1}{A_2} = \frac{1.1304 \, \text{cm}^2 \times 3.2 \, \text{cm/s}}{0.1018 \, \text{cm}^2} = \frac{3.6173 \, \text{cm}^3/\text{s}}{0.1018 \, \text{cm}^2} = 35.53 \, \text{cm/s}$$

Final Answer