Question

An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Suppose that , because of an aneurysm, the cross-sectional area A1 of the aorta increases to a value of A2 = 1.7A1. The speed of the blood (density =1060 kg/m3) through a normal portion of the aorta is v1 = 0.40 m/s. Assuming that their aorta is horizontal (the person is lying down), determine the amount by which the pressure P2 in the enlarged region exceeds the pressure P1 in the normal region.

Answer 1

The pressure difference (
`\( P_2 - P_1 \)`) in the enlarged region of the aorta exceeds the pressure in the normal region by approximately
`\(52.8 \, \text{Pa}\).`

The relationship between pressure, velocity, and cross-sectional area in fluid flow is described by Bernoulli's equation:

`\[ P + (1)/(2)\rho v^2 + \rho gh = \text{constant} \]`

where:

- P is pressure,

-
`\( \rho \)` is the fluid density,

- v is the fluid velocity,

- g is the acceleration due to gravity,

- h is the height above some reference point.

Since the aorta is horizontal, we can ignore the gravitational term (
`\( \rho gh \)`).

For a normal portion of the aorta (subscript 1), and an enlarged portion (subscript 2), Bernoulli's equation becomes:

`\[ P_1 + (1)/(2)\rho v_1^2 = P_2 + (1)/(2)\rho v_2^2 \]`

Given that
`\( A_2 = 1.7A_1 \)`, the relationship between velocities is
`\( v_2 = (A_1)/(A_2)v_1 \).`

Substitute this relationship into Bernoulli's equation:

`\[ P_1 + (1)/(2)\rho v_1^2 = P_2 + (1)/(2)\rho\left((A_1)/(A_2)v_1\right)^2 \]`

Now, we are interested in the difference in pressure (
`\( P_2 - P_1 \)`):

`\[ P_2 - P_1 = (1)/(2)\rho v_1^2 - (1)/(2)\rho\left((A_1)/(A_2)v_1\right)^2 \]`

`\[ P_2 - P_1 = (1)/(2)\rho v_1^2\left(1 - \left((A_1)/(A_2)\right)^2\right) \]`

`\[ P_2 - P_1 = (1)/(2)\rho v_1^2\left(1 - \left((A_1)/(1.7A_1)\right)^2\right) \]`

`\[ P_2 - P_1 = (1)/(2)\rho v_1^2\left(1 - \left((1)/(1.7)\right)^2\right) \]`

`\[ P_2 - P_1 = (1)/(2)\rho v_1^2\left(1 - (1)/(2.89)\right) \]`

`\[ P_2 - P_1 = (1)/(2)\rho v_1^2 * (1.89)/(2.89) \]`

Given that
`\( \rho = 1060 \, \text{kg/m}^3 \) and \( v_1 = 0.40 \, \text{m/s} \)`, substitute these values:

`\[ P_2 - P_1 = (1)/(2) * 1060 * (0.40)^2 * (1.89)/(2.89) \]`

`\[ P_2 - P_1 \approx 52.8 \, \text{Pa} \]`

Therefore, the pressure difference (
`\( P_2 - P_1 \)`) in the enlarged region of the aorta exceeds the pressure in the normal region by approximately
`\(52.8 \, \text{Pa}\).`