Question

Estimate the average blood pressure in a person’s foot if the foot is 1.23 m below the aorta, where the average blood pressure is 104 mmHg. For the purposes of this estimate, assume the blood isn’t flowing. The density of blood is 1060 kg/m3.(mmHg)

Answer 1

Given,

The height difference, h=1.23 m

The blood pressure at the aorta, P₁=104 mmHg

The density of the blood, ρ=1060 kg/m³

From Pascal's law, the pressure difference is given by,

`\begin{gathered} (P_1-P_2)=\rho gh \\ \Rightarrow P_2=P_1-\rho gh \end{gathered}`

Where P₂ is the required pressure and g is the acceleration due to gravity.

Let us calculate the value of ρgh as it will be in the pascal.

On substituting the known values,

`\begin{gathered} \rho gh=1060*9.8*1.23 \\ =12777.24\text{ Pa} \end{gathered}`

On converting it to mmHg,

`\rho gh=12777.24\text{ Pa}\approx96\text{ mmHg}`

Therefore the pressure in a person's foot is,

`\begin{gathered} P_2=104-96 \\ =8\text{ mmHg} \end{gathered}`

Therefore the average blood pressure in a person's foot is 8 mmHg