Giraffe bending to drink. In a giraffe with its head $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$ above its heart, and its heart $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$above its feet, the (hydrostatic) gauge pressure in the blood at its heart is250 $\mathit{t}\mathit{o}\mathit{r}\mathit{r}$. Assume that the giraffe stands upright and the blood density is $\mathbf{1}\mathbf{.}\mathbf{06}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$. (a) In $\mathit{t}\mathit{o}\mathit{r}\mathit{r}$ (or role="math" localid="1657260976786" $\mathit{m}\mathit{m}\mathbf{}\mathit{H}\mathit{g}$), find the (gauge) blood pressure at the brain (the pressure is enough to perfuse the brain with blood, to keep the giraffe from fainting). (b) $\mathit{t}\mathit{o}\mathit{r}\mathit{r}$In (or$\mathit{m}\mathit{m}\mathbf{}\mathit{H}\mathit{g}$), find the (gauge) blood pressure at the feet (the pressure must be countered by tight-fitting skin acting like a pressure stocking). (c) If the giraffe were to lower its head to drink from a pond without splaying its legs and moving slowly, what would be the increase in the blood pressure in the brain? (Such action would probably be lethal.)

1. Density of blood,$p=1.06\times {10}^{3}kg/m^3$
2. Heart pressure,$p=250torr$

Use the formula for gauge pressure, which depends on density, height, and acceleration due to gravity. Gauge pressure is that, which is exerted by a fluid at any point of time at equilibrium due to the force applied by gravity.

Formula:

Pressure applied on a fluid surface,$p=pgh$

Net pressure applied on a body, $∆p=p_2-p_1$

From equation (ii), the pressure at brain can be given as:

$$\begin{gathered} p_{brain}=p_{heart}-pgh \\ =250-1.06\times {10}^3\times 9.8\times 2\times \frac{1torr}{133.33Pa}(\because 1torr=133.33Pa) \\ =94.2torr \end{gathered}$$

The pressure at brain is 94.2 torr.

From equation (ii), the pressure at brain can be given as:

$$\begin{gathered} P_{feet}=P_{heart}+pgh(\because \mathrm{here}\mathrm{pressure}\mathrm{is}\mathrm{total}\mathrm{pressure}) \\ =250+1.06\times {10}^3\times 9.8\times 2\times \frac{1torr}{133.33Pa}(\because 1torr=133.33Pa) \\ =405.8torr \end{gathered}$$

The pressure at feet is 405.8 torr.

From equation (ii), the net pressure between brain and feet can be given as:

$$\begin{gathered} ∆p=p_{brain}-p_{feet} \\ =405.8-94.2 \\ =311.6torr \end{gathered}$$

Hence, the increased value of pressure is 311.6 torr.