Question

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.)

Short answer

1. Gauge pressure at the brain of giraffe is$94torr$
2. Gauge pressure at the feet of giraffe is $406torr$
3. Increment in pressure is 312$torr$

Step-by-step solution

The given data

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

Understanding the concept of gauge pressure

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, $∆p=p_2-p_1$

a) Calculation of pressure at brain

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.

b) Calculation of pressure at feet

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.

c) Calculation of increment in pressure

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.