Question

Blood pressure in Argentinosaurus. (a) If this longnecked, gigantic sauropod had a head height of 21 mand a heart height of9.0m,what (hydrostatic) gauge pressure in its blood was required at the heart such that the blood pressure at the brain was 80 torr (just enough to perfuse the brain with blood)? Assume the blood had a density of$\mathbf{1}\mathbf{.}\mathbf{06}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{K}\mathit{G}\mathbf{/}{\mathit{M}}^{\mathbf{3}}$. (b) What was the blood pressure (in torror mm ${\mathit{H}}_{\mathbf{g}}$) at the feet?

Short answer

1. Gauge pressure required in the blood is$1\times {10}^{3}torr$
2. The blood pressure at feet is$1.72\times {10}^{3}torr$

Step-by-step solution

The given data

1. Density of blood is $p=1060kg/m^3$
2. Pressure in brain, localid="1657265202495" $p=80torr$(to be used for required gauge pressure)
3. Head height, $h_2=21m$
4. Heart height, $h_1=9m$

Understanding the concept of hydrostatic pressure

To solve for pressure, we can use the formula for hydrostatic pressure difference, which depends on density and height. Hydrostatic pressure is that, which is exerted by a fluid at any point of time at equilibrium due to the force applied by gravity.

Formulae:

Pressure applied on a fluid surface,$p=pgh$ (i)

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

a) Calculation of gauge pressure in the blood

Using equation (i) and (ii), we can write the applied pressure on blood as:

$$\begin{gathered} p_{blood}=p_{brain}+pgh(\because heightatwhichpressureisaapiled,h=h_{2-}{h}_1) \\ =80torr+1060kg/m^3\times (21m-9m)\times \frac{1torr}{133.33pa}(\because 1torr=133.33pa) \\ =1\times {10}^{3}torr \\ Hence, \\ therequiredgaugepressureis1\times {10}^{3}torr \end{gathered}$$

b) Calculation of blood pressure at feet

Using equation (i) and (ii), we can write the applied pressure on blood as:

$$\begin{gathered} p_{feet}=p_{brain}+pgh(\because heightatwhichpressureisapplied,h=h_{2)} \\ =80torr+\left(1060kg/m^3\times 9.8m/s^2\times \left(21m\right)\times \frac{1torr}{133.33pa}\right)(\because 1torr=133.33pa) \\ =1.72\times {10}^{3}torr \\ Hence,thebloodpressireatthefeetis1.72\times {10}^{3}torr \end{gathered}$$