Question

Blood pressure in Argentinosaurus. If this long-necked, gigantic sauropod had a head height of 23.7 m and a heart height of 7.00 m, (a) what (hydrostatic) gauge pressure in its blood was required at the heart such that the blood pressure at the brain was 93.1 torr (just enough to perfuse the brain with blood)? Assume the blood had a density of 1.06 × 103 kg/m3. (b) What was the blood pressure at the feet?

Answer 1

Answer:

In a system the difference in pressure between 2 points is given by

$\Delta p=\rho \times g\times h\\\\P_{heart}-P_{brain}=1.06\times 10^{3}\times 9.81\times (23.7-7.0)\\\\P_{heart}-P_{brain}=173.656kPa$

Now it is given that pressure in the brain was 93.1 torr=12.41 kPa

Thus we have

$P_{heart}-12.41kPa=173.656kPa\\\\P_{heart}=186.066kPa$

Part 2)

Applying the same equation again we get

$P_{feet}-P_{brain}=1.06\times 10^{3}\times 9.81\times (23.7)\\\\P_{feet}-P_{brain}=246.44kPa\\\\P_{feet}=246.44kPa+12.41kPa=258.856kPa$

Answer 2

The gauge pressure in its blood that was required at the heart is equal to 186.07 kPa.

Given the following data:

Head height = 23.7 m.

Heart height of 7.00 m.

Blood pressure at brain = 93.1 torr.

Density of blood = $1.06 \times 10^3 \;kg/m^3$

Conversion:

1 torr = 0.1333 kPa.

93.1 torr = 12.41 kPa.

How to calculate the gauge pressure.

Mathematically, the change in pressure of a system is given by this formula:

$\Delta P = \rho g(h_2-h_1)\\\\P_{h}-P_{b}=\rho g(h_2-h_1)\\\\P_{h}=\rho g(h_2-h_1) + P_{b}$

Substituting the given parameters into the formula, we have;

$P_{h}=1.06 \times 10^3 \times 9.8 (23.7-7.00) + 12.41\\\\P_{h}= 173.66+12.41\\\\P_{h}=186.07\;kPa$

For the blood pressure at the feet, we have:

$P_{f}=1.06 \times 10^3 \times 9.8 \times (23.7) + 12.41\\\\P_{f}= 246.44+12.41\\\\P_{f}=258.85\;kPa$

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