
When a pilot takes a tight turn at high speed in a modern fighter airplane, the blood pressure at the brain level decreases, blood no longer perfuses the brain, and the blood in the brain drains. If the heart maintains the (hydrostatic) gauge pressure in the aorta at \[120torr{\text{ }}\left( {or{\text{ }}mm{\text{ }}Hg} \right)\]when the pilot undergoes a horizontal centripetal acceleration of \[4g\] , what is the blood pressure (in torr) at the brain, \[30cm\] radially inward from the heart? The perfusion in the brain is small enough that the vision switches to black and white and narrows to “tunnel vision” and the pilot can undergo g-LOC (“g-induced loss of consciousness”).Blood density is \[1.06 \times 10{\text{ }}3{\text{ }}kg/m\].
Answer
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Hint: Blood pressure is the blood circulating pressure on the blood vessel walls. The heart blood pumping through the circulatory system results in most of this pressure. The word "blood pressure" refers to the pressure of large arteries if it is used without qualification.
Complete step by step answer:
Given, ${P_{Heart}} = 120\,torr$.
Horizontal centripetal acceleration $\left( a \right) = 4g$
Blood density = \[1.06 \times 10{\text{ }}3{\text{ }}kg/m\]
$g = 9.8m/{s^2}$
To Find :- Blood pressure at brain which is \[30{\text{ }}cm\] radially inwards from heart
$a = 4 \times g \\
\Rightarrow a = 4 \times 9.8m/{s^2} \\
\Rightarrow a = 39.2\,m/{s^2} \\ $
Now, Blood pressure at Brain $\left( {{P_{Brain}}} \right)$ is :
Pressure at Brain = Pressure at Heart - ( Blood Density × Acceleration × Radial Distance of Brain from Heart )
\[{P_{Brain}} = {P_{Heart}} - \rho \times a \times r \\
\Rightarrow {P_{Brain}} = 120torr - \left( {1.06 \times {{10}^3}kg/{m^3}} \right)\left( {39.2m/{s^2}} \right)\left( {0.30m} \right) \\
\Rightarrow {P_{Brain}}= \left( {120torr} \right) - \left( {12465.6kg/m{s^2}} \right) \\
\Rightarrow {P_{Brain}}= \left( {120torr - 94torr} \right)\,\,\,\,\,\left( {\because 1kg/m{s^2} = \dfrac{1}{{133}}torr} \right) \\
\therefore {P_{Brain}} = 26torr \\ \]
Hence, Blood pressure at Brain is \[26{\text{ }}torr\] .
Note:Hydrostatic gauges (e.g., the mercury column gauge) measure the pressure by the unit area at the basis of a fluid column against the hydrostatic force. Hydrostatic gauge measurements can be designed to achieve a very linear calibration regardless of the type of gas being measured.
Complete step by step answer:
Given, ${P_{Heart}} = 120\,torr$.
Horizontal centripetal acceleration $\left( a \right) = 4g$
Blood density = \[1.06 \times 10{\text{ }}3{\text{ }}kg/m\]
$g = 9.8m/{s^2}$
To Find :- Blood pressure at brain which is \[30{\text{ }}cm\] radially inwards from heart
$a = 4 \times g \\
\Rightarrow a = 4 \times 9.8m/{s^2} \\
\Rightarrow a = 39.2\,m/{s^2} \\ $
Now, Blood pressure at Brain $\left( {{P_{Brain}}} \right)$ is :
Pressure at Brain = Pressure at Heart - ( Blood Density × Acceleration × Radial Distance of Brain from Heart )
\[{P_{Brain}} = {P_{Heart}} - \rho \times a \times r \\
\Rightarrow {P_{Brain}} = 120torr - \left( {1.06 \times {{10}^3}kg/{m^3}} \right)\left( {39.2m/{s^2}} \right)\left( {0.30m} \right) \\
\Rightarrow {P_{Brain}}= \left( {120torr} \right) - \left( {12465.6kg/m{s^2}} \right) \\
\Rightarrow {P_{Brain}}= \left( {120torr - 94torr} \right)\,\,\,\,\,\left( {\because 1kg/m{s^2} = \dfrac{1}{{133}}torr} \right) \\
\therefore {P_{Brain}} = 26torr \\ \]
Hence, Blood pressure at Brain is \[26{\text{ }}torr\] .
Note:Hydrostatic gauges (e.g., the mercury column gauge) measure the pressure by the unit area at the basis of a fluid column against the hydrostatic force. Hydrostatic gauge measurements can be designed to achieve a very linear calibration regardless of the type of gas being measured.
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