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Chapter 10: Q74GP (page 260)

Intravenous transfusions are often made under gravity, as shown in Fig. 10–55. Assuming the fluid has a density of \({\bf{1}}{\bf{.00\;g/c}}{{\bf{m}}^{\bf{3}}}\)at what height h should the bottle be placed so the liquid pressure is (a) 52 mm-Hg, and (b) \({\bf{680\;mm - }}{{\bf{H}}_{\bf{2}}}{\bf{O?}}\)(c) If the blood pressure is 75 mm-Hg above atmospheric pressure, how high should the bottle be placed so that the fluid just barely enters the vein?

Short Answer

Expert verified

(a) The value of required height is\(70.72\;{\rm{cm}}\).

(b) The value of required height is\(68\;{\rm{cm}}\).

(c) The required height of bottle is \(102{\rm{ cm}}\).

Step by step solution

01

Understanding the hydrostatic pressure

Whenever a specific liquid contains in a container, then there would be a pressure exerted by the liquid on the bottom surface of the container, referred to as hydrostatic pressure.

The relation between the hydrostatic pressure and liquid height is a linear one.

02

Identification of given data

The density of fluid is\(\rho = 1.0\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}\).

The liquid pressure in terms of height is\({h_1} = \left( {52\;{\rm{mm}} - {\rm{Hg}}\; \times \frac{{{{10}^{ - 1}}\;{\rm{cm}} - {\rm{Hg}}}}{{1\;{\rm{mm}} - {\rm{Hg}}}}} \right) = 5.2\;{\rm{cm}} - {\rm{Hg}}\).

The other liquid pressure in terms of height is\({h_2} = \left( {680\;{\rm{mm}} - {{\rm{H}}_{\rm{2}}}{\rm{O}} \times \frac{{{{10}^{ - 1}}\;{\rm{cm}} - {{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{1\;{\rm{mm}} - {{\rm{H}}_{\rm{2}}}{\rm{O}}}}} \right) = 68\;{\rm{cm}} - {{\rm{H}}_{\rm{2}}}{\rm{O}}\).

The height blood pressure height is\({h_b} = \left( {75\;{\rm{mm}} - {\rm{Hg}}\, \times \frac{{{{10}^{ - 1}}\;{\rm{cm}} - {\rm{Hg}}}}{{1\;{\rm{mm}} - {\rm{Hg}}}}} \right) = 7.5\;{\rm{cm}} - {\rm{Hg}}\).

03

Determination of the liquid height

(a)

The expression to calculate the liquid pressure can be represented as,

\({P_1} = {\rho _{Hg}}g{h_1}\)

Here, \({P_1}\) is the liquid pressure, \({\rho _{Hg}}\) is the density of mercury \(\left( {{\rho _{Hg}} = 13.6\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}\;} \right)\).

Substitute all the known values in the above expression,

\(\begin{array}{c}{P_1} = \left( {13.6\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}} \right)\left( {5.2\;{\rm{cm}} - {\rm{Hg}}} \right)g\\ = 70.72g\;{\rm{g/c}}{{\rm{m}}^{\rm{2}}}\end{array}\)

The expression of the height of the liquid pressure is given by,

\(\begin{array}{c}{P_1} = \rho gh\\h = \frac{{{P_1}}}{{\rho g}}\end{array}\)

Substitute all the known values in the above expression,

\(\begin{array}{c}h = \frac{{\left( {70.72g} \right)\;{\rm{g/c}}{{\rm{m}}^{\rm{2}}}}}{{\left( {1\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}} \right)g}}\\ = 70.72\;{\rm{cm}}\end{array}\)

So, the value of required height is \(70.72\;{\rm{cm}}\).

(b)

The expression to calculate the liquid pressure can be represented as,

\({P_2} = {\rho _{{H_2}O}}g{h_2}\).

Here, \({P_2}\) is the liquid pressure, \({\rho _{{H_2}O}}\) is the density of water \(\left( {{\rho _{{H_2}O}} = 1\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}\;} \right)\).

Substitute all the known values in the above expression.

\(\begin{array}{c}{P_2} = \left( {1\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}} \right)\left( {68\;{\rm{cm}} - {H_2}O} \right)g\\ = 68g\;{\rm{g/c}}{{\rm{m}}^{\rm{2}}}\end{array}\)

The expression of the height of the liquid pressure is given by,

\(\begin{array}{c}{P_2} = \rho gh\\h = \frac{{{P_2}}}{{\rho g}}\end{array}\)

Substitute all the known values in the above expression.

\(\begin{array}{c}h = \frac{{\left( {68g} \right)\;{\rm{g/c}}{{\rm{m}}^{\rm{2}}}}}{{\left( {1\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}} \right)g}}\\ = 68\;{\rm{cm}}\end{array}\)

So, the value of required height is \(68\;{\rm{cm}}\).

04

Determine the required height of bottle

The expression to calculate the height of bottle can be represented as,

\({P_b} = {\rho _{Hg}}g{h_b}\)

Here, \({P_b}\) is the blood pressure, \({\rho _{Hg}}\) is the density of mercury \(\left( {{\rho _{Hg}} = 13.6\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}\;} \right)\)

Substitute all the known values in the above expression.

\(\begin{array}{c}{P_b} = \left( {13.6\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}} \right)\left( {7.5\;{\rm{cm}} - Hg} \right)g\\ = 102g\;{\rm{g/c}}{{\rm{m}}^{\rm{2}}}\end{array}\)

The expression of the bottle height is given by,

\(h = \frac{{{P_b}}}{{\rho g}}\)

Substitute all the known values in the above expression.

\(\begin{array}{c}h = \frac{{\left( {102g} \right)\;{\rm{g/c}}{{\rm{m}}^{\rm{2}}}}}{{\left( {1\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}} \right)g}}\\ = 102\;{\rm{cm}}\end{array}\)

Thus, the required height of bottle is \(102{\rm{ cm}}\).

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Most popular questions from this chapter

Water at a gauge pressure of \({\bf{3}}{\bf{.8}}\;{\bf{atm}}\) at street level flows into an office building at a speed of \({\bf{0}}{\bf{.78}}\;{\bf{m/s}}\) through a pipe \({\bf{5}}{\bf{.0}}\;{\bf{cm}}\)in diameter. The pipe tapers down to \({\bf{2}}{\bf{.8}}\;{\bf{cm}}\) in diameter by the top floor, \({\bf{16}}\;{\bf{m}}\) above (Fig. 10–53), where the faucet has been left open. Calculate the flow velocity and the gauge pressure in the pipe on the top floor. Assume no branch pipes and ignore viscosity.

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