Question

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

(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

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.

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

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 \(68\;{\rm{cm}}\).

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 \(102{\rm{ cm}}\).