Q66P (III) A patient is to be given a... [FREE SOLUTION]

Chapter 10: Q66P (page 260)

(III) A patient is to be given a blood transfusion. The blood is to flow through a tube from a raised bottle to a needle inserted in the vein (Fig. 10鈥�55). The inside diameter of the 25-mm-long needle is 0.80 mm, and the required flow rate is \(2.0\;{\rm{c}}{{\rm{m}}^3}\) of blood per minute. How high h should the bottle be placed above the needle? Obtain \(\rho \) and \(\eta \) from the Tables. Assume the blood pressure is 78 torr above atmospheric pressure.
Figure 10-55

Short Answer

Expert verified

The bottle should be placed \(1.04\;{\rm{m}}\) above the needle.

Step by step solution

Concept

When a fluid is in static equilibrium condition, the pressure drop in the fluid relies upon the value of fluid鈥檚 density and its change in elevation point from the datum.

The relation for the pressure change is given by,

\(\Delta P = \rho gh\)

Given data

The length of the needle is, \(L = 25\;{\rm{mm}} \times \frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}} = 0.025\;{\rm{m}}\).

The inside diameter of the needle is, \(D = 0.80\;{\rm{mm}} \times \frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}} = 0.0008\;{\rm{m}}\).

The blood flow rate is, \(Q = 2.0\;{\rm{c}}{{\rm{m}}^3}{\rm{/min}} \times \frac{{1\;{{\rm{m}}^3}{\rm{/s}}}}{{{{10}^6} \times 60\;{\rm{c}}{{\rm{m}}^3}{\rm{/min}}}} = 3.33 \times {10^{ - 8}}\;{{\rm{m}}^3}{\rm{/s}}\).

The blood pressure is, \({P_{blood}} = 78\;{\rm{torr}} \times \frac{{133.3\;{\rm{Pa}}}}{{1\;{\rm{torr}}}} + {P_{atm}} = 10397.4\;{\rm{Pa}} + {P_{atm}}\)

Calculation

Using the property table,

The density of the blood is given as,

\(\rho = 1050\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\)

The viscosity of the blood is given as,

\(\eta = 4.0 \times {10^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}}\)

Applying the Poiseuille鈥檚 law, the expression for the pressure drops between the inlet and exit of the needle is given as,

\(\begin{array}{c}Q = \frac{{\pi {D^4}\Delta P}}{{128\eta L}}\\3.33 \times {10^{ - 8}}\;{{\rm{m}}^3}{\rm{/s}} = \frac{{\pi \times {{\left( {0.0008\;{\rm{m}}} \right)}^4} \times \Delta P}}{{128 \times 4.0 \times {{10}^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}} \times 0.025\;{\rm{m}}}}\\\Delta P = 331.24\;{\rm{Pa}}\\{P_{exit}} - {P_{blood}} = 331.24\;{\rm{Pa}}\end{array}\)

Substituting the value in the above equation,

\(\begin{array}{c}{P_{exit}} = 331.24\;{\rm{Pa}} + {P_{blood}}\\{P_{exit}} = 331.24\;{\rm{Pa}} + 10397.4\;{\rm{Pa}} + {P_{atm}}\\{P_{exit}} = 10728.64\;{\rm{Pa}} + {P_{atm}}\end{array}\)

Using hydrostatic law, the change in pressure between the top of the bottle and the exit of the needle is given as,

\(\begin{array}{c}{P_{top}} + \rho gh = {P_{exit}}\\{P_{atm}} + 1050\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}} \times 9.81\;{\rm{m/}}{{\rm{s}}^2} \times h = 10728.64\;{\rm{Pa}} + {P_{atm}}\\10300.5\;{\rm{kg/}}{{\rm{m}}^2} \cdot {{\rm{s}}^2} \times \frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}} \times h = 10728.64\;{\rm{Pa}} \times \frac{{1\;{\rm{N/}}{{\rm{m}}^2}}}{{1\;{\rm{Pa}}}}\\h = 1.04\;{\rm{m}}\end{array}\)

Thus, the bottle should be placed \(1.04\;{\rm{m}}\) above the needle.