Chapter 10, Problem 58P

To determine

The height at which the bottle should be placed above the needle is to be calculated under the given conditions.

Answer to Problem 58P

The height at which the bottle should be placed above the needle is $1.88m$ .

Explanation of Solution

Given info:

Diameter,

  $\begin{array}{l}d=0.40mm=0.40\times {10}^{-3}m\\ \Rightarrow r=0.20\times {10}^{-3}m\end{array}$

Blood rate flow, $Q=4c{m}^3/\min =\frac{4\times {10}^{-6}}{60}{m}^3/s$

Needle length, $l=4.0cm=0.04m$

The density of blood is, $\rho \text{=1050 }kg/m^3$

Viscosity coefficient, $\eta =4\times {10}^{-3}Pas$

Pressure, $P=18torr=2392Pa$

Formula used:

The continuity equation is,

  $P_1+\frac{1}{2}\rho v^2+\rho gh$

And,

The volume flow rate is, $Q=\frac{\pi r^4\left(P_2-P_1\right)}{8\eta l}$

Calculation:

From the continuity equation,

  $P_1+\frac{1}{2}\rho v_1{}^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

Here the volume is same. So, the pressure becomes,

  $P_2=P_1+\rho g\left(h_2-h_1\right)=P_1+\rho gh$

now consider the rate flow equation,

  $\begin{array}{l}d=0.40mm=0.40\times {10}^{-3}m\\ \Rightarrow r=0.20\times {10}^{-3}m\\ Q=4c{m}^3/\min =\frac{4\times {10}^{-6}}{60}{m}^3/s\\ l=4.0cm=0.04m\\ \rho \text{=1050 }kg/m^3\\ Q=\frac{\pi r^4\left(P_2-P_1\right)}{8\eta l}\\ \frac{4\times {10}^{-6}}{60}{m}^3/s=\frac{3.14\times {\left(0.20\times {10}^{-3}m\right)}^4\left(\rho gh-2392Pa\right)}{8\times 4\times {10}^{-3}Pas\times 0.04m\times }\\ \left(\rho gh-2392Pa\right)=\frac{4\times {10}^{-6}\times 8\times 4\times {10}^{-3}\times 0.04m}{60\times 3.14\times {\left(0.20\times {10}^{-3}m\right)}^4}\\ \left(10290h-2392\right)=\frac{5.12\times {10}^{-9}}{0.30144}=16985\\ 10290h=16985+2392=19377\\ h=\frac{19377}{10290}=1.88m\end{array}$

Conclusion:

The height at which the bottle should be placed above the needle is $1.88m$ .