Question

wide tube that extends down from the bag of solution, which hangs from a pole so that the fluid level is 90.0 cm above the needle. The inner radius of the needle is 0.200 mm. The top of the fluid is exposed to the atmosphere, and the flow rate of the fluid (which has a density of 1025 kg/m^3 and a viscosity of 0.0010 Pass) through the needle is 0.200 L/h. What is the average gauge pressure inside the vein where the needle is? Use g = 9.8 m/s^2. _______ Pa

Answer 1

Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

$P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa$

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

$S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}$

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

$P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa$