Answer :
Answer:
Flow in liters per minute is 4 L/min
Explanation:
For this case, we have an aorta of 0.9 cm of diameter (D). Let's suppose an uniform and constant diameter for calculation purposes.
D = (0.9 cm)(1m/100cm)
D = 0.009 m
It is required to calculate the blood flow in liters per minute when the flow regime changes from laminar to turbulent. Laminar flow is usually less than 2500 for Reynolds value, and Turbulent flow when Re is higher than 2500. So, we need to study the phenomenon for
Re = 2500.
Using the definition of Reynolds we can find the velocity average of the blood, and use it to find flow. Where blood density is [tex]\rho[/tex], aorta diameter is D, average velocity is v and blood viscosity is [tex]\mu[/tex]
[tex]Re = \frac{\rho v D}{\mu } \\v = \frac{Re*\mu}{\rho D}[/tex]
From data problem, we have Re, D values. As we need blood density and blood viscosity we can find them in medical studies. For example: in this online document: Blood flow analysis of the aortic arch using computational fluid dynamics †
Satoshi Numata, Keiichi Itatani, Keiichi Kanda, Kiyoshi Doi, Sachiko Yamazaki, Kazuki Morimoto, Kaichiro Manabe, Koki Ikemoto, Hitoshi Yaku Author Notes
European Journal of Cardio-Thoracic Surgery, Volume 49, Issue 6, June 2016, Pages 1578–1585, https://doi.org/10.1093/ejcts/ezv459
Published: 20 January 2016
[tex]\rho = 1060 kg/m^3\\\mu = 0.0004 kg/(ms)\\[/tex]
[tex]v = \frac{Re*\mu}{\rho D}\\v = \frac{2500 * 0.004 kg/(ms)}{(1060 kg/m3)(0.009 m)}\\v = 1.04 m/s\\[/tex]
Average blood velocity is 1.04 m/s. After that, we can calculate the flow (Q) using the flow are (Ao) of aorta.
Q = v*Ao
[tex]Q = (1.4m/s)*(\pi*(0.009m/2)^2)\\Q = 6.67 * 10^-5 \frac{m^3}{s}\\Q = (6.67 * 10^-5 \frac{m^3}{s})(\frac{1000 L}{1 m^3} ) (\frac{60 s}{1 min} )\\Q = 4 L/min[/tex]
Finally, the blood flow in liters per minute is 4 L/min when the flow regime in the aorta changes from laminar to turbulent.