Question
(a) What is the largest average velocity of blood flow in an artery of radius 2 × 10–3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s).
Solution 1
a)
Radius of the artery, r = 2 × 10–3 m
Diameter of the artery, d = 2 × 2 × 10–3 m = 4 × 10– 3 m
Viscosity of blood, eta = `2.084 xx 10^(-3)` Pas
Density of blood, ρ = 1.06 × 103 kg/m3
Reynolds’ number for laminar flow, NR = 2000
The largest average velocity of blood is given by the relation:
`V_"arg" = (N_Reta)/(rhod)`
= `(2000xx2.084xx10^(-3))/(1.06xx10^3xx4xx10^(-3))`
= 0.983 m/s
Therefore, the largest average velocity of blood is 0.983 m/s.
b) Flow rate is given by the relation:
`R = pir^2V_"arg"`
`= 3.14 xx (2xx10^(-3))^2 xx 0.983`
`= 1.235 xx 10^(-5) m^3 s^(-1)`
Therefore, the corresponding flow rate is `1.235 xx 10^(-5) m^3 s^(-1)`
Solution 2
Here `r = 2 xx 10^(-3) m; D = 2r = 2xx2xx10^(-3) = 4xx 10^(-3) m`
`eta = 2.084 xx 10^(-3) Pa-s; rho = 1.06 xx 10^3 kg m^(-3)`
For flow to be laminar , `N_R = 2000`
a) Now `v_c = `(N_Reta)/rho_D` = (2000xx(2.084xx10^(-3)))/((1.06xx10^3)xx(4xx10^(-3))) = 0.98 "m/s"`
b)Volume flowering per second =`pir^2v_c = 22/7 xx (2xx10^(-3))^2 xx 0.98 = 1.23 xx 10^(-5) m^3s^(-1)`
