#### 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 × 10^{3} kg/m^{3}

Reynolds’ number for laminar flow, *N*_{R} = 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)`