Question #200424

Consider the blood flow in an artery following Poiseuille’s law. If the length of the

artery is 3 cm, radius is 7×10-3 cm and driving force is 5×103 dynes/cm2 then using blood viscosity, μ = 0 × 027 poise, find the

(i) velocity u( y) and the maximum peak velocity of blood, and

(ii) shear stress at the wall of the artery.


Expert's answer

(i) Here, viscosity of blood is given as

 μ=0.027 poise=0.0027Nsm2\mu=0.027\ poise=0.0027\dfrac{N\cdot s}{m^2} ,

length of artery l=3cm=3×102m,l=3cm=3\times10^{-2}m,

radius r=7×103cm=7×105m,r=7\times10^{-3} cm=7\times10^{-5}m, and P1P2=5×103dynecm2=5×102Nm2P_1-P_2= 5\times10^{3} \dfrac{dyne}{cm^2}=5\times 10^2\dfrac{N}{m^2}

We know that maximum velocity in case of fluid flow



Vmax=ΔPr24μl=5×102×49×10104×0.0027×3×102=0.756 cm/sV_{max}=\frac{\Delta P r^2}{4 \mu l}= \frac{5\times10^{2}\times49\times10^{-10}}{4\times0.0027\times3\times10^{-2}}=0.756\ cm/su=Vmax2=0.756cm/s2=0.378cm/su=\dfrac{V_{max}}{2}=\dfrac{0.756cm/s}{2}=0.378cm/s



(ii) For shear stress we know that,



τ=ΔPr2l=5×102×7×1052×3×102m=0.583Nm2=5.83dynecm2\tau=\frac{\Delta P r}{2l}= \frac{5\times10^{2}\times7\times10^{-5}}{2\times3\times10^{-2}m}=0.583\dfrac{N}{m^2}=5.83 \dfrac{dyne}{cm^2}

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