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.


1
Expert's answer
2021-06-01T09:07:43-0400

(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}

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS