Answer
a)
$v(0)=0.925$$cm^2/s$
$v(0.005)=0.694$ $cm^2/s$
$v(0.01)=0$$cm^2/s$
b)
$v'(0)=0$
$v'(0.005)=-92.592$ cm/s/cm
$v'(0.01)=-185.185 $ $cm^2/s$
c)
The velocity is greater at $r=0$ and changing most at $R=0.01$
Work Step by Step
(a) Find the velocity of the blood along the center-line $r − 0$,
at radius $r − 0.005$ cm, and at the wall $r − R − 0.01$ cm.
solution:-
By using $v=\frac{p}{4{\eta}l}(R^2-r^2)$
Where $R=0.01, l=3,p=3000, {\eta}=0.027$
Now we have a function,
$v(r)=\frac{p}{4{\eta}l}(R^2-r^2)$
$v(r)=\frac{3000}{4(0.027)3}((0.01)^2-r^2)$
$v(0)=\frac{1000}{4(0.027)3}((0.01)^2-r^2)$
$v(0)=0.925$cm/s/cm
$v(0)=0.925$$cm^2/s$
$v(0.005)=0.694$ cm/s/cm
$v(0.005)=0.694$ $cm^2/s$
$v(0.01)=0$ cm/s/cm
$v(0.01)=0$$cm^2/s$
b) As we know we have a function ; $v(r)=\frac{p}{4{\eta}l}(R^2-r^2)$
Taking differentiate w.r.t $r$ on both sides
$v'(r)=\frac{d}{dr}(\frac{p}{4{\eta}l}(R^2-r^2))$
$v'(r)=(\frac{P}{4{\eta}l})\frac{d}{dr}(R^2-r^2)$
$v'(r)=(\frac{P}{4{\eta}l)})(\frac{d}{dr}(R^2)-\frac{d}{dr}(r^2))$
$v'(r)=(\frac{P}{4{\eta}l)})(-2r)$
$v'(r)=-\frac{Pr}{2{\eta}l}$
We have $R=0.01, l=3,p=3000, {\eta}=0.027$ so,
$v'(r)=\frac{3000r}{2(0.027)3}$
$v'(0)=\frac{3000{\times}(0)}{2(0.027)3}$
$v'(0)=0$
$v'(0.005)=\frac{3000{\times}(0.005)}{2(0.027)3}$
$v'(0.005)=-92.592$ cm/s/cm
$v'(0)=-92.592$ $cm^2/s$
$v'(0.01)=\frac{3000{\times}(0.01)}{2(0.027)3}$
$v'(0.01)=-185.185 $ cm/s/cm
$v'(0.01)=-185.185 $ $cm^2/s$
c)
The velocity is greater at $r=0$ and changing most at $R=0.01$
