Page: Velocity distribution in smooth and rough pipes
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Velocity distribution in smooth pipes

$u=\dfrac{u_*}{K}\log_e y+C$

It is seen that at (y=0), the velocity ‘u’ at wall is ($-\infty$).

This means ‘u’ is positive at some distance far from the wall.

Hence, at some finite distance from wall, (‘u’=0).

Let the distance from pipe wall b y’. Now, constant C is determined from the boundary condition, i.e., at (y=y’), (u’=0),

$\therefore 0=\dfrac{u_*}K\log_e y’+C$

$\therefore C=-\dfrac{u_*}K\log_e y’$

Putting the value of ‘C’ in the above equation, we get

$\therefore u=\dfrac{u_*}K\log_e y-\dfrac{u_*}K\log_e y’$

$\therefore u=\dfrac{u_*}K\log_e \left(\dfrac y{y’}\right)$

Substituting the value of K = 0.4, we get

$\therefore u=\dfrac{u_*}{0.4}\log_e \left(\dfrac y{y’}\right)$

$\therefore u=2.5 u_*\log_e \left(\dfrac y{y’}\right)$

$\therefore \dfrac u{u_*}=2.5 \times2.3\log_{10} \left(\dfrac y{y’}\right)$

$\left[\because \log_e \left(\dfrac y{y’}\right)= 2.3\log_{10} \left(\dfrac y{y’}\right)\right]$

$\therefore \dfrac u{u_*}=5.75\log_{10} \left(\dfrac y{y’}\right)………………(1)$

For smooth boundary, there exist a laminar sub-layer.

The velocity distribution in laminar sub-layer is parabolic in nature.

Thus, in laminar sub-layer, velocity distribution does not hold.

Therefore assume y’ proportional to $\delta'$

$\therefore y’=\dfrac {\delta'}{107}$

Where, $\delta'=\dfrac{11.6v}{u_*}$, where v = kinematic viscosity.

$\therefore y’=\dfrac {\dfrac{11.6v}{u_*}}{107}$

$\therefore y’=\dfrac {0.108v}{u_*}$

Substituting this value in equation (1), we get

$\therefore \dfrac u{u_*}=5.75\log_{10} \left(\dfrac y{\dfrac {0.108v}{u_*}}\right)$

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac {yu_*}{0.108v}\right)$

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac {yu_*}{v}\times9.259\right)$

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac {yu_*}{v}\right)+ 5.75\log_{10}(9.259)$

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac {yu_*}{v}\right)+ 5.55…………..(2)$

Velocity distribution in rough pipes:-

In case of rough boundaries, the thickness of laminar sub-layer is very small.

The surface irregularities are above the laminar sub-layer, hence the laminar sub-layer is completely destroyed.

Therefore y’ is proportional to ‘K’.

$\therefore y’=\dfrac {K}{30}$

Substituting this value of y’ from equation $\left[ \dfrac u{u_*}=5.75\log_{10} \left(\dfrac y{y’}\right)\right]$ , we get

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac y{\dfrac {K}{30}}\right)$

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac {y}{K}\times30\right)$

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac {y}{K}\right)+ 5.75\log_{10}(30.0)$

$\dfrac u{u_*}=5.75\log_{10} \left(\dfrac {y}{K}\right)+ 8.5…………..(3)$

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modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 270 written 7 months ago by gravatar for Syedahina Mohi Syedahina Mohi0
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