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Numericals related to boundary

Q1) A pipeline carrying water has average height of irregularities projecting from the surface of the boundary of the pipe as 0.15 mm. What type of boundary is it? The shear stress developed is $4.9N/m^2$. The kinematics viscosity of water is 0.01 Stokes.

Solution:- Given:-

$K=0.15mm=0.15\times10^{-3}$

$\tau_0=4.9N/m^2$

v=0.01 stokes

$v=0.01\times10^{-4}m^2/s$

Density of water, $\rho=1000kg/m^3$

Shear velocity,

$u_*=\sqrt{\dfrac{\tau_0}{\rho}}$

$u_*=\sqrt{\dfrac{4.9}{1000}}=0.07m/s$

Roughness Reynolds number,

$R_e=\dfrac{u_*\times K}v$

$R_e=\dfrac{0.07\times0.15\times 10^{-3}}{0.01\times10^{-4}}$

$\dfrac{u_*\times K}v=10.5$

Since $\left(\dfrac{u_*\times K}v\right)$ lies between 4 and 100, so the pipe behaves as in transition.


Q2) A rough pipe is of diameter 8.0 cm. The velocity at a point 3.0 cm from the wall is 30% more than the velocity at a point 1 cm from the pipe wall. Determine the average height of the roughness.

Solution:- Given:-

D=8m=0.08m

Let the velocity of flow at 1 cm from the pipe wall = u

Then the velocity of flow at 3 cm from pipe wall = 1.3u

$\dfrac u{u_*}$ = Velocity distribution for rough pipe

$\dfrac u{u_*}=5.75\log_{10}\left(\dfrac yK\right)+8.5$

a) For a point, 1 cm from pipe wall

$\dfrac u{u_*}=5.75\log_{10}\left(\dfrac 1K\right)+8.5………..(1)$

(b) For a point, 3 cm from pipe wall, velocity is 1.3 u, and hence

$\dfrac {1.3u}{u_*}=5.75\log_{10}\left(\dfrac 3K\right)+8.5…………(2)$

Dividing (1) and (2) equation, we get

$1.3=\dfrac{5.75\log_{10}\left(\dfrac 3K\right)+8.5}{5.75\log_{10}\left(\dfrac 1K\right)+8.5}$

$1.3\left[5.75\log_{10}\left(\dfrac 1K\right)+8.5\right]= 5.75\log_{10}\left(\dfrac 3K\right)+8.5$

$7.475\log_{10}\left(\dfrac 1K\right)-5.75\log_{10}\left(\dfrac 3K\right)=8.5-11.05$

$7.475\log_{10}\left(\dfrac 1K\right)-5.75\log_{10}\left(\dfrac 3K\right)=-2.55$

$7.475[\log_{10} (1.0)-\log_{10}(K)]-5.75[\log_{10}(3.0)-\log_{10}(K)]=-2.55$

$7.475[0-log_{10}(K)]-5.75[0.4771-\log_{10}(K)]=-2.55$

$-7.475log_{10}(K)-2.7433+5.75\log_{10}(K)]=-2.55$

$-1.725log_{10}(K)=2.7433-2.55$

$-1.725log_{10}(K)=0.1933$

$log_{10}(K)=\dfrac{0.1933}{-1.725}$

$log_{10}(K)=-0.1120=\overline{1}.888$

$K=0.7726 \ cm$

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