A smooth pipe of diameter 400mm and length 800m carries water at the rate of 0.04m3/s0.04m3/s. Determine the head lost due to friction, wall shear stress, center-line velocity and thickness of laminar sub-layer. Take the kinematics viscosity of water as 0.018 stokes.

 A smooth pipe of diameter 400mm and length 800m carries water at the rate of 

0.04m3/s$0.04m^3/s$. Determine the head lost due to friction, wall shear stress, center-line velocity and thickness of laminar sub-layer. Take the kinematics viscosity of water as 0.018 stokes.

Solution:-

 Given:-

Diameter of pipe, D=400mm=0.40m

Radius, R=D2=0.42=0.2m$R={\displaystyle \frac{D}{2}}={\displaystyle \frac{0.4}{2}}=0.2m$

Length of pipe, L=800m

Discharge, Q=0.04m3/s$Q=0.04m^3/s$

Kinematics viscosity, v=0.018stokes

v=0.018cm2/s=0.018×10−4m2/s$v=0.018c{m}^2/s=0.018\times {10}^{-4}{m}^2/s$

Average velocity, V¯¯¯¯=QArea=0.04π4(0.4)2=0.318m/s$\overline{V}={\displaystyle \frac{Q}{\text{Area}}}={\displaystyle \frac{0.04}{{\displaystyle \frac{\pi }{4}}(0.4{)}^2}}=0.318m/s$

Reynolds number

Re=V×Dv$R_e={\displaystyle \frac{V\times D}{v}}$

Re=V¯¯¯¯×Dv$R_e={\displaystyle \frac{\overline{V}\times D}{v}}$

Re=0.3183×0.40.018×10−4=7.073×10−4$R_e={\displaystyle \frac{0.3183\times 0.4}{0.018\times {10}^{-4}}}=7.073\times {10}^{-4}$

The flow is turbulent.

Step 1: The coefficient viscosity of friction ‘f’ is obtained from

f=0.0791(Re)1/4$f={\displaystyle \frac{0.0791}{(R_e{)}^{1/4}}}$

f=0.0791(7.073×10−4)1/4$f={\displaystyle \frac{0.0791}{(7.073\times {10}^{-4}{)}^{1/4}}}$

f=0.0791(16.30)=0.00485$f={\displaystyle \frac{0.0791}{(16.30)}}=0.00485$

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Step 2: Head lost due to friction,

hf=4fLV2D×2g$hf={\displaystyle \frac{4fL{V}^2}{D\times 2g}}$

hf=4fLV¯¯¯¯2D×2g$hf={\displaystyle \frac{4fL{\overline{V}}^2}{D\times 2g}}$

hf=4×0.00485×800×(0.3183)20.40×2×9.81=0.2m$hf={\displaystyle \frac{4\times 0.00485\times 800\times (0.3183{)}^2}{0.40\times 2\times 9.81}}=0.2m$

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Step 3: Wall shear stress (τ0${\tau }_0$)

τ0=fρV22${\tau }_0={\displaystyle \frac{f\rho V^2}{2}}$

τ0=fρV¯¯¯¯22${\tau }_0={\displaystyle \frac{f\rho {\overline{V}}^2}{2}}$

τ0=0.00485×1000×(0.3184)22${\tau }_0={\displaystyle \frac{0.00485\times 1000\times (0.3184{)}^2}{2}}$

τ0=0.245Nm2${\tau }_0=0.245N{m}^2$

${\mathrm{<span\; style="color:\; \#351c75;\; font-family:\; georgia;\; font-size:\; medium;"><i><br></i></span>}}^{}$

Step 4: Center-line viscosity (umax$u_{max}$)for smooth pipe,

umaxu∗=5.75log10u∗Rv+5.55${\displaystyle \frac{u_{max}}{u_{\ast }}}=5.75{\log }_{10}{\displaystyle \frac{u_{\ast }R}{v}}+5.55$

[put u=umax$u=u_{max}$ at y=R]

u∗=τ0ρ−−−√$u_{\ast }=\sqrt{{\displaystyle \frac{{\tau }_0}{\rho }}}$

u∗=0.2451000−−−−−√=0.0156m/s$u_{\ast }=\sqrt{{\displaystyle \frac{0.245}{1000}}}=0.0156m/s$

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Substituting the value of u∗$u_{\ast }$, R and v in the above equation,

umax0.0156=5.75log100.0156×0.200.018×10−4+5.55${\displaystyle \frac{u_{max}}{0.0156}}=5.75{\log }_{10}{\displaystyle \frac{0.0156\times 0.20}{0.018\times {10}^{-4}}}+5.55$

umax0.0156=24.173${\displaystyle \frac{u_{max}}{0.0156}}=24.173$

∴umax=0.0156×24.173=0.377m/s$\therefore u_{max}=0.0156\times 24.173=0.377m/s$

Step 5: Thickness of laminar sublayer (δ′${\delta }^{\prime }$)

δ′=11.6×vu∗${\delta }^{\prime }={\displaystyle \frac{11.6\times v}{u_{\ast }}}$

δ′=11.6×0.018×10−40.0156=0.001338m${\delta }^{\prime }={\displaystyle \frac{11.6\times 0.018\times {10}^{-4}}{0.0156}}=0.001338m$

δ′=1.338mm