A smooth pipe line of 100 mm diameter carries 2.27m32.27m3 per minute of water at 20∘C20∘C with kinematic viscosity of 0.0098 Stokes. Calculate the friction factor maximum velocity as well as shear stress at the boundary.

  A smooth pipe line of 100 mm diameter carries 

2.27m3$2.27m^3$ per minute of water at 20∘C${20}^{\circ }C$ with kinematic viscosity of 0.0098 Stokes. Calculate the friction factor maximum velocity as well as shear stress at the boundary.

Solution:- 

Given:-

Diameter of pipe, D=100mm=0.1m

Radius, R=0.12=0.05m$R={\displaystyle \frac{0.1}{2}}=0.05m$

Discharge, Q=2.27m3/min=2.2760=0.0378m3/s$Q=2.27m^3/min={\displaystyle \frac{2.27}{60}}=0.0378m^3/s$

Kinematic viscosity, v=0.0098stokes

v=0.0098×10−4m2/s$v=0.0098\times {10}^{-4}{m}^2/s$

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Therefore Average velocity,

U¯¯¯¯=QA=0.0378π4(0.1)2=4.817m/s$\overline{U}={\displaystyle \frac{Q}{A}}={\displaystyle \frac{0.0378}{{\displaystyle \frac{\pi }{4}}(0.1{)}^2}}=4.817m/s$

Therefore Reynolds number is given by,

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

Re=4.817×0.10.0098×10−4=4.9154×105$R_e={\displaystyle \frac{4.817\times 0.1}{0.0098\times {10}^{-4}}}=4.9154\times {10}^5$

The flow is turbulent and Re$R_e$ is more than 105${10}^5$.

Hence for smooth pipe, the coefficient of friction ‘f’ is obtained by,

14f−−√=2.0log10(Re4f−−√)−0.8${\displaystyle \frac{1}{\sqrt{4f}}}=2.0{\log }_{10}(R_e\sqrt{4f})-0.8$

14f−−√=2.0log10(4.9154×1054f−−√)−0.8${\displaystyle \frac{1}{\sqrt{4f}}}=2.0{\log }_{10}(4.9154\times {10}^5\sqrt{4f})-0.8$

14f−−√=2.0[log10(4.9154×105)+log104f−−√]−0.8${\displaystyle \frac{1}{\sqrt{4f}}}=2.0[{\log }_{10}(4.9154\times {10}^5)+{\log }_{10}\sqrt{4f}]-0.8$

14f−−√=2.0[5.6915+log104f−−√]−0.8${\displaystyle \frac{1}{\sqrt{4f}}}=2.0[5.6915+{\log }_{10}\sqrt{4f}]-0.8$

14f−−√=2×5.6915+2log104f−−√−0.8${\displaystyle \frac{1}{\sqrt{4f}}}=2\times 5.6915+2{\log }_{10}\sqrt{4f}-0.8$

14f−−√=11.3830+log10(4f−−√)2−0.8${\displaystyle \frac{1}{\sqrt{4f}}}=11.3830+{\log }_{10}(\sqrt{4f}{)}^2-0.8$

14f−−√=11.383+log10(4f)−0.8${\displaystyle \frac{1}{\sqrt{4f}}}=11.383+{\log }_{10}(4f)-0.8$

14f−−√−log10(4f)=11.383−0.8=10.583……………….(1)${\displaystyle \frac{1}{\sqrt{4f}}}-{\log }_{10}(4f)=11.383-0.8=10.583\dots \dots \dots \dots \dots \dots .(1)$

Step 1: Friction factor

(f*)=4×$\times$ coefficient of friction=4f

substitute the value in equation (1),

1f∗−−√−log10(f∗)=10.583………………(2)${\displaystyle \frac{1}{\sqrt{f\ast }}}-{\log }_{10}(f\ast )=10.583\dots \dots \dots \dots \dots \dots (2)$

The above equation is solved by hit and trial method,

Therefore, let f*=0.1, then LHS of equation (2), becomes

LHS=10.1−−−√−log10(0.1)=3.16−(−1.0)=4.16${\displaystyle \frac{1}{\sqrt{0.1}}}-{\log }_{10}(0.1)=3.16-(-1.0)=4.16$

Let f*=0.01, then LHS becomes,

LHS=10.01−−−−√−log10(0.01)=10−(−2)=12${\displaystyle \frac{1}{\sqrt{0.01}}}-{\log }_{10}(0.01)=10-(-2)=12$

But for exact solution, LHS should be (10.583). Hence value of f* lies between 0.1 and 0.01.

Let f*=0.013, then LHS becomes,

LHS=10.013−−−−√−log10(0.013)=8.77−(−1.886)=10.656${\displaystyle \frac{1}{\sqrt{0.013}}}-{\log }_{10}(0.013)=8.77-(-1.886)=10.656$

which is approximately (10.583).

Hence the value of (f*=0.013).

Therefore friction factor, f*=0.013.

Step 2: Maximum velocity (umax$u_{max}$)

We know, f*=4f

∴f=f∗4$\therefore f={\displaystyle \frac{f\ast }{4}}$

∴f=0.0134=0.00325$\therefore f={\displaystyle \frac{0.013}{4}}=0.00325$

Now,

u∗=U¯¯¯¯f2−−√$u_{\ast }=\overline{U}\sqrt{{\displaystyle \frac{f}{2}}}$

u∗=4.817×0.003252−−−−−−−√$u_{\ast }=4.817\times \sqrt{{\displaystyle \frac{0.00325}{2}}}$

u∗=4.817×0.04303=0.194$u_{\ast }=4.817\times 0.04303=0.194$

For smooth,

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

The velocity will be maximum at the centre of pipe.

Where, y=R=0.05,

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

umax0.194=5.75log100.194×0.050.0098×10−4+5.55${\displaystyle \frac{u_{max}}{0.194}}=5.75{\log }_{10}{\displaystyle \frac{0.194\times 0.05}{0.0098\times {10}^{-4}}}+5.55$

umax=0.194[22.974+5.55]=5.528m/s$u_{max}=0.194[22.974+5.55]=5.528m/s$

Step 3: Shear stress at boundary (τ0${\tau }_0$)

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

∴τ0=ρu2∗$\therefore {\tau }_0=\rho u{2}_{\ast }$

∴τ0=1000×(0.194)2=37.63 N/m2