The loss of head, due to friction in pipes is given by,

$hf=\dfrac{4fLV^2}{d\times2g}$

In this equation, the value of coefficient of friction (f), should be known accurately for predicting the loss of head due to friction in pipes.

On the basis of dimensional analysis, it can be shown that the pressure loss in a straight pipe of diameter (D), length (L), roughness (K), average velocity ($\overline{V}$), viscosity (‘$\mu$’) and density of fluid (‘$\rho$’) is

$\Delta P=\dfrac{\rho\overline{V}^2}2\phi\left[R_e,\dfrac KD, \dfrac LD\right]$

$\dfrac{\Delta P}{\dfrac{\rho\overline{V}^2}2}=\phi\left[R_e,\dfrac KD, \dfrac LD\right]$

Experimentally it was found that pressure drop is a function of $\dfrac LD$ to the first power.

$\dfrac{\Delta P}{\dfrac{\rho\overline{V}^2}2}=\dfrac LD\phi\left[R_e,\dfrac KD\right]$

$\dfrac{\Delta P\times D}{L\dfrac{\rho\overline{V}^2}2}=\phi\left[R_e,\dfrac KD\right]$

$f=\phi\left[R_e,\dfrac KD\right]$

Numericals

Q1) A smooth pipe of diameter 400mm and length 800m carries water at the rate of $0.04m^3/s$. 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, $R=\dfrac D2=\dfrac {0.4}2=0.2m$

Length of pipe, L=800m

Discharge, $Q=0.04m^3/s$

Kinematics viscosity, v=0.018stokes

$v=0.018cm^2/s=0.018\times10^{-4}m^2/s$

Average velocity, $\overline{V}=\dfrac Q{\text{Area}}=\dfrac {0.04}{\dfrac\pi4(0.4)^2}=0.318m/s$

Reynolds number

$R_e=\dfrac {V\times D}v$

$R_e=\dfrac {\overline{V}\times D}v$

$R_e=\dfrac {0.3183\times 0.4}{0.018\times10^{-4}}=7.073\times10^{-4}$

The flow is turbulent.

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

$f=\dfrac{0.0791}{(R_e)^{1/4}}$

$f=\dfrac{0.0791}{(7.073\times10^{-4})^{1/4}}$

$f=\dfrac{0.0791}{(16.30)}=0.00485$

Step 2: Head lost due to friction,

$hf=\dfrac{4fLV^2}{D\times2g}$

$hf=\dfrac{4fL\overline{V}^2}{D\times2g}$

$hf=\dfrac{4\times0.00485\times800\times(0.3183)^2}{0.40\times2\times9.81}=0.2m$

Step 3: Wall shear stress ($\tau_0$)

$\tau_0=\dfrac{f\rho V^2}{2}$

$\tau_0=\dfrac{f\rho\overline{V}^2}{2}$

$\tau_0=\dfrac{0.00485\times1000\times(0.3184)^2}{2}$

$\tau_0=0.245Nm^2$

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

$\dfrac {u_{max}}{u_*}=5.75\log_{10}\dfrac{u_*R}v+5.55$

[put $u=u_{max}$ at y=R]

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

$u_*=\sqrt{\dfrac{0.245}{1000}}=0.0156m/s$

Substituting the value of $u_*$, R and v in the above equation,

$\dfrac {u_{max}}{0.0156}=5.75\log_{10}\dfrac{0.0156\times0.20}{0.018\times10^{-4}}+5.55$

$\dfrac {u_{max}}{0.0156}=24.173$

$\therefore u_{max}=0.0156\times24.173=0.377m/s$

Step 5: Thickness of laminar sublayer ($\delta'$)

$\delta'=\dfrac{11.6\times v}{u_*}$

$\delta'=\dfrac{11.6\times 0.018\times10^{-4}}{0.0156}=0.001338m$

$\delta'=1.338mm$

Q2) A smooth pipe line of 100 mm diameter carries $2.27m^3$ per minute of water at $20^\circ C$ 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, $R=\dfrac{0.1}2=0.05m$

Discharge, $Q=2.27m^3/min=\dfrac{2.27}{60}=0.0378m^3/s$

Kinematic viscosity, v=0.0098stokes

$v=0.0098\times10^{-4}m^2/s$

Therefore Average velocity,

$\overline{U}=\dfrac QA=\dfrac{0.0378}{\dfrac\pi4(0.1)^2}=4.817m/s$

Therefore Reynolds number is given by,

$R_e=\dfrac{\overline{U}\times D}v$

$R_e=\dfrac{4.817\times 0.1}{0.0098\times10^{-4}}=4.9154\times10^5$

The flow is turbulent and $R_e$ is more than $10 ^ 5$.

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

$\dfrac1{\sqrt{4f}}=2.0\log_{10}(R_e\sqrt{4f})-0.8$

$\dfrac1{\sqrt{4f}}=2.0\log_{10}(4.9154\times10^5\sqrt{4f})-0.8$

$\dfrac1{\sqrt{4f}}=2.0[\log_{10}(4.9154\times10^5)+\log_{10}\sqrt{4f}]-0.8$

$\dfrac1{\sqrt{4f}}=2.0[5.6915+\log_{10}\sqrt{4f}]-0.8$

$\dfrac1{\sqrt{4f}}=2\times5.6915+2\log_{10}\sqrt{4f}-0.8$

$\dfrac1{\sqrt{4f}}=11.3830+\log_{10}(\sqrt{4f})^2-0.8$

$\dfrac1{\sqrt{4f}}=11.383+\log_{10}(4f)-0.8$

$\dfrac1{\sqrt{4f}}-\log_{10}(4f)=11.383-0.8=10.583……………….(1)$

Step 1: Friction factor

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

substitute the value in equation (1),

$\dfrac1{\sqrt{f*}}-\log_{10}(f*)=10.583………………(2)$

The above equation is solved by hit and trial method,

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

LHS=$\dfrac1{\sqrt{0.1}}-\log_{10}(0.1)=3.16-(-1.0)=4.16$

Let f*=0.01, then LHS becomes,

LHS=$\dfrac1{\sqrt{0.01}}-\log_{10}(0.01)=10-(-2)=12$

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=$\dfrac1{\sqrt{0.013}}-\log_{10}(0.013)=8.77-(-1.886)=10.656$

which is approximately (10.583).

Hence the value of (f*=0.013).

Therefore friction factor, f*=0.013.

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

We know, f*=4f

$\therefore f=\dfrac{f*}4$

$\therefore f=\dfrac{0.013}4=0.00325$

Now,

$u_*=\overline{U}\sqrt{\dfrac{f}2}$

$u_*=4.817\times\sqrt{\dfrac{0.00325}2}$

$u_*=4.817\times0.04303=0.194$

For smooth,

$\dfrac {u_{max}}{u_*}=5.75\log_{10}\dfrac{u_*y}v+5.55$

The velocity will be maximum at the centre of pipe.

Where, y=R=0.05,

$\dfrac {u_{max}}{u_*}=5.75\log_{10}\dfrac{u_*R}v+5.55$

$\dfrac {u_{max}}{0.194}=5.75\log_{10}\dfrac{0.194\times0.05}{0.0098\times10^{-4}}+5.55$

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

Step 3: Shear stress at boundary ($τ_0$)

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

$\therefore \tau_0=\rho u2_*$

$\therefore \tau_0=1000\times(0.194)^2=37.63 \ N/m^2$