Chagen – Poiseville Law

For this, consider or determine the velocity distribution across a section, the ratio of maximum velocity to average velocity, and the shear stress distribution and drop of pressure for a given length.

The flow of the flow would be laminar, if ($R_e$) is less than 2000.

Use $R_e=\dfrac{\rho VD}\mu$

Where $\rho$ is the density of fluid flowing through pipe

V = average velocity of fluid

D= Diameter of pipe

$\mu$ = viscosity of fluid

Consider a horizontal pipe of radius ‘R’. The fluid is flowing from left to right in the pipe as shown.

Consider fluid element of radius ‘r’, sliding in a cylindrical fluid element of radius (r+dr). Let the length of fluid element be ‘dx’. If ‘P’ is intensity of pressure on the face AB, then the intensity of pressure on face CD will be $\left(P+\dfrac{\delta P}{\delta x}dx\right)$.

Forces acting on fluid elements are:-

1) The pressure force, $P\times \mu r^2$ on face AB

2) The pressure force, $\left(P+\dfrac{\delta P}{\delta x}dx\right)\pi r^2$ on face CD

3) The shear force, $\tau \times2\pi rdx$ on the surface of fluid element. As there is no acceleration, hence the summation of all the forces in the direction of flow must be zero, i.e.,

$P\mu r^2-\left(P+\dfrac{\delta P}{\delta x}dx\right)\pi r^2- \tau\times2\pi rdx =0$

$-\left(\dfrac{\delta P}{\delta x} \right)dx\pi r^2-\tau \times2\pi rdx =0$

$-\left(\dfrac{\delta P}{\delta x} \right) r- 2\tau =0$

$\left[\tau=-\dfrac{\delta P}{\delta x}.\dfrac r2 \right]$

(a) Velocity distribution is given by equation,

$u=-\dfrac1{4\mu}.\dfrac{\delta P}{\delta x}.[R^2-r^2]$

(b) Ratio of maximum velocity to average velocity:-

$\dfrac{U_{max}}{\overline{u}}=\dfrac{-\dfrac1{4\mu}.\dfrac{\delta P}{\delta x}.R^2}{\dfrac1{8\mu}.-\dfrac{\delta P}{\delta x}.R^2}=2.0$

$\text {(c) Hagen Poiseville formula: - (drop of pressure)}$

$hf=\dfrac {32\mu\overline{u}L}{\rho gD^2}$

Numericals

Q1) An oil of viscosity $0.1NS/m^2$ and relative density 0.9 is flowing through a circular pipe of diameter 50 mm and of length 300 m. The rate of flow of fluid through the pipe is 3.5 litres per second. Find the pressure drop in a length of 300 m and also the shear stress at the pipe wall.

Solution:- Given:-

Viscosity, $\mu=0.1NS/m^2$

Relative density = 0.9

$\rho_o=0.9\times1000=900kg/m^3$ (As density of water =$1000kg/m^3$)

D=50mm=0.05m$ L=300m Q=3.5lit/s=0.0035m^3/s$

1) Pressure Drop ($P_1-P_2$)

$=\dfrac {32\mu\overline{u}L}{D^2}$

$\therefore \overline{u}=\dfrac QA$

$=\dfrac{0.0035}{\dfrac\pi4D^2}$

$=\dfrac{0.0035}{\dfrac\pi4(0.05)^2}=1.782m/s$

$R_e=\dfrac{\rho VD}{\mu}$ ($\rho=900kg/m^3, V=\overline{u}=1.782$)

$R_e=\dfrac{900\times1.782\times0.05}{0.1}=801.9$

As $R_e\lt2000$, the flow is laminar,

$\therefore P_1-P_2=\dfrac{32\times0.1\times1.782\times300}{(0.05)^2}$

$=684288N/m^2=684288\times10^{-4}N/cm^2$

$\therefore P_1-P-2=68.43N/cm^2$

2) Shear stress at the pipe wall ($τ_0$)

$\therefore \tau=-\dfrac{\delta P}{\delta x}.\dfrac r2$ (r=R)

$\therefore \tau=-\dfrac{\delta P}{\delta x}.\dfrac R2$

$-\dfrac{\delta P}{\delta x}=\dfrac{-(P_2-P_1)}{x_2-x_1}$

$-\dfrac{\delta P}{\delta x}=\dfrac{P_1-P_2}{x_2-x_1}$

$-\dfrac{\delta P}{\delta x}=\dfrac{P_1-P_2}{L}=\dfrac{684288}{300}=2280.96N/m^2$

$R=\dfrac D2=\dfrac{0.05}2$

R=0.025m

$\therefore \tau_0=2280.96\times\dfrac{0.025}2$

$\tau_0=28.512N/m^2$

Q2) A crude oil of viscosity 0.97 poise and relative density 0.9 is flowing through a horizontal circular pipe of diameter 100 mm and length 10m. Calculate the difference of pressure at the two ends of the pipe, if 100 kg of the oil is collected in a tank in 30 seconds.

Solution:- Given:-

$\mu=0.97$ poise

$\mu=\dfrac{0.97}{10}=0.097NS/m^2$

$\rho_o=0.9\times1000=900kg/m^3$ (As density of water =$1000kg/m^3$)

D=100mm=0.1m

L=10m

Mass, m=100kg

T=30seconds

(i) Pressure difference ($P_1-P_2$)

$=\dfrac {32\mu\overline{u}L}{D^2}$

$\therefore \overline{u}=\dfrac QA$

Mass of oil/sec = $\dfrac{100}{30}kg/s…………………(1)$

$=\rho_o\times Q$

$=900\times Q……………..(2)$

Equate (1) and (2), we get

$\dfrac{100}{30}=900\times Q$

$\therefore Q=\dfrac{100}{30}\times\dfrac1{900}$

$Q=0.0037m^3/s$

$\therefore \overline{u}=\dfrac QA$

$\overline{u}=\dfrac{0.0037}{\dfrac\pi4D^2}$

$\overline{u}=\dfrac{0.0037}{\dfrac\pi4(0.1)^2}$

$\overline{u}=0.471m/s$

Reynolds Number,

$R_e=\dfrac{\rho VD}{\mu}$ ($\rho=900kg/m^3, V=\overline{u}=0.471$)

$R_e=\dfrac{900\times0.471\times0.1}{0.097}=436.91$

As $R_e\lt2000$, the flow is laminar,

$\therefore P_1-P_2=\dfrac{32\times0.097\times0.471\times10}{(0.1)^2}$

$=1462.28NS/m^2=1462.28\times10^{-4}N/cm^2$

$\therefore P_1-P-2=0.1462N/cm^2$