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.

(10 Marks) May-2018

Subject Fluid Mechanics 2

Topic Laminar Flow

Difficulty Medium

Data : $\mu = 0.97 poise = \frac{0.97}{10} = 0.097 Ns/m^2$

Relative density = 0.9 (of 0:1)

$\therefore Density \rho_o = 0.9 \times 1000 = 900 kg/m^3$ (of oil)

Dia. of pipe D = 100 mm = 0.1 m

Length L = 10 m

Mass of Oil = m = 100 kg.

and time t = 30 seconds.

$P_1 - P_2 = ? $

The difference of pressure for viscous flow is

$p_1 - p_2 = \frac{32mŪL}{D^2}$ Avg. velocity $Ū = \frac{Q}{Area}$

Now, mass of oil per second = $\frac{100}{30}$ kg/s

= $\rho \times Q = 900 \times Q$

$\therefore \frac{100}{30} \times Q$

$\therefore Q = 0.0037 m^3/s$

Now, $Ū = \frac{Q}{A} = \frac{0.0037}{\pi \times D^2} = \frac{0.0037}{\pi \times 0.1^2} = 0.471 m/s$

For Laminar/viscous flow, Re < 2000

Now, $Re = \frac{\rho VD}{\mu}$

$\rho = \rho_o = 900, v = Ū = 0.471 m/s$

D = 0.1m, $\mu$ = 0.097

$Re = 900 \times \frac{0.471 \times 0.1}{0.097} = 436.91$

$\because$ Value of Re < 2000, hence flow is laminar.

$\therefore p_1 - p_2 = \frac{32 \mu Ū L}{D^2} = \frac{32 \times 0.097 \times 0.471 \times 10}{(0.1)^2}$

$p_1 - p_2 = 1462.28 N/m^2$

= $1462.28 \times 10^-4 N/cm^2$

= $p_1 - p_2 = 0.1462 N/cm^2$