Question

8) A horizontal pipe of 50 cm2 cross-sectional area carries a fluid of unknown density at a speed of 3.0 m/s. This pipe feeds a smaller pipe of cross section 25 cm2 . If the pressure difference between the two sections of pipe is 9000 Pa, find the density of the fluid.

Answer 1

Answer:

fluid density, $\rho = 666.67 kg/m^{3}$

Given:

Area of cross section of larger pipe, $A_{l} = 50 cm^{2}$

Area of cross section of smaller pipe, $A_{s} = 25 cm^{2}$

Velocity of fluid, $v_{i}$ = 3.0 m/s[/tex]

Pressure difference, p = 9000 Pa

Solution:

Using Bernoulli's equation:

$\frac{1}{2}\rho v_{i}^{2} + \rho g h_{a} + P_{i} = \frac{1}{2}\rho v_{o}^{2} + \rho g h_{b} + P_{o}$                (1)

where

$v_{i}$ = inlet velocity

$v_{o}$ = velocity at outlet

$h_{a}$ and $h_{b}$ = vertical heights

$\rho$ = density of the fluid

$P_{i}$ and $P_{o}$ = Pressure

Since, the pipe is horizontal, thus:

$h_{a}$ = $h_{b}$

Thus eqn (1) reduces to :

$\frac{1}{2}\rho v_{i}^{2} + P_{i} = \frac{1}{2}\rho v_{o}^{2} + P_{o}$

Now,

$P_{i} - P_{o} = \frac{1}{2}\rho v_{o}^{2} - \frac{1}{2}\rho v_{i}^{2}$

$9000 = \frac{1}{2}\rho (v_{o}^{2} - v_{i}^{2})$            (2)

Also, from the eqn of continuity:

$A_{l}v_{i} = A_{s}v_{o}$

$v_{o} = \frac{A_{l}}{A_{s}}v_{i}$

$v_{o} = \frac{50}{25}\times 3.0 = 6.0 m/s$

Now, using the value of   $v_{o} = 6.0 m/s$ in eqn (2) for fluid density:

$9000 = \frac{1}{2}\rho (6.0^{2} - 3.0^{2})$

$\rho = 666.67 kg/m^{3}$