Question

Air at 105 kPa and 37°C flows upward through a 6-cm-diameter inclined duct at a rate of 65 L/s. The duct diameter is then reduced to 4 cm through a reducer. The pressure change across the reducer is measured by a water manometer. The elevation difference between the two points on the pipe where the two arms of the manometer are attached is 0.20 m. Determine the differential height between the fluid levels of the two arms of the manometer.

Answer 1

Answer:

$h=2.5456\ m$

Explanation:

Given:

• pressure of air in the duct, $P_1=105\ kPa$

• temperature of air in the duct, $T_1=37^{\circ}C$

• diameter of the duct, $d_d=0.06\ m$

• flow rate through the duct, $\dot V_1=65\ L.s^{-1}=0.065\ m^3.s^{-1}$

• diameter after reduction, $d_r=0.04\ m$

• difference in the datum of the two points, $z_2=0.2\ m$

Using continuity equation:

$\dot V_1=\dot V_2$

$\therefore \dot V_2=0.065\ m^3.s^{-1}$

As we know,

$\dot V_2=A_2.v_2$

$A_2=\pi\times \frac{d_r^2}{4}$

$A_2=\pi\times \frac{0.04^2}{4}$

$A_2=\pi\times 10^{-4}\ m^3.s^{-1}$

Hence:

$v_2=\frac{\dot V}{A_2}$

$v_2=\frac{0.065}{\pi\times 10^{-4}}$

$v_2=206.9\ m.s^{-1}$

And:

$v_1=\frac{\dot V_1}{A_1}$

$v_1=0.065\div \pi \times \frac{0.06^2}{4}$

$v_1=22.99\ m.s^{-1}$

According to Bernoulli's principle we have for an incomprehensible flow:

$\frac{P}{\rho.g} +\frac{v^2}{2g} +z=constt.$

Now we're concerned about the density of air:

$P.V=m.R.T$

$\rho=\frac{P}{R.T}$

$\rho=\frac{105000}{0.167\times 310}$

$\rho=1.18\ kg.m^{-3}$

Apply Bernoulli's principle assuming it an in-compressible flow:

$\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} +z_1=\frac{P_2}{\rho.g} +\frac{v_2^2}{2g} +z_2$

$\frac{105000}{1.18\times 9.8} +\frac{22.99^2}{2\times 9.8} +0=\frac{P_2}{1.18\times 9.8} +\frac{206.9^2}{2\times 9.8} +0.2$

$P_2=80053.04\ Pa$

Now the difference in pressure:

$\Delta P=105000-80053.04$

$\Delta P=24946.96\ Pa$

∴The corresponding difference in the water levels of manometer:

$\Delta P=\rho.g.h$

$24946.96=1000\times9.8\times h$

$h=2.5456\ m$