Question

A 6-cm diameter horizontal pipe expands gradually to a 9-cm
diameter pipe (Fig below). The walls of the expansion section are
angled 10° from the axis. The average velocity and pressure of
water before the expansion section are 7 m/s and 150 kPa,
respectively. Determine the head loss in the expansion section an
the pressure in the larger-diameter pipe.
Properties: density of water = 1000 kg/m3
, total angle inclination
20°, diameter ratio d/D = 6/9, KL = 0.133.

Answer 1

Explanations & Calculations

• Assume the fluid is incompressible.

• Therefore, the flow rate is constant hence the velocity of the fluid at the enlargement can be calculated to be,

$\qquad\qquad \begin{aligned} \small v_2 &= \small \frac{A_1v_1}{A_2}= \frac{d_1^2v_1}{d_2^2} = \frac{6^2\times 7ms^{-1}}{9^2}\\ \small &= \small \bold{3.11ms^{-1}} \end{aligned}$

• With the head loss due to similar variations in pipe lines (as discussed here) the Bernoulli equation (energy equation) can be written as,

$\qquad\qquad \begin{aligned} \small h_1+\frac{v_1^2}{2g}+\frac{P_1}{\rho g} &= \small h_2+\frac{v_2^2}{2g}+\frac{P_2}{\rho g} +h_{\text{loss}}\\ \small \frac{v_1^2}{2g}+\frac{P_1}{\rho g} &= \small \frac{v_2^2}{2g}+\frac{P_2}{\rho g} +h_{\text{loss}}\cdots(1)\\ \end{aligned}$

• Head loss could occur due to several reasons & that due to these kinds of variations is calculated as follows.

$\qquad\qquad \begin{aligned} \small h_{\text{loss}}&= \small K_L\times\frac{v_1^2-v_2^2}{2g}\\ &= \small 0.133\times \frac{(7ms^{-1})^2-(3.11ms^{-1})^2}{2\times9.8ms^{-2}}\\ &= \small \bold{0.2669m} \end{aligned}$

• Now using equation (1) the pressure can be calculated to be,

$\qquad\qquad \begin{aligned} \small \frac{7^2}{2(9.8)}+\frac{150kPa}{1000\times9.8}&= \small \frac{3.11^2}{2(9.8)}+\frac{P_2}{1000\times9.8}+0.2669\\ \small P_2 &= \small \bold{17198.33kPa} \end{aligned}$