Question

At a certain position in a piping system, a solution with a specific gravity of 1.50 passes through an 8-cm pipe with an average speed of 1.2 m/s. At some position downstream, the elevation of the pipe has increased 15.0 m and the pipe size has decreased to 5 cm. The temperature of the fluid is assumed constant at 30 C, and a heat loss of 25 Nm/kg occurs. Determine the change in pressure, in bars and megapascals.

Answer 1

Answer:

$\Delta p=-2.60bar\\ \\\Delta p=-0.260MPa$

Explanation:

Hello,

In this case, by using the Bernoulli equation:

$p_1+\frac{1}{2}\rho v^2_1+\rho gh_1= p_2+\frac{1}{2}\rho v^2_2+\rho gh_2+\rho h_L$

Whereas $h_L$ accounts for the heat loss, so we can compute the change of pressure by:

$p_1 +\frac{1}{2}\rho v^2_1+\rho gh_1= p_2+\frac{1}{2}\rho v^2_2+\rho gh_2+\rho h_L\\\\p_2-p_1=\frac{1}{2}\rho v^2_1+\rho gh_1-\frac{1}{2}\rho v^2_2-\rho gh_2-\rho h_L\\\\\Delta p=\rho *[\frac{1}{2}(v_1^2-v_2^2)+g(h_1-h_2)-h_L ]$

Thus, we must first compute the velocity inside the 5-cm section by using the continuity equation:

$v_2=v_1\frac{A_1}{A_2}=v_1*(\frac{\pi d_1^2/4}{\pi d_2^2/4} ) \\\\v_2=1.2m/s*\frac{8cm}{5cm}=1.92m/s$

It means that the change in pressure in Pa turns out (density is 1500 kg/m³ given the specific gravity of the fluid):

$\Delta p=1500\frac{kg}{m^3} *[\frac{1}{2}((1.2\frac{m}{s})^2-(1.92\frac{m}{s})^2)+9.8m/s^2(0m-15.0m)-25\frac{m^2}{s^2} ]\\\\\Delta p=1500\frac{kg}{m^3}*(-1.1232\frac{m^2}{s^2} -147\frac{m^2}{s^2} -25\frac{m^2}{s^2} )\\\\\Delta p=-259684.8Pa$

Therefore, the change in pressure in bar and MPa turns out:

$\Delta p=-259684.8Pa*\frac{1bar}{1x10^5Pa}=-2.60bar\\ \\\Delta p=-259684.8Pa*\frac{1MPa}{1x10^6Pa}=-0.260MPa$

Such negative sign means that the pressure at the 5-cm section is lower than the pressure at the 8-cm section.

Best regards.