Answer:
Explanation:
From the given information:
We can estimate the cross-sectional area of the pipe by using the formula:
[tex]A_c = \dfrac{\pi}{4} D^2[/tex]
[tex]A_c = \dfrac{\pi}{4} (6 \times 10^{-2} \ m)^2[/tex]
[tex]A_c = 0.0028274 \ m^2[/tex]
To find the average velocity of the water flowing through the pipe by using the formula;
[tex]V = \dfrac{\mathtt{V}}{A_c}[/tex]
[tex]V = \dfrac{10 \times 10^{-3} \ m^3/s}{0.0028274}[/tex]
[tex]V = 3.537 \ m/s[/tex]
However, the Reynolds number can be determined by using the equation
[tex]Re= \dfrac{\rho \times V \times D}{\mu}[/tex]
[tex]Re= \dfrac{(999.1 \ kg/m^3) \times (3.537 \ m/s) \times (6 \times 10^{-3} \ m)}{1.138 \times 10^{-3} \ kg/m.s}[/tex]
Re = 18631.7225
From above, Re> 4000, thus, the flow will be turbulent.
By using Colebrook's equation in a turbulent flow; we have the following expression for Colebrook's equation:
[tex]\dfrac{1}{\sqrt{f}}= -2.0 \ log \bigg ( \dfrac{\epsilon/D_h}{3.7}+ \dfrac{2.51}{Re \sqrt{f}} \bigg )[/tex]
where;
[tex]f[/tex] = frictional force = ???
[tex]\epsilon[/tex] = roughness of steel= 0.002 mm
Replacing our values into the above equation, we have:
[tex]\dfrac{1}{\sqrt{f}}= -2.0 \ log \bigg ( \dfrac{ 2\times 10^{-6 \ m }/6 \times 10^{2 }\ m}{3.7}+ \dfrac{2.51}{18631.7225 \sqrt{f}} \bigg )[/tex]
By solving the above equation for f; we get
f = 0.026409
The pressure drop in flow;
[tex]\Delta P_L = f \dfrac{L}{D} \dfrac{\rho V^2}{2}[/tex]
[tex]=0.026409(\dfrac{29}{0.06})(\dfrac{999.1*3.537^2}{2})[/tex]
= 79771.50524 Pa
= 79.772 kPa
The pressure drop = 79.772 kPa
The head loss in the pipe:
[tex]h_L = \dfrac{\Delta P_L}{\rho g}[/tex]
= [tex]\dfrac{79771.50524 }{999.1 * 9.81}[/tex]
= 8.14 m
The power input
[tex]W_{pump} = V\Delta P_L[/tex]
= 10 * 10^{-3} * 79771.50524
= 797.72 W
