Water at 15°C (rho = 999.1 kg/m³ and μ = 1.138 × 10⁻³ kg/m·s) is flowing steadily in a 29-m-long and 6-cm-diameter horizontal pipe made of stainless steel at a rate of 10 L/s. Determine the pressure drop, the head loss, and the pumping power requirement to overcome this pressure drop. The roughness of stainless steel is 0.002 mm.

Explanation:

From the given information:

We can estimate the  cross-sectional area of the pipe by using the formula:

To find the average velocity of the water flowing through the pipe by using the formula;

However, the Reynolds number can be determined by using the equation

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:

where;

= frictional force = ???

= roughness of steel= 0.002 mm

Replacing our values into the above equation, we have:

By solving the above equation for f; we get

f = 0.026409

The pressure drop in flow;

= 79771.50524 Pa

= 79.772 kPa

The pressure drop = 79.772 kPa

The head loss in the pipe:

=

= 8.14 m

The power input

=  10 * 10^{-3} * 79771.50524

= 797.72 W