Jasmine W. answered • 06/20/21

Civil Engineer with Extensive Background in Applying Geometry Concepts

**GIVEN:**

L = 200 m ; D = 30 mm = 0.03 m

ɛ = 0.1 m = 0.0001 m ; Q = 1.5 L / s = 0.0015 m^{3}/s

**DETERMINE:** J [unit head loss, in m/m]

**ASSUMPTIONS:**

- The fluid is flowing under steady state conditions.
- The density of the fluid is uniform throughout the fluid's streamline.

**SOLUTION PROCEDURE:**

V = velocity = Q / A = Q / [0.25*πD^{2}] = (0.0015 m^{3}/s) / [0.25*π*(0.03 m)^{2}] = 2.122 m/s

To determine f, the friction factor, the relative roughness and Reynolds number must be known.

Relative roughness = ε / D = 0.0001 m / 0.03 m

Re = Reynolds number = (ρVD) / μ where ρ = density of fluid, V = velocity of fluid, D = diameter of pipe, μ = dynamic viscosity of fluid

At 20 °C, ρ_{Water} = 998.21 kg/m^{3}

[Source: https://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html]

At 20 °C, μ_{Water} = 1.00160 cP = 0.0010016 kg / (m•s)

[Source: https://www.engineeringtoolbox.com/water-dynamic-kinematic-viscosity-d_596.html]

Re = (ρVD) / μ = [998.21 kg/m^{3} • 2.122 m/s • 0.03 m] / [0.0010016 kg / (m•s)] = 6.34 x 10^{4}

By observation of a Moody diagram, f = 0.026

[Source: https://www.engineersedge.com/graphics/moodys-diagram.png]

Darcy-Weisbach equation: h_{f} = f • (L/D) • [v^{2}/(2g)], where L = length of pipe, 2 = acceleration due to gravity = 9.81 m/s^{2}

[Source: http://fluid.itcmp.pwr.wroc.pl/~znmp/dydaktyka/fundam_FM/Lecture11_12.pdf ; Equation (4)]

h_{f} = (0.026) • (200 m / 0.03 m) • [(2.122 m/s)^{2} / (2 • 9.81 m/s^{2})] = 39.78 m

J = h_{f} / L = 39.78 m / 200 m = 0.1989 m/m

*J = 0.2 m/m*