Answer:
The friction factor [tex]f[/tex] is approximately 0.0179.
Explanation:
At first we need to know what flow regime water flow is found in. Reynolds number offers an appropriate dimensionless indicator for flow in pipes and whose formula is:
[tex]Re_{D} = \frac{\rho \cdot v\cdot D}{\mu}[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density of water, measured in kilograms per cubic meter.
[tex]\mu[/tex] - Dynamic viscosity, measured in kilograms per meter-second.
[tex]D[/tex] - Inner diameter of pipe, measured in meters.
[tex]v[/tex] - Flow average speed, measured in meters per second.
[tex]Re_{D}[/tex] - Reynolds number, dimensionless.
The average speed of water is determined by the following expression:
[tex]v = \frac{4 \cdot \dot V}{\pi \cdot D^{2}}[/tex] (Eq. 2)
Where:
[tex]\dot V[/tex] - Volume flow, measured in cubic meters per second.
[tex]D[/tex] - Inner diameter of pipe, measured in meters.
If we know that [tex]\dot V = 0.05\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.5\,m[/tex], the flow average speed is:
[tex]v = \frac{4\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.5\,m)^{2}}[/tex]
[tex]v\approx 0.255\,\frac{m}{s}[/tex]
The properties of water at given conditions ([tex]T = 10\,^{\circ}C[/tex]) are, respectively:
[tex]\rho = 999.7\,\frac{kg}{m^{3}}[/tex]
[tex]\mu = 1.307\times 10^{-3}\,\frac{kg}{m\cdot s}[/tex]
And the Reynolds Number is:
[tex]Re_{D} = \frac{\left(999.7\,\frac{kg}{m^{3}} \right)\cdot \left(0.255\,\frac{m}{s} \right)\cdot (0.5\,m)}{1.307\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re_{D} = 97522.380[/tex]
Which means that water is in turbulent flow. There are several empirical and semi-empirical expression to estimate friction factor, we decided to use the Haaland approximation due to its exactness and simplicity:
[tex]\frac{1}{\sqrt{f}} = -1.8\cdot \log_{10}\left[\frac{6.9}{Re}+\left(\frac{\epsilon}{3.7\cdot D}\right)^{1.11} \right][/tex] (Eq. 3)
Where:
[tex]f[/tex] - Friction factor, dimensionless.
[tex]\epsilon[/tex] - Smoothness factor, dimensionless.
If we know that [tex]\epsilon = 0[/tex] and [tex]Re_{D} = 97522.380[/tex], then we get that:
[tex]\frac{1}{\sqrt{f}}=-1.8\cdot \log_{10}\left[\frac{6.9}{97522.380} \right][/tex]
[tex]\frac{1}{\sqrt{f}} = 7.470[/tex]
[tex]f = \left(\frac{1}{7.470} \right)^{2}[/tex]
[tex]f = 0.0179[/tex]
The friction factor [tex]f[/tex] is approximately 0.0179.
