Answer:
The friction factor of a 60-mm-diameter galvanized iron pipe is 0.045.
Explanation:
Losses due to friction flowing through iron pipe is determined by the Darcy-Weisbach model:
[tex]\Delta p = \rho \cdot f \cdot \frac{L}{D}\cdot \frac{v^{2}}{2}[/tex]
Where:
[tex]\Delta p[/tex] - Pressure drop, measured in pascals.
[tex]\rho[/tex] - Density of water, measured in kilograms per cubic meter.
[tex]f[/tex] - Friction factor, dimensionless.
[tex]L[/tex] - Length of the pipe, measured in meters.
[tex]D[/tex] - Diameter of the pipe, measured in meters.
[tex]v[/tex] - Velocity of the flow, measured in meters per second.
The friction factor is now cleared:
[tex]f = \frac{2 \cdot \Delta p \cdot D}{\rho \cdot L \cdot v^{2}}[/tex]
The flow velocity is equal to the volume flow divided by the cross area of the iron pipe. That is:
[tex]v = \frac{4 \cdot \dot V}{\pi \cdot D^{2}}[/tex]
Given that [tex]\dot V = 0.017\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.06\,m[/tex], the velocity of the flow is:
[tex]v = \frac{4\cdot \left(0.017\,\frac{m^{3}}{s} \right)}{\pi \cdot (0.06\,m)^{2}}[/tex]
[tex]v \approx 6.013\,\frac{m}{s}[/tex]
Now, if [tex]\Delta p = 135000\,Pa[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex] and [tex]L = 10\,m[/tex]. The friction factor is:
[tex]f = \frac{2\cdot (135000\,Pa)\cdot (0.06\,m)}{\left(1000\,\frac{kg}{m^{3}} \right)\cdot (10\,m)\cdot \left(6.013\,\frac{m}{s} \right)^{2}}[/tex]
[tex]f = 0.045[/tex]
The friction factor of a 60-mm-diameter galvanized iron pipe is 0.045.
