Answer: a) f = 0.0203, b) Re = 338999.254, c) [tex]\epsilon_{r} \approx 0.0008[/tex]
Explanation:
a) The Friction Factor is isolated in this formula:
[tex]h_{L} = \frac{L}{D} \cdot f \cdot \frac{v^{2}} {2\cdot g}[/tex]
[tex]f = \frac{2 \cdot g \cdot D \cdot h_{L}}{v^{2} \cdot L}[/tex]
The flowrate unit is converted to cubic feet per second:
[tex]\dot V = 1200\frac{gal}{min} \cdot (\frac{0.134 ft^{3}}{1 gal})\cdot(\frac{1 min}{60 sec}) \\ \dot V= 2.68 \frac{ft^{3}}{s}[/tex]
The diameter unit is converted to feet:
[tex]D = 10 in \cdot (\frac{1 ft}{12 in} )\\D = 0.833 ft[/tex]
The velocity is found by using this formula:
[tex]v = \frac{4 \cdot \dot V}{\pi \cdot D^{2}}[/tex]
[tex]v = \frac{4 \cdot (2.68 \frac{ft^{3}}{s} )}{\pi \cdot (0.833 ft)^{2}} \\v \approx 4.918 \frac{ft}{s}[/tex]
The friction factor is found after replacing the known variables:
[tex]f = \frac{2 \cdot (32.714 \frac{ft}{s^{2}}) \cdot (0.833 ft) \cdot (11.25 ft) )}{(4.918 ft)^{2} \cdot (1250 ft)} \\f = 0.0203[/tex]
b) The Reynolds number is given by this formula:
[tex]Re_{D} = \frac{\rho_{water} \cdot v \cdot D}{\mu_{water}}[/tex]
The density and dynamic viscosity of water at 60 °F are [tex]62.36 \frac{lbm}{ft^{3}}[/tex] and [tex]7.536 \times 10^{-4} \frac{lbm}{ft \cdot s}[/tex], respectively. The Reynolds number is:
[tex]Re = 338999.254[/tex]
Since [tex]Re > 4000[/tex], the flow has a turbulent regime.
c) A relatively approach to estimate relative roughness of the steel pipe is using the Moody diagram in terms of friction factor and Reynolds number. The relative roughness of the steel pipe is approximately 0.0008.
