Question 5.1: A pipeline connecting two tanks contains four standard elbow...

Cross-sectional area of pipe =\frac{\pi}{4}\left(25 \times 10^{-3}\right)^{2}=0.491 \times 10^{-3} m ^{2}

Fluid velocity, u =\frac{3500}{3600} \times \frac{1}{0.491 \times 10^{-3}} \times \frac{1}{998}=1.98 m / s

Reynolds number, R e=\left(998 \times 1.98 \times 25 \times 10^{-3}\right) / 0.99 \times 10^{-3} =49,900=5 \times 10^{4} (5.4)

Absolute roughness commercial steel pipe, Table 5.2 = 0.046 mm

Relative roughness =0.046 /\left(25 \times 10^{-3}\right)=0.0018, \text { round to } 0.002

From friction factor chart, Figure 5.7, f = 0.0032

Miscellaneous losses

Method 1, velocity heads

A velocity head =u^{2} / 2 g=1.98^{2} /(2 \times 9.8)=0.20 m \text { of liquid. }

Head loss =0.20 \times 14.7=2.94 m

as pressure =2.94 \times 998 \times 9.8=28,754 N / m ^{2}

Friction loss in pipe, \Delta P_{f}=8 \times 0.0032 \frac{(120)}{\left(25 \times 10^{-3}\right)} 998 \times \frac{1.98^{2}}{2} =240,388 N / m ^{2} (5.3)

Total pressure =28,754+240,388=269,142 N / m ^{2}=270 kN / m ^{2}

Method 2, equivalent pipe diameters

Extra length of pipe to allow for miscellaneous losses =735 \times 25 \times 10^{-3}=18.4 m

So, total length for \Delta P calculation = 120 + 18.4 = 138.4 m

\begin{aligned}\Delta P_{f} &=8 \times 0.0032 \frac{(138.4)}{\left(25 \times 10^{-3}\right)} 998 \times \frac{1.98^{2}}{2}=277,247 N / m ^{2} \\&=277 kN / m ^{2}\end{aligned} (5.3)

Note: the two methods will not give exactly the same result. The method using velocity heads is the more fundamentally correct approach, but the use of equivalent diameters is easier to apply and sufficiently accurate for use in design calculations.