Explanation:
First, calculate v = 500 kg/min x 1 min/60 s x 1 m3/780 kg x (1/.025x2.54)2 x 1/3.14 =0.84 m/s. Re = 780 x 0.84 x 0.05 x 2.54/ 0.0022 = 38,000. Now refer to a Moody plot, such as the one on the Roymech resource on "Fluid Flow in Pipes" in Unit 4, to obtain f = 0.034 for that plot. We can then calculate directly the head of that fluid or the pressure from the definition of f. In this case, we just take ΔP = (100m/(0.05x2.54m)) x (0.84 m/s)2/2 x 0.034 x 780 kg /m3= 7.4 kPa. Turbulent flow does not necessarily imply a large pressure drop.
