The pipe is schedule 40 and has a nominal diameter of 2.5 inches. It is 50 meters long. Then pressure drop is 29.1kpa,Head loss is 2.967m and power is 176.986 watt.
a)Pressure drop = 32μVL/D²
=32*1.307*10⁻³*2*50/0.0119882
=29102.6 pascal
=29.1 Kpa
b)Head loss=p1-p2/pg=29.1*10³/999.7*9.81
=2.967 m
c)power=(p1-P2)q
=29.1*10³*6.082*10⁻³
Power =176.986 watt //watt is the unit of power
The difference in total pressure between two points on a network used to transport fluids is known as a pressure drop. When frictional forces acting on a fluid as it flows through the tube, brought on by the resistance to flow, occur, a pressure drop happens.
Learn more about pressure drop here:
https://brainly.com/question/15851784
#SPJ4
