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water (10 oc) flows with a speed of 2 m/s through a horizontal run of pvc pipe. the length of the pipe is 50 m, and the pipe is schedule 40 with a nominal diameter of 2.5 inch. calculate: a) the pressure drop in kpa b) the head loss in meters c) the power in watts needed to overcome the head loss

Sagot :

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

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