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Consider a pipe in which flow is taking place.A textbook says,when there is a valve or heat exchanger or resistance to flow between two points in this pipe,pressure decreases from one point to another.It occurts to me that when there is resistance to flow, velocity decreases and hence kinetic head decreases which implies pressure head increases (according to Bernoulli's equation).Could someone clarify?

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  • $\begingroup$ velocity doesn drop. same mass gos in as gos out. mass flow is velocitiy x area x density. $\endgroup$ – mart Jun 20 '18 at 9:15
  • $\begingroup$ If you restrict a pipe you think the velocity decreases? $\endgroup$ – paparazzo Jun 22 '18 at 0:12
  • $\begingroup$ I got confused with the term 'resistance to the flow.'Got it now@paparazzo $\endgroup$ – Shwetha Jun 22 '18 at 3:25
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The Bernoulli Equation can be considered to be a statement of the conservation of energy, i.e. the internal energy of the fluid remains constant. This can be useful for certain calculations e.g. the Venturi effect

In a long pipe run, however, the internal energy is not constant, as some is lost. Any frictional effects (due to surface roughness of the pipe, going around bends etc. as noted by Solar Mike) will cause the pipework to warm up slightly, and consequently that energy is lost to the environment.

If the diameter of the pipe is equal at the two points that you're considering, and the internal energy has been reduced, then the fluid must be at a reduced pressure, since the constant mass-flow rate fixes it to being at the same velocity.

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The pressure losses in a pipe are due to

The velocity squared

The length, material and surface roughness ie pipe friction (Thanks Fred)

And any losses due to bends, restrictions etc

This has a good explanation of the phenomena : Losses in pipes - Queens University

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    $\begingroup$ You could also add pipe friction (roughness) to that list $\endgroup$ – Fred Jun 20 '18 at 4:40

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