# Fluid mechanics problems?

Water flows through a pipeline 50m long at a velocity of 2 m/s. What increase in preassure difference between the inlet and outlet is required to accelerate the water in the pipe at the rate 0.02 m/s2?

Pressure difference in a pipe is directly proportional to the square of the velocity.

delta p ~ v^2 .

You specify a 50 m pipe with acceleration of 0.02 m/s^2 with initial velocity of 2 m/s. After 50 m, the velocity increase is 1 m/s which is a 50% increase so that the output velocity is 1.5x the input velocity. So as a first approximation, the pressure is 1.5^2 or 2.25x, more than double. The actual is more complex due to flow friction and other factors all of which tends to require an increase in the pressure. The following is a link to an engineering approach, including a software package.

http://www.coolit.co.za/pipeflow/tech01.

Traditional municipal water calculations use tables. The low velocity flow and the numerous angles make for a high resistance flow circuit. Tables are used because it requires the solution of a pretty complex system of differential equations.
I made the assumption reading the problem that the velocity was constant at 2m/s throughout the full 50m of pipe...a bit different than kyq's response

Using Bernoulli, we can look at the inlet and outlets

If H20 density = rho= 1kg/liter= 1000kg/cubic meter then taking just the KE/Velocity component of Bernoulli will give you

1/2(rho)v^2 = 0.5(1000)(2^2) = 2000 kg/(m*s^2) = 2000 Pa

Assuming your water is traveling a 2m/s throughout the full length, then a pressure gauge at the inlet and outlet would have 2000 Pa as a reading.(this assumes that the pipe is horizontal, i.e.. no elevation difference between inlet and outlet, and the water is in fact flowing at 2 m/s before the inlet and after the outlet ...ie. not open to atm))

We know that velocity is the derivative of position and that acceleration is the derivative of velocity ...so

A graph of the distance traveled (from inlet to outlet) in the pipe vs time would give us a straight line with a slope of 2 m/s or

d=2t + 0

0m = inlet

So we know that it takes 25 seconds at 2m/s for a molecule of water to get from inlet to outlet.

If acceleration is the derivitive of velocity then a = (delta v)/t

with an initial velocity of 2m/s and a desired increase in acceleration of 0.02 m/s^2 and keeping out travel time at 25sec

We have

0.02 = (v2-v1)/t = (v2 - 2)/t

and

v2 = 0.02t + 2

@ 25 seconds v2 = 2.5 m/s

Plug this back into

1/2(rho)v^2 using v2 and you get

0.5(1000)(2.5^2) = 3125 Pa

So we increase our Pressure from 2000 to 3125 or 156%

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