# 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%

More Questions and Answers:

More Questions and Answers:

How much does an elevator cost?
Who invented the wheel?
Why does a boat made of steel float in water when a flat piece of steel with the same mass sinks?
Need Diode bridge help?
Is there a Fiber Optic Modem or phone Protection Invention?
What are the problems and challenges faced in implementation of cost of quality program?
plz can you suggest any projects for final year be student?
list of engineering colleges of jaipur?
were dose the water go in a sauna?

**Answer:**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%

The answers post by the user, for information only, FunQA.com does not guarantee the right.

More Questions and Answers:

More Questions and Answers: