Bernoulli's Equation, Find gauge pressure at 2nd point

AI Thread Summary
The discussion revolves around applying Bernoulli's equation to determine the gauge pressure at a second point in a pipeline. Given the initial conditions of velocity and gauge pressure, the velocity at the second point is calculated to be 2 m/s due to a reduced cross-section. Bernoulli's principle states that the total energy per unit volume remains constant, leading to the equation P1 + ρgh1 + 1/2 ρv1^2 = P2 + ρgh2 + 1/2 ρv2^2. After substituting the known values, the gauge pressure at the second point is found to be 4.95 x 10^5 N/m2. This calculation demonstrates the application of fluid dynamics principles in real-world scenarios.
Ishraq Begum
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Homework Statement


At a certain point in a pipeline, the velocity is 1 m/s and the gauge pressure is 3 x 105 N/m2. Find the gauge pressure at a second point in the line 20 m lower than the first if the cross-section at the second point is one half that at the first. The liquid in the pipe is water.

Homework Equations


Bernoulli's equation and Equation of continuity.

The Attempt at a Solution


I don't really know where to start
 
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Ishraq Begum said:

Homework Statement


At a certain point in a pipeline, the velocity is 1 m/s and the gauge pressure is 3 x 105 N/m2. Find the gauge pressure at a second point in the line 20 m lower than the first if the cross-section at the second point is one half that at the first. The liquid in the pipe is water.

Homework Equations


Bernoulli's equation and Equation of continuity.

The Attempt at a Solution


I don't really know where to start

Try to start writing down the equations with symbols. What will be the velocity of the water at the second point?
 
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stockzahn said:
Try to start writing down the equations with symbols. What will be the velocity of the water at the second point?
The velocity of the water at the second point will be double of that at the first point, i.e., 2 m/s
How do you find the gauge pressure??
 
Ishraq Begum said:
The velocity of the water at the second point will be double of that at the first point, i.e., 2 m/s
How do you find the gauge pressure??

What does Bernoulli's law state (in case of a water stream without frictional pressure losses)?
 
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stockzahn said:
What does Bernoulli's law state (in case of a water stream without frictional pressure losses)?
It states that total energy per unit volume remains constant throughout the flow.
Mathematically,
P + ρgh + 1/2 ρv2 = constant
P = pressure, ρ = density of fluid, h = height of fluid, v = velocity with fluid is flowing
 
Ishraq Begum said:
It states that total energy per unit volume remains constant throughout the flow.
Mathematically,
P + ρgh + 1/2 ρv2 = constant
P = pressure, ρ = density of fluid, h = height of fluid, v = velocity with fluid is flowing

Well, that means, that the sum of static, dynamic and hydrostatic pressure at both points are equal. How could you describe this mathematically (e.g. by using subscripts ##_1## and ##_2## for the two different points)?
 
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stockzahn said:
Well, that means, that the sum of static, dynamic and hydrostatic pressure at both points are equal. How could you describe this mathematically (e.g. by using subscripts ##_1## and ##_2## for the two different points)?
or P1 + ρgh1 + 1/2 ρv12 = P2 + ρgh2 + 1/2 ρv22

Here, P1 = 3 x 105
ρgh1 = 9.8 x 103 x h
1/2 ρv12 = 1/2 x 13 x 1 = 500
P2 = ? (to be found)
ρgh2 = 9.8 x 103 x (h - 20)
1/2 ρv22 = 2 x 102

⇒ (3x105) + 9.8 x 103 (h - h +20) + (500 - 2 x 103) = P2
⇒ (3 x 105) + (1.96 x 105) + (-1500) = P2
Therefore, P2 = 4.95 x 105 N/m2

That's the answer! Thanks a lot!
 
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