At one poinn a pipeline the water's spced

Munaza Awan

Munaza Awan

Answered question

2022-07-31

At one poinn a pipeline the water's spced is 3.00 ms and the gange pressure is

Sx l0 Pa. Find the gauge peessure al a seoond point in the line, 11.0 m bower thas

the first, if the pipe dianeter at the seccond point is twice that at the first.

Answer & Explanation

nick1337

nick1337

Expert2023-06-17Added 777 answers

To solve this problem, we can use Bernoulli's equation, which relates the pressure, speed, and height of a fluid in a horizontal flow.
The equation can be expressed as:
P1+12ρv12+ρgh1=P2+12ρv22+ρgh2
where:
- P1 and P2 are the gauge pressures at the first and second points, respectively.
- v1 and v2 are the speeds of water at the first and second points, respectively.
- ρ is the density of water (assumed constant).
- g is the acceleration due to gravity.
- h1 and h2 are the heights of the first and second points, respectively.
Given that the second point is 11.0 m lower than the first point, we have h2=h111.0m.
We are given the speed at the first point, v1=3.00m/s, and the gauge pressure at the first point, P1=10Pa.
To find the gauge pressure at the second point, P2, we need to solve Bernoulli's equation.
The equation and solution can be written as follows:
Bernoulli's equation:
P1+12ρv12+ρgh1=P2+12ρv22+ρgh2
Given values:
v1=3.00m/s, P1=10Pa, h2=h111.0m
Solving this equation will give us the gauge pressure at the second point, P2.

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