Munaza Awan

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.

nick1337

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:
${P}_{1}+\frac{1}{2}\rho {v}_{1}^{2}+\rho g{h}_{1}={P}_{2}+\frac{1}{2}\rho {v}_{2}^{2}+\rho g{h}_{2}$
where:
- ${P}_{1}$ and ${P}_{2}$ are the gauge pressures at the first and second points, respectively.
- ${v}_{1}$ and ${v}_{2}$ are the speeds of water at the first and second points, respectively.
- $\rho$ is the density of water (assumed constant).
- $g$ is the acceleration due to gravity.
- ${h}_{1}$ and ${h}_{2}$ 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 ${h}_{2}={h}_{1}-11.0\phantom{\rule{0.167em}{0ex}}\text{m}$.
We are given the speed at the first point, ${v}_{1}=3.00\phantom{\rule{0.167em}{0ex}}\text{m/s}$, and the gauge pressure at the first point, ${P}_{1}=10\phantom{\rule{0.167em}{0ex}}\text{Pa}$.
To find the gauge pressure at the second point, ${P}_{2}$, we need to solve Bernoulli's equation.
The equation and solution can be written as follows:
Bernoulli's equation:
${P}_{1}+\frac{1}{2}\rho {v}_{1}^{2}+\rho g{h}_{1}={P}_{2}+\frac{1}{2}\rho {v}_{2}^{2}+\rho g{h}_{2}$
Given values:
${v}_{1}=3.00\phantom{\rule{0.167em}{0ex}}\text{m/s}$, ${P}_{1}=10\phantom{\rule{0.167em}{0ex}}\text{Pa}$, ${h}_{2}={h}_{1}-11.0\phantom{\rule{0.167em}{0ex}}\text{m}$
Solving this equation will give us the gauge pressure at the second point, ${P}_{2}$.

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