# At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure

At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure is $5.00×{10}^{4}Pa$. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.
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Tuthornt

Step 1
The diameters of the two pipe cross-sections are ${d}_{1}$ and ${d}_{2}=2{d}_{1}$. The area of the first is ${A}_{1}=\pi {d}_{1}^{2}$ and of the the second ${A}_{2}=\pi {d}_{2}^{2}=\pi 4{d}_{1}^{2}=4A1$
Step 2
Let's first determine the relation between the velocities based on the continuity equation ${v}_{1}{A}_{1}={v}_{2}{A}_{2}$ and the fact that ${A}_{2}=4{A}_{1}$ from which we can write
${v}_{2}=\frac{{v}_{1}{A}_{1}}{{A}_{2}}$
${v}_{2}=\frac{{v}_{1}}{4}$
Step 3
Since the atmospheric pressures are the same we can write Bernoulli's equation as
${p}_{1}+\rho g{y}_{1}+\frac{1}{2}\rho {v}_{1}^{2}={p}_{2}+\rho g{y}_{2}+\frac{1}{2}\rho {v}_{2}^{2}$ where ${p}_{1}$ and ${p}_{2}$ are gauge pressures. We can now insert ${v}_{2}=\frac{{v}_{1}}{4}$ to get
${p}_{2}={p}_{1}+\rho g\left({y}_{1}-{y}_{2}\right)+\frac{1}{2}\rho \left({v}_{1}^{2}-\frac{{v}_{1}^{2}}{16}\right)$
${p}_{2}={p}_{1}+\rho g\left({y}_{1}-{y}_{2}\right)+\frac{15}{32}\rho {v}_{1}^{2}$
Step 4
At this point we can insert all the given values to get the following
${p}_{2}=5×{10}^{4}+{10}^{3}×9.81×11+{10}^{3}×\frac{15}{32}×{3}^{2}$
${p}_{2}=162kPa$