Step 1

The diameters of the two pipe cross-sections are \(\displaystyle{d}_{{{1}}}\) and \(\displaystyle{d}_{{{2}}}={2}{d}_{{{1}}}\). The area of the first is \(\displaystyle{A}_{{{1}}}=\pi{{d}_{{{1}}}^{{{2}}}}\) and of the the second \(\displaystyle{A}_{{{2}}}=\pi{{d}_{{{2}}}^{{{2}}}}=\pi{4}{{d}_{{{1}}}^{{{2}}}}={4}{A}{1}\)

Step 2

Let's first determine the relation between the velocities based on the continuity equation \(\displaystyle{v}_{{{1}}}{A}_{{{1}}}={v}_{{{2}}}{A}_{{{2}}}\) and the fact that \(\displaystyle{A}_{{{2}}}={4}{A}_{{{1}}}\) from which we can write

\(\displaystyle{v}_{{{2}}}={\frac{{{v}_{{{1}}}{A}_{{{1}}}}}{{{A}_{{{2}}}}}}\)

\(v_{2}=\frac{v_{1}}{4}\)

Step 3

Since the atmospheric pressures are the same we can write Bernoulli's equation as

\(\displaystyle{p}_{{{1}}}+\rho{g}{y}_{{{1}}}+{\frac{{{1}}}{{{2}}}}\rho{{v}_{{{1}}}^{{{2}}}}={p}_{{{2}}}+\rho{g}{y}_{{{2}}}+{\frac{{{1}}}{{{2}}}}\rho{{v}_{{{2}}}^{{{2}}}}\) where \(\displaystyle{p}_{{{1}}}\) and \(\displaystyle{p}_{{{2}}}\) are gauge pressures. We can now insert \(\displaystyle{v}_{{{2}}}=\frac{{v}_{{{1}}}}{{4}}\) to get

\(\displaystyle{p}_{{{2}}}={p}_{{{1}}}+\rho{g}{\left({y}_{{{1}}}-{y}_{{{2}}}\right)}+{\frac{{{1}}}{{{2}}}}\rho{\left({{v}_{{{1}}}^{{{2}}}}-\frac{{{v}_{{{1}}}^{{{2}}}}}{{16}}\right)}\)

\(\displaystyle{p}_{{{2}}}={p}_{{{1}}}+\rho{g}{\left({y}_{{{1}}}-{y}_{{{2}}}\right)}+{\frac{{{15}}}{{{32}}}}\rho{{v}_{{{1}}}^{{{2}}}}\)

Step 4

At this point we can insert all the given values to get the following

\(\displaystyle{p}_{{{2}}}={5}\times{10}^{{{4}}}+{10}^{{{3}}}\times{9.81}\times{11}+{10}^{{{3}}}\times{\frac{{{15}}}{{{32}}}}\times{3}^{{{2}}}\)

\(\displaystyle{p}_{{{2}}}={162}{k}{P}{a}\)