Water flows horizontally in a pipe of diameter $12.1 \mathrm{~cm}$ at a rate of $0.783 \mathrm{~m} / \mathrm{s}$ and a gauge pressure of $5.94 \times 10^{\wedge} 3 \mathrm{~Pa}$. The fluid (still in the pipe) then descends a $6.4 \mathrm{~m}$ high hill, passes through a reducer and into a smaller $2.2 \mathrm{~cm}$ diameter pipe, and then flows horizontally again. Determine the speed of the water in the lower section of pipe.

Step 1 ：Given: \(P1 = 5.94 \times 10^3 \, Pa\), \(v1 = 0.783 \, m/s\), \(h1 = 0 \, m\), \(P2 = ?\), \(v2 = ?\), \(h2 = -6.4 \, m\), \(\rho = 1000 \, kg/m^3\), \(g = 9.81 \, m/s^2\)

Step 2 ：Bernoulli's equation is given by: \(P1/\rho + v1^2/2 + g*h1 = P2/\rho + v2^2/2 + g*h2\)

Step 3 ：Substitute the given values into Bernoulli's equation: \(5.94 \times 10^3 / 1000 + 0.783^2/2 + 9.81*0 = P2/1000 + v2^2/2 + 9.81*(-6.4)\)

Step 4 ：Solve for \(v2^2\): \(v2^2 = 2 * [(P1 - P2)/\rho + v1^2/2 + g*h1 - g*h2]\)

Step 5 ：Substitute the values: \(v2^2 = 2 * [(5.94 \times 10^3 - P2)/1000 + 0.783^2/2 + 9.81*6.4]\)

Step 6 ：Simplify: \(v2^2 = 2 * [(5.94 \times 10^3 - 5.94 \times 10^3)/1000 + 0.783^2/2 + 9.81*6.4]\)

Step 7 ：Further simplify: \(v2^2 = 2 * [0 + 0.306289/2 + 62.784]\)

Step 8 ：Simplify again: \(v2^2 = 2 * [0.1531445 + 62.784]\)

Step 9 ：Simplify further: \(v2^2 = 2 * 62.9371445\)

Step 10 ：Calculate: \(v2^2 = 125.874289\)

Step 11 ：Take the square root: \(v2 = \sqrt{125.874289}\)

Step 12 ：\(\boxed{v2 = 11.22 \, m/s}\)