Water flows in a vertical pipe of 0.15 -m diameter at a rate of \(0.2 \mathrm{m}^{3} / \mathrm{s}\) and a pressure of \(200 \mathrm{kPa}\) at an elevation of \(25 \mathrm{m}\). Determine the velocity head and pressure head at elevations of 20 end 55 m.

The fluid velocity \(V\) can be determined using the equation \(Q = AV\), where \(Q\) is the flow rate and \(A\) is the cross sectional area of the pipe. In this case, \(Q = 0.2 m^3/s\) and \(A = \pi(D^2)/4\), where \(D\) is the diameter of the pipe, given as 0.15 m. Hence, \(V = Q/A = 0.2m^3/s / \pi*(0.15m)^2/4 = 22.62 m/s\).

The velocity head \(h_v\) can be calculated using the formula \(h_v = V^2 / 2g\), where \(g\) is the acceleration due to gravity, which we'll take as 9.81 m/s^2. Hence, \(h_v = 22.62 m/s / (2 * 9.81 m/s^2) = 26.06 m\). Since the fluid velocity remains constant along the pipe, the velocity head remains the same at all elevations.

The pressure head \(h_p\) can be calculated using the formula \(h_p = P / ρg\), where \(P\) is the fluid pressure, \(ρ\) is the fluid density, and \(g\) is the acceleration due to gravity. The fluid is water, so we'll take \(ρ = 1000 kg/m^3\) and \(g = 9.81 m/s^2\). Given the initial pressure \(P=200 kPa = 200000 Pa\), then the initial pressure head \(h_P=200000 Pa / (1000 kg/m^3 * 9.81 m/s^2) = 20.39 m\). According to Bernoulli's principle, the sum of the pressure head, velocity head and elevation remains constant, which we can take as \(h_1 = h_p1 + h_v + z1 = 20.39 m + 26.06 m + 25 m = 71.45 m\). At elevation \(z_2 = 20 m\), the pressure head \(h_p2\) is \(h_1 - h_v - z_2 = 71.45 m - 26.06 m - 20 m = 25.39 m\), and at elevation \(z_3 = 55 m\), the pressure head \(h_p2\) is \(h_1 - h_v - z_3 = 71.45 m - 26.06 m - 55 m = -9.61 m\). Note that the negative pressure head points to a decrease in absolute pressure below the reference pressure.