Pipe passing over barrier between reservoirs

Data. $D = 0.1$ m, $L = 36$ m, $f = 0.04$. Upper water surface is 1.8 m above the inlet; lower water surface is 1.2 m above the outlet; ends differ in level by 4 m, so the difference of the water surfaces is $H = 4 + 1.8 - 1.2 = 4.6$ m. Barrier at 7.5 m horizontally from the upper reservoir. Minimum absolute pressure 40 kPa; $p_a = 101.3$ kPa. All losses: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$, friction.

Step 1: velocity in the pipe

$$H = \left(0.5 + f\frac{L}{D} + 1.0\right)\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.04\left(\frac{36}{0.1}\right) = 14.4$$ $$4.6 = 15.9\,\frac{V^2}{2g} \;\Rightarrow\; \frac{V^2}{2g} = 0.2893\ \text{m},\quad V = 2.382\ \text{m/s}\quad(Q = 18.71\ \text{l/s})$$

Step 2: energy equation from the upper surface to the barrier X

Take the pipe inlet level as datum. Let the barrier be at height $h_X$ above the inlet. Since the pipe is straight from the inlet to the barrier, its length is $l = \sqrt{7.5^2 + h_X^2}$.

Absolute pressure heads: $\dfrac{p_a}{\gamma} = \dfrac{101.3}{9.81} = 10.326$ m; minimum $\dfrac{p_X}{\gamma} = \dfrac{40}{9.81} = 4.077$ m.

$$\frac{p_a}{\gamma} + 1.8 = h_X + \frac{p_X}{\gamma} + \frac{V^2}{2g}\left(1 + 0.5 + f\frac{l}{D}\right)$$ $$h_X = 10.326 + 1.8 - 4.077 - 0.2893\left(1.5 + 0.4\,l\right)$$

Iterating ($h_X$ then $l$): $l = 9.90$ m, so the bracket is $5.46$ and

$$h_X = 6.47\ \text{m above the pipe inlet}$$

Answer: The pipe may rise to about 6.47 m above its inlet at the barrier (about 4.67 m above the upper reservoir water level).