Hose siphon discharge and minimum pressure

Data. s.g. 0.6 so $\rho = 600$ kg/m³, $\nu = 5.0\times10^{-7}$ m²/s, $D = 0.025$ m, $k/D = 0.0004$, total hose length $L = 9$ m, length to the highest point A $= 3.25$ m. From the figure: A is 1.5 m above the liquid surface; the outlet is 5 m below A, so it is $5 - 1.5 = 3.5$ m below the liquid surface. The hose discharges freely to the atmosphere.

Volumetric flow

Energy equation from the tank surface to the outlet (entrance loss neglected, velocity head at the outlet kept):

$$3.5 = \frac{V^2}{2g}\left(1 + f\frac{L}{D}\right) = \frac{V^2}{2g}\left(1 + f\frac{9}{0.025}\right) = \frac{V^2}{2g}\left(1 + 360f\right)$$

$f$ depends on $Re = VD/\nu$ and $k/D = 0.0004$ (Colebrook-White), so iterate:

1. Assume $f = 0.02$: $\dfrac{V^2}{2g} = \dfrac{3.5}{8.2} = 0.427$ m, $V = 2.89$ m/s, $Re = 1.4\times10^5$.
2. Colebrook: $f \approx 0.0189$ and the result converges (3 iterations) to

$$f = 0.01891,\qquad \frac{V^2}{2g} = \frac{3.5}{1 + 360(0.01891)} = 0.4483\ \text{m},\quad V = 2.966\ \text{m/s},\quad Re = 148284$$ $$Q = \frac{\pi}{4}(0.025)^2(2.966) = 0.001456\ \text{m}^3/\text{s} = 1.456\ \text{l/s} \;(87.3\ \text{l/min})$$

Minimum pressure (at the top point A)

Energy equation from the tank surface (gauge pressure 0) to A, 1.5 m above the surface, with friction over 3.25 m:

$$0 = 1.5 + \frac{p_A}{\gamma} + \frac{V^2}{2g}\left(1 + f\frac{3.25}{0.025}\right), \qquad f\frac{3.25}{0.025} = 2.458$$ $$\frac{p_A}{\gamma} = -1.5 - 0.4483(3.458) = -3.050\ \text{m of liquid}$$ $$p_A = \rho g\left(\frac{p_A}{\gamma}\right) = 600(9.81)(-3.050) = -17954\ \text{Pa} \approx -17.95\ \text{kPa (gauge)}$$

This is about 83.3 kPa absolute, above the vapour pressure of the liquid unless it is very volatile.

Answer: $Q = 1.46$ l/s ($0.001456$ m³/s, $V = 2.97$ m/s); minimum pressure at A $= -17.95$ kPa gauge.