Venturimeter as suction device, minimum discharge

Given: throat area $a_t = 0.00025$ m², outlet area $a_e = 0.001$ m², venturi vertical, discharging to atmosphere.

The figure values (2.0 m, 1.9 m, 1.0 m) are not fully readable, so assume: the throat is $z_t = 1.0$ m above the outlet plane, and the water surface in the suction tank is $H_s = 1.9$ m below the throat.

        inlet
        ||
        ||        <- 
      throat ---- suction pipe (to tank)
        /\          1.9 m lift
       /  \
      outlet (atmosphere) ... 1.0 m below throat

Condition for flow up the suction pipe: the throat pressure must fall at least to $p_t = -\gamma H_s$ (gauge). The minimum discharge is when $p_t/\gamma = -1.9$ m.

Continuity: $V_t = Q/a_t$, $V_e = Q/a_e = V_t/4$.

Bernoulli, throat to outlet (exit pressure atmospheric):

$$\frac{p_t}{\gamma} + \frac{V_t^2}{2g} + z_t = 0 + \frac{V_e^2}{2g} + 0$$ $$\begin{aligned} -1.9 + \frac{V_t^2}{2g} + 1.0 &= \frac{V_t^2}{16\times 2g}\\ \frac{V_t^2}{2g}\left(1 - \frac{1}{16}\right) &= 0.9\\ \frac{V_t^2}{2g} &= 0.96\ \text{m}\\ V_t &= \sqrt{2\times 9.81\times 0.96} = 4.340\ \text{m/s} \end{aligned}$$ $$Q_{min} = a_tV_t = 0.00025\times 4.340 = 1.085\times 10^{-3}\ \text{m}^3/\text{s}$$

Answer: minimum discharge $\approx 1.085\times 10^{-3}$ m³/s (about 1.09 L/s) for the assumed dimensions. General result: $Q_{min} = a_t\sqrt{2g\,(H_s - z_t)\big/\left(1 - (a_t/a_e)^2\right)}$.