Question 2.9: Tank Drainage Consider the efflux of a liquid through a smoo......

At descending point ➀ and exit point ➁, most of the information is given. Thus, Eq. (2.28) now reads:

${\frac{\mathrm{v}_{1}^{2}}{2}}+{\frac{\mathrm{p}_{1}}{\mathrm{\rho }}}+{\mathrm{g}}\mathrm{z_1}={\frac{\mathrm{v}_{2}^{2}}{2}}+{\frac{\mathrm{p}_{2}}{\mathrm{\rho }}}+{{{\mathrm{gz_2}}}}$                          (2.28)

The velocities are related via continuity (see Eq. (2.7)):

$\Sigma{\rm Q}_{\mathrm{in}}=\Sigma{\rm Q}_{\mathrm{out}}$                          (2.7)

so that with $\mathrm{v_1=v_0\,A_0/A_T},$

Comment: Toricelli’s law, $\mathrm{v=\sqrt{2gh} }$ can be directly recovered when Δp ≈ 0 and $\mathrm{A_T >> A_O}$ . Clearly, the liquid level y and pressure drop $\mathrm{p_T − p_A}$ are the key driving forces. The solution breaks down when $\mathrm{A_O → A_T}$.