Question 2.1: Volumetric Flow Rate Consider a liquid-filled tank (depth H)......

The u(z)-equation will be derived in Sect. 2.3.2. Here, we apply a reduced version of Eq. (2.3). Specifically, with \rm\frac{\partial}{\partial t} \iiint{ρd∀} =0 (steady-state because no system parameter changes with time), we have with ρ =¢ (incompressible fluid):

\rm 0=\frac{\partial}{\partial t} \iiint\limits_{C.∀.} ρd∀+\iint\limits_{C.S.}ρ\vec{v} \cdot d\vec{A}                         (2.3)

\rm 0=\iint \limits_{c.s.}\vec{v}\cdot d\vec{A}                               (E.2.1.1)

Fluid mass crosses the control surface at two locations (see graph). Recalling that “inflow” is negative and “outflow” positive (see Fig. 2.2b), Eq. (E.2.1.1) reads with \rm\frac{\dot m _{in}}{ρ} =Q, ~Q+\iint\limits_{A_{slot}} \vec{v}\cdot d\vec{A}=0 or \rm  Q=\int\limits_{A} v_ndA               (E.2.1.2)

Here, \rm Q\equiv Q_{outlet},\,v_n=u(z)=\sqrt{2g(H-z)} and dA = wdz so that

which yields

\rm Q_{outlet}=K[(H+h)^{3/2}-(H-h)^{3/2}]                       (E.2.1.3)

where \rm K\equiv \frac{2}{3}w\sqrt{2g} .

For h << H, as it is often the case, Eq. (E.2.1.3) can be simplified to (see HW Assignment):

\rm{Q_{{outlet}}\approx2~w h\sqrt{2g H} }                                  (E.2.1.4)

Comment: Equation (E.2.1.4) is known as Toricelli’s law.