Question 8.6: Determine the time tf it takes for a cylindrical container w...

Referring to Fig. 8.13, let us consider a streamline from the free surface, at 1, to the drain, at 2. Assuming an atmospheric pressure at 1 and 2, Bernoulli’s equation reduces to

                                                                 gh+\frac{1}{2}\nu ^{2}_{1}=\frac{1}{2}\nu ^{2}_{2},
or
                                                                  \nu ^{2}_{2}-\nu ^{2}_{1}=2gh(t),

where we emphasize that h varies with time t. Mass balance gives \nu _{1}A_{1}=\nu _{2}A_{2}; thus,

                                                \nu _{2}=\nu _{1}\frac{\pi D^{2}/4}{\pi d^{2}/4}=\nu _{1}\frac{D^{2}}{d^{2}}
and, therefore,
                                        \nu ^{2}_{1}\left(\frac{D^{4}}{d^{4}}-1 \right)=2gh(t)\rightarrow \nu _{1}=\sqrt{\frac{d^{4}2gh(t)}{D^{4}-d^{4}} }.

Now, we recognize that \nu _{1},=-dh/dt and, therefore,

Integrating with respect to time,

                                                \int_{H}^{0}{\frac{1}{\sqrt{h} }\frac{dh}{dt} }dt=\int_{0}^{t_{f}}{\frac{-d^{2}\sqrt{2g} }{\sqrt{D^{4}-d^{4}} } }dt,
or
                                                      -2\sqrt{H}=\frac{-d^{2}\sqrt{2g} }{\sqrt{D^{4}-d^{4}} }t_{f};
thus,
                                         t_{f}=2\sqrt{\frac{H(D^{4}-d^{4})}{2gd^{4}} }=\sqrt{\frac{2H(D^{4}-d^{4})}{gd^{4}} }.