Question 2.3: Draining of a Tank: A “deformable C.∀.” because the fluid le......
Draining of a Tank: A “deformable C.∀.” because the fluid level decreases and hence we have a shrinking C.∀. with moving C.S
Consider a relatively small tank of diameter D and initially filled to height \rm h_0. The fluid drains through a pipe of radius \rm(r_0) according to:
\rm u(r)=2\bar{u} [1-(\frac{r}{r_0} )^2] (see Example 2.10)
where \rm\bar{u} =\sqrt{2gh} (see Example 2.1), \rm r_0 is the outlet pipe radius, and r is its variable radius, \rm 0 ≤ r ≤ r_0 . The fluid depth was \rm h_0 at time t = 0.
Find h(t) for a limited observation time Δt.
| Control Volume | Assumptions | Sketch |
 |
• Transient incompressib le flow |  |
| • Stationary C.∀. (tank) but deforming liquid volume |
Equation (2.8b) can be expanded for this problem with ∀_{C.∀.}=\frac{D^2\pi }{4}h to
\rm\frac{d∀}{dt} \big|_{C.∀.}=-\iint\limits_{C.S.} \vec{v}\cdot d\vec{A} (2.8b)
\rm(\frac{{ D}^{2}\pi}{4})\frac{{dh}}{{dt}}=-\int_{0}^{r_0}{u}({r})\,{dA} (E.2.3.1)
where dA = 2πrdr is the variable ring element as part of the crosssectional area of the outlet pipe. Thus,
{\frac{\mathrm{dh}}{\mathrm{dt}}}=-{\frac{16\,{\sqrt{2\mathrm{gh}}}}{\mathrm{D}^{2}}}\int_{0}^{r_0}\left[1-({\frac{r}{r_{0}}})^{2}\right]\mathrm{rdr} (E.2.3.2a)
or
\rm\frac{{d} h}{{d}t}=-4\sqrt{2{g}}\,(\frac{{r_{0}}}{ D})^{2}\sqrt{{h}} (E.2.3.2b)
subject to h (t = 0) = \rm h_0. Separation of variables and integration yields:
\rm\sqrt{h_{0}}\,-\,\sqrt{h}\,=\,\sqrt{8g}(\frac{r_{0}}{D})^{2}\,{t}or
\rm{\frac{\ h({ t})}{ h_0}}=\left[1-\sqrt{\frac{8{g}}{{h}_{0}}}(\frac{{r}_{0}}{{{{{D}}}}})^{2}{t}\right]^{2} (E.2.3.3)
Graph:
Comments:
• The standard assumption that h = const. <reservoir> is only approximately true when \rm r_0 / D < 0.1
• A variable speed dh/dt, i.e., accelerated tank-draining, occurs when \rm r_0/ D > 0.2
