Answer:
[tex]f_{in} = 476544.862\,\frac{cm^{3}}{min}[/tex]
Step-by-step explanation:
The conical tank is modelled by the Principle of Mass Conservation:
[tex]\dot m_{in} - \dot m_{out} = \frac{dm_{tank}}{dt}[/tex]
As water is an incompressible fluid, the equation can be simplified into this:
[tex]f_{in} - f_{out} = \frac{dV_{tank}}{dt}[/tex]
The rate at which water is being pumped into the tank is:
[tex]f_{in} = \frac{dV_{tank}}{dt} + f_{out}[/tex]
[tex]f_{in} = \frac{2}{3}\cdot \pi \cdot r\cdot h \cdot \frac{dr}{dt} + \frac{1}{3}\cdot \pi \cdot r^{2}\cdot \frac{dh}{dt} + f_{out}[/tex]
The cone obeys the following relationship:
[tex]\frac{r}{h} = \frac{2.75\,m}{14\,m}[/tex]
[tex]r = 0.196\cdot h[/tex]
[tex]r = 0.196\cdot (450\,cm)[/tex]
[tex]r = 88.2\,cm[/tex]
By deriving the expression:
[tex]\frac{dr}{dt} = 0.196\cdot \frac{dh}{dt}[/tex]
[tex]\frac{dr}{dt} = 0.196\cdot (19\,\frac{cm}{min} )[/tex]
[tex]\frac{dr}{dt} = 3.724\,\frac{cm}{min}[/tex]
The flow required to be pumped is:
[tex]f_{in} = \frac{2}{3}\cdot \pi \cdot (88.2\,cm)\cdot (450\,cm) \cdot (3.724\,\frac{cm}{min} )+\frac{1}{3}\cdot \pi \cdot (88.2\,cm)^{2}\cdot (19\,\frac{cm}{min} )+12200\,\frac{cm^{3}}{min}[/tex]
[tex]f_{in} = 476544.862\,\frac{cm^{3}}{min}[/tex]
