Question

Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flowrate (m3/sec) from the tank is 0.01 times of the liquid height (in meter) in the tank. If the cross section area of thank is 0.01 m2 how long does it take to fill the tank up to 0.6 meter from the bottom? how long does it take to fill the tank up to 2 meter from the bottom?

Answer 1

Explanation:

Apply the mass of balance as follows.

   Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

         $\frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)$

      $\rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h$

   $\frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}$

          [/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = $\frac{10}{\rho A_{c}}$

                       $A_{c} = 0.01 m^{2}$

              $\frac{dh}{dt}$ + h = 1

                  $\frac{dh}{dt}$ = 1 - h

               $\frac{dh}{1 - h}$ = dt  

                $\frac{ln(1 - h)}{-1}$ = t + C      

Given at t = 0 and V = 0  

                         $A \times h$ = 0  

 or,                     h = 0

                 -ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

                                C = 0  

                So,   -ln(1 - h) = t

or,                      t = $ln (\frac{1}{1 - h})$  ........... (1)

(a)    Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

                    t = $ln (\frac{1}{1 - h})$  

                     t = $ln (\frac{1}{1 - 0.6})$  

                        = $ln (\frac{1}{0.4})$

                        = 0.916 seconds

(b)   As maximum height of water level in the tank is achieved at steady state that is, t = $\infty$.  

                    1 - h = exp (-t)

                    1 - h = 0  

                         h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.