Respuesta :
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
[tex]\frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)[/tex]
[tex]\rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h[/tex]
[tex]\frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}[/tex]
[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = [tex]\frac{10}{\rho A_{c}}[/tex]
[tex]A_{c} = 0.01 m^{2}[/tex]
[tex]\frac{dh}{dt}[/tex] + h = 1
[tex]\frac{dh}{dt}[/tex] = 1 - h
[tex]\frac{dh}{1 - h}[/tex] = dt
[tex]\frac{ln(1 - h)}{-1}[/tex] = t + C
Given at t = 0 and V = 0
[tex]A \times h[/tex] = 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 = [tex]ln (\frac{1}{1 - h})[/tex] ........... (1)
(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.
t = [tex]ln (\frac{1}{1 - h})[/tex]
t = [tex]ln (\frac{1}{1 - 0.6})[/tex]
= [tex]ln (\frac{1}{0.4})[/tex]
= 0.916 seconds
(b) As maximum height of water level in the tank is achieved at steady state that is, t = [tex]\infty[/tex].
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.
