A cylindrical tank with crossectional area, S_c, initially contains V_o liters o

Question

A cylindrical tank with crossectional area, S_c, initially contains V_o liters of fluid in which A_o grams of salt are dissolved.  At the bottom of the tank is an orifice with crossectional area S_o from which the fluid flows out.  By Bernoulli's Law, the velocity of the fluid coming out is approximately given by: sqrt(2gh), where g is the acceleration of gravity and h is the height of the fluid in the tank at any time, t.

a) show that the rate of volume leaving the tank through the orifice is given by R_o= a*sqrt(V(t)), where a = (S_o*sqrt(2g))/(sqrt(S_c)) and V(t) is the volume of the fluid in the tank at any t

b) if brine containing c grams of salt per liter is being pumped into the tank at a constant rate of R_in liters per minute, show that the ODE (ordinary differential equation) for the rate of change of the volume, V(t) in the tank is given by:  dV/dt = R_in - a*sqrt(V).  For what value of V will equilibrium be reached (i.e. for what value of V is dV/dt = 0)

c)  Solve the above ODE assuming that at t=0 , V=V_o.  You cannot explicitly express V as a function of t, so express it as a function of V (and V_o, a, and R_in).  At what time will the tank reach equilibrium?

d) Show that the ODE for the rate of change of the amount of salt in the tank, A(t), is given by dA/dt = cR_in - a*(A/(sqrt(V)))

e)  Since we cannot explicitly express V as a function of t, to substitute into the above ODE, we cannot solve the ODE in part d to find A as a function of t.  But we can find A as a function of V by noting that dA/dt = dA/dV * dV/dt.  Use this and the result from part b to get an ODE in A and V and then solve it, noting that when V=V_o,  A=A_o

A cylindrical tank with crossectional area, S_c, initially contains V_o liters of fluid in which A_o grams of salt are dissolved. At the bottom of the tank is an orifice with crossectional area S_o from which the fluid flows out. By Bernoulli's Law, the velocity of the fluid coming out is approximately given by: sqrt(2gh), where g is the acceleration of gravity and h is the height of the fluid in the tank at any time, t. show that the rate of volume leaving the tank through the orifice is given by R_o= a*sqrt(V(t)), where a = (S_o*sqrt(2g))/(sqrt(S_c)) and V(t) is the volume of the fluid in the tank at any t if brine containing c grams of salt per liter is being pumped into the tank at a constant rate of R_in liters per minute, show that the ODE (ordinary differential equation) for the rate of change of the volume, V(t) in the tank is given by: dV/dt = R_in - a*sqrt(V). For what value of V will equilibrium be reached (i.e. for what value of V is dV/dt = 0) Solve the above ODE assuming that at t=0 , V=V_o. You cannot explicitly express V as a function of t, so express it as a function of V (and V_o, a, and R_in). At what time will the tank reach equilibrium? Show that the ODE for the rate of change of the amount of salt in the tank, A(t), is given by dA/dt = cR_in - a*(A/(sqrt(V))) Since we cannot explicitly express V as a function of t, to substitute into the above ODE, we cannot solve the ODE in part d to find A as a function of t. But we can find A as a function of V by noting that dA/dt = dA/dV * dV/dt. Use this and the result from part b to get an ODE in A and V and then solve it, noting that when V=V_o, A=A_o

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