Answer:
A) 8 l/s - 4 l/s = dv/dt
initial conditions : t = 0 , v = 400 L
B) Vt = 4t + 400
Explanation:
Given data :
volume of tank = 2000 liter
volume of water in tank = 400 liters
aqueous solution contains = 1 g/l potassium chloride
flow rate of solution = 8.00 L/s ( v1 )
outlet stream flow rate = 4.00 L/s ( v2 )
t = 0
density of the whole content of the tank ( feed stream and tank solution )= rho (g/l )
assuming
v (t) liters = volume of tank contents
C (t) g/l = concentration of potassium chloride in tank contents and outlet stream
A ) Total mass balance on the tank contents
mass inlet - mass outlet = mass accumulation ------ 1
note : mass accumulation = dm/dt
mass = density * v
hence equation 1 becomes
( D1 * v1 ) - (D2 *v2) = dM/dt
D1 * (8 l/s) - D2 * (4 l/s) = dM/dt ----- 2
since D1 = D2 = Dm and M = ( D*v )
hence equation 2 becomes
8 l/s - 4 l/s = dv/dt
initial conditions : t = 0 , v = 400 l
B) attached below is the required sketch
solving the mass balance equation to obtain an expression for V(t)
8 l/s - 4 l/s = dv/dt
= 8 - 4 = dv/dt
= 4 = dv/dt
hence dv = 4dt
integrating
Vt - vi = 4*t
input the initial volume ( vi ) = 400
Vt = 4t + 400
The reason behind the sketch is that the volume increase because the inflow is greater than the outflows
also the concentration increases because of the addition of a concentrated mixture
The mass balance equation for V(t) has been given as V(t) = 4t + 400.
(A) The flow volume of the tank has been considered as:
Mass inlet - Mass outlet = Mass accumulated
The mass can be defined by: mass = Density × Volume
The mass accumulated has been given by: [tex]\rm \dfrac{dM}{dt}[/tex]
Thus, the equation can be written as:
(Density of inlet × Volume of inlet) - (Density of outlet × Volume of outlet) = [tex]\rm \dfrac{dM}{dt}[/tex]
The given tank has an inlet flow rate = 8 L/s
Outlet flow rate = 4 L/s
Substituting the values for volume:
(Density of inlet × 8 L/s) - (Density of outlet ×4 L/s = [tex]\rm \dfrac{dM}{dt}[/tex]
Since the Potassium chloride has been inlet and outlet, the density has been the same.
The differentiation of mass with the same density cancels the term density in the equation:
8 L/s - 4 L/s = [tex]\rm \dfrac{dV}{dt}[/tex]
4 = [tex]\rm \dfrac{dV}{dt}[/tex]
(B) At the initial condition of the tank, at time t = 0 sec,
The initial volume of the tank has been 400 L.
Solving the equation for V(t):
4 dt = dV
Integrating with volume Vi to Vt:
V (at t) - V (initial) = 4 dt
V(t) = 4t + 400
The plot for V versus t has been a straight line, as there has been a direct relationship between the Volume and time, and with an increase in time, there has been an increase in the volume.
The plot of C versus t has been a hyperbolic growth curve, as there has been an exponential relationship between the C and t.
For more information about the mass balance, refer to the link:
https://brainly.com/question/24860461
