Question 10.4.1: Transient Balances on a Semibatch Reactor An aqueous solutio......
Transient Balances on a Semibatch Reactor
An aqueous solution containing 0.015 mol/L of species A is fed into a holding tank that initially contains 75 liters of pure water. The reactant decomposes at a rate
r[mol A/(L·s)] = 0.0375C_{A}
where C_{A}(mol A/L) is the concentration of A in the tank. The volumetric feed rate of the solution, \dot{V}(t), increases linearly over a 10-second period from 0 to 25 L/s and stays constant at that rate thereafter until the tank is filled to the desired level. The density of the feed stream is constant.
1. Write transient balances for the total mass of the tank contents and the mass of A in the tank. Convert the equations to differential equations for V(t) (the volume of the tank contents) and C_{A}(t) (the concentration of A in the tank) that have the form of Equations 10.4-1 and 10.4-2, and provide initial conditions.
\begin{matrix}\frac{dy_{1}}{dt} = f_{1}(y_{1},y_{2},…,y_{n},t)\\ y_{1}(t_{0})=y_{10} &\pmb{(10.4-1)} \\\frac{dy_{2}}{dt} = f_{2}(y_{1},y_{2},…,y_{n},t)\\ y_{2}(t_{0})=y_{20}& \pmb{(10.4-2)}\end{matrix}
2. Sketch the shapes of the plots you would expect for the volume of the tank contents, V(L), and the concentration of A in the tank, C_{A}(mol/L), versus time.
3. Outline how the equations would be solved to derive an expression for C_{A}(t) for the period from t = 0 to t = 60 s
A flowchart of the process is as follows:
The concentration of A in the tank changes with time because both the moles of A in the tank and the volume of the tank contents are changing.
1. Total mass balance: accumulation (kg/s) = input (kg/s). The total mass of the tank contents (kg) at any time is ρ(kg/L)V(L), and the mass flow rate of the feed stream (kg/s) is ρ(kg/L)\dot{V}(L/s). The mass balance thus becomes d(ρV)/dt = ρ\dot{V} , or, bringing ρ (which we are told is constant) out of the derivative and canceling it,
\boxed{\begin{matrix} \frac{dV}{dt}=\dot{V}\\ V(0)=75.0 L \end{matrix} } (1)
Balance on A: accumulation(mol A/s) = input – consumption. The number of moles of A in the tank at any time equals V(L)C_{A}(mol A/L). The balance equation therefore becomes
\boxed{\begin{matrix} \frac{dC_{A}}{dt} = \frac{\dot{V}}{V} (0.015 mol A/L – C_{A}) – 0.0375C_{A} \\ C_{A}(0)=0 mol A/L \end{matrix} } (2)
The initial condition in Equation 2 follows from the statement that the tank initially contains pure water. In Equations 1 and 2,
\begin{matrix} \dot{V}(t) =2.5t 0\leq t\leq 10 s& \pmb{(3a)} \\ = 25 L/s t \gt 10 s & \pmb{(3b)} \end{matrix}
(Verify Equation 3a.)
Equations 1 and 2 are two differential equations in two dependent variables that have the form of Equations 10.4-1 and 10.4-2, where V and C_{A} correspond to y_{1} and y_{2}, respectively. The equations may therefore be solved with any of the computer programs mentioned at the beginning of this section.³
2. To predict the shape of the curve on a plot of V versus t, we need only remember that the slope of the curve is dV/dt, which in turn equals \dot{V}(t) (from Equation 1). Try to follow this chain of reasoning:
• A point on the plot of V versus t is the initial condition (t = 0, V = 75 L).
• During the first 10 seconds, dV/dt = 2.5t (from Equations 1 and 3a). The slope of the curve therefore equals zero at t = 0 (so that the curve is horizontal at the V axis) and increases over the first 10 seconds (so that the curve is concave up).
• At t = 10 seconds, dV/dt reaches a value of 25 L/s and thereafter remains constant at that value.
A curve with a constant slope is a straight line. The plot of V versust for t ≥ 10 s must therefore be a straight line with a slope of 25 L/s.
• Putting the preceding observations together, we conclude that the plot of V versus t starts horizontally at (t = 0, V = 75 L), curves up for 10 seconds, and then becomes a straight line with a slope of 25 L/s. It should have the following appearance:
• The plot of C_{A} versus t must begin at (t = 0, C_{A} = 0), since the tank initially contains pure water.
• At t = 0, the expression of Equation 2 for dC_{A}/dt equals zero since both t and C_{A} are zero at this point. (Verify.) The plot of C_{A} versus t is therefore horizontal at the C_{A} axis. Since we are adding A to the tank, its concentration must increase, and so the curve must be concave up.
• As time proceeds, more and more of the tank volume is occupied by fluid in which the A has had a long time to react. We could anticipate that at a very long time, the tank would contain a huge volume with very little A in it, and the A being added would be diluted down to a concentration approaching zero. C_{A} should therefore increase near t = 0, rise to a maximum, start decreasing, and approach zero at long times.
• Furthermore, the concentration in the tank can never be greater than that in the feed stream (0.015 mol/L) and, in fact, must always be less than this amount since (a) the feed is diluted by the water initially in the tank and (b) some of the A in the feed reacts once it is in the tank. The maximum value of C_{A} must therefore be less than 0.015 mol A/L.
• All of these observations combine to predict a plot with the following shape:
3. The system of equations must be solved in two stages—the first from t = 0 to t = 10 s (when \dot{V} = 2.5t) and the second for t > 10 s, when \dot{V} = 25 L/s. The procedure is as follows:
• Substitute 2.5t for \dot{V} (t) in Equations 1 and 2.
\begin{matrix}\frac{dV}{dt} = 2.5t \\ V(0) = 75.0 L&\pmb{(1a)} \end{matrix}
\begin{matrix}\frac{dC_{A}}{dt} = \frac{2.5t}{V} (0.015 – C_{A}) – 0.0375 C_{A} \\ C_{A}(0)=0 &\pmb{(2a)}\end{matrix}
When this pair of equations is solved for V(t) and C_{A}(t) (we will omit details of the solution procedure), we determine that V(10 s) =200 L and C_{A}(10 s) =0.00831 mol A/L.
• Substitute \dot{V}(t) = 25 L/s in Equations 1 and 2 and substitute the dependent variable values at t = 10 s for the initial conditions:
\begin{matrix}\frac{dV}{dt} = 25 L/s \\ V(10) = 200 L&\pmb{(1b)}\end{matrix}
\begin{matrix}\frac{dC_{A}}{dt} = \frac{25}{V} (0.015 – C_{A}) – 0.0375 C_{A} \\ C_{A}(10)=0.0831 mol A/L &\pmb{(2b)}\end{matrix}
These equations may be solved for V(t) and C_{A}(t) for t > 10 s. These solutions along with the previous solutions for t ≤ 10 s are shown in the following plots:
³It would be easier in this particular problem to solve Equation 1 analytically and substitute for V(t) in Equation 2. The methods we are illustrating now would work even if an analytical solution to Equation 1 could not be found.
