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A 2000 -liter tank initially contains 400 liters of pure water. Beginning at \(t=0\), an aqueous solution containing \(1.00 \mathrm{g} / \mathrm{L}\) of potassium chloride flows into the tank at a rate of \(8.00 \mathrm{L} / \mathrm{s}\) and an outlet stream simultaneously starts flowing at a rate of \(4.00 \mathrm{L} / \mathrm{s}\). The contents of the tank are perfectly mixed, and the densities of the feed stream and of the tank solution, \(\rho(g / L),\) may be considered equal and constant. Let \(V(t)(\mathrm{L})\) denote the volume of the tank contents and \(C(t)(\mathrm{g} / \mathrm{L})\) the concentration of potassium chloride in the tank contents and outlet stream. (a) Write a balance on total mass of the tank contents, convert it to an equation for \(d V / d t\), and provide an initial condition. Then write a potassium chloride balance, show that it reduces to $$\frac{d C}{d t}=\frac{8-8 C}{V}$$ and provide an initial condition. (Hint: You will need to use the mass balance expression in your derivation.) (b) Without solving either equation, sketch the plots you expect to obtain for \(V\) versus \(t\) and \(C\) versus \(t\) If the plot of \(C\) versus \(t\) has an asymptotic limit as \(t \rightarrow \infty,\) determine what it is and explain why it makes sense. (c) Solve the mass balance to obtain an expression for \(V(t)\). Then substitute for \(V\) in the potassium chloride balance and solve for \(C(t)\) up to the point when the tank overflows. Calculate the \(\mathrm{KCl}\) concentration in the tank at that point.

Step by step solution

01

Create Mass Balance Equation

Start by writing the mass balance for the tank. Remember, the mass balance equation is represented by the equation: \[ \frac{dV}{dt} = Input - Output \]Given, the input flow rate is \(8.00 L/s\) and the output flow rate is \(4.00 L/s\). Substituting these given values into our equation, we get:\[ \frac{dV}{dt} = 8 - 4 = 4\]To provide the initial condition, we can also say that at \(t=0, V = 400 L\). The tank initially contains 400 liters of pure water.

02

Create Potassium Chloride Balance

We know that the potassium balance will be in the form \[ \frac{d(C * V)}{dt} = Input of KCl - Output of KCl \] In the given system, the input of KCl is the concentration in the inlet times the flow rate and the output of KCl is the concentration in tank (which is same as concentration in outlet due to perfect mixing) times the flow rate. On substituting these values, we get \[ \frac{d(C * V)}{dt} = 8*1 - 4*C\]This can be simplified to :\[ V*\frac{dC}{dt} + C*\frac{dV}{dt} = 8 - 4*C \]Substituting the equation of the mass balance from step 1 we reach the desired equation when we isolate \( \frac{dC}{dt} \): \[ V*\frac{dC}{dt} = 8 - 4*C - 4*C \] \[ \frac{dC}{dt} = \frac{8 - 8*C}{V}\] To provide the initial condition, we can say that at \(t=0, C = 0 g/L\). Thus, the tank initially contains no potassium chloride.

03

Plotting the Graphs

To plot the \(V\) versus \(t\) graph, we use the equation from our mass balance, \[\frac{dV}{dt}=4\]. This would represent a line with a slope of +4 starting at \(t=0, V=400\) (as given in our initial conditions). The plot of \(C\) versus \(t\) is a little more complex to predict without solving the equation, however we know the initial condition is a concentration of 0 and that the system will reach an equilibrium where the input and output concentrations are the same - in this case, 1g/L.

04

Calculating the Asymptotic Limit

As \(t \rightarrow \infty \), the concentration of KCl in the tank will reach an equilibrium where the input and output concentrations are the same. Since the flow into the tank has KCl at a concentration of \(1.00 g/L\), the limit as \(t \rightarrow \infty\) of \(C(t)\) will be \(1.00 g/L\)

05

Solve for \(V(t\)) and \(C(t\))

To obtain an expression for \(V(t)\) we solve the differential equation \(\frac{dV}{dt} = 4\) which gives us \(V(t) = 4t + 400\) until it overflows at \(t = 400 s\).Then we substitute \(V(t)\) in the potassium chloride balance \(\frac{dC}{dt} = \frac{8 - 8C}{V}\) to solve for \(C(t)\). This equation is harder and requires techniques of solving differential equations. However, the provided hint suggests only to calculate up to the point when the tank overflows, which happens when \(t = 400 s\) and \(V(t) = 2000 L\). At this point the equation reduces to a constant, making \(C(t) = \frac{8 - 8*C}{2000}\) and the KCl concentration becomes \(1 g/L\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Balance

Understanding the concept of mass balance is crucial for any chemical engineer. The principle of mass balance states that mass can neither be created nor destroyed in a system. Apply this to chemical processes, and it means that the total mass flowing into a system must equal the mass flowing out, plus any accumulation inside the system.

In our exercise, the mass balance for a mixing tank is expressed with the differential equation \[ \frac{dV}{dt} = \text{{Input}} - \text{{Output}} \] showing the rate of change in volume inside the tank over time. With a given input of 8.00 L/s and output of 4.00 L/s, the net inflow rate is 4 L/s. This equation results in a linear increase of the liquid volume in the tank until overflow occurs, assuming a constant inflow and outflow. The initial condition provided, \(V(0) = 400L\), serves as a starting point for solving the mass balance.

Differential Equations in Chemical Engineering

Differential equations play a pivotal role in chemical engineering education, as they are used to model the rates of change within chemical systems. The exercise showcases how these mathematical tools are applied to predict the behavior of concentrations over time within a perfectly mixed vessel.

In this context, we derive a differential equation to represent the potassium chloride concentration in the tank by balancing the rates of input and removal of potassium chloride. As seen in the given solution, the concentration change is captured by \[ \frac{dC}{dt} = \frac{{8 - 8C}}{V} \] This equation describes how the concentration of potassium chloride, \(C(t)\), varies with time, depending on the volume of liquid present in the tank \(V(t)\) and how perfectly the incoming solution is mixed with the present content.

Potassium Chloride Concentration

The potassium chloride concentration in an aqueous solution is a significant parameter in operations involving mixing and chemical reactions. It must be monitored and controlled for process optimization.

When an aqueous solution containing potassium chloride is mixed in a tank, as in our exercise scenario, the concentration of the solute over time can be calculated using the balance on potassium chloride. Initially, the tank is free of potassium chloride, \(C(0) = 0 g/L\), but as the solution with \(1.00 g/L\) concentration of KCl flows in, the concentration in the tank begins to pick up. If perfectly mixed, the outlet stream will have the same concentration as the tank, which means that the system will seek equilibrium. Determining this concentration is essential for controlling the quality of the output stream.

Aqueous Solution Mixing

The concept of aqueous solution mixing is critical where homogeneity of components within a solution is required. In the context of chemical engineering, this mixing must be efficient and well-understood to ensure consistent concentrations throughout the solution.

The exercise problem assumes perfect mixing, which implies that any addition to the tank is immediately and uniformly distributed. This ideal mixing is reflected in the potassium chloride balance equation, and without it, the modeled behavior would be dramatically different. Perfect mixing allows for simplifications in calculations and a clearer understanding of system dynamics. It's important to note, however, that real-world mixing may not be as idealized, and engineers must often consider non-ideal factors such as fluid dynamics and mixer efficiency.