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Chapter 8: Problem 4

The following equation pertains to the concentration of a chemical in a completely mixed reactor: \\[c=c_{\text {in }}\left(1-e^{-0.04 t}\right)+c_{0} e^{-0.04 t}\\] If the initial. consentration \(c_{0}=5\) and the inflow concentration \(c_{\text {in }}=12,\) compute the time required for \(c\) to be 85 percent of \(c_{\text {in }}\).

Short Answer

Expert verified
The time required for the concentration \(c\) to be 85 percent of the inflow concentration \(c_\text{in}\) is approximately 15.86 time units.

Step by step solution

01

Find 85 percent of the inflow concentration

We need to find 85 percent of the inflow concentration, \(c_{\text{in}}\), which will be the value of \(c\) we are looking for. To find the 85 percent of inflow concentration, multiply the inflow concentration by 0.85. \[c = 0.85 \times c_{\text{in}}\] Given that \(c_{\text{in}} = 12\). Therefore, \[ c = 0.85 \times 12 = 10.2\]
02

Substitute known values and solve for \(t\)

Now our concentration \(c\) is 10.2 when it reaches 85 percent of the inflow concentration. We need to find the time \(t\) for which this occurs. Using the given equation, \[10.2 = 12\left(1-e^{-0.04t}\right) + 5e^{-0.04t}\] Now, our goal is to solve this equation for \(t\).
03

Solve the equation for \(t\)

First, isolate the exponential terms by moving the constant term to the other side: \[10.2 - 12 = -12e^{-0.04t} + 5e^{-0.04t}\] Now combine the exponential terms: \[-1.8 = -7e^{-0.04t}\] Now, divide both sides by -7 and obtain: \[0.2571 = e^{-0.04t}\] Now take the natural logarithm of both sides: \[ln(0.2571) = -0.04t\] Finally, divide by -0.04 to obtain the time \(t\): \[t = \frac{ln(0.2571)}{-0.04} \approx 15.86\]
04

Interpret the result

The time required for the concentration \(c\) to be 85 percent of the inflow concentration \(c_\text{in}\) is approximately 15.86 time units.

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Key Concepts

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

Mixed Reactor
A mixed reactor, also known as a continuously stirred tank reactor (CSTR), is a common type of chemical reactor. In a mixed reactor, the contents are well-mixed and uniform in composition throughout. This is achieved through continuous stirring or agitation. The reactor allows the inflow and outflow of substances to maintain a steady state. In practical scenarios, achieving perfect mixing can be challenging, but the model helps in simplifying calculations.

Key characteristics of a mixed reactor include:
  • Uniform temperature and concentration.
  • Continuous input and output of reactants and products.
  • Typically modeled mathematically for steady-state conditions.
Engineers often use mixed reactors in industries such as chemical production, wastewater treatment, and pharmaceuticals. Understanding mixed reactors is essential for chemical engineers tasked with designing and optimizing them for efficient chemical processing.
Concentration Calculation
Concentration in chemical reaction engineering refers to the amount of a substance within a given volume. In our exercise, we are dealing with the concentration of a chemical in a mixed reactor influenced by both the initial concentration and the inflow concentration.

To compute concentrations in mixed reactors, use the provided equation:
\[c = c_{ ext{in}}(1-e^{-0.04t}) + c_{0}e^{-0.04t}\]
Where:
  • \(c\) is the final concentration.
  • \(c_{ ext{in}}\) is the inflow concentration.
  • \(c_{0}\) is the initial concentration.
Through these calculations, you determine the concentration at any given time. Here, you observe how factors such as inflow concentration, initial concentration, and time affect the chemical mix in reactors.
Exponential Decay
Exponential decay describes the process where a quantity decreases at a rate proportional to its current value. It is a common mathematical model in chemical reactions, indicating how reactant concentrations decline over time. In our context, the concentration term with time, \(e^{-0.04t}\), represents exponential decay.

Exponential decay in the reactor means:
  • The rate of change of concentration is not constant but decreases over time.
  • It forms a significant part of the alternative terms in the equation for concentration.
  • The decrease is rapid initially and slows down as time progresses.
Understanding exponential decay is pivotal for analyzing and predicting the time-dependent behavior of chemical reactants in reactors.
Time Calculation
Time calculation in the context of chemical reactions is crucial when predicting how long a reaction will take to reach a certain concentration. In the exercise, we calculated the time required for the concentration to reach 85% of the inflow concentration using a series of algebraic steps.

Steps to calculate time:
  • Set the desired concentration as a percentage of the inflow concentration.
  • Substitute known values into the concentration equation.
  • Rearrange the equation to isolate the exponential term.
  • Use natural logarithms to solve for \(t\).
  • Calculate \(t\) to find the approximate time needed.
These steps help understand the dynamics between reaction time and concentration levels, crucial for efficient chemical processing.

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Most popular questions from this chapter

The pressure drop in a section of pipe can be calculated as \\[\Delta p=f \frac{L \rho V^{2}}{2 D}\\] where \(\Delta p=\) the pressure drop (Pa), \(f=\) the friction factor, \(L=\) the length of pipe \([\mathrm{m}], \rho=\) density \(\left(\mathrm{kg} / \mathrm{m}^{3}\right), V=\) velocity \((\mathrm{m} / \mathrm{s})\) and \(D=\) diameter (m). For turbulent flow, the Colebrook equation provides a means to calculate the friction factor, \\[\frac{1}{\sqrt{f}}=-2.0 \log \left(\frac{\varepsilon}{3.7 D}+\frac{2.51}{\operatorname{Re} \sqrt{f}}\right)\\] where \(\varepsilon=\) the roughness \((\mathrm{m}),\) and \(\mathrm{Re}=\) the Reynolds number \\[\mathrm{Re}=\frac{\rho V D}{\mu}\\] where \(\mu=\) dynamic viscosity \(\left(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\right)\) (a) Determine \(\Delta p\) for a 0.2 -m-long horizontal stretch of smooth drawn tubing given \(\rho=1.23 \mathrm{kg} / \mathrm{m}^{3}, \mu=1.79 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) \(D=0.005 \mathrm{m}, V=40 \mathrm{m} / \mathrm{s},\) and \(\varepsilon=0.0015 \mathrm{mm} .\) Use a numer- ical method to determine the friction factor. Note that smooth pipes with \(\mathrm{Re}<10^{5},\) a good initial guess can be obtained using the Blasitus formula, \(f=0.316 / \mathrm{Re}^{0.25}\) (b) Repeat the computation but for a rougher commercial steel pipe \((\varepsilon=0.045 \mathrm{mm})\)

For fluid flow in pipes, friction is described by a dimensionless number, the Fanning friction factor \(f\). The Fanning friction factor is dependent on a number of parameters related to the size of the pipe and the fluid, which can all be represented by another dimensionless quantity, the Reynolds number Re. A formula that predicts \(f\) given \(\operatorname{Re}\) is the von Karman equation, \\[\frac{1}{\sqrt{f}}=4 \log _{10}(\operatorname{Re} \sqrt{f})-0.4\\] Typical values for the Reynolds number for turbulent flow are 10,000 to 500,000 and for the Fanning friction factor are 0.001 to \(0.01 .\) Develop a function that uses bisection to solve for \(f\) given a user-supplied value of Re between 2,500 and 1,000,000 . Design the function so that it ensures that the absolute error in the result is \(E_{a, d}<0.000005\).

The space shuttle, at lift-off from the launch pad, has four forces acting on it, which are shown on the free-body diagram (Fig. P8.46). The combined weight of the two solid rocket boosters and external fuel tank is \(W_{B}=1.663 \times 10^{6} \mathrm{lb}\). The weight of the orbiter with a full payload is \(W_{s}=0.23 \times 10^{6} \mathrm{Jb}\). The combined thrust of the two solid roctet boosters is \(T_{B}=5.30 \times 10^{6}\) lb. The combined thrust of the three liquid fuel orbiter engines is \(T_{S}=1.125 \times 10^{6} \mathrm{lb}\) At liftoff, the orbiter engine thrust is directed at angle \(\theta\) to make the resultant moment acting on the entire craft assembly (external tank, solid rocket boosters, and orbiter) equal to zero. With the resultant moment equal to zero, the craft will not rotate about its mass center \(G\) at liftoff. With these forces, the craft will have a resultant force with components in both the vertical and horizontal direction. The vertical resultant force component is what allows the craft to lift off from the launch pad and fly vertically. The horizontal resultant force component causes the craft to fly horizontally. The resultant mornent acting an the craft will be zero when \(\theta\) is adjusted to the proper value. If this angle is not adjusted properly, and there is some resultant moment acting on the craft, the craft will tend to rotate about it mass center. (a) Resolve the orbiter thrust \(T_{S}\) into horizontal and vertical components, and then sum moments about point \(G,\) the craft mass center. Set the resulting moment equation equal to zero. This equation can now be solived for the value of \(\theta\) required for liftoff. (b) Derive an equation for the resutent moment acting on the craft in terms of the angle \(\theta\). Plot the resultant moment as a function of the angle \(\theta\) over a range of -5 radians to +5 radians. (c) Write a computer program to solve for the angle \(\theta\) using Newton's method to find the root of the resultant moment equation. Make an initial first guess at the root of interest using the plot. Terminate your iterations when the value of \(\theta\) has better than five significant figures. (d) Repeat the program for the minimum payload weight of the orbiter of \(W_{S}=195,000 \mathrm{lb}\).

The operation of a constant density plug flow reactor for the production of a substance via an enzymatic reaction is described by the equation below, where \(V\) is the volume of the reactor, \(F\) is the flow rate of reactant \(C, C_{\text {in }}\) and \(C_{\text {out }}\) are the concentrations of reactant entering and leaving the reactor, respectively, and \(K\) and \(k_{\max }\) are constants. For a 500 -L reactor, with an inlet concentration of \(C_{\text {in }}=0.5 \mathrm{M},\) an inlet flow rate of \(40 \mathrm{L} / \mathrm{s}, k_{\max }=5 \times 10^{-3} \mathrm{s}^{-1},\) and \(K=0.1 \mathrm{M},\) find the concentration of \(C\) at the outlet of the reactor \\[\frac{V}{F}=-\int_{C_{i n}}^{C_{m}} \frac{K}{k_{\max } C}+\frac{1}{k_{\max }} d C\\]

The concentration of pollutant bacteria \(c\) in a lake decreases according to \\[c=75 e^{-1.5 t}+20 e^{-0.075 t}\\] Determine the time required for the bacteria concentration to be reduced to 15 using (a) the graphical method and (b) using the Newton-Raphson method with an initial guess of \(t=6\) and a stopping criterion of \(0.5 \% .\) Check your result.
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