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Chapter 10: Problem 1

A solution containing hydrogen peroxide with a mass fraction \(x_{\mathrm{p} 0}\) \(\left(\mathrm{kg} \mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{kg} \text { solution }\right)\) is added to a storage tank at a steady rate \(\dot{m}_{0}(\mathrm{kg} / \mathrm{h})\). During this process, the liquid level reaches a corroded spot in the tank wall and a leak develops. As the filling continues, the leak rate \(\dot{m}_{1}(\mathrm{kg} / \mathrm{h})\) becomes progressively worse. Moreover, once it is in the tank the peroxide begins to decompose at a rate $$r_{\mathrm{d}}(\mathrm{kg} / \mathrm{h})=k M_{\mathrm{p}}$$ where \(M_{\mathrm{p}}(\mathrm{kg})\) is the mass of peroxide in the tank. The tank contents are well mixed, so that the peroxide concentration is the same at all positions. At a time \(t=0\) the liquid level reaches the corroded spot. Let \(M_{0}\) and \(M_{\mathrm{p} 0}\) be the total liquid mass and mass of peroxide, respectively, in the tank at \(t=0,\) and let \(M(t)\) be the total mass of liquid in the tank at any time thereafter. (a) Show that the leakage rate of hydrogen peroxide at any time is \(\dot{m}_{1} M_{\mathrm{p}} / M\) (b) Write differential balances on the total tank contents and on the peroxide in the tank, and provide initial conditions. Your solution should involve only the quantities \(\dot{m}_{0}, \dot{m}_{1}, x_{\mathrm{p} 0}, k, M, M_{0}, M_{\mathrm{p}}\) \(M_{\mathrm{p} 0},\) and \(t\)

Short Answer

Expert verified
The leakage rate of the hydrogen peroxide at the time 't' is given by \(\dot{m}_{1} M_{\mathrm{p}} / M\). The total mass balance equation is \(dM/dt = \dot{m}_{0} - \dot{m}_{1}\) with initial condition \(M(0) = M_{0}\). The peroxide mass balance equation is \(dM_{\mathrm{p}}/dt = \dot{m}_{0}x_{\mathrm{p} 0}- (\dot{m}_{1} M_{\mathrm{p}} / M) - k M_{\mathrm{p}}\) with an initial condition of \(M_{\mathrm{p}}(0) = M_{\mathrm{p}0}\).

Step by step solution

01

Derive the Leakage Rate

To show that the leakage rate of hydrogen peroxide at any time is \(\dot{m}_{1} M_{\mathrm{p}} / M\), begin by considering that the rate of mass leaving the tank per unit time is proportional to the concentration of the species in the tank. The concentration is given by mass of the species divided by total mass, hence for peroxide, the concentration \(C_{\mathrm{p}}\) is given by \(M_{\mathrm{p}} / M\). Thus, the leakage rate of hydrogen peroxide \(\dot{m}_{\mathrm{p}1}\) is given by \(C_{\mathrm{p}} \cdot \dot{m}_{1} = (M_{\mathrm{p}} / M) \cdot \dot{m}_{1}\).
02

Derive the Total Mass Balance Equation

Start by writing the total mass balance equation, which states that the rate of change of mass in the tank is equal to the rate of mass entering minus rate of mass leaving plus rate of mass produced. However, as there is no production of mass (only decomposition of peroxide), the mass balance is: \(dM/dt = \dot{m}_{0} - \dot{m}_{1}\). The initial condition for this is \(M(0) = M_{0}\).
03

Derive the Peroxide Mass Balance Equation

Now write the mass balance for the peroxide. Using a similar approach, the rate of change of peroxide mass in the tank is equal to the rate of peroxide mass entering minus (rate of peroxide mass leaking + rate of peroxide mass decomposing). Therefore, \(dM_{\mathrm{p}}/dt = \dot{m}_{0}x_{\mathrm{p} 0}- (\dot{m}_{1} M_{\mathrm{p}} / M) - k M_{\mathrm{p}}\). The initial condition for this is \(M_{\mathrm{p}}(0) = M_{\mathrm{p}0}\).

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Key Concepts

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

Mass Balance Equations
Mass balance equations are fundamental to understanding chemical process dynamics. They represent the principle of conservation of mass, stating that mass cannot be created or destroyed in a system. In the context of a storage tank receiving a hydrogen peroxide solution, the mass balance equation calculates the change in mass within the tank over time. It considers the mass rate of the solution entering, as well as the mass leaking out of the tank, and reacts by decomposing.

For a well-mixed tank, the general mass balance equation is written as: \[\frac{dM}{dt} = \text{rate in} - \text{rate out} + \text{rate of production}\]. In the case of our storage tank with no production, this simplifies to \[\frac{dM}{dt} = \. \]. This equation is essential as it forms the basis for determining the leakage and decomposition rates, which affect the system's dynamics. It captures the idea that the rate at which the total mass of the solution in the tank changes is simply the difference between the mass flow in and the mass flow out.
Hydrogen Peroxide Leakage
When dealing with chemical storage, such as in the case of hydrogen peroxide, leakage can significantly impact the system's stability and safety. Understanding how to determine the leakage rate is crucial in managing the integrity of storage facilities. In our scenario, the mass balance approach helps ascertain the rate at which hydrogen peroxide leaks from a corroded storage tank as a function of time.

To calculate the specific leakage rate of the peroxide \(\. \), one must regard both the leakage rate \(\dot{m}_{1}\) and the fraction of peroxide in the total mixture, provided by \(C_{\text{p}} = \frac{M_{\text{p}}}{M}\). Combining these factors gives the leakage rate of hydrogen peroxide specifically as \(\. \) This understanding is vital for effective management and safety assessments of the chemical storage process.

It is also noteworthy to consider the practical aspect: by monitoring leakage rates, one can predict and prevent large-scale failures, ensuring a high level of operational safety.
Peroxide Decomposition Rate
Peroxide decomposition is a reaction in which hydrogen peroxide breaks down into water and oxygen. The decomposition rate is of particular interest in storage and handling, as it can contribute to changes in concentration and potential pressure build-up in closed containers. In our mass balance analysis, we quantify this rate with the expression \[\]. Here, \(k\) is the rate constant for the decomposition reaction, and \(M_{\text{p}}\) denotes the mass of hydrogen peroxide in the tank.

The formula illustrates that the peroxide decomposition rate is directly proportional to its mass in the tank. This dynamic behavior, typically exhibited by first-order reactions where the rate is proportional to the concentration of the reactant, needs to be carefully managed due to the potential for rapid changes in reaction rates.

In summary, the peroxide decomposition rate gives us insight into how quickly the chemical reaction proceeds, which is vital for safely maintaining chemical storage systems and preventing unintended escalation of the decomposition reaction within the tank.

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Most popular questions from this chapter

A \(40.0-\mathrm{ft}^{3}\) oxygen tent initially contains air at \(68^{\circ} \mathrm{F}\) and 14.7 psia. At a time \(t=0\) an enriched air mixture containing \(35.0 \%\) v/v \(\mathrm{O}_{2}\) and the balance \(\mathrm{N}_{2}\) is fed to the tent at \(68^{\circ} \mathrm{F}\) and 1.3 psig at a rate of \(60.0 \mathrm{ft}^{3} / \mathrm{min},\) and gas is withdrawn from the tent at \(68^{\circ} \mathrm{F}\) and 14.7 psia at a molar flow rate equal to that of the feed gas. (a) Calculate the total Ib-moles of gas \(\left(\mathrm{O}_{2}+\mathrm{N}_{2}\right)\) in the tent at any time. (b) Let \(x(t)\) equal the mole fraction of oxygen in the outlet stream. Write a differential mole balance on oxygen, assuming that the tent contents are perfectly mixed (so that the temperature, pressure, and composition of the contents are the same as those propertics of the exit stream). Convert the balance into an equation for \(d x / d t\) and provide an initial condition. (c) Integrate the equation to obtain an expression for \(x(t)\). How long will it take for the mole fraction of oxygen in the tent to reach 0.33 ? Sketch a plot of \(x\) versus \(t,\) labeling the value of \(x\) at \(t=0\) and the asymptotic value at \(t \rightarrow \infty\)

A tracer is used to characterize the degree of mixing in a continuous stirred tank. Water enters and leaves the mixer at a rate of \(\dot{V}\left(\mathrm{m}^{3} / \mathrm{min}\right) .\) Scale has built up on the inside walls of the tank, so that the effective volume \(V\left(\mathrm{m}^{3}\right)\) of the tank is unknown. At time \(t=0,\) a mass \(m_{0}(\mathrm{kg})\) of the tracer is injected into the tank and the tracer concentration in the outlet stream, \(C\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) is monitored. (a) Write a differential balance on the tracer in the tank in terms of \(V, C,\) and \(\dot{V},\) assuming that the tank contents are perfectly mixed, and convert the balance into an equation for \(d C / d t\). Provide an initial condition, assuming that the injection is rapid enough so that all of the tracer may be considered to be in the tank at \(t=0 .\) Without doing any calculations, sketch a plot of \(C\) versus \(t\) labeling the value of \(C\) at \(t=0\) and the asymptotic value at \(t \rightarrow \infty\) (b) Integrate the balance to prove that $$C(t)=\left(m_{0} / V\right) \exp (-\dot{V} t / V)$$ (c) Suppose the flow rate through the mixer is \(\dot{V}=30.0 \mathrm{m}^{3} / \mathrm{min}\) and that the following data are taken: (For example, at \(t=1\) min, \(C=0.223 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{3}\).) Verify graphically that the tank is functioning as a perfect mixer- -that is, that the expression of Part (b) fits the data- -and determine the effective volume \(V\left(\mathrm{m}^{3}\right)\) from the slope of your plot. (d) A solution of a radioactive element with a fairly short half-life (see Problem 10.16 ) is often used as a tracer for applications like the one in this problem. The advantage of doing so is that the concentration of the tracer at the outlet can be measured with a sensitive radiation detector mounted outside the exit pipe rather than having to draw fluid samples from the pipe and analyze them. What is a potential drawback of radiotracers? Why is it important that the half-life of the tracer be neither too short nor too long?

Water is added at varying rates to a 300 -liter holding tank. When a valve in a discharge line is opened, water flows out at a rate proportional to the height and hence to the volume \(V\) of water in the tank. The flow of water into the tank is slowly increased and the level rises in consequence, until at a steady input rate of \(60.0 \mathrm{L} / \mathrm{min}\) the level just reaches the top but does not spill over. The input rate is then abruptly decreased to \(40.0 \mathrm{L} / \mathrm{min}\). (a) Write the equation that relates the discharge rate, \(\dot{V}_{\text {out }}(\mathrm{L} / \mathrm{min}),\) to the volume of water in the tank, \(V(\mathrm{L}),\) and use it to calculate the steady-state volume when the input rate is \(40 \mathrm{L} / \mathrm{min}\). (b) Write a differential balance on the water in the tank for the period from the moment the input rate is decreased \((t=0)\) to the attainment of steady state \((t \rightarrow \infty),\) expressing it in the form \(d V / d t=\cdots \cdot\) Provide an initial condition. (c) Without integrating the equation, use it to confirm the steady-state value of \(V\) calculated in Part (a) and then to predict the shape you would anticipate for a plot of \(V\) versus \(t\). Explain your reasoning. (d) Separate variables and integrate the balance equation to derive an expression for \(V(t)\). Calculate the time in minutes required for the volume to decrease to within \(1 \%\) of its steady-state value.

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.

An immersed electrical heater is used to raise the temperature of a liquid from \(20^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 20.0 min. The combined mass of the liquid and the container is \(250 \mathrm{kg}\), and the mean heat capacity of the system is 4.00 kJ/(kg.'C). The liquid decomposes explosively at 85"C. At 10: 00 a.m. a batch of liquid is poured into the vessel, and the operator turns on the heater and answers a call on his cell phone. Ten minutes later, his supervisor walks by and looks at the computer display of the power input. This what she sees. The supervisor immediately shuts off the heater and charges off to pass on to the operator several brief observations that come to her mind. (a) Calculate the required constant power input \(\dot{Q}(\mathrm{k} \mathrm{W})\), neglecting energy losses from the container. (b) Write and integrate using Simpson's rule (Appendix A.3) an energy balance on the system to estimate the system temperature at the moment the heater is shut off. Use the following data from the recorder chart: $$\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|} \hline t(\mathrm{s}) & 0 & 30 & 60 & 90 & 120 & 150 & 180 & 210 & 240 & 270 & 300 \\ \hline \dot{Q}(\mathrm{kW}) & 33 & 33 & 34 & 35 & 37 & 39 & 41 & 44 & 47 & 50 & 54 \\ \hline \end{array}$$ $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline t(\mathrm{s}) & 330 & 360 & 390 & 420 & 450 & 480 & 510 & 540 & 570 & 600 \\ \hline \dot{Q}(\mathrm{kW}) & 58 & 62 & 66 & 70 & 75 & 80 & 85 & 90 & 95 & 100 \\\ \hline \end{array}$$ (c) Suppose that if the heat had not been shut off, \(\dot{Q}\) would have continued to increase linearly at a rate of \(10 \mathrm{kW} / \mathrm{min}\). At what time would everyone in the plant realize that something was wrong?
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