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Chapter 10: Problem 8

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.

Short Answer

Expert verified
The steady-state volume of the tank when the input rate is decreased to \( 40 \, L/min \) is \( 200 \,L \). The differential balance for the system is \( dV/dt = r_{in} - kV \). The volume of water in the tank decreases over time, reaching the steady-state value, and follows an exponential decay curve. The volume decreases to within \(1 \%\) of its steady-state value in approximately \( 4.6 \, mi

Step by step solution

01

Model Discharge Rate

Given the input rate \( r_{in} = 60.0 \, L/min \), and no water is spilling means, the output rate \(r_{out}\) is equal to the input rate at steady state. As the output rate is proportional to the volume of water in the tank, we set proportional constant \(k\) and write it as \(r_{out} = kV\). Here the proportionality constant \(k\) is \(60.0 \, L/min / 300 \, L = 0.2 \, min^{-1}\).
02

Calculation at New Steady State

The input rate decreases to \(40.0 \, L/min\). At the new steady state, the output rate equals the input rate \(40.0 \, L/min = kV_{s.s}\). Solving it for \(V_{s.s}\), we find \(V_{s.s} = 40.0 \, L/min / 0.2 \, min^{-1} = 200 \, L\).
03

Writing and Analyzing the Differential Balance

The rate of change of volume \(V\) with respect to time \(t\) is given by the difference of the input and output rates \(dV/dt = r_{in} - r_{out} = r_{in} - kV\). Initially at \(t=0\), the volume is \(300 \, L\). As \(t \rightarrow \infty\), the volume strives to reach the steady-state value of \(200 \, L\). The differential equation confirms the steady-state value, and anticipates that the volume decreases with time
04

Solving for V(t)

Separating the variable and integrating, we find \( \int_{V(0)}^{V(t)} dV / (V_{s.s} - V) = - \int_0^t k dt \), on solving which, we obtain \( - \ln |V_{s.s} - V(t)| = -kt + \ln|V_{s.s} - V(0)| \). Hence, \( V(t) = V_{s.s} + (V(0) - V_{s.s}) e^{-kt} \), where \(V(0)\) is the initial volume, \(V_{s.s}\) is the steady-state volume and \(k\) is the proportionality constant. To find the time when volume reaches within \(1 \%\) of its steady state, we set \( V(t) = 0.99 V_{s.s} \) and solve for \( t \), which gives \( t = -\ln(0.01) / k \approx 4.6 \, minutes \).

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Key Concepts

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

Differential Equation Modeling
Modeling a chemical process using differential equations involves understanding how variables like volume or concentration change over time. When water flows into and out of a tank, the situation can be captured with a differential equation. Here, the input rate, which is the flow of water into the tank, is described as a function of time, while the output rate is dependent on the tank's volume. This results in an equation expressing how the volume of water in the tank changes with respect to time.For this specific tank problem, we model the rate of change of water volume, denoted as \( \frac{dV}{dt} \), based on the difference between the input and output rates: \[\frac{dV}{dt} = r_{in} - r_{out} = r_{in} - kV\]Here, \(r_{in}\) is the rate at which water enters the tank, and \(r_{out} = kV\), where \(k\) is a constant of proportionality. This equation accounts for how water is both entering and draining from the tank, illustrating the dynamic nature of such processes.
Steady State Analysis
Steady state analysis helps in determining conditions where the system parameters become constant over time. In a chemical process, this means finding the point where the input and output rates are balanced, leading to a constant volume in the tank.In this exercise, initially, the tank is filled to the top at a rate of \(60.0 \, \text{L/min}\). Assuming no spillage means that this input matches an equal output rate defining its first steady state. When the input rate changes to \(40.0 \, \text{L/min}\), the system seeks a new balance.By setting the differential change in volume, \( \frac{dV}{dt} \), to zero, we solve for the steady state volume:\[0 = 40.0 - kV_{s.s}\]Solving for \(V_{s.s}\), we find it to be \(200 \, \text{L}\). The importance of steady state analysis lies in its ability to predict system behavior without time dependency, simplifying complex dynamic systems into manageable forms.
Rate of Change in Chemical Processes
In chemical processes, the rate of change is crucial for understanding how quickly a system evolves from one state to another. This concept revolves around how factors like volume or concentration change over time.For the water tank problem, considering the initial starting volume \(V(0) = 300 \, \text{L}\) and a different input rate, we utilize the differential equation:\[\frac{dV}{dt} = 40.0 - 0.2V\]This indicates the loss of volume over time as the tank adjusts to the new input rate. Solving and integrating this differential equation gives:\[V(t) = V_{s.s} + (V(0) - V_{s.s}) e^{-kt}\]This expression reveals how the volume approaches its steady state value. It incorporates both the initial condition and the rate constant, providing insight into the dynamic adjustments.By understanding this, students can predict how fast a system reaches equilibrium, like determining that it takes approximately \(4.6\) minutes for the volume to be within \(1\%\) of the steady state, thus, appreciating the pace of chemical adjustments.

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Most popular questions from this chapter

A kettle containing 3.00 liters of water at a temperature of \(18^{\circ} \mathrm{C}\) is placed on an electric stove and begins to boil in three minutes. (a) Write an energy balance on the water and determine an expression for \(d T / d t,\) neglecting evaporation of water before the boiling point is reached, and provide an initial condition. Sketch a plot of \(T\) versus \(t\) from \(t=0\) to \(t=4\) minutes. (b) Calculate the average rate (W) at which heat is being added to the water. Then calculate the rate (g/s) at which water vaporizes once boiling begins. (c) The rate of heat output from the stove element differs significantly from the heating rate calculated in Part (b). In which direction, and why?

Ninety kilograms of sodium nitrate is dissolved in \(110 \mathrm{kg}\) of water. When the dissolution is complete (at time \(t=0\) ), pure water is fed to the tank at a constant rate \(\dot{m}(\mathrm{kg} / \mathrm{min}),\) and solution is withdrawn from the tank at the same rate. The tank may be considered perfectly mixed. (a) Write a total mass balance on the tank and use it to prove that the total mass of liquid in the tank remains constant at its initial value. (b) Write a balance on sodium nitrate, letting \(x(t, \dot{m})\) equal the mass fraction of \(\mathrm{NaNO}_{3}\) in the tank and outlet stream. Convert the balance into an equation for \(d x / d t\) and provide an initial condition. (c) On a single graph of \(x\) versus \(t,\) sketch the shapes of the plots you would expect to obtain for \(\dot{m}=50 \mathrm{kg} / \mathrm{min}, 100 \mathrm{kg} / \mathrm{min},\) and \(200 \mathrm{kg} / \mathrm{min} .\) (Don't do any calculations.) Explain your reason- ing, using the equation of Part (b) in your explanation. (d) Separate variables and integrate the balance to obtain an expression for \(x(t, \dot{m})\). Check your solution. Then generate the plots of \(x\) versus \(t\) for \(\dot{m}=50 \mathrm{kg} / \mathrm{min}, 100 \mathrm{kg} / \mathrm{min},\) and \(200 \mathrm{kg} / \mathrm{min}\) and show them on a single graph. (A spreadsheet is a convenient tool for carrying out this step.) (e) If \(\dot{m}=100 \mathrm{kg} / \mathrm{min}\), how long will it take to flush out \(90 \%\) of the sodium nitrate originally in the tank? How long to flush out 99\%? 99.9\%? (f) The stream of water enters the tank at a point near the top, and the exit pipe from the tank is located on the opposite side toward the bottom. One day the plant technician forgot to turn on the mixing impeller in the tank. On the same chart, sketch the shapes of the plots of \(x\) versus \(t\) you would expect to see with the impeller on and off, clearly showing the differences between the two curves at small values and large values of \(t .\) Explain your reasoning.

A stirred tank contains \(1500 \mathrm{lb}_{\mathrm{m}}\) of pure water at \(70^{\circ} \mathrm{F}\). At time \(t=0,\) two streams begin to flow into the tank and one is withdrawn. One input stream is a \(20.0 \mathrm{wt} \%\) aqueous solution of \(\mathrm{NaCl}\) at \(85^{\circ} \mathrm{F}\) flowing at a rate of \(15 \mathrm{lb}_{\mathrm{m}} / \mathrm{min},\) and the other is pure water at \(70^{\circ} \mathrm{F}\) flowing at \(10 \mathrm{lb}_{\mathrm{m}} / \mathrm{min} .\) The mass of liquid in the tank is held constant at \(1500 \mathrm{lb}_{\mathrm{m}}\). Perfect mixing in the tank may be assumed, so that the outlet stream has the same \(\mathrm{NaCl}\) mass fraction \((x)\) and temperature \((T)\) as the tank contents. Also assume that the heat of mixing is zero and the heat capacity of all fluids is \(C_{p}=1 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) (a) Write differential material and energy balances and use them to derive expressions for \(d x / d t\) and \(d T / d t\) (b) Without solving the equations derived in Part (a), sketch plots of \(T\) and \(x\) as a function of time \((t)\) Clearly identify values at time zero and as \(t \rightarrow \infty\)

A radioactive isotope decays at a rate proportional to its concentration. If the concentration of an isotope is \(C(\mathrm{mg} / \mathrm{L}),\) then its rate of decay may be expressed as $$r_{\mathrm{d}}[\mathrm{mg} /(\mathrm{L} \cdot \mathrm{s})]=k C$$ where \(k\) is a constant. (a) A volume \(V(\mathrm{L})\) of a solution of a radioisotope whose concentration is \(C_{0}(\mathrm{mg} / \mathrm{L})\) is placed in a closed vessel. Write a balance on the isotope in the vessel and integrate it to prove that the half-life \(t_{1 / 2}\) of the isotope \(-\) by definition, the time required for the isotope concentration to decrease to half of its initial value- equals ( \(\ln 2\) )/ \(k\). (b) The half-life of \(^{56} \mathrm{Mn}\) is \(2.6 \mathrm{h}\). A batch of this isotope that was used in a radiotracing experiment has been collected in a holding tank. The radiation safety officer declares that the activity (which is proportional to the isotope concentration) must decay to \(1 \%\) of its present value before the solution can be discarded. How long will this take?

The flow rate of a process stream has tended to fluctuate considerably, creating problems in the process unit to which the stream is flowing. A horizontal surge drum has been inserted in the line to maintain a constant downstream flow rate even when the upstream flow rate varies. A cross-section of the drum, which has length \(L\) and radius \(r,\) is shown below. The level of liquid in the drum is \(h\), and the expression for liquid volume in the drum is $$V=L\left[r^{2} \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{r^{2}-(r-h)^{2}}\right]$$ Here is how the drum works. The rate of drainage of a liquid from a container varies with the height of the liquid in the container: the greater the height, the faster the drainage rate. The drum is initially charged with enough liquid so that when the input rate has its desired value, the liquid level is such that the drainage rate from the drum has the same value. A sensor in the drum sends a signal proportional to the liquid level to a control valve in the downstream line. If the input flow rate increases, the liquid level starts to rise; the control valve detects the rise from the transmitted signal and opens to increase the drainage rate, stopping when the level comes back down to its set-point value. Similarly, if the input flow rate drops, the control valve closes enough to bring the level back up to its set point. (a) The drum is to be charged initially with benzene (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3}\) ) at a constant rate \(\dot{m}(\mathrm{kg} / \mathrm{min})\) until the tank is half full. If \(L=5 \mathrm{m}, r=1 \mathrm{m},\) and \(\dot{m}=10 \mathrm{kg} / \mathrm{min},\) how long should it take to reach that point? (b) Now suppose the flow rate into the tank is unknown. A sight gauge on the tank allows determination of the liquid level, and instructions are to stop the flow when the tank contains 3000 kg. At what value of \(h\) should this be done? (c) After the tank has been charged, the flow rate into the drum, \(\dot{m}_{1}\), varies with upstream operations, and the flow rate out is \(10 \mathrm{kg} / \mathrm{min}\). Write a mass balance around the drum so that you obtain a relationship between \(\dot{m}_{1}\) and the rate of change in the height of liquid in the tank \((d h / d t)\) as a function of \(h .\) Estimate the flow rate into the tank when \(h\) has an approximate value of \(50 \mathrm{cm}\), and \(d h / d t=1 \mathrm{cm} / \mathrm{min} .\) (Hint: Although an analytical solution is feasible, you may find it easier to create plots of \(V\) and \(d V / d h\) at \(0.1 \mathrm{m}\) increments in \(h,\) which can be used in obtaining an approximate solution to the problem.) (d) Speculate on why the drum would provide better performance than feeding a signal proportional to the flow rate directly to the control valve that would cause the valve to close if the flow rate drops below the set point and to open if the flow rate rises above that point.
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