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Water enters a \(2.00-\mathrm{m}^{3}\) tank at a rate of \(6.00 \mathrm{kg} / \mathrm{s}\) and is withdrawn at a rate of \(3.00 \mathrm{kg} / \mathrm{s}\). The tank is initially half full. (a) Is this process continuous, batch, or semibatch? Is it transient or steady state? (b) Write a mass balance for the process (see Example 4.2-1). Identify the terms of the general balance equation (Equation 4.2-1) present in your equation and state the reason for omitting any terms. (c) How long will the tank take to overflow?

Short Answer

Expert verified
a) This is a semi-batch transient process. b) The mass balance equation is dV/dt = (6.00 kg/s - 3.00 kg/s) / 1000 kg/m3. c) The tank will overflow after 250 seconds.

Step by step solution

01

Identifying the process

Usually, we have three types of processes, namely continuous process, batch process, or semi-batch process. That depends on whether any material is added or removed from the system over time. Since in this case, both adding and removing of water are simultaneously happening, it is a semi-batch process. Regarding its state, because there's a change in the system with time (as the amount of water in the tank is changing due to the difference in inflow and outflow), the process is transient.
02

Formulating the mass balance equation

The general form of a mass balance equation includes terms for accumulation, in flow, out flow and generation. In this case, the generation term is zero because water is neither being created nor destroyed inside the tank. Thus, the mass balance equation becomes: Accumulation = Inflow - Outflow.
03

Setting up the equation

The flow rates given are in terms of mass (kg/s). Therefore, we'll need to convert them to a volume flow rate by dividing by the density of water, which is about 1000 kg/m3. Then we can apply the mass balance equation developed in Step 2. Thus the equation becomes: change in volume with respect to time = Inflow rate - Outflow rate. This further simplifies to: dV/dt = (6.00 kg/s - 3.00 kg/s) / 1000 kg/m3 = 0.003 m3/s.
04

Solving for time

The time it will take for the tank to be filled up can be calculated by dividing the volume left to be filled by the net volume flow rate (the difference between the inflow and outflow). This can be obtained by integrating the equation from Step 3 over the volume from \(2.00 m3/2\) to \(2.00 m3\), and finding the corresponding time it takes. So, t = ∫(dV/((6 - 3)/1000 kg/m3) = (2 m3 - 2 m3 / 2) / (0.003 m3/s) = 250 seconds. Thus, it will take 250 seconds for the tank to be completely filled.

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Key Concepts

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

Semi-Batch Process
In chemical process engineering, a semi-batch process is a blend between batch and continuous processes. It involves both the continuous addition and removal of materials from the system.
For instance, imagine you're filling a bathtub while also draining water out. If the tap's flow is faster than the drain, eventually, the bathtub will overflow. This is similar to the tank scenario mentioned in the exercise. The tank is the 'bathtub,' with water constantly entering and leaving. Because there's simultaneous input and output, it's called a semi-batch process.
In real-world applications, semi-batch processes are quite beneficial. They offer flexibility, allowing for controlled reactions, and can be adapted to optimize product yield or purity. Think of it as having a more sophisticated control on your bathtub, where you tweak the inflow and outflow to maintain the right amount of water you need. This is particularly useful in chemical reactions needing precise ingredient additions or in waste treatment processes.
Mass Balance Equation
Moving on, we encounter the mass balance equation—an essential tool for engineers to track the amount of mass moving in and out of a system.
Using our previous bathtub analogy, it's like having a ledger that notes every cup of water added and every cup drained, to keep track of how much water is in the tub at any time.
In more technical terms, the mass balance equation is formulated as: Accumulation = Inflow - Outflow + Generation - Consumption. Here, generation and consumption refer to chemical reactions creating or using up the substance, but since our tank example involves plain water without any reactions, these terms are zero.
  • Accumulation: Represents the change in mass within the system over time.
  • Inflow: The incoming mass to the system (6 kg/s of water entering the tank).
  • Outflow: The outgoing mass from the system (3 kg/s of water leaving the tank).
The difference between inflow and outflow shows whether the mass in the system accumulates or decreases. It's fundamental in designing all sorts of processes, from megastructures like dams to miniature chemical reactors.
Transient State Analysis
Finally, transient state analysis is the examination of non-steady-state conditions, where variables such as concentration or temperature change with time within the system.
Consider when you first turn on the heat under a pot of water. For a while, the water temperature will keep rising—it's not constant. This period, until the water starts boiling at a steady temperature, is akin to a transient state. In the tank scenario, since the water level changes over time (due to different inflow and outflow rates), we’re also dealing with a transient system.
Engineers must understand these changing conditions to design systems that can cope with fluctuations without failure. In our case, knowing how long before the tank overflows is crucial to prevent spillage. Transient analysis involves a lot of differential equations, as seen in the step-by-step solution, to track how the system evolves over time.
This insight goes beyond just tanks—it's used in environmental engineering to predict pollutant spread in a lake, in mechanical engineering to understand heat transfer in car engines, and in nearly every corner where change occurs over time.

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Most popular questions from this chapter

A stream consisting of 44.6 mole \(\%\) benzene and \(55.4 \%\) toluene is fed at a constant rate to a process unit that produces two product streams, one a vapor and the other a liquid. The vapor flow rate is initially zero and asymptotically approaches half of the molar flow rate of the feed stream. Throughout this entire period, no material accumulates in the unit. When the vapor flow rate has become constant, the liquid is analyzed and found to be 28.0 mole\% benzene. (a) Sketch a plot of liquid and vapor flow rates versus time from startup to when the flow rates become constant. (b) Is this process batch or continuous? Is it transient or steady-state before the vapor flow rate reaches its asymptotic limit? What about after it becomes constant? (c) For a feed rate of 100 mol/min, draw and fully label a flowchart for the process after the vapor flow rate has reached its limiting value, and then use balances to calculate the molar flow rate of the liquid and the composition of the vapor in mole fractions.

In an absorption tower (or absorber), a gas is contacted with a liquid under conditions such that one or more species in the gas dissolve in the liquid. A stripping tower (or stripper) also involves a gas contacting a liquid, but under conditions such that one or more components of the feed liquid come out of solution and exit in the gas leaving the tower. A process consisting of an absorption tower and a stripping tower is used to separate the components of a gas containing 30.0 mole \(\%\) carbon dioxide and the balance methane. A stream of this gas is fed to the bottom of the absorber. A liquid containing 0.500 mole\% dissolved \(\mathrm{CO}_{2}\) and the balance methanol is recycled from the bottom of the stripper and fed to the top of the absorber. The product gas leaving the top of the absorber contains 1.00 mole \(\% \mathrm{CO}_{2}\) and essentially all of the methane fed to the unit. The CO_-rich liquid solvent leaving the bottom of the absorber is fed to the top of the stripper and a stream of nitrogen gas is fed to the bottom. Ninety percent of the \(\mathrm{CO}_{2}\) in the liquid feed to the stripper comes out of solution in the column, and the nitrogen/CO_stream leaving the column passes out to the atmosphere through a stack. The liquid stream leaving the stripping tower is the \(0.500 \% \mathrm{CO}_{2}\) solution recycled to the absorber. The absorber operates at temperature \(T_{\mathrm{a}}\) and pressure \(P_{\mathrm{a}}\) and the stripper operates at \(T_{\mathrm{s}}\) and \(P_{\mathrm{s}}\) Methanol may be assumed to be nonvolatile- -that is, none enters the vapor phase in either column and \(\mathrm{N}_{2}\), may be assumed insoluble in methanol. (a) In your own words, explain the overall objective of this two-unit process and the functions of the absorber and stripper in the process. (b) The streams fed to the tops of each tower have something in common, as do the streams fed to the bottoms of each tower. What are these commonalities and what is the probable reason for them? (c) Taking a basis of 100 mol/h of gas fed to the absorber, draw and label a flowchart of the process. For the stripper outlet gas, label the component molar flow rates rather than the total flow rate and mole fractions. Do the degree-of-freedom analysis and write in order the equations you would solve to determine all unknown stream variables except the nitrogen flow rate entering and leaving the stripper. Circle the variable(s) for which you would solve each equation (or set of simultaneous equations), but don't do any of the calculations yet. (d) Calculate the fractional \(\mathrm{CO}_{2}\) removal in the absorber (moles absorbed/mole in gas feed) and the molar flow rate and composition of the liquid feed to the stripping tower. (e) Calculate the molar feed rate of gas to the absorber required to produce an absorber product gas flow rate of \(1000 \mathrm{kg} / \mathrm{h}\). (f) Would you guess that \(T_{\mathrm{s}}\) would be higher or lower than \(T_{\mathrm{a}} ?\) Explain. (Hint: Think about what happens when you heat a carbonated soft drink and what you want to happen in the stripper.) What about the relationship of \(P_{\mathrm{s}}\) to \(P_{\mathrm{a}} ?\) (g) What properties of methanol would you guess make it the solvent of choice for this process? (In more general terms, what would you look for when choosing a solvent for an absorption-stripping process to separate one gas from another?)

A variation of the indicator-dilution method (see preceding problem) is used to measure total blood volume. A known amount of a tracer is injected into the bloodstream and disperses uniformly throughout the circulatory system. A blood sample is then withdrawn, the tracer concentration in the sample is measured, and the measured concentration [which equals (tracer injected)/(total blood volume) if no tracer is lost through blood vessel walls] is used to determine the total blood volume. In one such experiment, \(0.60 \mathrm{cm}^{3}\) of a solution containing \(5.00 \mathrm{mg} / \mathrm{L}\) of a dye is injected into an artery of a grown man. About 10 minutes later, after the tracer has had time to distribute itself uniformly throughout the bloodstream, a blood sample is withdrawn and placed in the sample chamber of a spectrophotometer. A beam of light passes through the chamber, and the spectrophotometer measures the intensity of the transmitted beam and displays the value of the solution absorbance (a quantity that increases with the amount of light absorbed by the sample). The value displayed is 0.18. A calibration curve of absorbance \(A\) versus tracer concentration \(C\) (micrograms dye/liter blood) is a straight line through the origin and the point \((A=0.9, C=3 \mu \mathrm{g} / \mathrm{L}) .\) Estimate the patient's total blood volume from these data.

In the production of a bean oil, beans containing 13.0 wt\% oil and \(87.0 \%\) solids are ground and fed to a stirred tank (the extractor) along with a recycled stream of liquid \(n\) -hexane. The feed ratio is \(3 \mathrm{kg}\) hexane/kg beans. The ground beans are suspended in the liquid, and essentially all of the oil in the beans is extracted into the hexane. The extractor effluent passes to a filter where the solids are collected and form a filter cake. The filter cake contains 75.0 wt\% bean solids and the balance bean oil and hexane, the latter two in the same ratio in which they emerge from the extractor. The filter cake is discarded and the liquid filtrate is fed to a heated evaporator in which the hexane is vaporized and the oil remains as a liquid. The oil is stored in drums and shipped. The hexane vapor is subsequently cooled and condensed, and the liquid hexane condensate is recycled to the extractor. (a) Draw and label a flowchart of the process, do the degree-of-freedom analysis, and write in an efficient order the equations you would solve to determine all unknown stream variables, circling the variables for which you would solve. (b) Calculate the yield of bean oil product (kg oil/kg beans fed), the required fresh hexane feed \(\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{14} / \mathrm{kg} \text { beans fed }\right),\) and the recycle to fresh feed ratio (kg hexane recycled/kg fresh feed). (c) It has been suggested that a heat exchanger might be added to the process. This process unit would consist of a bundle of parallel metal tubes contained in an outer shell. The liquid filtrate would pass from the filter through the inside of the tubes and then go on to the evaporator. The hot hexane vapor on its way from the evaporator to the extractor would flow through the shell, passing over the outside of the tubes and heating the filtrate. How might the inclusion of this unit lead to a reduction in the operating cost of the process? (d) Suggest additional steps that might improve the process economics.

A liquid mixture containing 30.0 mole \(\%\) benzene \((\mathrm{B}), 25.0 \%\) toluene \((\mathrm{T}),\) and the balance xylene \((\mathrm{X})\) is fed to a distillation column. The bottoms product contains 98.0 mole \(\% \mathrm{X}\) and no \(\mathrm{B},\) and \(96.0 \%\) of the \(\mathrm{X}\) in the feed is recovered in this stream. The overhead product is fed to a second column. The overhead product from the second column contains \(97.0 \%\) of the \(\mathrm{B}\) in the feed to this column. The composition of this stream is 94.0 mole\% B and the balance T. (a) Draw and label a flowchart of this process and do the degree-of-freedom analysis to prove that for an assumed basis of calculation, molar flow rates and compositions of all process streams can be calculated from the given information. Write in order the equations you would solve to calculate unknown process variables. In each equation (or pair of simultaneous equations), circle the variable(s) for which you would solve. Do not do the calculations. (b) Calculate (i) the percentage of the benzene in the process feed (i.e., the feed to the first column) that emerges in the overhead product from the second column and (ii) the percentage of toluene in the process feed that emerges in the bottom product from the second column.

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