/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A liquid stream containing 50.0 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A liquid stream containing 50.0 mole \(\%\) benzene and the balance toluene at \(25^{\circ} \mathrm{C}\) is fed to a continuous single-stage evaporator at a rate of \(1320 \mathrm{mol} / \mathrm{s}\). The liquid and vapor streams leaving the evaporator are both at \(95.0^{\circ} \mathrm{C}\). The liquid contains 42.5 mole \(\%\) benzene and the vapor contains 73.5 mole\% benzene. (a) Calculate the heating requirement for this process in \(\mathrm{kW}\). (b) Using Raoult's law (Section 6.4b) to describe the equilibrium between the vapor and liquid outlet streams, determine whether or not the given benzene analyses are consistent with each other. If they are, calculate the pressure (torr) at which the evaporator must be operating; if they are not, give several possible explanations for the inconsistency.

Short Answer

Expert verified
The detailed calculation of the heating requirement will depend on the heat of vaporization values at the given temperature, which are not provided in the problem statement. The check for consistency depends on these values as well as the vapor pressures of each compound at the system temperature. If these are known, the calculation can be completed as outlined in the solution steps. If the benzene analyses are consistent, the operating pressure will depend on the benzene's vapor pressure at the system temperature.

Step by step solution

01

Determine molar fractions of the input

In order to find the heat requirement, we first need to break down the input feed. We know it consists of 50% benzene and 50% toluene, and the total rate is 1320 mol/s. Therefore, the rate of benzene is \(0.5 \times 1320\) mol/s and the rate of toluene is also \(0.5 \times 1320\) mol/s.
02

Calculate heat requirement

The heating requirement can be obtained by using the principle of conservation of energy, with the understanding that all the heat provided is used to vaporize the mixture. We know that the total heat energy supplied will be equal to the total enthalpy change of the system. Enthalpy change can be calculated by multiplying the molar flow rate with the heat of vaporization for each component and adding them up. We will need to find the heat of vaporization for both compounds at the given temperature (95°C) and then calculate the heat requirement.
03

Check consistency of benzene analysis

The consistency of the benzene analyses can be verified using Raoult's law, which describes the partial pressure contribution of each component in a mixture. We know that the benzene content in the liquid phase after evaporation is 42.5% and in the vapour phase is 73.5%. According to Raoult's Law, the ratio of these two should be equal to the ratio of their vapor pressures at the system temperature. We will also need the vapor pressure of benzene and toluene at the given temperature to calculate this. If the ratios are not equal, some possible explanations could be non-ideal behaviour or interactions between the components.
04

Calculate operating pressure

If the benzene analyses are consistent, we can use the Raoult's law to calculate the total pressure at which the evaporator must be operating. Raoult's law states that the partial pressure of a component in a mixture is equal to its mole fraction times the total pressure. Given that the benzene content in the vapour phase is 73.5%, we can multiply this by the benzene's vapour pressure at the system temperature to find the total pressure.

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Key Concepts

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

Raoult's Law
Understanding Raoult's Law is crucial for comprehending vapor-liquid equilibrium in mixtures. This law states that the partial pressure of each component in an ideal mixture of liquids is directly proportional to its mole fraction in the liquid phase times the pure component's vapor pressure. Essentially, Raoult’s Law provides a method to predict how different substances will behave when mixed and heated.

For a solution with two components, A and B, Raoult’s Law can be written as: \[ P_A = X_A \times P_A^* \] and \[ P_B = X_B \times P_B^* \] where \(P_A\) and \(P_B\) are the partial pressures of components A and B, \(X_A\) and \(X_B\) are the mole fractions of the components in the liquid phase, and \(P_A^*\) and \(P_B^*\) are the vapor pressures of the pure components at the same temperature.

Applying Raoult's Law helps in understanding if the benzene-toluene mixture in our exercise behaves ideally and thus verifies the consistencies in vapor and liquid phases. It's a fundamental principle for chemical engineers when analyzing separation processes like distillation.
Vapor-Liquid Equilibrium
The concept of vapor-liquid equilibrium (VLE) is pivotal in the analysis of separation processes. It describes the condition under which a liquid and its vapor phase are in equilibrium with each other at a given temperature and pressure, meaning that the rate at which molecules vaporize from the liquid equals the rate at which they condense from the vapor.

In the exercise, the benzene-toluene mixture is brought to equilibrium at 95°C. At this equilibrium state, each component of the mixture will have a specific composition in the liquid and vapor phases. The equilibrium compositions are often read from a phase diagram or calculated using thermodynamic principles, such as Raoult's Law, to ensure a proper understanding of the separation process.

Ensuring the correct calculation of the VLE is essential for designing and operating evaporators and distillation columns efficiently. A thorough understanding of this concept would allow us to determine if the given analyses of benzene are consistent, as it affects the overall process design and energy requirements.
Enthalpy Change
The enthalpy change of a system is the heat absorbed or released during a process at constant pressure. In chemical process calculations, it is often connected to the amount of energy required to break intermolecular forces during phase changes, such as vaporization.

In our exercise scenario, the enthalpy change is crucial for determining the heating requirement of the evaporator. For an isobaric (constant pressure) process, the enthalpy change is calculated by summing the products of the molar flow rates and the respective heats of vaporization for each component, which can be expressed as: \[ \text{Enthalpy Change} = (\text{Flow Rate of Benzene} \times \text{Heat of Vaporization of Benzene}) + (\text{Flow Rate of Toluene} \times \text{Heat of Vaporization of Toluene}) \] Understanding enthalpy change allows students to grasp how much energy is needed for the vaporization process, which is vital for optimal energy usage and equipment sizing in the industry.
Heat of Vaporization
The heat of vaporization is defined as the amount of energy required to convert a unit amount of a substance from its liquid phase to the vapor phase without a temperature change. This physical property is intrinsic to each substance and varies with temperature.

In practical terms, for the single-stage evaporator in our exercise, the heat of vaporization is a vital factor for calculating the total energy we must supply to vaporize benzene and toluene at 95°C. It’s typically measured in kJ/mol or kJ/kg. The correct values for the heat of vaporization are important to ensure accurate calculation of the heating requirement and the feasibility of the process. Without precise data on the heats of vaporization, the energy balance would be off, leading to inefficient process operation or even design failure.

Familiarity with the concept and the ability to calculate the heat of vaporization correctly are, therefore, essential skills for engineers working with thermodynamic and phase change processes.

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Most popular questions from this chapter

On a cold winter day the temperature is \(2^{\circ} \mathrm{C}\) and the relative humidity is \(15 \% .\) You inhale air at an average rate of \(5500 \mathrm{mL} / \mathrm{min}\) and exhale a gas saturated with water at body temperature, roughly \(37^{\circ} \mathrm{C} .\) If the mass flow rates of the inhaled and exhaled air (excluding water) are the same, the heat capacities \(\left(C_{p}\right)\) of the water-free gases are each \(1.05 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right),\) and water is ingested into the body as a liquid at \(22^{\circ} \mathrm{C},\) at what rate in \(\mathrm{J} /\) day do you lose energy by breathing? Treat breathing as a continuous process (inhaled air and liquid water enter, exhaled breath exits) and neglect work done by the lungs.

Your next-door neighbor, Josephine Rackstraw, surprised her husband last January by having a hot tub installed in their back yard while he was away on an ice-fishing trip. It surprised him, all right, but instead of being pleased he was horrified. "Have you lost your mind, Josephine?" he sputtered. "It will cost a fortune to keep this thing hot, and you know what the President said about conserving energy." "Don't be silly, Ralph," she replied. "It can't cost more than a few pennies a day, even in the dead of winter." "No way -just because you have a PhD, you think you're an expert on everything!" They argued for a while, bringing up several issues that each had been storing for just such an occasion. After calming down and using the tub for a week, they remembered their neighbor (i.e., you) had a chemical engineering education and came to ask if you could settle their argument. You asked a few questions, made several observations, converted everything to metric units, and arrived at the following data, all corresponding to an average outside temperature of \(5^{\circ} \mathrm{C}\). \- The tub holds 1230 liters of water. \- Ralph normally keeps the tub temperature at \(29^{\circ} \mathrm{C}\), raises it to \(40^{\circ} \mathrm{C}\) when he plans to use it, keeps it at \(40^{\circ} \mathrm{C}\) for about one hour, and drops it back to \(29^{\circ} \mathrm{C}\) when he is finished. \- During heating, it takes about three hours for the water temperature to rise from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). When the heat is shut off, it takes eight hours for the water temperature to drop back to \(29^{\circ} \mathrm{C}\). \- Electricity costs 10 cents per kilowatt-hour. Taking the heat capacity of the tub contents to be that of pure liquid water and neglecting evaporation, answer the following questions. (a) What is the average rate of heat loss ( \(k W\) ) from the tub to the outside air? (Hint: Consider the period when the tub temperature is dropping from \(40^{\circ} \mathrm{C}\) to \(29^{\circ} \mathrm{C}\).) (b) At what average rate ( \(\mathrm{kW}\) ) does the tub heater deliver energy to the water when raising the water temperature? What is the total quantity of electricity (kW\cdoth) that the heater must deliver during this period? [Consider the result of Part (a) when performing the calculation.] (c) (These answers should settle the argument.) Consider a day in which the tub is used once. Use the results of Parts (a) and (b) to estimate the cost (S) of heating the tub from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) and the cost \((\mathrm{S})\) of keeping the tub at a constant temperature. (There is no cost for the period in which \(T\) is dropping.) What is the total daily cost of running the tub? Assume the rate of heat loss is independent of the tub temperature. (d) The tub lid, which is an insulator, is removed when the tub is in use. Explain how this fact would probably affect your cost estimates in Part (c).

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The off-gas from a reactor in a process plant in the heart of Freedonia has been condensing and plugging up the vent line, causing a dangerous pressure buildup in the reactor. Plans have been made to send the gas directly from the reactor into a cooling condenser in which the gas and liquid condensate will be brought to \(25^{\circ} \mathrm{C}.\) (a) You have been called in as a consultant to aid in the design of this unit. Unfortunately, the chief (and only) plant engineer has disappeared and nobody else in the plant can tell you what the offgas is (or what anything else is, for that matter). However, a job is a job, and you set out to do what you can. You find an elemental analysis in the engineer's notebook indicating that the gas formula is \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\). On another page of the notebook, the off-gas flow rate is given as \(235 \mathrm{m}^{3} / \mathrm{h}\) at \(116^{\circ} \mathrm{C}\) and 1 atm. You take a sample of the gas and cool it to \(25^{\circ} \mathrm{C}\), where it proves to be a solid. You then heat the solidified sample at 1 atm and note that it melts at \(52^{\circ} \mathrm{C}\) and boils at \(113^{\circ} \mathrm{C}\). Finally, you make several assumptions and estimate the heat removal rate in \(\mathrm{kW}\) required to bring the off-gas from \(116^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\). What is your result? (b) If you had the right equipment, what might you have done to get a better estimate of the cooling rate?

The flow of groundwater often plagues construction of tunnels and other underground systems. Dne way of preventing it is with an ice seal-freezing the water in the soil so that the ice formed is a barrier to water movement. Such a structure was planned for the Fukushima TEPCO nuclear power plant, which was severely damaged by a 2011 tsunami that created a tremendous environmental challenge. A major concern was potential contamination with radioactive isotopes of groundwater flowing under the plant and into the ocean. A proposal under consideration was to channel the flow around the plant by forming an ice dam with a 1,400 -meter perimeter, a depth of \(30 \mathrm{m},\) and a thickness of approximately \(2 \mathrm{m}\). This was to be done by pumping a brine solution at a temperature of \(-40^{\circ} \mathrm{C}\) though vertical pipes spaced at \(1-\mathrm{m}\) intervals. The brine would exit at a temperature no greater than \(-25^{\circ} \mathrm{C}\). To keep ambient temperature fluctuations from causing occasional melting, the dam was to have a mean temperature of about \(-20^{\circ} \mathrm{C}\). (a) Estimate the average cooling rate (kW) and associated flow rate of brine (L/min) required to complete formation of the dam within 60 days of starting the refrigeration system. State and give your rationale for each of the assumptions and/or approximations necessary to obtain your result. (b) From a suitable reference, for which you must provide a citation, find an estimate of the ratio of the heat removed to the work done by a refrigeration system. Use the value to estimate the power usage during the time the dam is being created. (c) It is expected that substantially less power will be used once the dam has been formed. Explain. (d) Identify the primary radioactive species that were of greatest concern regarding contamination of the groundwater. (e) Explosions of hydrogen occurred in the power plants after the cooling water system was shut down upon being flooded by the tsunami. What was the source of the hydrogen? Describe the scenario that led to hydrogen formation.

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