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Chapter 6: Problem 90

An ore containing \(90 \mathrm{wt} \% \mathrm{MgSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) and the balance insoluble minerals is fed to a dissolution tank at a rate of \(60,000 \mathrm{lb}_{\mathrm{m}} / \mathrm{h}\) along with fresh water and a recycle stream. The tank contents are heated to \(120^{\circ} \mathrm{F}\), causing all of the magnesium sulfate monohydrate in the ore to dissolve, forming a solution 10^0 F above saturation. The resulting slurry of the insoluble minerals in MgSO_solution is pumped to a heated filter, where a wet filter cake is separated from a solids-free filtrate. The filter cake retains \(5 \mathrm{lb}_{\mathrm{m}}\) of solution per \(100 \mathrm{lb}_{\mathrm{m}}\) of solids. The filtrate is sent to a crystallizer in which the temperature is reduced to \(50^{\circ} \mathrm{F},\) producing a slurry of \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) crystals in a saturated solution that is sent to another filler. The product filter cake contains all of the prea entrained solution in a ratio of \(5 \mathrm{Ib}_{\mathrm{m}}\) solution per \(100 \mathrm{lb}_{\mathrm{m}}\) crystals. The filtrate from this filter is returned to the dissolution tank as the recycle stream.Solubility data: Saturated magnesium sulfate solutions at \(110^{\circ} \mathrm{F}\) and \(50^{\circ} \mathrm{F}\) contain \(32 \mathrm{wt} \%\) \(\mathrm{MgSO}_{4}\) and \(23 \mathrm{wt} \% \mathrm{MgSO}_{4},\) respectively.(a) Explain why the solution is first heated (in the dissolution tank) and filtered and then cooled (in the crystallizer) and filtered. (b) Calculate the production rate of crystals and the required feed rate of fresh water to the dissolution tank. (Note: Don't forget to include water of hydration when you write a mass balance on water.)(c) Calculate the ratio \(\mathrm{lb}_{\mathrm{m}}\) recycle/lb \(_{\mathrm{m}}\) makeup water.

Short Answer

Expert verified
The production rate of MgSO4.7H2O crystals is 54000 lbm/h. The required feed rate of fresh water to the dissolution tank is 114750 lbm/h. The ratio of recycle to makeup water is 1.

Step by step solution

01

Understanding the Process

The first part of our production process is the dissolution tank, where the ore (containing 90 wt% of MgSO4.H2O) is mixed with fresh water. During this phase, all the MgSO4.H2O in the ore dissolves, forming a supersaturated solution as all the dissolved MgSO4.H2O is above saturation. This solution is then filtered to separate the insoluble solids. The filtrate is fed into a crystallizer, where the temperature is reduced to form MgSO4.7H2O crystals in a saturated solution. The slurry, in which the MgSO4.7H2O is suspended, is filtered to separate the crystals from the solution, and the remaining solution is recycled back to the dissolution tank.
02

Calculate mass of MgSO4.H2O and Insoluble Mass

From the problem, it's given that the feed rate to the dissolution tank is 60,000 lbm/h, of which 90% is MgSO4.H2O. Therefore, the mass of MgSO4.H2O = \(0.90 \times 60000 = 54000 lbm/h\). The remaining 10% of the feed is insoluble minerals, therefore, the insoluble mass in the feed = \(0.10 \times 60000 = 6000 lbm/h\).
03

Calculate production rate of crystals

Since the MgSO4.H2O is converted into MgSO4.7H2O in the crystallizer, and based on the principle of mass balance, the mass of MgSO4.H2O should be equal to the mass of MgSO4.7H20. Hence, in this case the production rate of MgSO4.7H20 crystals is the same as the mass of MgSO4.H2O = 54000 lbm/h.
04

Calculate mass of fresh water needed

From the solubility data, it's known that at \(110^{\circ} F\), the concentration of a saturated solution of magnesium sulfate is 32 wt%. Therefore, in the filtrate leaving the dissolution tank, the mass of MgSO4 (from MgSO4.H2O) is 32% of the total mass. So the total mass = \( Mass of MgSO4 /0.32 = 54000 lb_m /0.32 = 168750 lb_m/h \). Since the filtrate contains only the MgSO4 from the MgSO4.H2O and the fresh water, the mass of fresh water needed = \( total mass - mass of MgSO4.H2O = 168750 - 54000= 114750 lb_m/h\).
05

Calculate ratio of recycle to makeup water

In addition to fresh water, the dissolution tank also receives the recycle stream from the crystallizer filter. To calculate the mass of the recycle stream, since all of the magnesium sulfate monohydrate in the filtrate crystallizes in the crystallizer, the mass of the remaining solution in the slurry after filtration is \( total mass from step 4 鈥� mass of crystals = 168750 - 54000 = 114750 lb_m/h \). Therefore, the ratio lb_m recycle/lb_m makeup water becomes \( 114750 / 114750 = 1 \)

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Key Concepts

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

Understanding Mass Balance Calculations
Mass balance calculations are a fundamental aspect of chemical engineering processes, critical for designing efficient and cost-effective systems. They are based on the principle of conservation of mass, which dictates that mass cannot be created or destroyed in a chemical process, only transformed. In the process described in our textbook exercise, mass balance is applied to assess the movement and transformation of materials through the chemical process, which involves dissolving magnesium sulfate from ore in a dissolution tank and subsequently crystallizing it.

When solving mass balance problems, the first step is to create a basis of calculation. In the exercise, the known feed rate of ore is the starting point. With the given feed rate of 60,000 lbm/h and the concentration of MgSO鈧�.H鈧侽 in the ore, we calculated the mass of MgSO鈧�.H鈧侽 entering the dissolution tank. Understanding these relationships and the movement of mass through the system is critical for the next steps, such as determining the amount of fresh water needed and the rate of production of crystals. The mass of materials at each stage must account for all inputs, transformations, and outputs, ensuring the 'mass in equals mass out' rule is satisfied.

To make the concept more digestible, consider the following points to guide you through typical mass balance problems:
Designing Chemical Processes
The design of a chemical process, like the operation described in the exercise, involves meticulously planning each step to ensure that the desired transformation of materials is achieved efficiently and safely. Key steps in chemical process design include identifying the sequence of chemical and physical operations, selecting the appropriate equipment, and controlling process conditions such as temperature and pressure.

In this exercise, the ore dissolution and crystallization stages are carefully designed to optimize the yield of magnesium sulfate crystals. The temperature of the process is controlled to enhance solubility in the dissolution tank, and later, reduced in the crystallizer to facilitate crystallization. Additionally, filtration steps are included to separate insoluble minerals and to recover the desired product.

The design also cleverly incorporates a recycle stream, which minimizes waste and improves the sustainability of the overall process. This stream is analyzed as part of the mass balance, showing how the same principles apply whether in the initial design phase or during a problem-solving exercise.
Solubility and Crystallization
Solubility and crystallization are key concepts in many chemical engineering processes, including the exercise's purification of magnesium sulfate. Solubility refers to the maximum amount of solute that can be dissolved in a solvent at a given temperature and pressure. In our exercise, the solubility of magnesium sulfate is different at the two temperatures provided, which directly influences the design and operation of the dissolution tank and crystallizer.

During the dissolution phase, the solution is heated to a temperature where the solubility of magnesium sulfate increases, allowing for more of the compound to dissolve. This is indicated by the solution being 10掳F above saturation. In contrast, the crystallization phase involves cooling the solution, decreasing solubility and encouraging the formation of magnesium sulfate heptahydrate crystals. The process steps of heating and cooling are thus critical to achieve the desired solubility before filtration.

It is important for students to grasp that temperature manipulation is a powerful tool in controlling solubility and crystallization. This understanding is not only vital to tackling exercises but also to the practical application in real-world scenarios where precise predictions of solubility behavior influence process outcomes.

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Most popular questions from this chapter

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at \(58.3^{\circ} \mathrm{C}\), the height of the mercury in the arm of the manometer connected to the flask is \(747 \mathrm{mm}\) and that in the arm open to the atmosphere is \(52 \mathrm{mm} . \mathrm{At} 110^{\circ} \mathrm{C},\) the mercury level is \(577 \mathrm{mm}\) in the arm connected to the flask and \(222 \mathrm{mm}\) in the other arm. Atmospheric pressure is \(755 \mathrm{mm} \mathrm{Hg}\). (a) Extrapolate the data using the Clausius-Clapeyron equation to estimate the vapor pressure of chlorobenzene at \(130^{\circ} \mathrm{C}\). (b) Air saturated with chlorobenzene at \(130^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) is cooled to \(58.3^{\circ} \mathrm{C}\) at constant pressure. Estimate the percentage of the chlorobenzene originally in the vapor that condenses. (See Example 6.3-2.)(c) Summarize the assumptions you made in doing the calculation of Part (b).

Sulfur trioxide (SO \(_{3}\) ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) The vapor in equilibrium with the solution contains both \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). If enough \(\mathrm{SO}_{3}\) is added, all of the water reacts and the solution becomes pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If still more \(\mathrm{SO}_{3}\) is added, it dissolves to form a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure \(\mathrm{SO}_{3}\). Twenty percent oleum by definition contains \(20 \mathrm{kg}\) of dissolved \(\mathrm{SO}_{3}\) and \(80 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as \(\% \mathrm{SO}_{3}\) by mass, with the constituents of the oleum considered to be \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Prove that a \(15.0 \%\) oleum contains \(84.4 \% \mathrm{SO}_{3}\) (b) Suppose a gas stream at \(40^{\circ} \mathrm{C}\) and 1.2 atm containing 90 mole \(\% \mathrm{SO}_{3}\) and \(10 \% \mathrm{N}_{2}\) contacts a liquid stream of 98 wt\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (aq), producing \(15 \%\) oleum. Tabulated equilibrium data indicate that the partial pressure of \(S O_{3}\) in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of \(S O_{3}\) in the outlet gas if this gas is in equilibrium with the liquid product at \(40^{\circ} \mathrm{C}\) and 1 atm, and (ii) the ratio ( \(\mathrm{m}^{3}\) gas feed) \(/\) (kg liquid feed).

Using Raoult's law or Henry's law for each substance (whichever one you think appropriate), calculate the pressure and gas-phase composition (mole fractions) in a system containing a liquid that is 0.3 mole \(\% \mathrm{N}_{2}\) and 99.7 mole \(\%\) water in equilibrium with nitrogen gas and water vapor at \(80^{\circ} \mathrm{C}\).

Nitric acid is a chemical intermediate primarily used in the synthesis of ammonium nitrate, which is used in the manufacture of fertilizers. The acid also is important in the production of other nitrates and in the separation of metals from ores. Nitric acid may be produced by oxidizing ammonia to nitric oxide over a platinum-rhodium catalyst, then oxidizing the nitric oxide to nitrogen dioxide in a separate unit where it is absorbed in water to form an aqueous solution of nitric acid.The reaction sequence is as follows:$$\begin{aligned} 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \\\4 \mathrm{NO}+2 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}_{2} \\\4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2} & \rightarrow 4 \mathrm{HNO}_{3}(\mathrm{aq}) \end{aligned}$$.Ammonia vapor produced by vaporizing pure liquid ammonia at 820 kPa absolute is mixed with air, and the combined stream enters the ammonia oxidation unit. Air at \(30^{\circ} \mathrm{C}, 1\) atm absolute, and \(50 \%\) relative humidity is compressed and fed to the process. A fraction of the air is sent to the cooling and hydration units, while the remainder is passed through a heat exchanger and mixed with the ammonia. The total oxygen fed to the process is the amount stoichiometrically required to convert all of the ammonia to HNO \(_{3},\) while the fraction sent to the ammonia oxidizer corresponds to the stoichiometric amount required to convert ammonia to NO.The ammonia reacts completely in the oxidizer, with \(97 \%\) forming NO and the rest forming \(\mathrm{N}_{2}\). Only a negligible amount of \(\mathrm{NO}_{2}\) is formed in the oxidizer. However, the gas leaving the oxidizer is subjected to a series of cooling and hydration steps in which the NO is completely oxidized to \(\mathrm{NO}_{2}\) which in turn combines with water (some of which is present in the gas from the oxidizer and the rest is added) to form a 55 wt\% aqueous solution of nitric acid. The product gas from the process may be taken to contain only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). (a) Taking a basis of \(100 \mathrm{kmol}\) of ammonia fed to the process, calculate (i) the volumes \(\left(\mathrm{m}^{3}\right)\) of the ammonia vapor and air fed to the process using the compressibility-factor equation of state; (ii) the amount (kmol) and composition (in mole fractions) of the gas leaving the oxidation unit; (iii) the required volume of liquid water \(\left(\mathrm{m}^{3}\right)\) that must be fed to the cooling and hydration units; and (iv) the fraction of the air fed to the ammonia oxidizer. (b) Scale the results from Part (a) to a new basis of 100 metric tons per hour of 55\% nitric acid solution.(c) Nitrogen oxides (collectively referred to as \(\mathrm{NO}_{x}\) ) are a category of pollutants that are formed in many ways, including processes like that described in this problem. List the annual emission rates of the three largest sources of \(\mathrm{NO}_{x}\) emissions in your home region. What are the effects of exposure to excessive concentrations of \(\mathrm{NO}_{x} ?\) (d) A platinum-rhodium catalyst is used in ammonia oxidation. Fxplain the function of the catalyst, describe its structure, and explain the relationship of the structure to the function.

When a flammable liquid (e.g.. gasoline) ignites, the substance actually buming is vapor generated from the liquid. If the concentration of the vapor in the air above the liquid exceeds a certain level (the lower flammability limit), the vapor will ignite if it is exposed to a spark or another ignition source. Once ignited, the heat released is likely to cause additional vaporization of the liquid, and the resulting fire may continue until all combustible material has been consumed.(a) The flash point is defined as the minimum temperature at which a flammable liquid or volatile solid gives off sufficient vapor to form an ignitable mixture with air near the surface of the liquid or within a vessel (page \(2-515,\) Perry's Chemical Engineers' Handbook, see Footnote 1 ). For example, the flash point of \(n\) -octane at 1.0 atm is \(13^{\circ} \mathrm{C}\left(55^{\circ} \mathrm{F}\right)\), which means that dropping a match into an open container of octane is likely to start a fire in a laboratory, but not outside on a cold winter day. (Do not try it! One reference- -L. Bretherick, Bretherick's Handbook of Reactive Chemical Hazards, 4th Edition, Butterworths, London, 1990, p. 1596 - points out there is "usually a fair [our emphasis] correlation between flash point and probability of involvement in fire.")Suppose you are keeping two solvents in your laboratory, one with a flash point of \(15^{\circ} \mathrm{C}\) and the other with a flash point of \(75^{\circ} \mathrm{C}\). How do these solvents differ from the standpoint of safety? What differences, if any, should there be in how you treat them?(b) The lower flammability limit (LFL) of methanol in air is 6.0 mole \(\%\). Calculate the temperature at which a saturated methanol-air mixture at 1 atm would have a composition corresponding to the LFL. What is the relationship of this value to the flash point, and what value would you assign the flash point of methanol?(c) Give reasons why it would be unsafe to maintain an open container of methanol in an environment below the LFL (i.e., the value calculated in Part (b)) if there are ignition sources nearby. List common ignition sources that may be found in a laboratory.
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