/*! 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 43 The popularity of orange juice, ... [FREE SOLUTION] | 91Ó°ÊÓ

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The popularity of orange juice, especially as a breakfast drink, makes this beverage an important factor in the economy of orange-growing regions. Most marketed juice is concentrated and frozen and then reconstituted before consumption, and some is "not-from-concentrate." Although concentrated juices are less popular in the United States than they were at one time, they still have a major segment of the market for orange juice. The approaches to concentrating orange juice include evaporation, freeze concentration, and reverse osmosis. Here we examine the evaporation process by focusing only on two constituents in the juice: solids and water. Fresh orange juice contains approximately 10 wt\% solids (sugar, citric acid, and other ingredients) and frozen concentrate contains approximately 42 wt\% solids. The frozen concentrate is obtained by evaporating water from the fresh juice to produce a mixture that is approximately 65 wt\% solids. However, so that the flavor of the concentrate will closely approximate that of fresh juice, the concentrate from the evaporator is blended with fresh orange juice (and other additives) to produce a final concentrate that is approximately 42 wt\% solids. (a) Draw and label a flowchart of this process, neglecting the vaporization of everything in the juice but water. First prove that the subsystem containing the point where the bypass stream splits off from the evaporator feed has one degree of freedom. (If you think it has zero degrees, try determining the unknown variables associated with this system.) Then perform the degree- offreedom analysis for the overall system, the evaporator, and the bypass- evaporator product mixing point, and write in order the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don't do any calculations. (b) Calculate the amount of product (42\% concentrate) produced per 100 kg fresh juice fed to the process and the fraction of the feed that bypasses the evaporator. (c) Most of the volatile ingredients that provide the taste of the concentrate are contained in the fresh juice that bypasses the evaporator. You could get more of these ingredients in the final product by evaporating to (say) 90\% solids instead of 65\%; you could then bypass a greater fraction of the fresh juice and thereby obtain an even better tasting product. Suggest possible drawbacks to this proposal.

Short Answer

Expert verified
The solution includes creating a clear flowchart representation of the process. Then, using these visual aids and the information provided, a degree-of-freedom analysis is performed to determine the unknown parameters of the system and the system's boundaries. Finally, a calculation is performed to find the quantity of product produced per 100 kg of juice and the fraction that bypasses the evaporator. Potential drawbacks of increasing the evaporator solids concentration are also discussed. Please note, that without specific data and exact calculations, the exact quantities and reactions can't be determined.

Step by step solution

01

- Understand the Process Flow and System

Make sure to familiarize yourself with the process of converting fresh orange juice into concentrated juice as described in the problem. Create a flowchart showcasing the process flow, and label each step properly.
02

- Degree-of-Freedom Analysis

Identify the subsystem where the bypass stream splits off from the evaporator feed. This subsystem should have one degree of freedom, meaning only one independent variable can change without changing the others. Next, perform a degree-of-freedom analysis for the larger systems: the whole process, the evaporator, and the bypass-evaporator product mixing point.
03

- Formulation of Equations

According to the degree-of-freedom analysis, write down the equations which would need to be solved to determine all unknown stream variables. For each equation, circle the variable which would be solved for.
04

- Calculate the Product Concentrate and Bypass

Apply the mass balance principle to calculate the amount of 42% concentrate produced per 100 kg fresh juice. Also, using the given percentages of solids in fresh juice, evaporator output, and the final concentrate, calculate the fraction of fresh juice that bypasses the evaporator.
05

- Analyse Potential Drawbacks

Consider the proposal to increase the blend ratios by increasing the solids concentration to 90% solids instead of 65% through evaporation. Discuss potential issues regarding factors like production cost, flavor concentration, and product stability.

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Key Concepts

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

Degree of Freedom Analysis
Degree of freedom (DOF) analysis is a vital step in solving material balance problems. It helps determine whether the system has enough information to find unknowns without ambiguity. In this context, DOF analysis is applied to understand how the system behaves under certain variables and conditions.

To conduct a DOF analysis, the degree of freedom is defined as the difference between the number of independent equations and the number of unknown variables in a particular system. Specifically:
  • Identify known variables, typically starting from the mass fractions or flow rates provided, such as the solids content in orange juice and concentrate.
  • Determine the independent equations, which often include mass balance equations for both the total mass and the individual components (like water and solids).
  • Calculate the degrees of freedom by subtracting the total number of independent equations from the number of unknowns.
In the case of the subsystem where a bypass stream splits off from the evaporator feed, this analysis helps to verify that there is one degree of freedom. It guides which stream variables can independently change without affecting others, aiding in identifying and solving the correct equations for the desired outputs.
Evaporation Process
The evaporation process is crucial in concentrating orange juice by removing water while retaining the essential solids that contribute to flavor and nutritional content.

In this context, fresh orange juice initially consists of 10 weight percent solids, including sugars and acids. Evaporation aims to reach a concentrate with a higher solids content, typically around 42% for frozen concentrate products. This requires removing enough water to yield a product of 65% solids before it is diluted or blended with fresh juice for final consumption.

Here’s how the evaporation process ties into the overall concentration techniques:
  • Evaporating enough water from the juice increases the weight percent of solids, thereby enhancing the concentration of taste and aroma compounds.
  • This process requires careful balance to ensure that the evaporated juice retains a quality and flavor profile similar to fresh juice, achieved by blending it back with some of the original fresh juice.
  • The efficiency of the evaporator is crucial, as it impacts not only the concentration level but also the energy consumption of the process.
Optimizing the evaporation process, therefore, involves not only the technical aspects of heat and mass transfer but also an understanding of the desired product characteristics and the market demands.
Orange Juice Concentration
Orange juice concentration is an essential practice in the orange juice industry, aiming to enhance shelf-life and reduce transportation costs. It involves reducing the water content while maximizing the retention of solid components that produce the coveted orange juice flavor.

Methods to achieve orange juice concentration, such as the evaporation process, focus on improving solids concentration, including:
  • Evaporation, to reduce water content and increase solids concentration to around 65% by weight before final adjustment to 42%.
  • Freeze concentration and reverse osmosis, as alternatives to evaporation, but less commonly used.
  • The blending process post-evaporation to ensure that the concentrate's flavor closely resembles that of fresh juice.
Thus, orange juice concentration is a balance between technological processes and product quality, ensuring that these methods effectively retain flavor while providing an economical and practical product for consumers.
Mass Balance Calculations
Mass balance calculations are foundational in determining material flow and composition in the orange juice concentration process. This involves accounting for all mass entering and leaving each part of the system to ensure consistency and accuracy.

Here's how mass balance equations are typically applied in this scenario:
  • Begin by establishing the total mass balance equation: the total mass entering the system must equal the total mass leaving the system.
  • Next, focus on component balances, specifically for solids and water, given their significance in juice concentration.
  • Apply these equations to individual process steps, like the evaporator and the bypass system, to identify the flow rates and concentrations of each stream.
By following these steps, mass balance calculations help to determine specific figures such as the amount of 42% concentrate produced per kilogram of fresh juice and the fraction of juice bypassing the evaporator.

This analytical approach is critical for optimizing the process, ensuring that resources are used efficiently and that the final product meets desired specifications and quality standards.

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Most popular questions from this chapter

A stream containing \(\mathrm{H}_{2} \mathrm{S}\) and inert gases and a second stream of pure \(\mathrm{SO}_{2}\) are fed to a sulfur recovery reactor, where the reaction $$2 \mathrm{H}_{2} \mathrm{S}+\mathrm{SO}_{2} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$$ takes place. The feed rates are adjusted so that the ratio of \(\mathrm{H}_{2} \mathrm{S}\) to \(\mathrm{SO}_{2}\) in the combined feed is always stoichiometric. In the normal operation of the reactor the flow rate and composition of the \(\mathrm{H}_{2} \mathrm{S}\) feed stream both fluctuate. In the past, each time either variable changed the required \(\mathrm{SO}_{2}\) feed rate had to be reset by adjusting a valve in the feed line. A control system has been installed to automate this process. The \(\mathrm{H}_{2} \mathrm{S}\) feed stream passes through an electronic flowmeter that transmits a signal \(R_{\mathrm{f}}\) directly proportional to the molar flow rate of the stream, \(\dot{n}_{\mathrm{f}}\). When \(\dot{n}_{\mathrm{f}}=100 \mathrm{kmol} / \mathrm{h}\), the transmitted signal \(R_{\mathrm{f}}=15 \mathrm{mV}\). The mole fraction of \(\mathrm{H}_{2} \mathrm{S}\) in this stream is measured with a thermal conductivity detector, which transmits a signal \(R_{\mathrm{a}} .\) Analyzer calibration data are as follows: $$\begin{array}{|l|c|c|c|c|c|c|}\hline R_{\mathrm{a}}(\mathrm{mV}) & 0 & 25.4 & 42.8 & 58.0 & 71.9 & 85.1 \\ \hline x\left(\mathrm{mol} \mathrm{H}_{2} \mathrm{S} / \mathrm{mol}\right) & 0.00 & 0.20 & 0.40 & 0.60 &0.80 & 1.00 \\\\\hline\end{array}$$ The controller takes as input the transmitted values of \(R_{\mathrm{f}}\) and \(R_{\mathrm{a}}\) and calculates and transmits a voltage signal \(R_{\mathrm{c}}\) to a flow control valve in the \(\mathrm{SO}_{2}\) line, which opens and closes to an extent dependent on the value of \(R_{c} .\) A plot of the \(S O_{2}\) flow rate, \(\dot{n}_{c},\) versus \(R_{c}\) on rectangular coordinates is a straight line through the points \(\left(R_{c}=10.0 \mathrm{mV}, \dot{n}_{c}=25.0 \mathrm{kmol} / \mathrm{h}\right)\) and \(\left(R_{c}=25.0 \mathrm{mV}, \dot{n}_{c}=60.0 \mathrm{kmol} / \mathrm{h}\right)\) (a) Why would it be important to feed the reactants in stoichiometric proportion? (Hint: \(\mathrm{SO}_{2}\) and especially \(\mathrm{H}_{2} \mathrm{S}\) are serious pollutants.) What are several likely reasons for wanting to automate the \(\mathrm{SO}_{2}\) feed rate adjustment? (b) If the first stream contains 85.0 mole \(\% \mathrm{H}_{2} \mathrm{S}\) and enters the unit at a rate of \(\dot{n}_{\mathrm{f}}=3.00 \times 10^{2} \mathrm{kmol} / \mathrm{h}\) what must the value of \(\dot{n}_{c}\left(\mathrm{kmol} \mathrm{SO}_{2} / \mathrm{h}\right)\) be? (c) Fit a function to the \(\mathrm{H}_{2} \mathrm{S}\) analyzer calibration data to derive an expression for \(x\) as a function of \(R_{\mathrm{a}}\) Check the fit by plotting both the function and the calibration data on the same graph. (d) Derive a formula for \(R_{\mathrm{c}}\) from specified values of \(R_{\mathrm{f}}\) and \(R_{\mathrm{a}},\) using the result of Part (c) in the derivation. (This formula would be built into the controller.) Test the formula using the flow rate and composition data of Part (a). (e) The system has been installed and made operational, and at some point the concentration of \(\mathrm{H}_{2} \mathrm{S}\) in the feed stream suddenly changes. A sample of the blended gas is collected and analyzed a short time later and the mole ratio of \(\mathrm{H}_{2} \mathrm{S}\) to \(\mathrm{SO}_{2}\) is not the required 2: 1 . List as many possible reasons as you can think of for this apparent failure of the control system.

Ammonia is oxidized to nitric oxide in the following reaction: $$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$$ (a) Calculate the ratio (lb-mole \(\mathrm{O}_{2}\) react/lb-mole NO formed). (b) If ammonia is fed to a continuous reactor at a rate of \(100.0 \mathrm{kmol} \mathrm{NH}_{3} / \mathrm{h}\), what oxygen feed rate (kmol/h) would correspond to 40.0\% excess O_? (c) If \(50.0 \mathrm{kg}\) of ammonia and \(100.0 \mathrm{kg}\) of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

The indicator-dilution method is a technique used to determine flow rates of fluids in channels for which devices like rotameters and orifice meters cannot be used (e.g., rivers, blood vessels, and largediameter pipelines). A stream of an easily measured substance (the tracer) is injected into the channel at a known rate, and the tracer concentration is measured at a point far enough downstream of the injection point for the tracer to be completely mixed with the flowing fluid. The larger the flow rate of the fluid, the lower the tracer concentration at the measurement point. A gas stream that contains 1.50 mole \(\% \mathrm{CO}_{2}\) flows through a pipeline. Twenty (20.0) kilograms of \(\mathrm{CO}_{2}\) per minute is injected into the line. A sample of the gas is drawn from a point in the line 150 meters (a) Estimate the gas flow rate (kmol/min) upstream of the injection point. (b) Eighteen seconds elapse from the instant the additional \(\mathrm{CO}_{2}\) is first injected to the time the \(\mathrm{CO}_{2}\) concentration at the measurement point begins to rise. Assuming that the tracer travels at the average velocity of the gas in the pipeline (i.e., neglecting diffusion of \(\mathrm{CO}_{2}\) ), estimate the average velocity (m/s). If the molar gas density is \(0.123 \mathrm{kmol} / \mathrm{m}^{3}\), what is the pipe diameter?

Effluents from metal-finishing plants have the potential of discharging undesirable quantities of metals, such as cadmium, nickel, lead, manganese, and chromium, in forms that are detrimental to water and air quality. A local metal-finishing plant has identified a wastewater stream that contains 5.15 wt\% chromium (Cr) and devised the following approach to lowering risk and recovering the valuable metal. The wastewater stream is fed to a treatment unit that removes \(95 \%\) of the chromium in the feed and recycles it to the plant. The residual liquid stream leaving the treatment unit is sent to a waste lagoon. The treatment unit has a maximum capacity of 4500 kg wastewater/h. If wastewater leaves the finishing plant at a rate higher than the capacity of the treatment unit, the excess (anything above \(4500 \mathrm{kg} / \mathrm{h}\) ) bypasses the unit and combines with the residual liquid leaving the unit, and the combined stream goes to the waste lagoon. (a) Without assuming a basis of calculation, draw and label a flowchart of the process. (b) Wastewater leaves the finishing plant at a rate \(\dot{m}_{1}=6000 \mathrm{kg} / \mathrm{h}\). Calculate the flow rate of liquid to the waste lagoon, \(\dot{m}_{6}(\mathrm{kg} / \mathrm{h}),\) and the mass fraction of \(\mathrm{Cr}\) in this liquid, \(x_{6}(\mathrm{kg} \mathrm{Cr} / \mathrm{kg})\) (c) Calculate the flow rate of liquid to the waste lagoon and the mass fraction of Crin this liquid for \(\dot{m}_{1}\) varying from \(1000 \mathrm{kg} / \mathrm{h}\) to \(10,000 \mathrm{kg} / \mathrm{h}\) in \(1000 \mathrm{kg} / \mathrm{h}\) increments. Generate a plot of \(x_{6}\) versus \(\dot{m}_{1}\). (Suggestion: Use a spreadsheet for these calculations.) (d) The company has hired you as a consultant to help them determine whether or not to add capacity to the treatment unit to increase the recovery of chromium. What would you need to know to make this determination? (e) What concerns might need to be addressed regarding the waste lagoon?

Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor. The fresh feed to the process contains 32.0 mole \(\%\) CO, \(64.0 \%\) H \(_{2}\), and \(4.0 \%\) Ne. This stream is mixed with a recycle stream in a ratio 5 mol recycle/ 1 mol fresh feed to produce the feed to the reactor, which contains 13.0 mole\% \(\mathrm{N}_{2}\). A low single-pass conversion is attained in the reactor. The reactor effluent goes to a condenser from which two streams emerge: a liquid product stream containing essentially all the methanol formed in the reactor, and a gas stream containing all the \(\mathrm{CO}, \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) leaving the reactor. The gas stream is split into two fractions: one is removed from the process as a purge stream, and the other is the recycle stream that combines with the fresh feed to the reactor. (a) Assume a methanol production rate of \(100 \mathrm{kmol} / \mathrm{h}\). Perform the DOF for the overall system and all subsystems to prove that there is insufficient information to solve for all unknowns. (b) Briefly explain in your own words the reasons for including (i) the recycle stream and (ii) the purge stream in the process design.

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