/*! 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 12 Eggs are sorted into two sizes (... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 12

Eggs are sorted into two sizes (large and extra large) at the Cheerful Chicken Coop. Recently, the economic downturn has not allowed Cheerful Chicken to repair the egg-sorting machine bought in 2000. Instead, the company has Chick Poulet, one of the firm's sharper-eyed employees, stamp the big eggs with a "Large" rubber stamp in her right hand and the really big eggs with an "X-large" stamp in her left as the eggs go by on a conveyor belt. Down the line, another employee puts the eggs into one of two hoppers, each egg according to its stamp. On average Chick breaks \(8 \%\) of the 120 eggs that pass by her each minute. At the same time, a check of the "X-large" stream reveals a flow rate of 70 eggs/min, of which 8 eggs/min are broken. (a) Draw and label a flowchart for this process. (b) Write and solve balances about the egg sorter on total eggs and broken eggs. (c) How many "large" eggs leave the plant each minute, and what fraction of them are broken? (d) Is Chick right- or left-handed?

Short Answer

Expert verified
There are 50 'Large' eggs leaving the plant each minute, with a broken fraction of 3.2%. Furthermore, Chick is inferred to be right-handed.

Step by step solution

01

Flowchart

Drawing a flowchart will give a graphical representation of the process. At the beginning, 120 eggs are passing by Chick per minute. Chick then sorts them into 'Large' and 'Extra Large' categories with her right and left hand, respectively. 'Large' and 'Extra large' eggs are then placed in two hoppers. During the process, some eggs get broken, which is represented in the flowchart as well. However, the drawing of the flowchart here is not possible as this is a text-based service.
02

Solve Balances

The balance for the entire system can be written as: Input = Output + Accumulation - Depletion. Since the system is in steady state, Accumulation and Depletion are both zero, leaving us with Input = Output. So, Total eggs = 'Large' eggs + 'X-large' eggs; 120 = Large eggs + 70. Solving for 'Large' eggs, we get 'Large' eggs = 50 per minute. For broken eggs: Total broken eggs = Broken 'Large' eggs + Broken 'X-large' eggs. 8% of 120 total eggs equals 9.6 eggs are broken in total. So, the equation becomes 9.6 = Broken 'Large' eggs + 8. Solving, we find that 1.6 'Large' eggs are broken every minute.
03

Determine Large Egg Output and Broken Fraction

From our calculations, 50 'Large' eggs are leaving the plant each minute, and 1.6 of them are broken. To find the fraction of 'Large' eggs which are broken, we divide the number of broken 'large' eggs by the total number of 'large' eggs. So, the fraction of 'Large' eggs which are broken equals 1.6 / 50 = 0.032 or 3.2%.
04

Determine Chick's Handedness

Typically, a person is more skilled and less likely to make mistakes with their dominant hand. In this case, 鈥楲arge鈥� eggs have fewer broken ones (1.6/min) than the 'X-large' category (8/min). Therefore, it can be inferred that Chick is right-handed as she uses her right hand for 'Large' eggs.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Flowchart Interpretation
Understanding a flowchart is like reading a roadmap for the process at hand. A flowchart in the context of the Cheerful Chicken Coop's egg-sorting scenario visually breaks down each step that eggs go through as they are sorted by Chick Poulet. The flowchart will show the initial input of 120 eggs per minute on a conveyor belt. Then, it separates into two branches representing "Large" and "X-large" eggs based on Chick's sorting.
  • The input arrow leads to two outcomes, marking where Chick applies the respective stamps with either her right or left hand.
  • Each category of eggs then flows into designated hoppers.
  • Additionally, broken eggs are accounted for as a separate path or note on the flowchart.
Drawing the flowchart helps in clearly visualizing the operations, indicating where in the process eggs may be broken and showing the distribution into two distinct categories. This greatly aids in comprehending the material balance problem by using a systematic approach.
Steady-State Assumptions
The steady-state assumption is crucial in simplifying the analysis of many process systems. In this egg-sorting process, the steady-state means that the number of eggs entering the system is consistently balanced by the number of eggs exiting, with no eggs accumulating at any point.
  • This assumption allows us to set the inflow equal to the outflow for both total eggs and broken eggs, which simplifies the calculations considerably.
  • Under this condition, changes over time can be ignored, making it easier to focus on the calculations of current flow rates and breaks.
Having no accumulation or depletion makes it possible to solve for unknown variables straightforwardly. For instance, when knowing the total egg inflow and the counts of sorted and broken eggs, this assumption helps deduce that the remaining eggs are accounted as output, simplifying the problem to a simple equation: Input = Output.
Process Analysis
Process analysis takes the flowchart and the steady-state assumptions and applies them to solve for missing information in a systematic way. This involves looking at each process unit and stating the mass balance around these units.
  • The total mass balance for the egg process involves the equation: Total eggs = 'Large' eggs + 'X-large' eggs, allowing us to solve for the 'Large' eggs.
  • By creating a specific balance for broken eggs: Total broken eggs = Broken 'Large' eggs + Broken 'X-large' eggs, we can find out the specifics of how many eggs are broken in each category.
Through analyzing each step in this detailed manner, we can not only find out the amount and type of eggs leaving the system but also evaluate the efficiency and skill (or lack thereof) in the process, such as determining that Chick Poulet is likely right-handed due to fewer breaks occurring with her dominant hand. This deeper understanding through process analysis allows us to make informed conclusions about the operation and performance of the egg-sorting task.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Inside a distillation column (see Problem 4.8), a downward-flowing liquid and an upward-flowing vapor maintain contact with each other. For reasons we will discuss in greater detail in Chapter \(6,\) the vapor stream becomes increasingly rich in the more volatile components of the mixture as it moves up the column, and the liquid stream is enriched in the less volatile components as it moves down. The vapor leaving the top of the column goes to a condenser. A portion of the condensate is taken off as a product (the overhead product), and the remainder (the reflux) is returned to the top of the column to begin its downward journey as the liquid stream. The condensation process can be represented as shown below: A distillation column is being used to separate a liquid mixture of ethanol (more volatile) and water (less volatile). A vapor mixture containing 89.0 mole \(\%\) ethanol and the balance water enters the overhead condenser at a rate of \(100 \mathrm{lb}\) -mole/h. The liquid condensate has a density of \(49.01 \mathrm{b}_{\mathrm{m}} / \mathrm{ft}^{3},\) and the reflux ratio is \(3 \mathrm{lb}_{\mathrm{m}}\) reflux/lb \(_{\mathrm{m}}\) overhead product. When the system is operating at steady state, the tank collecting the condensate is half full of liquid and the mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) is 10.0 minutes. Determine the overhead product volumetric flow rate (ft \(^{3}\) /min) and the condenser tank volume (gal).

The respiratory process involves hemoglobin (Hgb), an iron-containing compound found in red bloodcells. In the process, carbon dioxide diffuses from tissue cells as molecular \(\mathrm{CO}_{2}\), while \(\mathrm{O}_{2}\) simultaneously enters the tissue cells. A significant fraction of the \(\mathrm{CO}_{2}\) leaving the tissue cells enters red blood cells and reacts with hemoglobin; the \(\mathrm{CO}_{2}\) that does not enter the red blood cells ( \((\mathrm{D}\) in the figure below) remains dissolved in the blood and is transported to the lungs. Some of the \(\mathrm{CO}_{2}\) entering the red blood cells reacts with hemoglobin to form a compound (Hgb. \(\mathrm{CO}_{2} ;(\) 2) in the figure). When the red blood cells reach the lungs, the Hgb.CO_ dissociates, releasing free CO_ Meanwhile, the CO_ that enters the red blood cells but does not react with hemoglobin combines with water to form carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3},\) which then dissociates into hydrogen ions and bicarbonate ions ( (3) in the figure). The bicarbonate ions diffuse out of the cells ( (4) in the figure), and the ions are transported to the lungs via the bloodstream. For adult humans, every deciliter of blood transports a total of \(1.6 \times 10^{-4}\) mol of carbon dioxide in its various forms (dissolved \(\mathrm{CO}_{2}, \mathrm{Hgb} \cdot \mathrm{CO}_{2},\) and bicarbonate ions) from tissues to the lungs under normal, resting conditions. Of the total \(\mathrm{CO}_{2}, 1.1 \times 10^{-4}\) mol are transported as bicarbonate ions. In a typical resting adult human, the heart pumps approximately 5 liters of blood per minute. You have been asked to determine how many moles of \(\mathrm{CO}_{2}\) are dissolved in blood and how many moles of \(\mathrm{Hgb} \cdot \mathrm{CO}_{2}\) are transported to the lungs during an hour's worth of breathing. (a) Draw and fully label a flowchart and do a degree-of-freedom analysis. Write the chemical reactions that occur, and generate, but do not solve, a set of independent equations relating the unknown variables on the flowchart. (b) If you have enough information to obtain a unique numerical solution, do so. If you do not have enough information, identify a specific piece/pieces of information that (if known) would allow you to solve the problem, and show that you could solve the problem if that information were known. (c) When someone loses a great deal of blood due to an injury, they "go into shock": their total blood volume is low, and carbon dioxide is not efficiently transported away from tissues. The carbon dioxide reacts with water in the tissue cells to produce very high concentrations of carbonic acid, some of which can dissociate (as shown in this problem) to produce high levels of hydrogen ions. What is the likely effect of this occurrence on the blood pH near the tissue and the tissue cells? How is this likely to affect the injured person?

Draw and label the given streams and derive expressions for the indicated quantities in terms of labeled variables. The solution of Part (a) is given as an illustration. (a) A continuous stream contains 40.0 mole\% benzene and the balance toluene. Write expressions for the molar and mass flow rates of benzene, \(\dot{n}_{\mathrm{B}}\left(\operatorname{mol} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right)\) and \(\dot{m}_{\mathrm{B}}\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right),\) in terms of the total molar flow rate of the stream, \(\dot{n}(\mathrm{mol} / \mathrm{s})\) (b) The feed to a batch process contains equimolar quantities of nitrogen and methane. Write an expression for the kilograms of nitrogen in terms of the total moles \(n(\) mol) of this mixture. (c) A stream containing ethane, propane, and butane has a mass flow rate of \(100.0 \mathrm{g} / \mathrm{s}\). Write an expression for the molar flow rate of ethane, \(\dot{n}_{\mathrm{E}}\left(\text { Ib-mole } \mathrm{C}_{2} \mathrm{H}_{6} / \mathrm{h}\right)\), in terms of the mass fraction of this species, \(x_{\mathrm{E}}\). (d) A continuous stream of humid air contains water vapor and dry air, the latter containing approximately 21 mole \(\% \mathrm{O}_{2}\) and \(79 \% \mathrm{N}_{2}\). Write expressions for the molar flow rate of \(\mathrm{O}_{2}\) and for the mole fractions of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2}\) in the gas in terms of \(\dot{n}_{1}\left(\mathrm{lb}-\mathrm{mole} \mathrm{H}_{2} \mathrm{O} / \mathrm{s}\right)\) and \(\dot{n}_{2}(\text { lb- mole dry air/s })\) (e) The product from a batch reactor contains \(\mathrm{NO}, \mathrm{NO}_{2},\) and \(\mathrm{N}_{2} \mathrm{O}_{4} .\) The mole fraction of \(\mathrm{NO}\) is 0.400. Write an expression for the gram-moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) in terms of \(n(\mathrm{mol}\) mixture) and \(y_{\mathrm{NO}_{2}}\left(\operatorname{mol} \mathrm{NO}_{2} / \mathrm{mol}\right)\)

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.

A gas contains 75.0 wt\% methane, \(10.0 \%\) ethane, \(5.0 \%\) ethylene, and the balance water. (a) Calculate the molar composition of this gas on both a wet and a dry basis and the ratio (mol \(\mathrm{H}_{2} \mathrm{O} /\) mol dry gas). (b) If \(100 \mathrm{kg} / \mathrm{h}\) of this fuel is to be burned with \(30 \%\) excess air, what is the required air feed rate (kmol/ h)? How would the answer change if the combustion were only \(75 \%\) complete?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.