/*! 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 5 You have recently become interes... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You have recently become interested in collecting used classic vinyl recordings. The jackets for these disks often have no liners to protect the records from dust and other sources of scratches, so you seck to purchase some liners from online vendors. Two web sites offer 12 -in. inner plastic liners: site A advertises liners with a thickness of 50 microns, while the liners at site B are 3 mils thick. Which of the sites is selling the thicker liner? (Note: Neither of the given units of thickness is defined in the text, but definitions are available elsewhere.)

Short Answer

Expert verified
The liner from site B is thicker.

Step by step solution

01

Convert the thickness to common units.

First, find out the thickness of the liner from site B in microns. To do this, multiply the thickness in mils by 25.4. For this case, \(3 \, mils \times 25.4 = 76.2 \, microns\). Now both the thicknesses are in microns.
02

Compare the thicknesses.

Next, compare the thickness of the liner from site A with that of site B. For this case, the thickness of site A liner is 50 microns and site B liner is 76.2 microns. Thus, the liner from site B is thicker.

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 91Ó°ÊÓ!

Key Concepts

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

Unit Conversion
Understanding unit conversion is crucial for any student embarking on a journey through chemical engineering education or any technical field. The essence of unit conversion is the process of converting a given amount of a physical quantity from one unit of measurement to another. In our exercise, you have two different thickness measurements: microns and mils.

So, how do you compare thicknesses if they're not in the same unit? You find a common unit, which simplifies comparison. In the example, microns (a micron is one-millionth of a meter) and mils (one mil is one-thousandth of an inch) measure thickness. You can use the conversion factor that 1 mil equals 25.4 microns. This transformation allows you to perform a direct comparison, eliminating confusion and facilitating a streamlined decision-making process.
Materials Comparison
When comparing materials, understanding the properties is key, and thickness is a property that can affect performance, especially when dealing with protective linings like in our vinyl records scenario. Thicker materials often provide better protection but may also be less flexible or heavier.

In engineering and materials science, comparisons go beyond just measurements; they also consider the implications of variations in thickness, weight, durability, and suitability for the application. In our case, the thickness could imply better protection from scratches or dust for the vinyl records. Therefore, recognizing the utility of each characteristic helps to select the optimal material for a specific function. It’s not only about the raw numbers but also what those numbers mean in practical usage.
Thickness Measurement
Thickness measurement is a fundamental concept in material science and engineering. It can be measured in various units like inches, millimeters, microns, or mils, depending on the application and industry standards. In chemical engineering, material thickness can impact process efficiency, safety, and material costs.

Precision in measuring thickness is crucial, as incorrect measurements can lead to inadequate protection or unnecessary expense. Always ensure that the measuring tool or technique is suitable for the measurement's required resolution and accuracy. In the context of vinyl record protection, the correct thickness measurement ensures that the liner provides adequate protection without being too bulky or stiff, balancing protection with usability.

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

The relationship between the pressure \(P\) and volume \(V\) of the air in a cylinder during the upstroke of a piston in an air compressor can be expressed as \(P V^{k}=C\) where \(k\) and \(C\) are constants. During a compression test, the following data are taken: $$\begin{array}{|c|c|c|c|c|c|c|}\hline P(\mathrm{mm} \mathrm{Hg}) & 760 & 1140 & 1520 & 2280 & 3040 & 3800 \\ \hline V\left(\mathrm{cm}^{3}\right) & 48.3 & 37.4 & 31.3 & 24.1 & 20.0 & 17.4 \\\\\hline\end{array}$$ Determine the values of \(k\) and \(C\) that best fit the data. (Give both numerical values and units.)

The following empirical equation correlates the values of variables in a system in which solid particles are suspended in a flowing gas: $$\frac{k_{g} d_{p} y}{D}=2.00+0.600\left(\frac{\mu}{\rho D}\right)^{1 / 3}\left(\frac{d_{p} u \rho}{\mu}\right)^{1 / 2}$$ Both \((\mu / \rho D)\) and \(\left(d_{p} u \rho / \mu\right)\) are dimensionless groups; \(k_{g}\) is a coefficient that expresses the rate at which a particular species transfers from the gas to the solid particles; and the coefficients 2.00 and 0.600 are dimensionless constants obtained by fitting experimental data covering a wide range of values of the equation variables. The value of \(k_{g}\) is needed to design a catalytic reactor. since this coefficient is difficult to determine directly, values of the other variables are measured or estimated and \(k_{g}\) is calculated from the given correlation. The variable values are as follows: $$\begin{aligned}d_{p} &=5.00 \mathrm{mm} \\\y &=0.100 \quad(\text { dimensionless }) \\\D &=0.100 \mathrm{cm}^{2} / \mathrm{s} \\\\\mu &=1.00 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2} \\\\\rho &=1.00 \times 10^{-3} \mathrm{g} / \mathrm{cm}^{3} \\\u &=10.0 \mathrm{m} / \mathrm{s}\end{aligned}$$ (a) What is the estimated value of \(k_{g} ?\) (Give its value and units.) (b) Why might the true value of \(k_{g}\) in the reactor be significantly different from the value estimated in Part (a)? (Give several possible reasons.) (c) Create a spreadsheet in which up to five sets of values of the given variables ( \(d_{p}\) through \(u\) ) are entered in columns and the corresponding values of \(k_{g}\) are calculated. Test your program using the following variable sets: (i) the values given above; (ii) as above, only double the particle diameter \(d_{p}\) (making it \(10.00 \mathrm{mm}\) ); (iii) as above, only double the diffusivity \(D ;\) (iv) as above, only double the viscosity \(\mu ;(\mathrm{v})\) as above, only double the velocity \(u\). Report all five calculated values of \(k_{g}\).

On a website devoted to answering engineering questions, viewers were invited to determine how much power a 100 -MW power plant generates annually. The answer declared to be best was submitted by a civil engineering student, who stated, "It produces \(100 \mathrm{MW} / \mathrm{hr}\) so over the year that's \(100^{*} 24^{*} 365.25 \&\) do the math." (a) Carry out the calculation, showing all the units. (b) What is wrong with the statement of the question? (c) Why was the student wrong in saying that the plant produces \(100 \mathrm{MW} / \mathrm{hr} ?\)

During the early part of the 20 th century, sulfanilamide (an antibacterial drug) was only administered by injection or in a solid pill. In \(1937,\) a pharmaceutical company decided to market a liquid formulation of the drug. Since sulfanilamide was known to be highly insoluble in water and other common pharmaceutical solvents, a number of alternative solvents were tested and the drug was found to be soluble in diethylene glycol (DEG). After satisfactory results were obtained in tests of flavor, appearance, and fragrance, 240 gallons of sulfanilamide in DEG were manufactured and marketed as Elixir Sulfanilamide. After a number of deaths were determined to have been caused by the formulation, the Food and Drug Administration (FDA) mounted a campaign to recall the drug and recovered about 232 gallons. By this time, 107 people had died. The incident led to passage of the 1938 Federal Food, Drug, and cosmetic Act that significantly tightened FDA safety requirements. Not all of the quantities needed in solving the following problems can be found in the text. Give sources of such information and list all assumptions. (a) The dosage instructions for the elixir were to "take 2 to 3 teaspoons in water every four hours." Assume each teaspoon was pure DEG, and estimate the volume (mL) of DEG a patient would have consumed in a day. (b) The lethal oral dose of diethylene glycol has been estimated to be 1.4 mL DEG/kg body mass. Determine the maximum patient mass \(\left(1 \mathrm{b}_{\mathrm{m}}\right)\) for which the daily dose estimated in Part (a) would be fatal. If you need values of quantities you cannot find in this text, use the Internet. Suggest three reasons why that dose could be dangerous to a patient whose mass is well above the calculated value. (c) Estimate how many people would have been poisoned if the total production of the drug had been consumed. (d) List steps the company should have taken that would have prevented this tragedy.

Calculate (a) the weight in \(\mathrm{Ib}_{\mathrm{f}}\) of a \(25.0-\mathrm{lb}_{\mathrm{m}}\) object. (b) the mass in \(\mathrm{kg}\) of an object that weighs \(25 \mathrm{N}\). (c) the weight in dynes of a 10 -ton object (not metric tons).

See all solutions

Recommended explanations on Chemistry Textbooks

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.