/*! 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 14 Limestone (calcium carbonate) pa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Limestone (calcium carbonate) particles are stored in \(50-\mathrm{L}\) bags. The void fraction of the particulate matter is 0.30 (liter of void space per liter of total volume) and the specific gravity of solid calcium carbonate is 2.93. (a) Estimate the bulk density of the bag contents ( \(\mathbf{k g}\) CaCO \(_{3}\) /iter of total volume). (b) Estimate the weight ( \(W\) ) of the filled bags. State what you are neglecting in your estimate. (c) The contents of three bags are fed to a ball mill, a device something like a rotating clothes dryer containing steel balls. The tumbling action of the balls crushes the limestone particles and turns them into a powder. (See pp. \(21-64\) of Perry's Chemical Engineers' Handbook, 8 th ed.) The limestone coming out of the mill is put back into \(50-\mathrm{L}\) bags. Would the limestone (i) just fill three bags, (ii) fall short of filling three bags, or (iii) fill more than three bags? Briefly explain your answer.

Short Answer

Expert verified
The bulk density of the bag contents is 2.051 kg/L, the weight of the filled bags is approximately 102.55 kg, and the contents post-grinding should still be sufficient to fill three 50-L bags.

Step by step solution

01

Calculate the bulk density

Bulk density is the ratio of the total mass to the total volume, including the void. The mass of calcium carbonate in 1 liter can be calculated from its specific gravity (2.93 g/cm³). We know: 1 cm³ = 1 mL = 0.001 L, so the mass of calcium carbonate = 2.93 g/cm³ * 1000 cm³/L = 2930 g/L = 2.93 kg/L. Therefore, the bulk density = mass * (1 - void fraction) = 2.93 kg/L * (1 - 0.3) = 2.051 kg/L.
02

Estimate the weight of the filled bags

The weight of the filled bags can be calculated by multiplying the bulk density by the volume of the bag. Therefore, the weight = volume * bulk density = 50 L * 2.051 kg/L = 102.55 kg. The weight is an estimate because it does not include the weight of the bag itself or any possible packing materials.
03

Estimate the number of filled bags post-grinding

The total volume of limestone remains constant even after grinding. Therefore, the contents should still fill three bags because while the individual particles may be smaller, the total volume of material (limestone) hasn't changed.

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.

Specific Gravity
Understanding the specific gravity is essential when dealing with the bulk density calculation of a substance in chemical processes. Specific gravity, often termed as relative density, is the ratio of the density of a substance to the density of a reference substance, usually water at 4°C for liquids and solids, and air at room temperature for gases. It is a dimensionless unit because it is a ratio of two densities and can provide quick insights about the substance. For instance, a specific gravity greater than one, as in the case of calcium carbonate (2.93), indicates that the substance is denser than water.

When calculating the bulk density, one can quickly determine the mass per unit volume by using the specific gravity. The conversion factor from specific gravity to mass density, under standard conditions, is typically the density of water, which is approximately 1 g/cm³ or 1000 kg/m³. Hence, for calcium carbonate with a specific gravity of 2.93, its density can be calculated as 2.93 times the density of water, giving us a mass density of 2.93 g/cm³ or 2930 kg/m³.
Void Fraction
Void fraction is a measure of the amount of empty space in a material. In the context of chemical engineering and particulate solids, particularly when looking at storage or handling of bulk materials, the void fraction is critical. It represents the fraction of the total volume taken up by the voids (spaces) between particles. This means that in a container filled with particles, not every bit of space is occupied by the material itself; some of it is empty space.

The void fraction has implications for numerous properties, like the flowability of a material, its ability to compact, and of course, its bulk density. Here, it is given that the void fraction for the calcium carbonate particles is 0.30. When calculating bulk density, one must account for this space to get an accurate measure of the density of the material in a container, which includes both the particles and the voids. The calculation for bulk density effectively adjusts the solid density by a factor of \(1 - \text{void fraction}\), as seen in the step-by-step solution.
Chemical Engineering Principles in Bulk Density Calculation
Applying chemical engineering principles to the calculation of bulk density allows for a more accurate understanding of the material's behavior in various processes. Bulk density is not just an intrinsic property of a material; it is contextual and depends on both the material's specific gravity and the void fraction. Chemical engineers need to consider these variables, along with others such as particle size distribution, shape, and cohesiveness, when designing processes involving powders and particulates.

In the given problem, dealing with storage and material transformation (grinding to powder), the conservation of mass is an underlying principle used to solve part (b) and (c) of the exercise. The total mass of the limestone remains unchanged after grinding, indicating that, barring any losses, the same volume will be occupied. These principles guide the handling, transportation, and transformation of materials in industries ranging from pharmaceuticals, food processing, to mining.

A cornerstone of chemical engineering is the understanding of physical principles that govern the behavior of materials. By combining knowledge on specific gravity, void fraction, and mass conservation, engineers can predict the outcome of process changes, such as the one described involving the ball mill, and ensure effective and efficient designs.

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

A thermocouple is a temperature-measurement device that consists of two dissimilar metal wires joined at one end. An oversimplified diagram follows. A voltage generated at the metal junction is read on a potentiometer or millivoltmeter. When certain metals are used, the voltage varies linearly with the temperature at the junction of the two metals: $$V(\mathrm{mV})=a T\left(^{\circ} \mathrm{C}\right)+b$$ An iron-constantan thermocouple (constantan is an alloy of copper and nickel) is calibrated by inserting its junction in boiling water and measuring a voltage \(V=5.27 \mathrm{mV}\), and then inserting the junction in silver chloride at its melting point and measuring \(V=24.88 \mathrm{mV}\). (a) Derivethelinear equation for \(V(\mathrm{mV})\) in terms of \(T\left(^{\circ} \mathrm{C}\right)\). Then convert it to an equation for \(T\) in terms of \(V\). (b) If the thermocouple is mounted in a chemical reactor and the voltage is observed to go from 10.0 mV to \(13.6 \mathrm{mV}\) in \(20 \mathrm{s}\), what is the average value of the rate of change of temperature, the the the terme the are a renter \(d T / d t,\) during the measurement period? (c) State the principal benefits and disadvantages of thermocouples.

A mixture of methanol (methyl alcohol) and water contains \(60.0 \%\) water by mass. (a) Assuming volume additivity of the components, estimate the specific gravity of the mixture at \(20^{\circ} \mathrm{C} .\) What volume (in liters) of this mixture is required to provide 150 mol of methanol? (b) Repeat Part (a) with the additional information that the specific gravity of the mixture at \(20^{\circ} \mathrm{C}\) is 0.9345 (making it unnecessary to assume volume additivity). What percentage error results from the volume- additivity assumption?

The following data have been obtained for the effect of solvent composition on the solubility of a serine, an amino acid, at \(10.0^{\circ} \mathrm{C}\) : $$\begin{array}{|l|c|c|c|c|c|c|c|c|}\hline \text { Volume \% Methanol } & 0 & 10 & 20 & 30 & 40 & 60 & 80 & 100 \\\\\hline \text { Solubility (g/100 mL solvent) } & 22.72 & 18.98 & 11.58 & 6.415 & 4.205 & 1.805 & 0.85 & 0.65 \\\\\hline \text { Solution Density (g/mL) } & 1.00 & 0.98 & 0.97 & 0.95 & 0.94 & 0.91 & 0.88 & 0.79 \\\\\hline\end{array}$$ The data were obtained by mixing known volumes of methanol and water to obtain the desired solvent compositions, and then slowly adding measured amounts of serine to each mixture until no more would go into solution. The temperature was held constant at \(10.0^{\circ} \mathrm{C}\). (a) Derive an expression for solvent composition expressed as mass fraction of methanol, \(x\), as a function of volume fraction of methanol, \(f\) (b) Prepare a table of solubility of serine (g serine/g solution) versus mass fraction of methanol.

As described in Problem 3.16 , a drilling mud is a slurry pumped into oil wells being drilled. The mud has several functions: It floats rock cuttings to the top of the well where they can easily be removed; Iubricates and cools the drill bit; and keeps loose solids and water from leaking into the borchole. A drilling mud is prepared by blending barite (SG \(=4.37\) ) with seawater (SG \(=1.03\) ). The seawater has a dissolved salt content of approximately 3.5 wt\%. You have been asked to determine the specific gravity of the mud and the wt\% barite. You collect a sample of the mud from a blending tank on an oil platform and make the following observations: (i) The mud appears homogeneous, even after standing for 2 days; (ii) the tare mass of the calibrated vessel into which you pour the sample of mud is 118 g: (iii) the volume of the collected sample is \(100 \mathrm{mL}\), and the mass of the collection vessel and sample is \(323 \mathrm{g}\); and (iv) the mass of the vessel and residue remaining after completely evaporating water from the sample is \(254 \mathrm{g}\). (a) Estimate the specific gravity of the mud and the wt\% barite. (b) What is the practical importance of Observation (i)?

An open-end mercury manometer is to be used to measure the pressure in an apparatus containing a vapor that reacts with mercury. A 10 -cm layer of silicon oil \((\mathrm{SG}=0.92)\) is placed on top of the mercury in the arm attached to the apparatus. Atmospheric pressure is \(765\) \(\mathrm{mm}\) Hg. (a) If the level of mercury in the open end is 365 mm below the mercury level in the other arm, what is the pressure (mm Hg) in the apparatus? (b) When the instrumentation specialist was deciding on a liquid to put in the manometer, she listed several properties the fluid should have and eventually selected silicon oil. What might the listed properties have been?

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.