/*! 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 3 When a liquid or a gas occupies ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 3

When a liquid or a gas occupies a volume, it may be assumed to fill the volume completely. On the other hand, when solid particles occupy a volume, there are always spaces (voids) among the particles. The porosity or void fraction of a bed of particles is the ratio (void volume)/(total bed volume). The bulk density of the solids is the ratio (mass of solids)/(total bed volume), and the absolute density of the solids has the usual definition (mass of solids)/(volume of solids). Suppose \(600.0 \mathrm{g}\) of a crushed ore is placed in a graduated cylinder, filling it to the \(184 \mathrm{cm}^{3}\) level. One hundred \(\mathrm{cm}^{3}\) of water is then added to the cylinder, whereupon the water level is observed to be at the \(233.5 \mathrm{cm}^{3}\) mark. Calculate the porosity of the dry particle bed, the bulk density of the ore in this bed, and the absolute density of the ore.

Short Answer

Expert verified
The porosity of the dry particle bed is approximately 73.1%, the bulk density of the ore in this bed is approximately 3.26 g/cm\u00b3 and the absolute density of the ore is approximately 12.12 g/cm\u00b3.

Step by step solution

01

Calculate the bulk density

The bulk density of the solids is given by the formula: \( Bulk Density = \frac{Mass of Solids}{Total Bed Volume} \). Substituting the given values: \( Bulk Density = \frac{600.0 g}{184 cm^{3}} = 3.26 g/cm^{3} \).
02

Calculate the volume of solids

The volume of solids can be determined from the volume of the liquid displaced when the solid is introduced into it. Here, the volume of the solids is equal to the rise in the water level when the ore was added. This is given by the formula: \( Volume of Solids = Final Water Level - Initial Water Level \). Substitute the provided values to find: \( Volume of Solids = 233.5 cm^{3} - 184 cm^{3} = 49.5 cm^{3} \).
03

Calculate the absolute density

The absolute density of the solids is given by the formula: \( Absolute Density = \frac{Mass of Solids}{Volume of Solids} \). Substituting the determined values: \( Absolute Density = \frac{600.0 g}{49.5 cm^{3}} = 12.12 g/cm^{3} \).
04

Calculate the porosity or void fraction

The porosity or void fraction of a bed of particles is given by the formula: \( Porosity = \frac{Void Volume}{Total Bed Volume} \). First, calculate the void volume. Void Volume = Total bed Volume - Volume of Solids = 184 cm^{3} - 49.5 cm^{3} = 134.5 cm^{3}. Then, using the formula, we have: \( Porosity = \frac{134.5 cm^{3}}{184 cm^{3}} = 0.731 or 73.1% \).

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.

Bulk Density
Bulk density is an important concept that helps us understand the overall density of a mixture, including both the solid particles and the spaces (voids) between them. It refers to the mass of the particles divided by the total volume they occupy, which includes the pores within the material as well as the spaces between individual grains.

To calculate bulk density, use the formula:
  • Bulk Density = \( \frac{\text{Mass of Solids}}{\text{Total Bed Volume}} \)
This formula accounts for the entire volume occupied by the particles in their container and is measured in grams per cubic centimeter (g/cm鲁).

For example, if you have 600 grams of ore occupying a 184 cm鲁 volume, the bulk density is \( \frac{600.0 \, g}{184 \, cm^{3}} = 3.26 \, g/cm^{3} \).

An understanding of bulk density is crucial for applications in industries such as agriculture, geology, and construction, where material properties need to be carefully considered.
Absolute Density
Absolute density provides a measure of how dense the particles themselves are, excluding any voids or spaces between them. It is calculated by dividing the mass of the solids by their actual volume, not including the space between particles.

The formula to find absolute density is:
  • Absolute Density = \( \frac{\text{Mass of Solids}}{\text{Volume of Solids}} \)
This property's typical unit is also grams per cubic centimeter (g/cm鲁).

Continuing with our example, if the volume of solids is determined by how much they displace the water (49.5 cm鲁 in this case), the absolute density will be \( \frac{600.0 \, g}{49.5 \, cm^{3}} = 12.12 \, g/cm^{3} \).

Absolute density is especially significant when considering material strength and compaction processes, as it reflects the true density of the material matter.
Void Fraction
The void fraction, often referred to as porosity, is the measure of the empty spaces in a material compared to its total volume. This ratio indicates how much of the volume is occupied by void space rather than the material itself.

To calculate the void fraction, you use the following relationship:
  • Void Fraction (Porosity) = \( \frac{\text{Void Volume}}{\text{Total Bed Volume}} \)
Void volume can be found by subtracting the volume of solids from the total volume of the bed.

For instance, with a total bed volume of 184 cm鲁 and a volume of solids calculated as 49.5 cm鲁, the void volume becomes 134.5 cm鲁. Thus, the porosity is \( \frac{134.5 \, cm^{3}}{184 \, cm^{3}} = 0.731 \) or 73.1%.

Understanding void fraction is vital in fields like civil engineering, hydrogeology, and material science, where the extent to which materials can absorb or allow passage of fluids is crucial.
Volume of Solids
Volume of solids is the actual space occupied by the material particles, with no inclusion of the air spaces or voids between them. Determining this is essential for assessing the absolute density of the material.

It is established by observing how much space the solid itself takes up, which is achieved through displacement methods when the solid is immersed in a fluid.

Using the displacement principle, if the initial water level is measured at 184 cm鲁 and rises to 233.5 cm鲁 after placing the solid in it, the volume occupied by the solid particles can be computed as:
  • Volume of Solids = Final Water Level - Initial Water Level
Therefore, the volume of solids would be \( 233.5 \, cm^3 - 184 \, cm^3 = 49.5 \, cm^3 \).

This concept is particularly valuable in mineral processing and building materials testing, where accurate volume and density are essential for quality control.

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

Hydrogen sulfide has the distinctive unpleasant odor associated with rotten eggs, and it is poisonous. It often must be removed from crude natural gas and is therefore a product of refining natural gas. In such instances, the Claus process provides a means of converting \(\mathrm{H}_{2} \mathrm{S}\) to elemental sulfur. Consider a feed stream to a Claus process that consists of 10.0 mole \(\% \mathrm{H}_{2} \mathrm{S}\) and \(90.0 \% \mathrm{CO}_{2}\). Onethird of the stream is sent to a furnace where the \(\mathrm{H}_{2} \mathrm{S}\) is burned completely with a stoichiometric amount of air fed at 1 atm and \(25^{\circ} \mathrm{C}\). The combustion reaction is $$\mathrm{H}_{2} \mathrm{S}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}$$ The product gases from this reaction are then mixed with the remaining two- thirds of the feed stream and sent to a reactor in which the following reaction goes to completion: $$2 \mathrm{H}_{2} \mathrm{S}+\mathrm{SO}_{2} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$$ The gases leave the reactor at \(10.0 \mathrm{m}^{3} / \mathrm{min}, 320^{\circ} \mathrm{C},\) and \(205 \mathrm{kPa}\) absolute. Assuming ideal-gas behavior, determine the feed rate of air in kmol/min. Provide a single balanced chemical equation reflecting the overall process stoichiometry. How much sulfur is produced in \(\mathrm{kg} / \mathrm{min} ?\)

The current global reliance on fossil fuels for heating, transportation, and electric power generation raises concems regarding the release of \(\mathrm{CO}_{2}\) and \(\mathrm{CH}_{4},\) which are greenhouse gases thought to lead to climate change, and NO, which contributes to smog. One potential solution to these problems is to produce transportation fuels from renewable biomass. You have been asked to evaluate a proposed process for converting forest residues to alcohols that may be used as transportation fuels. In the first stage of the process, steam and dry wood from hybrid poplar trees (which grow between five and eight feet a year and can be harvested roughly every five years) are fed to a gasifier in which the biomass is converted to light gases in the following reactions: $$\begin{aligned} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}+\mathrm{H}_{2} \\\ \mathrm{CO}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \\ \mathrm{C}+\mathrm{CO}_{2} & \rightarrow 2 \mathrm{CO} \\ \mathrm{C}+2 \mathrm{H}_{2} & \rightarrow \mathrm{CH}_{4} \\ \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}+3 \mathrm{H}_{2} \end{aligned}$$ The effluents from the reactor are a gas stream containing \(\mathrm{H}_{2}, \mathrm{CO}, \mathrm{CO}_{2}, \mathrm{CH}_{4},\) and \(\mathrm{H}_{2} \mathrm{O},\) and a solid char stream that contains only carbon and hydrogen. The char is discarded and the gases go through additional steps in which the hydrogen and carbon monoxide are converted to mixed alcohols. This problem only concerns the gasifier. \(\cdot\) Elemental composition of biomass: 51.9 mass \(\%\) C \(, 6.3 \%\) H, and \(41.8 \%\) O \(\cdot\) Pressure and temperature of entering steam: \(155^{\circ} \mathrm{C}, 4.4 \mathrm{atm}\) \(\cdot\) Feed ratio of steam to biomass: 1.1 kg steam/kg biomass \(\cdot\) Yield and dry-basis composition of product gas: 1.35 kg dry gas/kg biomass at \(700^{\circ} \mathrm{C}, 1.2\) atm; 50.7 mol\% \(\mathrm{H}_{2}, 23.8 \%\) CO, \(18.0 \% \mathrm{CO}_{2}, 7.5 \% \mathrm{CH}_{4}\) (a) Taking a basis of \(100 \mathrm{kg}\) of biomass fed, draw and completely label a flowchart for the gasifier incorporating the given data, labeling the volumes of the steam fed and the gases produced. Perform a degree-of-freedom analysis. (b) Calculate the mass and mass composition of the char and the volumes of the steam feed and product gas streams. (c) List advantages and possible drawbacks of using biomass rather than petroleum as a fuel source.

The demand for a particular hydrogenated compound, \(\mathrm{S}\), is \(5.00 \mathrm{kmol} / \mathrm{h}\). This chemical is synthesized in the gas-phase reaction $$A+H_{2}=S$$ The reaction equilibrium constant at the reactor operating temperature is $$K_{p}=\frac{p_{\mathrm{S}}}{p_{\Lambda} p_{\mathrm{H}_{2}}}=0.1 \mathrm{atm}^{-1}$$ The fresh feed to the process is a mixture of \(A\) and hydrogen that is mixed with a recycle stream consisting of the same two species. The resulting mixture, which contains \(3 \mathrm{kmol} \mathrm{A} / \mathrm{kmol} \mathrm{H}_{2},\) is fed to the reactor, which operates at an absolute pressure of 10.0 atm. The reaction products are in equilibrium. The effluent from the reactor is sent to a separation unit that recovers all of the \(S\) in essentially pure form. The A and hydrogen leaving the separation unit form the recycle that is mixed with fresh feed to the process. Calculate the feed rates of hydrogen and A to the process in kmol/h and the recycle stream flow rate in SCMH (standard cubic meters per hour).

A \(5.0-\mathrm{m}^{3}\) tank is charged with \(75.0 \mathrm{kg}\) of propane gas at \(25^{\circ} \mathrm{C}\). Use the SRK equation of state to estimate the pressure in the tank; then calculate the percentage error that would result from the use of the ideal-gas equation of state for the calculation.

Chemicals are stored in a laboratory with volume \(V\left(\mathrm{m}^{3}\right) .\) As a consequence of poor laboratory practices, a hazardous species, A, enters the room air (from inside the room) at a constant rate \(\dot{m}_{\mathrm{A}}(\mathrm{g} \mathrm{A} / \mathrm{h})\) The room is ventilated with clean air flowing at a constant rate \(\dot{V}_{\text {air }}\left(\mathrm{m}^{3} / \mathrm{h}\right) .\) The average concentration of A in the room air builds up until it reaches a steady-state value \(C_{\mathrm{A}, \mathrm{r}}\left(\mathrm{g} \mathrm{A} / \mathrm{m}^{3}\right)\) (a) List at least four situations that could lead to A getting into the room air. (b) Assume that the A is perfectly mixed with the room air and derive the formula $$\dot{m}_{\mathrm{A}}=\dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ (c) The assumption of perfect mixing is never justificd when the enclosed space is a room (as opposed to, say, a stirred reactor). In practice, the concentration of A varies from one point in the room to another: it is relatively high near the point where A enters the room air and relatively low in regions far from that point, including the ventilator outlet duct. If we say that \(C_{\mathrm{A}, \text { duct }}=k C_{\mathrm{A}}\) where \(k < 1\) is a nonideal mixing factor (generally between 0.1 and \(0.5,\) with the lowest value corresponding to the poorest mixing), then the equation of Part (b) becomes $$\dot{m}_{\mathrm{A}}=k \dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ Use this equation and the ideal-gas equation of state to derive the following expression for the average mole fraction of \(A\) in the room air: $$y_{\mathrm{A}}=\frac{\dot{m}_{\mathrm{A}}}{k \dot{V}_{\mathrm{air}}} \frac{R T}{M_{\mathrm{A}} P}$$ where \(M_{\mathrm{A}}\) is the molecular weight of \(\mathrm{A}\) (d) The permissible exposure level (PEL) for styrene \((M=104.14\) ) defined by the U.S. Occupational Safcty and Health Administration is 50 ppm (molar basis). \(^{21}\) An open storage tank in a polymerization laboratory contains styrene. The evaporation rate from this tank is estimated to be \(9.0 \mathrm{g} / \mathrm{h}\). Room temperature is \(20^{\circ} \mathrm{C}\). Assuming that the laboratory air is reasonably well mixed (so that \(k=0.5\) ), calculate the minimum ventilation rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) required to keep the average styrene concentration at or below the PEL. Then give several reasons why working in the laboratory might still be hazardous if the calculated minimum ventilation rate is used. (e) Would the hazard level in the situation described in Part (d) increase or decrease if the temperature in the room were to increase? (Increase, decrease, no way to tell.) Explain your answer, citing at least two effects of temperature in your explanation.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

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.