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Chapter 3: Problem 30

Coal being used in a power plant at a rate of \(8000 \mathrm{lb}_{\mathrm{m}} / \mathrm{min}\) has the following composition: $$\begin{array}{lc}\hline \text { Component } & \text { Weight } \% \text { (dry basis) } \\ \hline \text { Ash } & 7.2 \\\\\text { Sulfur } & 3.5 \\\\\text { Hydrogen } & 5.0 \\\\\text { Carbon } & 75.2 \\ \text { Nitrogen } & 1.6 \\\\\text { Oxygen } & 7.5 \\\\\hline\end{array}$$ In addition, there are \(4.58 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O}\) per \(\mathrm{lb}_{\mathrm{m}}\) of coal. Determine the molar flow rate of each element in the coal (including water) other than ash.

Short Answer

Expert verified
The molar flow rates of each element (C, H, O, N, S and H2O) other than ash in the coal are calculated by firstly deriving their mass flow rates and then converting these values into molar flow rates using the respective molar masses. The specific values are determined from the calculations described in the step-by-step solution.

Step by step solution

01

Identify the Provided Data

From the problem, it's given that the overall coal usage is 8000 lbm/min. The composition of the coal by weight (on a dry basis) is also given, as well as the additional 4.58 lbm H2O per lbm of coal.
02

Determine the Mass Flow Rate of Each Elemental Component

The mass flow rate of each element can be calculated by multiplying the total coal usage with the respective weight percent (converted into decimal form). For example, the mass flow rate of carbon (C) is \(8000 lbm/min \times 0.752 (or 75.2 / 100)\). Follow this procedure for each element in the coal.
03

Compute the Molar Flow Rate of Each Element

The molar flow rate of each element is obtained by dividing its mass flow rate by its molar mass. The molar mass of each element is: carbon (C): 12 lbm/lbmol, hydrogen (H): 1 lbm/lbmol, oxygen (O): 16 lbm/lbmol, nitrogen (N): 14 lbm/lbmol, and sulfur (S): 32 lbm/lbmol. Water (H2O) has a molar mass of 18 lbm/lbmol.
04

Include Water in the Calculation

Besides the elements in the coal, water is also present. Calculate the mass flow rate of water by multiplying the total amount of coal with the given ratio (4.58 lbm H2O per lbm of coal). This mass flow rate is then converted into molar flow rate using the molar mass of water.

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Key Concepts

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

Elemental Composition
Elemental composition is a fundamental concept in chemical process analysis, where we dissect a substance into its elemental constituents. In the context of coal, this means understanding the makeup of each element present in the coal. Knowing the composition is crucial for further calculations, such as determining mass and molar flow rates.
The given problem specifies the elemental composition of coal used in a power plant. Each element, on a dry basis, has a percentage weight:
  • Ash: 7.2%
  • Sulfur: 3.5%
  • Hydrogen: 5.0%
  • Carbon: 75.2%
  • Nitrogen: 1.6%
  • Oxygen: 7.5%
Understanding these values helps us calculate how much of each element is present, considering both the coal and the additional water content. This is a critical step before diving into mass and molar flow calculations.
Molar Flow Rate
The molar flow rate tells us how many moles of a chemical species flow through a system over a given time. Understanding molar flow rate is essential in processes where chemical reactions occur. The calculation involves two steps: first, identifying the mass flow rate of an element, and second, converting that to its molar equivalent by dividing the mass by its molar mass.
For instance, in the given problem, if the carbon mass flow rate is determined by the equation \[ \text{Mass Flow Rate of Carbon} = 8000 \text{ lbm/min} \times 0.752 \] we then calculate the molar flow rate using the molar mass of carbon (12 lbm/lbmol):\[ \text{Molar Flow Rate of Carbon} = \frac{\text{Mass Flow Rate of Carbon}}{12 \text{ lbm/lbmol}} \]
We perform a similar calculation for other elements, like sulfur, hydrogen, oxygen, and nitrogen, using their respective molar masses. This method ensures streamlined and systematic conversion from mass to mole units.
Coal Combustion Analysis
Coal combustion analysis involves studying how coal burns, where elemental composition plays a significant role. By understanding the composition, we can predict how efficiently the coal will burn, the heat released, and the emissions generated.
This problem highlights coal's interaction with additional water. Water changes combustion dynamics since it needs energy to evaporate, thus impacting the efficiency.
In combustion analysis, calculating the molar flow rates helps understand the stoichiometry of the combustion reactions. Knowing exact proportions of carbon, hydrogen, and other elements, including additional water, allows engineers to optimize combustion processes, improve efficiency, and reduce pollutants.
For example, precise molar flow rates of hydrogen and carbon allow the calculation of water and carbon dioxide produced during combustion, which is vital in designing emission control systems. Therefore, a meticulous approach to elemental analysis boosts the understanding and optimization of coal combustion.

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Most popular questions from this chapter

A useful measure of an individual's physical condition is the fraction of his or her body that consists of fat. This problem describes a simple technique for estimating this fraction by weighing the individual twice, once in air and once submerged in water. (a) A man has body mass \(m_{b}=122.5 \mathrm{kg} .\) If he stands on a scale calibrated to read in newtons, what would the reading be? If he then stands on a scale while he is totally submerged in water at \(30^{\circ} \mathrm{C}\) (specific gravity \(=0.996\) ) and the scale reads \(44.0 \mathrm{N},\) what is the volume of his body (liters)? (Hint: Recall from Archimedes' principle that the weight of a submerged object equals the weight in air minus the buoyant force on the object, which in turn equals the weight of water displaced by the object. Neglect the buoyant force of air.) What is his body density, \(\rho_{\mathrm{b}}(\mathrm{kg} / \mathrm{L}) ?\) (b) Suppose the body is divided into fat and nonfat components, and that \(x_{f}\) (kilograms of fat/kilogram of total body mass) is the fraction of the total body mass that is fat: \(x_{\mathrm{f}}=\frac{m_{\mathrm{f}}}{m_{\mathrm{b}}}\) Prove that \(x_{\mathrm{f}}=\frac{\frac{1}{\rho_{\mathrm{b}}}-\frac{1}{\rho_{\mathrm{nf}}}}{\frac{1}{\rho_{\mathrm{f}}}-\frac{1}{\rho_{\mathrm{nf}}}}\) where \(\rho_{\mathrm{b}}, \rho_{\mathrm{f}},\) and \(\rho_{\mathrm{nf}}\) are the average densities of the whole body, the fat component, and the nonfat component, respectively. [Suggestion: Start by labeling the masses ( \(m_{\mathrm{f}}\) and \(m_{\mathrm{b}}\) ) and volumes \(\left(V_{\mathrm{f}} \text { and } V_{\mathrm{b}}\right)\) of the fat component of the body and the whole body, and then write expressions for the three densities in terms of these quantities. Then eliminate volumes algebraically and obtain an expression for \(\left.m_{f} / m_{b} \text { in terms of the densities. }\right]\) (c) If the average specific gravity of body fat is 0.9 and that of nonfat tissue is \(1.1,\) what fraction of the man's body in Part (a) consists of fat? (d) The body volume calculated in Part (a) includes volumes occupied by gas in the digestive tract, sinuses, and lungs. The sum of the first two volumes is roughly \(100 \mathrm{mL}\) and the volume of the lungs is roughly 1.2 liters. The mass of the gas is negligible. Use this information to improve your estimate of \(x_{\mathrm{f}}\).

The reaction \(A \rightarrow B\) is carried out in a laboratory reactor. According to a published article the concentration of A should vary with time as follows: \(C_{\mathrm{A}}=C_{\mathrm{A} 0} \exp (-k t)\) where \(C_{\mathrm{A} 0}\) is the initial concentration of \(\mathrm{A}\) in the reactor and \(k\) is a constant. (a) If \(C_{\mathrm{A}}\) and \(C_{\mathrm{A} 0}\) are in \(\mathrm{Ib}-\) moles \(/ \mathrm{ft}^{3}\) and \(t\) is in minutes, what are the units of \(k ?\) (b) The following data are taken for \(C_{\mathrm{A}}(t):\) $$\begin{array}{cc}\hline t(\min ) & C_{\mathrm{A}}\left(\mathrm{lb}-\mathrm{mole} / \mathrm{ft}^{3}\right) \\\\\hline 0.5 & 1.02 \\\1.0 & 0.84 \\\1.5 & 0.69 \\\2.0 & 0.56 \\\3.0 & 0.38 \\\ 5.0 & 0.17 \\\10.0 & 0.02 \\\\\hline\end{array}$$ Verify the proposed rate law graphically (first determine what plot should yield a straight line), and calculate \(C_{\mathrm{A} 0}\) and \(k\) (c) Convert the formula with the calculated constants included to an expression for the molarity of A in the reaction mixture in terms of \(t\) (seconds). Calculate the molarity at \(t=265 \mathrm{s}\).

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spilled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of mercury in the open arm is \(900 \mathrm{mm}\) above the lowest part of the manometer. (a) When the gas is not flowing, the pressure is the same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connected to the pipe? (b) When gas is flowing, the mercury level in the visible arm drops by \(25 \mathrm{mm}\). What is the gas pressure (psig) at this moment?

The little-known rare earth element nauseum (atomic weight \(=172\) ) has the interesting property of being completely insoluble in everything but 25 -year- old single-malt Scotch. This curious fact was discovered in the laboratory of Professor Ludwig von Schlimazel, the eminent German chemist whose invention of the bathtub ring won him the Nobel Prize. Having unsuccessfully tried to dissolve nauseum in 7642 different solvents over a 10 -year period, Schlimazel finally came to the \(30 \mathrm{mL}\) of The Macsporran that was the only remaining liquid in his laboratory. Always willing to suffer personal loss in the name of science, Schlimazel calculated the amount of nauseum needed to make up a 0.03 molar solution, put the Macsporran bottle on the desk of his faithful technician Edgar P. Settera, weighed out the calculated amount of nauseum and put it next to the bottle, and then wrote the message that has become part of history: "Ed Settera. Add nauseum/" How many grams of nauseum did he weigh out? (Neglect the change in liquid volume resulting from the nauseum addition.)

In the manufacture of pharmaceuticals, most active pharmaceutical ingredients (APIs) are made in solution and then recovered by separation. Acetaminophen, a pain-killing drug commercially marketed as Tylenol", is synthesized in an aqueous solution and subsequently crystallized. The slurry of crystals is sent to a centrifuge from which two effluent streams emerge: ( 1 ) a wet cake containing 90.0 wt\% solid acetaminophen \((\mathrm{MW}=\) 151 g/mol) and 10.0 wt\% water (plus some acetaminophen and other dissolved substances, which we will neglect), and (2) a highly dilute aqueous solution of acetaminophen that is discharged from the process. The wet cake is fed to a dryer where the water is completely evaporated, leaving the residual acetaminophen solids bone dry. If the evaporated water were condensed, its volumetric flow rate would be \(50.0 \mathrm{Lh}\). Following is a flowchart of the process, which runs 24 h/day, 320 days/yr. A denotes acetaminophen. (a) Calculate the yearly production rate of solid acetaminophen (tonne/yr), using as few dimensional equations as possible. (b) A proposal has been made to subject the liquid solution leaving the centrifuge to further processing to recover more of the dissolved acetaminophen instead of disposing of the solution. On what would the decision depend?
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