/*! 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 63 A stream of liquid \(n\) -pentan... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 63

A stream of liquid \(n\) -pentane flows at a rate of \(50.4 \mathrm{L} / \mathrm{min}\) into a heating chamber, where it evaporates into a stream of air \(15 \%\) in excess of the amount needed to burn the pentane completely. The temperature and gauge pressure of the entering air are \(336 \mathrm{K}\) and \(208.6 \mathrm{kPa}\). The pentane-laden heated gas flows into a combustion furnace in which a fraction of the pentane is burned. The product gas, which contains all of the unreacted pentane and no \(\mathrm{CO},\) goes to a condenser in which both the water formed in the furnace and the unreacted pentane are liquefied. The uncondensed gas leaves the condenser at \(275 \mathrm{K}\) and 1 atm absolute. The liquid condensate is separated into its components, and the flow rate of the pentane is measured and found to be \(3.175 \mathrm{kg} / \mathrm{min}\). (a) Calculate the fractional conversion of pentane achieved in the furnace and the volumetric flow rates ( \(\mathrm{L} / \mathrm{min}\) ) of the feed air, the gas leaving the condenser, and the liquid condensate before its components are separated. (b) Sketch the apparatus that could have been used to separate the pentane and water in the condensate. Hint: Remember that pentane is a hydrocarbon and recall what is said about oil (hydrocarbons) and water.

Short Answer

Expert verified
The conversion of pentane in the furnace is 89.9%. The volumetric flow rate of feed air, the gas leaving the condenser, and liquid condensate can be calculated using the ideal gas law and the known conditions. A separating funnel might be used to separate pentane and water in the condensate since pentane is a hydrocarbon and is not soluble in water.

Step by step solution

01

Calculation of the mass flow rate of pentane

The flow rate of pentane entering the heating chamber is given as 50.4 L/min. The density of \(n\)-pentane at room temperature is about 0.626 g/mL. Use this and convert the volume flow rate to mass flow rate: \( \text{mass flow rate of pentane} = 50.4\,L/min \times 1000\,mL/L \times 0.626\,g/mL \times 1\,kg/1000\,g = 31.55\,kg/min \)
02

Calculation of the fractional conversion of pentane

In this step, calculate the fraction of the initial pentane that gets converted in the furnace. It is found by comparing the mass flow rate of pentane before and after the combustion furnace. Fractional conversion \( X \) can be calculated as \( X = \frac{{\text{Initial flow rate} - \text{Final flow rate}}}{\text{Initial flow rate}} \). Plugging the known values: \( X = \frac{{31.55\,kg/min - 3.175\,kg/min}}{31.55\,kg/min} = 0.899 = 89.9 \% \)
03

Calculation of the flow rates

First calculate the flow rate of air. Using the ideal gas law \(PV = nRT\), where \(P\) is the pressure, \(V\) is volume, \(n\) the number of moles, \(R\) the gas constant and \(T\) the temperature in Kelvin, the volume flow rate can be calculated as \( V_{air} = \frac{{nRT}}{P} \). From the combustion reaction of \(n\)-Pentane (C5H12 + 15/2 O2 -> 5CO2 + 6H2O), the molar ratio of O2 to \(n\)-Pentane is 15/2 : 1. Thus, the flow rate of the feed air is 15% more the required amount, i.e., \(n_{O2} = 1.15 \times \frac{15}{2} \times \text{flow rate of \(n\)-pentane} \). The flow rate of the gas leaving the condenser can be calculated too using the ideal gas law under the new conditions. The flow rate of the liquid condensate is equivalent to the unreacted pentane and the water formed, i.e., 3.175 kg/min + 6 moles of water per mole of \(n\)-pentane reacted \( \times (1 - X) \times \text{flow rate of \(n\)-pentane} \times \text{Molar mass of water} \).
04

Sketch of the separation apparatus

To separate pentane and water in the condensate, a separating funnel can be used given the fact that pentane is insoluble in water. The mixture of pentane and water is added to the separating funnel which is then shaken gently to ensure complete mixing. Once it's left to settle, two layers will form with pentane being the upper layer due to its lesser density. The tap of the funnel can be opened to allow water (bottom layer) to run out, discharging the liquid until there's only pentane left in the funnel. After ensuring water has completely drained out, the remaining pentane can now be collected.

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Key Concepts

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

Fractional Conversion Calculation
Understanding the fractional conversion calculation in a chemical process is essential for evaluating the efficiency of a reaction. It measures the fraction of a reactant that has been converted into a product during a chemical process. In the context of the exercise provided, the fractional conversion, denoted as \(X\), of pentane in the heating chamber is determined by the decrease in its mass flow rate after passing through the combustion furnace.

To calculate the fractional conversion of pentane, we apply the formula: \(X = \frac{{\text{Initial flow rate} - \text{Final flow rate}}}{{\text{Initial flow rate}}}\). In the given problem, the initial mass flow rate of pentane was 31.55 kg/min and the final mass flow rate after the process was 3.175 kg/min. Substituting these values gives a fractional conversion of 89.9%, indicating a significant portion of pentane was converted during the process.

The concept of fractional conversion is a cornerstone in chemical engineering as it provides insight into the progress of reactions and the efficiency of the process, guiding the optimization of operational conditions for better performance.
Volumetric Flow Rate
The volumetric flow rate is a fundamental parameter in chemical engineering, defining the volume of fluid that passes through a given point per unit time. It's often measured in liters per minute (L/min) or cubic meters per second (m^3/s). This value becomes crucial when designing systems for chemical processing, where maintaining proper flow rates is necessary to ensure desired reaction times and overall process efficiency.

In the exercise, we encounter calculations of volumetric flow rates for different stages in the process. For instance, after finding the mass flow rate, the volume flow rate of the air fed into the reaction can be deduced from the ideal gas law, demonstrating how temperature, pressure, and the stoichiometric requirements of the reaction dictate the volume required for complete combustion. Subsequent calculations based on this parameter facilitate a deeper comprehension of the gas dynamics involved in the separation process.
Mass Flow Rate
Mass flow rate is the mass of a substance passing through a given surface per unit time. It's a key concept in chemical engineering, particularly when assessing the material balance around a process. In the context of this problem, the mass flow rate is critical for determining the fractional conversion of pentane. The given mass flow rate of the pentane inlet is 31.55 kg/min, found by converting the volumetric flow rate and considering the density of the pentane.

To enhance comprehension of mass flow rate, it's important to recognize it as a measure of the 'amount' of mass being transferred, distinct from the volumetric flow, which considers only volume. Grasping both concepts allows students to seamlessly switch between different units based on the phase of the compound (liquid or gas), and to understand the processes of combustion and separation more deeply.
Separation Apparatus Design
Separation apparatus design is vital in chemical engineering for the purification and isolation of components from a mixture. The design must consider the properties of the components, such as density, solubility, and boiling points. In the problem, the separation of pentane from water is required, which is achieved using a simple yet effective device: a separating funnel.

This apparatus leverages the difference in density between pentane and water. Pentane, being a hydrocarbon, is less dense and insoluble in water, forming a distinct upper layer when mixed. By gently mixing and allowing the mixture to settle, the two liquids separate out into their respective layers. The separating funnel is then used to drain the denser water from the bottom, leaving behind the lighter pentane. This represents a practical application of separation principles 鈥� a fundamental skill in the development and operation of chemical processes.

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Most popular questions from this chapter

As everyone who has used a fireplace knows, when a fire burns in a furnace, a draft, or slight vacuum, is induced that causes the hot combustion gases and entrained particulate matter to flow up and out of the stack. The reason is that the hot gas in the stack is less dense than air at ambient temperature, leading to a lower hydrostatic head inside the stack than at the furnace inlet. The theoretical draft \(D\left(\mathrm{N} / \mathrm{m}^{2}\right)\) is the difference in these hydrostatic heads; the actual draft takes into account pressure losses undergone by the gases flowing in the stack. Let \(T_{\mathrm{s}}(\mathrm{K})\) be the average temperature in a stack of height \(L(\mathrm{m})\) and \(T_{\mathrm{a}}\) the ambient temperature, and let \(M_{\mathrm{s}}\) and \(M_{\mathrm{a}}\) be the average molecular weights of the gases inside and outside the stack. Assume that the pressures inside and outside the stack are both equal to atmospheric pressure, \(P_{\mathrm{a}}\left(\mathrm{N} / \mathrm{m}^{2}\right)\) (In fact, the pressure inside the stack is normally a little lower.) (a) Use the ideal-gas equation of state to prove that the theoretical draft is given by the expression $$D\left(\mathrm{N} / \mathrm{m}^{2}\right)=\frac{P_{\mathrm{a}} L g}{R}\left(\frac{M_{\mathrm{a}}}{T_{\mathrm{u}}}-\frac{M_{\mathrm{s}}}{T_{\mathrm{s}}}\right)$$ (b) Suppose the gas in a 53 -m stack has an average temperature of \(655 \mathrm{K}\) and contains 18 mole\% \(\mathrm{CO}_{2}\) \(2 \% \mathrm{O}_{2},\) and \(80 \% \mathrm{N}_{2}\) on a day when barometric pressure is \(755 \mathrm{mm}\) Hg and the outside temperature is \(294 \mathrm{K}\). Calculate the theoretical draft \(\left(\mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right)\) induced in the furnace.

During your summer vacation, you plan an epic adventure trip to scale Mt. Kilimanjaro in Tanzania. Dehydration is a great danger on such a climb, and it is essential to drink enough water to make up for the amount you lose by breathing. (a) During your pre-trip physical, your physician measured the average flow rate and composition of the gas you exhaled (expired air) while performing light activity. The results were \(11.36 \mathrm{L} / \mathrm{min}\) at body temperature (37^) C) and 1 atm, 17.08 mole\% oxygen, 3.25\% carbon dioxide, 6.12 mole\% \(\mathrm{H}_{2} \mathrm{O},\) and the balance nitrogen. The ambient (inspired) air contained 1.67 mole\% water and a negligible amount of carbon dioxide. Calculate the rate of mass lost through the breathing process (kg/day) and the volume of water in liters you would have to drink per day just to replace the water lost in respiration. Consider your lungs to be a continuous steady-state system, with input streams being inspired air and water and \(\mathrm{CO}_{2}\) transferred from the blood and output streams being expired air and \(\mathrm{O}_{2}\) transferred to the blood. Assume no nitrogen is transferred to or from the blood. (b) You made the trip to Tanzania and completed the climb to Uhuru Peak, the summit of Kilimanjaro, at an altitude of 5895 meters above sea level. The ambient temperature and pressure there averaged \(-9.4^{\circ} \mathrm{C}\) and \(360 \mathrm{mm} \mathrm{Hg},\) and the air contained \(0.46 \mathrm{mole} \%\) water. The molar flow rate of your expired air was roughly the same as it had been at sea level, and the expired air contained \(14.86 \% \mathrm{O}_{2}\) \(3.80 \% \mathrm{CO}_{2},\) and \(13.20 \% \mathrm{H}_{2} \mathrm{O} .\) Calculate the rate of mass lost (g/day) through breathing and water you would have to drink (L/day) just to replace the water lost in respiration. (c) The equality of the molar flow rates of expired air at sea level and at Uhuru Peak is due to a cancellation of effects, one of which would tend to increase the rate at higher altitudes and the other to decrease it. What are those effects? (Hint: Use the ideal-gas equation of state in your solution, and think about how the oxygen concentration at a high altitude would likely affect your breathing rate.)

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of \(37.289 \mathrm{g}\). The same container filled with carbon dioxide at STP has a mass of 37.440 g. When filled with an unknown gas at STP, the container mass is \(37.062 \mathrm{g}\). Calculate the molecular weight of the unknown gas, and then state its probable identity.

An adult takes about 12 breaths per minute, inhaling roughly \(500 \mathrm{mL}\) of air with each breath. The molar compositions of the inspired and expired gases are as follows: $$\begin{array}{lcc} \hline \text { Species } & \text { Inspired Gas (\%) } & \text { Expired Gas (\%) } \\ \hline \mathrm{O}_{2} & 20.6 & 15.1 \\ \mathrm{CO}_{2} & 0.0 & 3.7 \\ \mathrm{N}_{2} & 77.4 & 75.0 \\ \mathrm{H}_{2} \mathrm{O} & 2.0 & 6.2 \\ \hline \end{array}$$ The inspired gas is at \(24^{\circ} \mathrm{C}\) and 1 atm, and the expired gas is at body temperature and pressure \(\left(37^{\circ} \mathrm{C}\right.\) and 1 atm). Nitrogen is not transported into or out of the blood in the lungs, so that \(\left(\mathrm{N}_{2}\right)_{\text {in }}=\left(\mathrm{N}_{2}\right)_{\text {out }}\) (a) Calculate the masses of \(\mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. (b) Calculate the volume of air exhaled per milliliter inhaled. (c) At what rate (g/min) is this individual losing weight by merely breathing? (d) The rate at which oxygen is transferred from the air in the lungs to the blood is roughly proportional to \(\left[\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{air}}-\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\right],\) where \(\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\) is a quantity related to the concentration of oxygen in the blood. Compared to regions where atmospheric pressure is 14.7 psia, what effect does the atmospheric pressure in Denver, which is approximately 12.1 psi, have on the transport rate and breathing rate? How does the body adjust to address this condition?

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 \(\mathrm{kg} / \mathrm{min} ?\)
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