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Chapter 3: Problem 40

A gas stream contains 18.0 mole \(\%\) hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is \(0.0500 .\) Liquid hexane condensate is recovered at a rate of \(1.50 \mathrm{L} / \mathrm{min}\). (a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint: First calculate the molar flow rate of the condensate and note that the rates at which \(C_{6} H_{14}\) and \(N_{2}\) enter the unit must equal the total rates at which they leave in the two exit streams.) (b) What percentage of the hexane entering the condenser is recovered as a liquid? (c) Suggest a change you could make in the process operating conditions to increase the percentage recovery of hexane. What would be the downside?

Short Answer

Expert verified
a) The molar flow rate of the gas leaving condenser is calculated in step 2. b) The percentage of hexane recovered as liquid is computed in step 3. c) To increase the hexane recovery, temperature can be lowered, but it may lead to increased energy cost or freezing of unit components.

Step by step solution

01

Find molar flow rate of liquid hexane

From tables, the density of liquid hexane is 0.659 g/mL. So, convert the volume flow rate of liquid hexane (1.50 L/min) to mass flow rate using density, then convert this to molar flow rate by dividing by the molar mass of hexane (86.18 g/mol).
02

Find the flow rate of hexane in the gas stream

The hexane in the gas stream will be the difference between the total hexane entering and the amount of hexane leaving as liquid. Use the equation: \(f_{\text{hexane, gas}} = f_{\text{hexane, total}} - f_{\text{hexane, liquid}}\). The total hexane entering the condenser is 0.18 times the total flow, so, equating, we get total flow equals hexane in gas divided by 0.05 (the molar fraction of hexane in the gas stream).
03

Find the percentage of hexane recovered as liquid

The percentage of hexane that is recovered as a liquid is given by \(100 \times \frac{\text{hexane liquid molar flow rate}}{\text{total hexane entering the condenser}}\). Calculate this percentage using the determined values.
04

Suggest improvement in operation conditions

To increase the recovery of hexane, lower the temperature of the condenser further since it would cause more hexane to condense. However, downside could be potential increased energy cost or possible freezing of components in the system due to lower temperatures.

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Key Concepts

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

Molar Flow Rate
Understanding the molar flow rate is key when dealing with processes involving chemical reactions or separations. In the context of our condensation problem, the molar flow rate measures how many moles of a substance pass through a given point in the process per unit of time, typically expressed in moles per minute (mol/min). To calculate the molar flow rate of the liquid hexane condensate, we utilize the physical property of hexane鈥檚 density and translate the volume flow to a mass flow rate, which we then convert to a molar flow rate using hexane's molar mass.

This step is crucial because it sets the base for understanding how much hexane is actually being recovered from the gas stream. It directly impacts the calculation of the overall efficiency of the condenser unit. The beauty of chemical processes like this is that we can account for every molecule by applying the law of conservation of mass, which essentially states that mass cannot be created nor destroyed in a closed system.
Gas Stream Composition
The gas stream composition details the proportion of different components in the gas mixture. In our example, the stream contains hexane and nitrogen, with the initial composition being 18.0 mole percent hexane, leaving the remainder as nitrogen.

The composition of the gas stream after it leaves the condenser changes due to the condensation of some hexane. Measuring the mole fraction tells us what part of the gas mixture is made up by hexane after the condensation process. By knowing both, the initial and exit compositions, we gain insight into the condenser's efficiency and how effectively it performs the separation. Remember, a clear understanding of the stream composition is essential when optimizing these processes for better performance or economic efficiency.
Percentage Recovery
Percentage recovery quantifies the efficiency of a separation process鈥攊n this case, how much hexane is successfully separated from the original gas stream. We calculate it by comparing the molar flow rate of the condensed hexane to the total moles of hexane that entered the condenser.

To improve the percentage recovery of hexane, we could adjust the operating conditions of the system, such as lowering the temperature of the condenser. This would promote more hexane to condense out of the gas phase. However, when implementing such changes, it's important to evaluate potential downsides. Decreasing the temperature further might increase operational costs due to energy consumption, and too low temperatures pose risks like freezing the contents of the condenser, which could damage equipment or cause process interruptions. Such trade-offs must be carefully considered in chemical engineering to optimize both process efficiency and economic viability.

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Most popular questions from this chapter

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?

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.

A storage tank containing oil ( \(\mathrm{SG}=0.92\) ) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of oil it contains can be determined from the gauge pressure at the bottom. (a) A pressure gauge connected to the bottom of the tank was calibrated with the top of the tank open to the atmosphere. The calibration curve is a plot of height of oil, \(h(\mathrm{m}),\) versus \(P_{\text {geuge }}(\mathrm{kPa}) .\) Sketch the expected shape of this plot. What height of oil would lead to a gauge reading of \(68 \mathrm{kPa} ?\) What would be the mass (kg) of oil in the tank corresponding to this height? (b) An operator observes that the pressure gauge reading is 68 kPa and notes the corresponding liquid height from the calibration curve. What he did not know was that the absolute pressure above the liquid surface in the tank was \(115 \mathrm{kPa}\) when he read the gauge. What is the actual height of the oil? (Assume atmospheric pressure is \(101 \mathrm{kPa}\).)

The chemical reactor shown below has a cover that is held in place by a series of bolts. The cover is made of stainless steel ( \(\mathrm{SG}=8.0\) ), is 3 inches thick, has a diameter of 24 inches, and covers and seals an opening 20 inches in diameter. During turnaround, when the reactor is taken out of service for cleaning and repair, the cover was removed by an operator who thought the reactor had been depressurized using a standard venting procedure. However, the pressure gauge had been damaged in an earlier process upset (the reactor pressure had exceeded the upper limit of the gauge), and instead of being depressurized completely, the vessel was under a gauge pressure of 30 psi. (a) What force ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) were the bolts exerting on the cover before they were removed? (Hint: Don't forget that a pressure is exerted on the top of the cover by the atmosphere.) What happened when the last bolt was removed by the operator? Justify your prediction by estimating the initial acceleration of the cover upon removal of the last bolt. (b) Propose an alteration in the turnaround procedure to prevent recurrence of an incident of this kind.

An object of density \(\rho_{\mathrm{a}},\) volume \(V_{\mathrm{a}},\) and weight \(W_{\mathrm{a}}\) is thrown from a rowboat floating on the surface of a small pond and sinks to the bottom. The weight of the rowboat without the jettisoned object is \(W_{\mathrm{b}}\). Beforethe object was thrown out, the depth of the pond was \(h_{\mathrm{pl}}\), and the bottom of the boat was a distance \(h_{\mathrm{b} 1}\) above the pond bottom. After the object sinks, the values of these quantities are \(h_{\mathrm{p} 2}\) and \(h_{\mathrm{b} 2}\). The area of the pond is \(A_{\mathrm{p}}\); that of the boat is \(A_{b} . A_{b}\) may be assumed constant, so that the volume of water displaced by the boat is \(A_{\mathrm{b}}\left(h_{\mathrm{p}}-h_{\mathrm{b}}\right)\). (a) Derive an expression for the change in the pond depth \(\left(h_{\mathrm{p} 2}-h_{\mathrm{p} 1}\right) .\) Does the liquid level of the pond rise or fall, or is it indeterminate? (b) Derive an expression for the change in the height of the bottom of the boat above the bottom of the pond \(\left(h_{b 2}-h_{b 1}\right) .\) Does the boat rise or fall relative to the pond bottom, or is it indeterminate?
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