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Methanol is produced by reacting carbon monoxide and hydrogen at \(644 \mathrm{K}\) over a \(\mathrm{ZnO}-\mathrm{Cr}_{2} \mathrm{O}_{3}\) catalyst. A mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) in a ratio \(2 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol}\) CO is compressed and fed to the catalyst bed at \(644 \mathrm{K}\) and 34.5 MPa absolute. A single-pass conversion of 25\% is obtained. The space velocity, or ratio of the volumetric flow rate of the feed gas to the volume of the catalyst bed, is The product gases are passed through a condenser, in which the methanol is liquefied. (a) You are designing a reactor to produce \(54.5 \mathrm{kmol} \mathrm{CH}_{3} \mathrm{OH} / \mathrm{h}\). Estimate (i) the volumetric flow rate that the compressor must be capable of delivering if no gases are recycled, and (ii) the required volume of the catalyst bed. (Use Kay's rule for pressure-volume calculations.) (b) If (as is done in practice) the gases from the condenser are recycled to the reactor, the compressor is then required to deliver only the fresh feed. What volumetric flow rate must it deliver assuming that the methanol produced is completely recovered in the condenser? (In practice it is not; moreover, a purge stream must be taken off to prevent the buildup of impurities in the system.)

Short Answer

Expert verified
For a reactor to produce 54.5 kmol/h methanol, the required volumetric flow rate for the compressor would be 44.95 m^3/h without gas recycling, and the volume of the catalyst bed required would be 22.475 m^3. If the gases are recycled, the fresh feed volumetric flow rate would be 11.18 m^3/h.

Step by step solution

01

(a) (i) Calculate the required feed of CO and H2

Using the stoichiometry of the reaction, for 54.5 kmol/h of CH3OH, the required feed of CO is also 54.5 kmol/h. As the ratio of H2:CO given is 2:1, the required feed of H2 is 2 * 54.5 = 109 kmol/h. The total molar flow rate of gases will be 54.5 (CO) + 109 (H2) = 163.5 kmol/h.
02

(a) (ii) Apply Kay's rule to find the volumetric flow rate

Applying Kay's rule, assuming unit volume at 0°C and 1 atm (1.013 bar), the total volume of feed gases at that condition will be 163.5 * 22.414 (molar volume at NTP) = 3664.3 m^3/h. Using the Ideal Gas Law, the volumetric flow rate for the feed gases can be found by correcting for actual temperature and pressure. At 644K (371°C) and 34.5 MPa (345 bar), this becomes 3664.3 * (273.15 + 371) / 273.15 * 1.013 / 345 = 44.95 m^3/h.
03

(a) (iii) Calculate the required volume of the catalyst bed

The space velocity (SV) given as ratio of volumetric flow rate to the volume of the catalyst bed, is defined as SV = V_feed/V_catalyst. The catalyst bed volume then can be expressed as V_catalyst = V_feed/SV. Given that SV is 2 h-1 and V_feed is 44.95 m^3/h, the volume of the catalyst bed required is V_catalyst = 44.95 / 2 = 22.475 m^3.
04

(b) Determine the fresh feed volumetric flow rate.

The fresh feed which is required to produce the methanol and is not recycled, includes the amount of gas which actually reacted in the reactor, which is 25% of the initial feed entering the reactor. Therefore, the fresh flow rate will be 0.25 * 163.5 (total molar flow rate as calculated earlier) = 40.875 kmol/h. Following the same procedure as before, the volumetric flow rate is 40.875 * 22.414 * (273.15 + 371) / 273.15 * 1.013 / 345 = 11.18 m^3/h.

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Key Concepts

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

Methanol Synthesis
Methanol (CH3OH) is a chemical with a wide array of uses, from fuel to feedstock for various chemical syntheses. To understand how methanol is made, one should look at the basic chemical reaction involving carbon monoxide (CO) and hydrogen (H2), generally over a catalyst like zinc oxide-chromium oxide (ZnO-Cr2O3). The reaction proceeds at elevated temperatures and pressures, yielding methanol through the following balanced equation:
CO + 2 H2 → CH3OH
For students grappling with the synthesis of methanol, remember it’s crucial to consider the molar ratios of reactants and the conditions required to favor the reaction, such as the 644 K temperature and 34.5 MPa pressure outlined in your textbook exercise.
Stoichiometry
Stoichiometry is the math behind chemistry. It's all about proportions and balancing act. When assessing the methanol synthesis, you may refer to the stoichiometric coefficients in the balanced reaction to understand the molar relationships between reactants and products. For the production of methanol, a 1:2 molar ratio of CO to H2 is needed, reflecting how many moles of each gas are necessary to form a mole of methanol. In practice, like in the textbook problem, we calculate the required feeds of CO and H2 based on the desired output of methanol—1 mole of CH3OH requires 1 mole of CO and 2 moles of H2. Thus, it's the stoichiometry that guides us in determining how much of each reactant we need to feed into the reactor to get our desired methanol output.
Catalyst Bed Design
In chemical engineering, the catalyst bed is where the magic happens. This is where reactant molecules meet and transform, with the help of a catalyst—our middleman. For a successful methanol synthesis process, the design of this catalyst bed is paramount. It needs to accommodate the correct volume of reactants and provide enough contact time for the reaction to occur. The space velocity, mentioned in the textbook problem, is a measure of how fast the gas flows through the bed relative to the catalyst's volume. It affects conversion rate, pressure drop, and ultimately the reactor's efficiency. Engineers must design the catalyst bed to ensure the right balance between size and flow rates to optimize production while minimizing costs and maximizing safety.
Kay's Rule
When dealing with gases under various conditions of temperature and pressure, one useful shortcut is Kay's rule. It's a thumb rule employed to quickly estimate the volume of a gas mixture at high pressures assuming ideal behavior. It asserts that the molar volume of a gas mixture can be approximated by the sum of the molar contributions of each individual gas, each adjusted for its mole fraction. This rule is applied, as demonstrated in the textbook solution, to estimate the volumetric flow rate of gases fed to the catalyst bed for methanol synthesis. Given the high-pressure conditions used in the methanol process, Kay's rule simplifies the step from knowing the molar flow rates to determining the actual gas volumes needed.
Ideal Gas Law
The Ideal Gas Law is a cornerstone of chemical engineering and broader physical science, encapsulating the relationship between a gas's pressure (P), volume (V), temperature (T), and the amount of substance (n). The law is expressed in the equation PV=nRT, where R is the universal gas constant. This equation allows engineers to calculate any one of the variables if the others are known and is crucial for designing processes like methanol synthesis. In our textbook exercise, the Ideal Gas Law is used alongside Kay's rule to correct the volumetric flow rate for actual operating conditions of the process—after all, real-life systems rarely operate at standard conditions of temperature and pressure.

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Most popular questions from this chapter

An oxygen tank with a volume of \(2.5 \mathrm{ft}^{3}\) is kept in a room at \(50^{\circ} \mathrm{F}\). An engineer has used the idealgas equation of state to determine that if the tank is first evacuated and then charged with \(35.3 \mathrm{lb}_{\mathrm{m}}\) of pure oxygen, its rated maximum allowable working pressure (MAWP) will be attained. Operation at pressures above this value is considered unsafe. (a) What is the maximum allowable working pressure (psig) of the tank? (b) You suspect that at the conditions of the fully charged tank, ideal-gas behavior may not be a good assumption. Use the SRK equation of state to obtain a better estimate of the maximum mass of oxygen that may be charged into the tank. Did the ideal-gas assumption lead to a conservative estimate (on the safe side) or a nonconservative estimate of the amount of oxygen that could be charged? (c) Suppose the tank is charged and ruptures before the amount of oxygen calculated in Part (b) enters it. (It should have been able to withstand pressures up to four times the MAWP.) Think of at least five possible explanations for the failure of the tank below its rated pressure limit.

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.

Oxygen therapy uses various devices to provide oxygen to patients having difficulty getting sufficient amounts from air through normal breathing. Among the devices is a nasal cannula, which transports oxygen through small plastic tubes from a supply tank to prongs placed in the nostril. Consider a specific configuration in which the supply tank, whose volume is \(6.0 \mathrm{ft}^{3},\) is filled to a pressure of 2100 psig at a temperature of \(85^{\circ} \mathrm{F}\). The paticnt is in an environment where the ambicnt temperature is \(40^{\circ} \mathrm{F}\). When the cannula is put into use, the pressure in the tank begins to decrease as oxygen flows at \(10-15 \mathrm{L} / \mathrm{min}\) through a tube and the cannula into the nostrils. (a) Estimate the original mass of oxygen in the tank using the compressibility-factor equation of state. (b) What is the initial pressure when the temperature is 40 \(^{\circ} \mathrm{F} ?\) How much oxygen remains in the tank when application of the ideal-gas equation of state produces a result that is within \(3 \%\) of that predicted by the compressibility-factor equation of state (i.e., when \(0.97 \leq z \leq 1.03\) )? (c) How long will it take for the gauge on the tank to read 50 psig, assuming an average flow rate of \(12.5 \mathrm{L} / \mathrm{min} ?\)

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

Steam reforming is an important technology for converting refined natural gas, which we take here to be methane, into a synthesis gas that can be used to produce a varicty of other chemical compounds. For example, consider a reformer to which natural gas and steam are fed in a ratio of 3.5 moles of steam per mole of methane. The reformer operates at 18 atm, and the reaction products leave the reformer in chemical equilibrium at \(875^{\circ} \mathrm{C}\). The steam reforming reaction is $$\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}+3 \mathrm{H}_{2}$$ and the water-gas shift reaction also occurs in the reformer. $$\mathrm{CO}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{H}_{2}$$ The equilibrium constants for these two reactions are given by the expressions At \(875^{\circ} \mathrm{C}, K_{\mathrm{R}}=872.9 \mathrm{atm}^{2}\) and \(K \mathrm{w} \mathrm{G}=0.2482 .\) The process is to produce \(100.0 \mathrm{kmol} / \mathrm{h}\) of hydrogen. Calculate the feed rates (kmol/h) of methane and steam and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of gas leaving the reformer.

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