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Chapter 10: Problem 24

A gas that contains \(\mathrm{CO}_{2}\) is contacted with liquid water in an agitated batch absorber. The equilibrium solubility of \(\mathrm{CO}_{2}\) in water is given by Henry's law (Section \(6.4 \mathrm{b}\) ) $$C_{\mathrm{A}}=p_{\mathrm{A}} / H_{\mathrm{A}}$$ where \(C_{\mathrm{A}}\left(\mathrm{mol} / \mathrm{cm}^{3}\right)=\) concentration of \(\mathrm{CO}_{2}\) in solution, \(p_{\mathrm{A}}(\mathrm{atm})=\) partial pressure of \(\mathrm{CO}_{2}\) in the gas phase, and \(H_{\mathrm{A}}\left[\mathrm{atm} /\left(\mathrm{mol} / \mathrm{cm}^{3}\right)\right]=\) Henry's law constant. The rate of absorption of \(\mathrm{CO}_{2}\) (i.e., the rate of transfer of \(\mathrm{CO}_{2}\) from the gas to the liquid per unit area of gas-liquid interface) is given by the expression $$r_{\mathrm{A}}\left[\operatorname{mol} /\left(\mathrm{cm}^{2} \cdot \mathrm{s}\right)\right]=k\left(C_{\mathrm{A}}^{*}-C_{\mathrm{A}}\right)$$ where \(C_{A}=\) actual concentration of \(\mathrm{CO}_{2}\) in the liquid, \(C_{\mathrm{A}}^{*}=\) concentration of \(\mathrm{CO}_{2}\) in the liquid that would be in equilibrium with the \(\mathrm{CO}_{2}\) in the gas phase, and \(k(\mathrm{cm} / \mathrm{s})=\) a mass transfer coefficient. The gas phase is at a total pressure \(\mathrm{P}\left(\text { atm) and contains } y_{\mathrm{A}}\left(\mathrm{mol} \mathrm{CO}_{2} / \mathrm{mol}\text { gas), and the liquid }\right.\right.\) phase initially consists of \(V\left(\mathrm{cm}^{3}\right)\) of pure water. The agitation of the liquid phase is sufficient for the composition to be considered spatially uniform, and the amount of \(\mathrm{CO}_{2}\) absorbed is low enough for \(P, V,\) and \(y_{\mathrm{A}}\) to be considered constant throughout the process. (a) Derive an expression for \(d C_{\mathrm{A}} / d t\) and provide an initial condition. Without doing any calculations, sketch a plot of \(C_{\mathrm{A}}\) versus \(t,\) labeling the value of \(C_{\mathrm{A}}\) at \(t=0\) and the asymptotic value at \(t \rightarrow \infty\) Give a physical explanation for the asymptotic value of the concentration. (b) Prove that $$C_{\mathrm{A}}(t)=\frac{p_{\mathrm{A}}}{H_{\mathrm{A}}}[1-\exp (-k S t / V)]$$ where \(S\left(\mathrm{cm}^{2}\right)\) is the effective contact area between the gas and liquid phases. (c) Suppose the system pressure is 20.0 atm, the liquid volume is 5.00 liters, the tank diameter is \(10.0 \mathrm{cm},\) the gas contains 30.0 mole \(\% \mathrm{CO}_{2},\) the Henry's law constant is \(9230 \mathrm{atm} / \mathrm{mole} / \mathrm{cm}^{3}\) ), and the mass transfer coefficient is \(0.020 \mathrm{cm} / \mathrm{s}\). Calculate the time required for \(C_{\mathrm{A}}\) to reach \(0.620 \mathrm{mol} / \mathrm{L}\) if the gas-phase properties remain essentially constant. (d) If A were not \(\mathrm{CO}_{2}\) but instead a gas with a moderately high solubility in water, the expression for \(C_{\mathrm{A}}\) given in Part (b) would be incorrect. Explain where the derivation that led to it would break down.

Short Answer

Expert verified
Step 1: The derivative of \(C_A\) with respect to time, \(dC_A/dt\) is found to be \(k/V (p_A/H_A - C_A)\). Step 2: The asymptotic value of concentration is equal to the equilibrium concentration from Henry's Law \(C_A = p_A/H_A\). Step 3: After integration and solving, \(C_A(t) = p_A/H_A [1 - \exp(-k S t / V)]\). Step 4: With the given values, the time \(t\) can be calculated. Step 5: If gas A has high solubility, assumptions would break down.

Step by step solution

01

Derive an expression for \(d C_{A} / dt\)

Start by expressing the rate of absorption of \(\mathrm{CO}_{2}\) per unit area, \(r_{A}\), in terms of \(dC_{A} / dt\). We know, \(r_{A} =VA dC_A/dt\), where V is the volume of water. By using the given expression of \(r_A = k (C^{*}_A - C_A)\) and substitizing for \(C^{*}_A\) from Henry's law, we get the equation: \(VA dC_A/dt = k (p_{A}/H_{A} - C_A)\) which simplifies to \\ \(dC_A/dt = k/V (p_{A}/H_{A} - C_A)\).
02

Physical explanation for the asymptotic value of concentration

The asymptotic value of the concentration, as \(t \rightarrow \infty\) represents the equilibrium state of the system where the rate of gas entering the liquid phase equals the rate leaving it. So, \(C_A\) will become constant at that value. In terms of the equation derived, at large times \(t\), \(p_A/H_A - C_A\) tends to zero which implies that \(C_A = p_A/H_A\), the Henry's law equilibrium concentration.
03

Integration of rate equation

Integrate equation derived in step 1 over \(C_A\) and \(t\) between appropriate limits to solve for \(C_A(t)\). Remember, \(kS/V\) is a constant, so when you integrate from \(0\) to \(t\) on the left hand side and from \(0\) to \(C_A\) on the right hand side, you would get: \(- \ln(1 - \frac{C_A H_A}{P_A}) = \frac{k S t}{V}\). Finally, solve for \(C_A\) to get: \(C_A(t) = \frac{p_{A}}{H_{A}} [1 - \exp(-k S t / V)]\).
04

Calculate time

Substitute the given values into the formula derived. This results in an equation in a single unknown, \(t\), which can be solved by rearranging and using logarithm function to get time.
05

Explain derivation breakdown

If \(\mathrm{A}\) were a gas with a moderately high solubility in water, the main assumption - that \(P, V,\) and \(y_{\mathrm{A}}\) are considered constant throughout the process - would be invalidated. This is due to the fact that, as the gas dissolves in the liquid, its partial pressure \(p_A\) and mole fraction \(y_A\) in the gas phase would significantly reduce, affecting the rate of absorption and the equilibrium concentration.

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Key Concepts

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

Henry's Law
Henry's Law is an essential principle in understanding gas absorption in liquids. It describes how the gas's solubility is proportional to its partial pressure in the gas phase. The equation for Henry's Law is given by:\[C_A = \frac{p_A}{H_A}\]Here:
  • \(C_A\) is the concentration of the gas, \(\mathrm{CO}_2\) in this case, dissolved in the liquid.
  • \(p_A\) refers to the partial pressure of the gas in the gas phase.
  • \(H_A\) is the Henry's Law constant, unique to each gas-liquid pair and depends on temperature.
Henry's Law is crucial for predicting how much gas will dissolve in a liquid under given conditions. In our exercise, it helps determine the equilibrium concentration of \(\mathrm{CO}_2\) in water when the pressure and temperature are known. The law assumes that the system is in equilibrium, meaning the rates of gas dissolving into the liquid and coming out of it are equal. This balance ensures that the dissolved gas concentration stabilizes at a constant value.
Mass Transfer Coefficient
The mass transfer coefficient \(k\) is a key factor in quantifying the rate at which a gas moves from the gas phase to the liquid phase. It's a measure of the efficiency of the transfer process.The rate of absorption or mass transfer of \(\mathrm{CO}_2\) can be expressed as:\[r_A = k(C_A^* - C_A)\]In this equation:
  • \(r_A\) is the rate of mass transfer per unit area.
  • \(C_A^*\) represents the equilibrium concentration of the gas in the liquid, matching Henry's Law determined concentration.
  • \(C_A\) is the actual concentration at any time.
  • \(k\) is the mass transfer coefficient, indicating how quickly the equilibrium is reached.
The mass transfer coefficient is controlled by factors such as temperature, agitation, and the nature of the phases involved. It provides engineers with an essential parameter for designing and assessing processes like gas absorption, as it affects the speed at which equilibrium is approached. Understanding \(k\) allows for optimization of systems to ensure efficient removal of gases like \(\mathrm{CO}_2\).
Gas-Liquid Equilibrium
Gas-liquid equilibrium is reached when the concentration of a gas in the liquid no longer changes, meaning the gas is dissolving into and leaving the solution at the same rate. This balance is guided by both Henry's Law and the mass transfer process.In the system from our exercise, the equilibrium is characterized by the equation:\[C_A = \frac{p_A}{H_A}\]At equilibrium:
  • The concentration \(C_A\) matches \(C_A^*\), which is determined using Henry鈥檚 Law.
  • No net movement of gas molecules is occurring between the liquid and the gas phases.
Gas-liquid equilibrium is the state where the system stabilizes, and it is dependent on maintaining constant conditions like pressure, as assumed in the exercise. If a gas dissolves significantly, these assumptions might not hold, as further dissolution could change the pressure. Understanding this equilibrium helps in predicting the final outcome of gas absorption processes and in designing equipment like absorbers where gases need to be efficiently transferred to liquids.

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Most popular questions from this chapter

Phosgene (COCl_) is formed by CO and Cl_ reacting in the presence of activated charcoal: $$\mathrm{CO}+\mathrm{Cl}_{2} \rightarrow \mathrm{COCl}_{2}$$ At \(T=303.8 \mathrm{K}\) the rate of formation of phosgene in the presence of 1 gram of charcoal is $$R_{\mathrm{f}}(\mathrm{mol} / \mathrm{min})=\frac{8.75 C_{\mathrm{CO}} C_{\mathrm{C}_{2}}}{\left(1+58.6 C_{\mathrm{C}_{2}}+34.3 C_{\mathrm{COC}_{2}}\right)^{2}}$$ where C denotes concentration in mollL. (a) Suppose the charge to a 3.00 -liter batch reactor is \(1.00 \mathrm{g}\) of charcoal and a gas initially containing 60.0 mole\% CO and the balance \(\mathrm{Cl}_{2}\) at \(303.8 \mathrm{K}\) and 1 atm. Calculate the initial concentrations (mol/L) of both reactants, neglecting the volume occupied by the charcoal. Then, letting \(C_{\mathrm{P}}(t)\) be the concentration of phosgene at an arbitrary time \(t,\) derive relations for \(C_{\mathrm{P}}\) \(C_{\mathrm{CO}}\) and \(C_{\mathrm{C}_{2}}\) in terms of (b) Write a differential balance on phosgene and show that it simplifies to $$\frac{d C_{\mathrm{P}}}{d t}=\frac{2.92\left(0.02407-C_{\mathrm{P}}\right)\left(0.01605-C_{\mathrm{P}}\right)}{\left(1.941-24.3 C_{\mathrm{P}}\right)^{2}}$$ Provide an initial condition for this equation. (c) A plot of \(C_{\mathrm{P}}\) versus \(t\) starts at \(C_{\mathrm{P}}=0\) and asymptotically approaches a maximum value. Explain how you could predict that behavior from the form of the equation of Part (b). Without attempting to solve the differential equation, determine the maximum value of \(C_{\mathrm{P}}\) (d) Starting with the equation of Part (b), derive an expression for the time required to achieve a \(75 \%\) conversion of the limiting reactant. Your solution should have the form \(t=a\) definite integral. (e) The integral you derived in Part (d) can be evaluated analytically; however, more complex rate laws than the one given for the phosgene formation reaction would yield an integral that must be evaluated numerically. One procedure is to evaluate the integrand at a number of points between the limits of integration and to use a quadrature formula such as the trapezoidal rule or Simpson's rule (Appendix A.3) to estimate the value of the integral. Usea spreadsheet to evaluate the integrand of the integral of Part (c) at \(n_{p}\) equally spaced points between and including the limits of integration, where \(n_{p}\) is an odd number, and then to evaluate the integral using Simpson's rule. Perform the calculation for \(n_{p}=5,21,\) and \(51 .\) What can you conclude about the number of points needed to obtain a result accurate to three significant figures?

The flow rate of a process stream has tended to fluctuate considerably, creating problems in the process unit to which the stream is flowing. A horizontal surge drum has been inserted in the line to maintain a constant downstream flow rate even when the upstream flow rate varies. A cross-section of the drum, which has length \(L\) and radius \(r,\) is shown below. The level of liquid in the drum is \(h\), and the expression for liquid volume in the drum is $$V=L\left[r^{2} \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{r^{2}-(r-h)^{2}}\right]$$ Here is how the drum works. The rate of drainage of a liquid from a container varies with the height of the liquid in the container: the greater the height, the faster the drainage rate. The drum is initially charged with enough liquid so that when the input rate has its desired value, the liquid level is such that the drainage rate from the drum has the same value. A sensor in the drum sends a signal proportional to the liquid level to a control valve in the downstream line. If the input flow rate increases, the liquid level starts to rise; the control valve detects the rise from the transmitted signal and opens to increase the drainage rate, stopping when the level comes back down to its set-point value. Similarly, if the input flow rate drops, the control valve closes enough to bring the level back up to its set point. (a) The drum is to be charged initially with benzene (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3}\) ) at a constant rate \(\dot{m}(\mathrm{kg} / \mathrm{min})\) until the tank is half full. If \(L=5 \mathrm{m}, r=1 \mathrm{m},\) and \(\dot{m}=10 \mathrm{kg} / \mathrm{min},\) how long should it take to reach that point? (b) Now suppose the flow rate into the tank is unknown. A sight gauge on the tank allows determination of the liquid level, and instructions are to stop the flow when the tank contains 3000 kg. At what value of \(h\) should this be done? (c) After the tank has been charged, the flow rate into the drum, \(\dot{m}_{1}\), varies with upstream operations, and the flow rate out is \(10 \mathrm{kg} / \mathrm{min}\). Write a mass balance around the drum so that you obtain a relationship between \(\dot{m}_{1}\) and the rate of change in the height of liquid in the tank \((d h / d t)\) as a function of \(h .\) Estimate the flow rate into the tank when \(h\) has an approximate value of \(50 \mathrm{cm}\), and \(d h / d t=1 \mathrm{cm} / \mathrm{min} .\) (Hint: Although an analytical solution is feasible, you may find it easier to create plots of \(V\) and \(d V / d h\) at \(0.1 \mathrm{m}\) increments in \(h,\) which can be used in obtaining an approximate solution to the problem.) (d) Speculate on why the drum would provide better performance than feeding a signal proportional to the flow rate directly to the control valve that would cause the valve to close if the flow rate drops below the set point and to open if the flow rate rises above that point.

A steam radiator is used to heat a \(60-\mathrm{m}^{3}\) room. Saturated steam at 3.0 bar condenses in the radiator and emerges as a liquid at the saturation temperature. Heat is lost from the room to the outside at a rate $$\dot{Q}(\mathrm{kJ} / \mathrm{h})=30.0\left(T-T_{0}\right)$$ where \(T\left(^{\circ} \mathrm{C}\right)\) is the room temperature and \(T_{0}=0^{\circ} \mathrm{C}\) is the outside temperature. At the moment the radiator is turned on, the temperature in the room is \(10^{\circ} \mathrm{C}\). (a) Let \(\dot{m}_{\mathrm{s}}(\mathrm{kg} / \mathrm{h})\) denote the rate at which steam condenses in the radiator and \(n(\mathrm{kmol})\) the quantity of air in the room. Write a differential energy balance on the room air, assuming that \(n\) remains constant at its initial value, and evaluate all numerical coefficients. Take the heat capacity of air \(\left(C_{v}\right)\) to be constant at \(20.8 \mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\) (b) Write the steady-state energy balance on the room air and use it to calculate the steam condensation rate required to maintain a constant room temperature of \(24^{\circ} \mathrm{C}\). Without integrating the transient balance, sketch a plot of \(T\) versus \(t,\) labeling both the initial and maximum values of \(T\) (c) Integrate the transient balance to calculate the time required for the room temperature to rise by \(99 \%\) of the interval from its initial value to its steady-state value, assuming that the steam condensation rate is that calculated in Part (b).

A steam coil is immersed in a stirred tank. Saturated steam at 7.50 bar condenses within the coil, and the condensate emerges at its saturation temperature. A solvent with a heat capacity of \(2.30 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\cdot} \mathrm{C}\right)\) is fed to the tank at a steady rate of \(12.0 \mathrm{kg} / \mathrm{min}\) and a temperature of \(25^{\circ} \mathrm{C},\) and the heated solvent is discharged at the same flow rate. The tank is initially filled with \(760 \mathrm{kg}\) of solvent at \(25^{\circ} \mathrm{C},\) at which point the flows of both steam and solvent are commenced. The rate at which heat is transferred from the steam coil to the solvent is given by the expression $$\dot{Q}=U A\left(T_{\mathrm{steam}}-T\right)$$ where \(U A\) (the product of a heat transfer coefficient and the coil surface area through which the heat is transferred) equals \(11.5 \mathrm{kJ} /\left(\min \cdot^{\circ} \mathrm{C}\right) .\) The tank is well stirred, so that the temperature of the contents is spatially uniform and equals the outlet temperature. (a) Prove that an energy balance on the tank contents reduces to the equation given below and supply an initial condition. \frac{d T}{d t}=1.50^{\circ} \mathrm{C} / \mathrm{min}-0.0224 T (b) Without integrating the equation, calculate the steady-state value of \(T\) and sketch the expected plot of \(T\) versus \(t,\) labeling the values of \(T_{\mathrm{b}}\) at \(t=0\) and \(t \rightarrow \infty\) (c) Integrate the balance equation to obtain an expression for \(T(t)\) and calculate the solvent temperature after 40 minutes. (d) The tank is shut down for routine maintenance, and a technician notices that a thin mineral scale has formed on the outside of the steam coil. The coil is treated with a mild acid that removes the scale and reinstalled in the tank. The process described above is run again with the same steam conditions, solvent flow rate, and mass of solvent charged to the tank, and the temperature after 40 minutes is \(55^{\circ} \mathrm{C}\) instead of the value calculated in Part (c). One of the system variables listed in the problem statement must have changed as a result of the change in the stirrer. Which variable would you guess it to be, and by what percentage of its initial value did it change?

A 2000 -liter tank initially contains 400 liters of pure water. Beginning at \(t=0\), an aqueous solution containing \(1.00 \mathrm{g} / \mathrm{L}\) of potassium chloride flows into the tank at a rate of \(8.00 \mathrm{L} / \mathrm{s}\) and an outlet stream simultaneously starts flowing at a rate of \(4.00 \mathrm{L} / \mathrm{s}\). The contents of the tank are perfectly mixed, and the densities of the feed stream and of the tank solution, \(\rho(g / L),\) may be considered equal and constant. Let \(V(t)(\mathrm{L})\) denote the volume of the tank contents and \(C(t)(\mathrm{g} / \mathrm{L})\) the concentration of potassium chloride in the tank contents and outlet stream. (a) Write a balance on total mass of the tank contents, convert it to an equation for \(d V / d t\), and provide an initial condition. Then write a potassium chloride balance, show that it reduces to $$\frac{d C}{d t}=\frac{8-8 C}{V}$$ and provide an initial condition. (Hint: You will need to use the mass balance expression in your derivation.) (b) Without solving either equation, sketch the plots you expect to obtain for \(V\) versus \(t\) and \(C\) versus \(t\) If the plot of \(C\) versus \(t\) has an asymptotic limit as \(t \rightarrow \infty,\) determine what it is and explain why it makes sense. (c) Solve the mass balance to obtain an expression for \(V(t)\). Then substitute for \(V\) in the potassium chloride balance and solve for \(C(t)\) up to the point when the tank overflows. Calculate the \(\mathrm{KCl}\) concentration in the tank at that point.
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