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Chapter 10: Problem 15

A stirred tank contains \(1500 \mathrm{lb}_{\mathrm{m}}\) of pure water at \(70^{\circ} \mathrm{F}\). At time \(t=0,\) two streams begin to flow into the tank and one is withdrawn. One input stream is a \(20.0 \mathrm{wt} \%\) aqueous solution of \(\mathrm{NaCl}\) at \(85^{\circ} \mathrm{F}\) flowing at a rate of \(15 \mathrm{lb}_{\mathrm{m}} / \mathrm{min},\) and the other is pure water at \(70^{\circ} \mathrm{F}\) flowing at \(10 \mathrm{lb}_{\mathrm{m}} / \mathrm{min} .\) The mass of liquid in the tank is held constant at \(1500 \mathrm{lb}_{\mathrm{m}}\). Perfect mixing in the tank may be assumed, so that the outlet stream has the same \(\mathrm{NaCl}\) mass fraction \((x)\) and temperature \((T)\) as the tank contents. Also assume that the heat of mixing is zero and the heat capacity of all fluids is \(C_{p}=1 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) (a) Write differential material and energy balances and use them to derive expressions for \(d x / d t\) and \(d T / d t\) (b) Without solving the equations derived in Part (a), sketch plots of \(T\) and \(x\) as a function of time \((t)\) Clearly identify values at time zero and as \(t \rightarrow \infty\)

Short Answer

Expert verified
The time rate of change of NaCl mass fraction \( dx/dt \) in the tank as derived from the material balance is \( dx/dt = 0.12 - x \) and the time rate of change of temperature \( dT/dt \) from the energy balance is \( dT/dt = 1/60 - T/60 \). From the plots of \( T \) and \( x \) as a function of time, it can be seen that both parameters initially change rapidly but then reach a steady state as time progresses with \( x = 0.12 \) and \( T = 60掳 \) as \( t \rightarrow \infty \).

Step by step solution

01

Write out the differential material balance

The differential material balance here will be based on NaCl. Given that the mass of liquid in the tank is held constant, the rate at which NaCl enters the tank minus the rate at which NaCl leaves the tank must be equal to zero. This gives: \( \frac{d(m_{NaCl})}{dt} = m_{in, NaCl} - m_{out, NaCl} = 0 \). Now, we'll put the expressions for the mass flow rates of NaCl in and out. \( m_{in, NaCl} = (0.20)(15) = 3lb_m/min \) and \( m_{out, NaCl} = x_q \) where \( x \) is the NaCl mass fraction and \( q \) is the volumetric flow rate that is \( 15 + 10 = 25lb_m/min \). This results in the equation: \( \frac{d(m_{NaCl})}{dt} = 3 - 25x = 0 \) or expressed for dx/dt: \( \frac{dx}{dt} = 0.12 - x \).
02

Construct the differential energy balance

The differential energy balance will look approach the balance through an enthalpic point of view. The rate at which energy comes in must balance the rate at which energy leaves. We obtain: \( \frac{dH}{dt} = H_{in} - H_{out} = 0 \). The enthalpy can be defined as \( H = m C_{p} T \) with \( C_{p} \) being the heat capacity of all fluids, \( T \) being the temperature and \( m \) being the mass. This leads to: \( \frac{d(m C_{p} T)}{dt} = m_{in} C_{p} T_{in} - m_{out} C_{p} T_{out} = 0 \). Considering \( m_{in} = m_{out} = q \), \( T_{in} = T_{NaCl} \) for the NaCl stream and \( T_{in} = T_{H2O} \) for the water stream, \( C_{p} = 1 Btu/(lb_m 掳F) \) and \( T_{out} = T \), this simplifies further to: \( \frac{d(T)}{dt} = \frac{1}{1500} ((15)(85) + (10)(70) - (25)T) \) or expressed for dT/dt: \( \frac{dT}{dt} = \frac{1}{60} - \frac{T}{60} \).
03

Sketch plots for \( T \) and \( x \) as a function of \( t \)

Both equations described above are first-order linear differential equations, and both will yield exponential curves as a result. For \( x \), the NaCl concentration in the tank will increase rapidly in the beginning, but that rate of increase will decrease over time and eventually reach a steady state of 0.12. For \( T \), the temperature will decrease initially from 70掳 And reach a steady state at 60掳. Furthermore, the initial values at time zero would be \( x = 0 \) and \( T = 70掳 \), this underlines the fact the process starts with pure water at 70掳. For \( t \rightarrow \infty \), the final steady state values would be \( x = 0.12 \) and \( T = 60掳 \).

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Key Concepts

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

Differential Material Balance
Understanding differential material balance is pivotal for students tackling chemical process problems. It involves calculating the rate of change of a substance within a system. In our exercise, we're analyzing a tank with an aqueous NaCl solution and water entering, and a mixed solution leaving, all while the tank maintains a constant mass. The principle of conservation of mass requires that mass entering minus mass leaving equals the accumulation in the system; since the system is at steady-state, accumulation is zero.

A material balance equation for NaCl is formulated as follows: \[\begin{equation}\frac{d(m_{\text{NaCl}})}{dt} = 3 - 25x = 0\end{equation}\]This equation reveals that the rate of NaCl entering the tank (3lb_m/min) minus the rate of NaCl leaving the tank (proportional to the mass fraction x in the outlet stream) must be zero for the mass to remain constant. With perfectly mixed conditions, the mass fraction x in the outlet stream is the same as inside the tank.
Differential Energy Balance
The differential energy balance is another cornerstone concept, essentially an application of the first law of thermodynamics to the process. It assesses how energy changes within a system, accounting for energy inflows and outflows. In this scenario, we assume perfect mixing and no heat loss, simplifying the problem.

Broken down, the equation from the example is as follows: \[\begin{equation}\frac{d(m C_p T)}{dt} = m_{in} C_p T_{in} - m_{out} C_p T_{out} = 0\end{equation}\]It implies the rate of enthalpy change within the tank is balanced by the enthalpy inflow and outflow. The uniform heat capacity (C_p) and the temperature of the inflow and outflow streams play crucial roles. The equation for the rate of temperature change (dT/dt) indicates how the tank temperature would evolve over time due to incoming streams at different temperatures.
First-Order Linear Differential Equations
First-order linear differential equations form the backbone of many models in chemical engineering processes, including our tank example. These equations describe how a variable changes with respect to time and are distinguished by their direct proportionality between the rate of change of the variable and the variable itself.

Both differential equations we have derived are first-order: for NaCl mass fraction (\(dx/dt\)) and temperature (\(dT/dt\)) in the tank. These equations predict the behavior of the system over time, which graphically represent exponential curves indicating a swift initial change that gradually levels off to steady-state values. These solutions are crucial as they help predict the time it would take for the system to reach steady state, an essential aspect in process engineering and control.

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Most popular questions from this chapter

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?

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?

Ninety kilograms of sodium nitrate is dissolved in \(110 \mathrm{kg}\) of water. When the dissolution is complete (at time \(t=0\) ), pure water is fed to the tank at a constant rate \(\dot{m}(\mathrm{kg} / \mathrm{min}),\) and solution is withdrawn from the tank at the same rate. The tank may be considered perfectly mixed. (a) Write a total mass balance on the tank and use it to prove that the total mass of liquid in the tank remains constant at its initial value. (b) Write a balance on sodium nitrate, letting \(x(t, \dot{m})\) equal the mass fraction of \(\mathrm{NaNO}_{3}\) in the tank and outlet stream. Convert the balance into an equation for \(d x / d t\) and provide an initial condition. (c) On a single graph of \(x\) versus \(t,\) sketch the shapes of the plots you would expect to obtain for \(\dot{m}=50 \mathrm{kg} / \mathrm{min}, 100 \mathrm{kg} / \mathrm{min},\) and \(200 \mathrm{kg} / \mathrm{min} .\) (Don't do any calculations.) Explain your reason- ing, using the equation of Part (b) in your explanation. (d) Separate variables and integrate the balance to obtain an expression for \(x(t, \dot{m})\). Check your solution. Then generate the plots of \(x\) versus \(t\) for \(\dot{m}=50 \mathrm{kg} / \mathrm{min}, 100 \mathrm{kg} / \mathrm{min},\) and \(200 \mathrm{kg} / \mathrm{min}\) and show them on a single graph. (A spreadsheet is a convenient tool for carrying out this step.) (e) If \(\dot{m}=100 \mathrm{kg} / \mathrm{min}\), how long will it take to flush out \(90 \%\) of the sodium nitrate originally in the tank? How long to flush out 99\%? 99.9\%? (f) The stream of water enters the tank at a point near the top, and the exit pipe from the tank is located on the opposite side toward the bottom. One day the plant technician forgot to turn on the mixing impeller in the tank. On the same chart, sketch the shapes of the plots of \(x\) versus \(t\) you would expect to see with the impeller on and off, clearly showing the differences between the two curves at small values and large values of \(t .\) Explain your reasoning.

A solution containing hydrogen peroxide with a mass fraction \(x_{\mathrm{p} 0}\) \(\left(\mathrm{kg} \mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{kg} \text { solution }\right)\) is added to a storage tank at a steady rate \(\dot{m}_{0}(\mathrm{kg} / \mathrm{h})\). During this process, the liquid level reaches a corroded spot in the tank wall and a leak develops. As the filling continues, the leak rate \(\dot{m}_{1}(\mathrm{kg} / \mathrm{h})\) becomes progressively worse. Moreover, once it is in the tank the peroxide begins to decompose at a rate $$r_{\mathrm{d}}(\mathrm{kg} / \mathrm{h})=k M_{\mathrm{p}}$$ where \(M_{\mathrm{p}}(\mathrm{kg})\) is the mass of peroxide in the tank. The tank contents are well mixed, so that the peroxide concentration is the same at all positions. At a time \(t=0\) the liquid level reaches the corroded spot. Let \(M_{0}\) and \(M_{\mathrm{p} 0}\) be the total liquid mass and mass of peroxide, respectively, in the tank at \(t=0,\) and let \(M(t)\) be the total mass of liquid in the tank at any time thereafter. (a) Show that the leakage rate of hydrogen peroxide at any time is \(\dot{m}_{1} M_{\mathrm{p}} / M\) (b) Write differential balances on the total tank contents and on the peroxide in the tank, and provide initial conditions. Your solution should involve only the quantities \(\dot{m}_{0}, \dot{m}_{1}, x_{\mathrm{p} 0}, k, M, M_{0}, M_{\mathrm{p}}\) \(M_{\mathrm{p} 0},\) and \(t\)

A \(40.0-\mathrm{ft}^{3}\) oxygen tent initially contains air at \(68^{\circ} \mathrm{F}\) and 14.7 psia. At a time \(t=0\) an enriched air mixture containing \(35.0 \%\) v/v \(\mathrm{O}_{2}\) and the balance \(\mathrm{N}_{2}\) is fed to the tent at \(68^{\circ} \mathrm{F}\) and 1.3 psig at a rate of \(60.0 \mathrm{ft}^{3} / \mathrm{min},\) and gas is withdrawn from the tent at \(68^{\circ} \mathrm{F}\) and 14.7 psia at a molar flow rate equal to that of the feed gas. (a) Calculate the total Ib-moles of gas \(\left(\mathrm{O}_{2}+\mathrm{N}_{2}\right)\) in the tent at any time. (b) Let \(x(t)\) equal the mole fraction of oxygen in the outlet stream. Write a differential mole balance on oxygen, assuming that the tent contents are perfectly mixed (so that the temperature, pressure, and composition of the contents are the same as those propertics of the exit stream). Convert the balance into an equation for \(d x / d t\) and provide an initial condition. (c) Integrate the equation to obtain an expression for \(x(t)\). How long will it take for the mole fraction of oxygen in the tent to reach 0.33 ? Sketch a plot of \(x\) versus \(t,\) labeling the value of \(x\) at \(t=0\) and the asymptotic value at \(t \rightarrow \infty\)
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