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Chapter 10: Problem 3

Methanol is added to a storage tank at a rate of \(1200 \mathrm{kg} / \mathrm{h}\) and is simultaneously withdrawn at a rate \(\dot{m}_{w}(t)(\mathrm{kg} / \mathrm{h})\) that increases linearly with time. At \(t=0\) the tank contains \(750 \mathrm{kg}\) of the liquid and \(\dot{m}_{w}=750 \mathrm{kg} / \mathrm{h} .\) Five hours later \(\dot{m}_{\mathrm{w}}\) equals \(1000 \mathrm{kg} / \mathrm{h}\) (a) Calculate an expression for \(\dot{m}_{w}(t),\) letting \(t=0\) signify the time at which \(\dot{m}_{w}=750 \mathrm{kg} / \mathrm{h},\) and incorporate it into a differential methanol balance, letting \(M(\mathrm{kg})\) be the mass of methanol in the tank at any time. (b) Integrate the balance equation to obtain an expression for \(M(t)\) and check the solution two ways. (See Example 10.2-1.) For now, assume that the tank has an infinite capacity. (c) Calculate how long it will take for the mass of methanol in the tank to reach its maximum value, and calculate that value. Then calculate the time it will take to empty the tank. (d) Now suppose the tank volume is \(3.40 \mathrm{m}^{3}\). Draw a plot of \(M\) versus \(t\), covering the period from \(t=0\) to an hour after the tank is empty. Write expressions for \(M(t)\) in each time range when the function changes.

Short Answer

Expert verified
The expression for the withdrawal rate is \(\dot{m}_{w}(t) = 750 \, kg/h + 50 \, t \, kg/h^2\) and the equation for the mass of methanol in the tank at any time is \(M(t) = 450t - \frac{25}{2}t^2 + 750\). The maximum mass of methanol is reached after 18 hours, and the tank will be empty after 36 hours.

Step by step solution

01

Calculate The Expression For The Withdrawal Rate

Given the rate of increase of \(\dot{m}_{w}(t)\), we calculate the slope as \( (1000 - 750) \, kg/h / 5 \, h = 50 \, kg/h^2 \). Therefore, the function for \(\dot{m}_{w}(t)\) is: \(\dot{m}_{w}(t) = 750 \, kg/h + 50 \, t \, kg/h^2\).
02

Incorporate It Into A Differential Methanol Balance

The rate of change of mass in the tank with respect to time is given by: \(dM/dt = \dot{m}_{in} - \dot{m}_{w}(t)\). Substituting the given values we obtain: \(dM/dt = 1200 \, kg/h - (750 \, kg/h + 50 \, t \, kg/h^2)\).
03

Integrate The Balance Equation To Obtain An Expression For M(t)

Integrating the equation \(dM/dt = 1200 \, kg/h - (750 \, kg/h + 50 \, t \, kg/h^2)\) gives us: \(M(t) = 1200t - 750t - \frac{25}{2}t^2 + C\). Using the initial condition at \(t=0, M = 750\), we can solve for \(C\) to obtain: \(M(t) = 450t - \frac{25}{2}t^2 + 750\).
04

Calculate Maximum Mass of Methanol And Time To Reach Maximum Mass

To calculate the maximum mass and time, differentiate \(M(t) = 450t - \frac{25}{2}t^2 + 750\) and set equal to zero to determine the time at which the maximum occurs. Solving gives \(t = 18h\). Sub the value of \(t\) into \(M(t)\) to get the maximum mass.
05

Calculate The Time When The Mass Of Methanol In The Tank Will Be Zero

Setting \(M(t) = 0\) and solving the quadratic equation will provide the solution for \(t\)
06

Plot The Graph Of M Versus t

Plot the obtained function \(M(t) = 450t - \frac{25}{2}t^2 + 750\) from \(t=0\) to \(t=t_{empty}\), the time when the tank was empty

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Key Concepts

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

Differential Equations in Chemical Engineering
Differential equations are an essential tool in chemical engineering, allowing us to model the dynamics of systems that change over time. They help describe how variables like mass, energy, and concentration evolve in a process. For this exercise, we consider a system where methanol's mass in a storage tank changes due to influx and outflux at varying rates. The differential equation we use arises from the balance of methanol mass inside the tank: the rate of mass increase equals the inflow minus the outflow. Mathematically, this is represented as:\[ \frac{dM}{dt} = \dot{m}_{in} - \dot{m}_{w}(t) \]Here, \(M\) denotes the mass of methanol at any time \(t\), and the differential equation helps us predict how this mass changes over time. By understanding this principle, engineers can design and adjust processes to ensure effective material handling in industrial operations.
Material Balances
Material balances are fundamental in chemical engineering, involving accounting for all material entering, leaving, and accumulating in a process. In this scenario, the methanol in the tank is subject to a balance equation that considers its rate of addition and withdrawal.1. **Influx Rate:** Methanol is added at a constant rate of 1200 kg/h.2. **Outflux Rate:** The withdrawal rate, \(\dot{m}_{w}(t)\), starts at 750 kg/h and increases linearly with time.To perform the material balance, the differential methanol balance equation is formed based on these inputs and outputs. This equation lets us track the changing mass of methanol in the tank as a function of time, ensuring all entering and leaving quantities are accounted for. This knowledge is critical to the design, control, and optimization of chemical processes.
Process Dynamics
Process dynamics in chemical engineering explore how processes evolve over time under changing conditions. This involves analyzing how input and output rates affect the overall system. In this problem, methanol's mass in a tank is subject to these dynamics due to the varying withdrawal rate. The process starts with a known quantity of methanol and responds to an inflow and a changing outflow, which is linearly dependent on time. By studying these dynamics, engineers can predict system behavior, such as when the mass of methanol will peak and when the tank will be emptied. Understanding such dynamic behavior aids in preventing operational inefficiencies and potential safety issues in real-world chemical processes. This skillset is pivotal for chemical engineers when modeling, simulating, and controlling processes.
Integration Techniques
Integration techniques are employed to solve the balance equations, converting differential forms into explicit functions of time. This requires integrating the methanol balance equation:\[ M(t) = 450t - \frac{25}{2}t^2 + C \]By applying the initial condition, where the tank mass is 750 kg at \(t = 0\), we solve for the constant \(C\). This gives us an expression for mass \(M(t)\) as a function of time. By examining the endpoints of this function, students can determine the maximum mass of methanol in the tank over time and when it will be entirely empty.Integration allows engineers to convert dynamic rates into total changes over time, making it integral for designing systems that work efficiently and safely. This technique is not only used in calculating fluid dynamics but spans across all areas of chemical and process engineering.

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Most popular questions from this chapter

A 10.0 -ft compressed-air tank is being filled. Before the filling begins, the tank is open to the atmosphere. The reading on a Bourdon gauge mounted on the tank increases linearly from an initial value of 0.0 to 100 psi after 15 seconds. The temperature is constant at \(72^{\circ} \mathrm{F}\), and atmospheric pressure is 1 atm. (a) Calculate the rate \(\dot{n}\) (lb-mole/s) at which air is being added to the tank, assuming ideal-gas behavior. (Suggestion: Start by calculating how much is in the tank at \(t=0 .)\) (b) Let \(N(t)\) equal the number of Ib-moles of air in the tank at any time. Write a differential balance on the air in the tank in terms of \(N\) and provide an initial condition. (c) Integrate the balance to obtain an expression for \(N(t)\). Check your solution two ways. (d) Estimate the Ib-moles of oxygen in the tank after two minutes. List reasons your answer might be inaccurate, assuming there are no mistakes in your calculation.

The following chemical reactions take place in a liquid-phase batch reactor of constant volume \(V\). $$\begin{aligned} &\mathrm{A} \rightarrow 2 \mathrm{B} \quad r_{1}[\mathrm{mol} \mathrm{A} \text { consumed } /(\mathrm{L} \cdot \mathrm{s})]=0.100 C_{\mathrm{A}}\\\ &\mathbf{B} \rightarrow \mathbf{C} \quad r_{2}[\mathrm{mol} \mathbf{C} \text { generated } /(\mathbf{L} \cdot \mathbf{s})]=0.200 C_{\mathrm{B}}^{2} \end{aligned}$$ where the concentrations \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) are in mol/L. The reactor is initially charged with pure \(\mathrm{A}\) at a concentration of 1.00 mol/L. (a) Write expressions for ( \(i\) ) the rate of generation of \(\mathrm{B}\) in the first reaction and (ii) the rate of consumption of \(\mathrm{B}\) in the second reaction. (If this takes you more than about 10 seconds, you're missing the point.) (b) Write mole balances on A, B, and C, convert them into expressions for \(d C_{\mathrm{A}} / d t, d C_{\mathrm{B}} / d t\), and \(d C_{\mathrm{C}} / d t,\) and provide boundary conditions. (c) Without doing any calculations, sketch on a single graph the plots you would expect to obtain of \(C_{\mathrm{A}}\) versus \(t, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t .\) Clearly show the function values at \(t=0\) and \(t \rightarrow \infty\) and the curvature (concave up, concave down, or linear) in the vicinity of \(t=0 .\) Briefly explain your reasoning. (d) Solve the equations derived in Part (b) using a differential equation- solving program. On a single graph, show plots of \(C_{\mathrm{A} \text { versust }}, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t\) from \(t=0\) to \(t=50\) s. Verify that your predictions in Part (c) were correct. If they were not, change them and revise your explanation.

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?

An iron bar \(2.00 \mathrm{cm} \times 3.00 \mathrm{cm} \times 10.0 \mathrm{cm}\) at a temperature of \(95^{\circ} \mathrm{C}\) is dropped into a barrel of water at \(25^{\circ} \mathrm{C} .\) The barrel is large enough so that the water temperature rises negligibly as the bar cools. The rate at which heat is transferred from the bar to the water is given by the expression $$\dot{Q}(\mathrm{J} / \mathrm{min})=U A\left(T_{\mathrm{b}}-T_{\mathrm{w}}\right)$$ where \(U\left[=0.050 \mathrm{J} /\left(\mathrm{min} \cdot \mathrm{cm}^{2} \cdot^{\circ} \mathrm{C}\right)\right]\) is a heat transfer coefficient, \(A\left(\mathrm{cm}^{2}\right)\) is the exposed surface area of the bar, and \(T_{\mathrm{b}}\left(^{\circ} \mathrm{C}\right)\) and \(T_{\mathrm{w}}\left(^{\circ} \mathrm{C}\right)\) are the surface temperature of the bar and the water temperature, respectively. The heat capacity of the bar is \(0.460 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right) .\) Heat conduction in iron is rapid enough for the temperature \(T_{\mathrm{b}}(t)\) to be considered uniform throughout the bar. (a) Write an energy balance on the bar, assuming that all six sides are exposed. Your result should be an expression for \(d T_{\mathrm{b}} / d t\) and an initial condition. (b) Without integrating the equation, sketch the expected plot of \(T_{\mathrm{b}}\) versus \(t\), labeling the values of \(T_{\mathrm{b}}\) at \(t=0\) and \(t \rightarrow \infty\) (c) Derive an expression for \(T_{\mathrm{b}}(t)\) and check it three ways. How long will it take for the bar to cool to \(30^{\circ} \mathrm{C} ?\)

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).
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