/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Ninety kilograms of sodium nitra... [FREE SOLUTION] | 91Ó°ÊÓ

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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\(t=0\) ), pure water is fed to the tank at a constant rate \(\dot{m}(\mathrm{kg} / \mathrm{min}),\) and solution is withdrawn from the tank at the same rate.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.

Short Answer

Expert verified
To summarise, the essential steps followed were establishing mass balance equations for the entire system and sodium nitrate, subsequently visualising the balance over time for varying input rates, calculating the time necessary to flush out a certain fraction of the sodium nitrate, and assessing the effect of mixing on the rate of sodium nitrate flushing.

Step by step solution

01

Establishing Total Mass Balance

The total mass in the tank is made up of the initial mass plus any mass that enters minus any mass that leaves the tank. This gives us M(t) = Mi + ∫_{0}^{t} m_dot dt - ∫_{0}^{t} m_dot dt. Simplifying gives M(t) = Mi since the quantities that are added and removed are equal.
02

Sodium Nitrate Balance

The mass balance for sodium nitrate can be represented by an equation such as: N(t) = Ni + ∫_{0}^{t} ω * m_dot dt - ∫_{0}^{t} (x * m_dot) dt. ω represents the mass fraction of NaNO3 in the feed. As pure water is being fed to the tank, ω is 0. So the equation simplifies to: N(t) = Ni - ∫_{0}^{t} (x * m_dot) dt. Differentiating with respect to 't' gives: dN/dt = - x*m_dot, which can be rearranged to find dx/dt.
03

Illustrating the Sodium Nitrate Balance Over Time

Without specific values for the curves, for a lower feed rate, we expect a slow decrease of sodium nitrate concentration. For a higher feed rate, the concentration decreases faster. As the rate increases, the curve becomes steeper since the water enters and leaves the tank faster, replacing the sodium nitrate with pure water. Analyzing the derivative, dx/dt = - (m_dot / Mi) * x, it can be inferred that the graph must decrease exponentially because dx/dt is directly proportional to x.
04

Compute x(t, m_dot) Expression

By rearranging the equation dx/dt = - m_dot/Mi * x and integrating, we get: x(t, m_dot) = x(0, m_dot) * exp(- m_dot/Mi * t), where x(0, m_dot) = Ni/Mi is the initial mass fraction.
05

Computing Required Time

The equation from Step 4 can be used to compute the time necessary to flush out a certain fraction of the sodium nitrate. It has to be set equal to (1 - desired fraction) and then solve for 't'.
06

Effect of Mixing on Sodium Nitrate Removal

If the mixing impeller was off, the rate at which sodium nitrate is washed out would be slower. This is because without mixing, the water flowing in mainly interacts with water near the inlet and not much with the sodium nitrate solution set at the bottom. The curve for this would therefore be below the curve with the impeller on.

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Key Concepts

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

Mass Balance Equation
In chemical engineering, a mass balance equation helps us understand how the mass of a system changes over time. It's crucial for determining how reactants and products are distributed.
The mass balance is based on the principle of conservation of mass, and for this exercise, we consider a tank initially containing a sodium nitrate solution.
The important takeaway is that when the inlet and outlet flows are equal, the total mass remains constant. This means if you add water and draw the same amount of solution out simultaneously, the total mass inside the tank doesn't change.
  • Initial Mass (Mi): The mass of sodium nitrate and water initially in the tank.
  • Mass Flow Rate (m): The rate at which water is added and solution removed.
  • Time (t): The period over which the process occurs.
To prove this, we set up a mass balance as: \( M(t) = M_i + \int_{0}^{t} \dot{m} \, dt - \int_{0}^{t} \dot{m} \, dt \) simplifies to \( M(t) = M_i \), showing total mass stays constant as water entering equals solution leaving.
Sodium Nitrate Concentration
Determining sodium nitrate concentration within the mixture is essential for describing the chemical process over time. This could be visualized through how the concentration changes as water flows in and solution flows out.
Sodium nitrate concentration is crucial because it determines the purity and usability of the solution in various processes.
In this scenario, the mass fraction equation involves calculating the amount of sodium nitrate compared to the total mixture.
  • Initial Sodium Nitrate Mass (Ni): How much sodium nitrate was there initially.
  • Mass Fraction (x): The proportion of sodium nitrate in the mixture.
If no sodium nitrate enters with water, meaning pure water is used, the balance equation is \( N(t) = N_i - \int_{0}^{t} (x \cdot \dot{m}) \, dt \). Differentiating gives \( \frac{dN}{dt} = -x \cdot \dot{m}\), meaning \( \frac{dx}{dt} = - \frac{\dot{m}}{M_i} \cdot x\), indicating an exponential decrease in concentration.
Differential Equations
Understanding the role of differential equations is critical when looking at how variables change with respect to time in chemical processes.
In our scenario, differential equations help track the sodium nitrate mass fraction as it exits the tank. The rate of change of this fraction illustrates how quickly the solute is replaced with water.
For sodium nitrate, the differential equation derived is \( \frac{dx}{dt} = - \frac{\dot{m}}{M_i} \cdot x \). Here, the equation signifies that the decrease rate of sodium nitrate is proportional to its current concentration. This type of equation generally results in an exponential decay, common in chemical processes of dilution.
Graphical Analysis
Graphical analysis involves visually comparing how sodium nitrate concentration decreases over time with varying flow rates.
By plotting concentration (\( x \) versus time (\( t \)), we can better understand the process behavior for different conditions.
A steeper graph indicates quicker depletion of sodium nitrate, while a flatter one shows slower depletion.
No calculations are required to see this intuitive pattern. As \( \dot{m} \) increases (e.g., 50 kg/min, 100 kg/min, 200 kg/min), the influence of rapidly incoming and outgoing streams is visible on a graph:
  • Slow feed results in a gently sloping graph, gradual decrease in concentration.
  • Fast feed shows a steep drop, swift change in concentration.
Observing these concepts graphically allows easy visualization of the effect flow rates have on mixing efficiency, and it's a common method engineers use to analyze and communicate chemical processes.
Mixing Impeller Effect
The mixing impeller, when operational, significantly changes the efficiency of decreasing the sodium nitrate concentration in the solution.
An impeller ensures a homogeneous mix, meaning solution properties are uniform throughout, leading to predictable and regular depletion patterns.
However, without it, the water influent is less effective at thoroughly mixing, creating a layered concentration:
  • With the impeller on, mixing ensures a smooth, predictable concentration decrease over time.
  • Without mixing, the graph may show irregularities, slower decrease initially due to poor mixing.
Therefore, the impeller plays a pivotal role in process optimization, aiding in faster removal of sodium nitrate and resulting in a more uniform solution composition over the entire volume of the tank.

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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?

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.

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.

One hundred fifty kmol of an aqueous phosphoric acid solution contains 5.00 mole\% \(\mathrm{H}_{3} \mathrm{PO}_{4}\). The solution is concentrated by adding pure phosphoric acid at a rate of \(20.0 \mathrm{L} / \mathrm{min}\). (a) Write a differential mole balance on phosphoric acid and provide an initial condition. [Start by defining \(n_{\mathrm{p}}(\mathrm{kmol})\) to be the total quantity of phosphoric acid in the tank at any time.] Without solving the equation, sketch a plot of \(n_{\mathrm{p}}\) versus \(t\) and explain your reasoning. (b) Solve the balance to obtain an expression for \(n_{\mathrm{p}}(t) .\) Use the result to derive an expression for \(x_{\mathrm{p}}(t)\) the mole fraction of phosphoric acid in the solution. Without doing any numerical calculations, sketch a plot of \(x_{\mathrm{p}}\) versus \(t\) from \(t=0\) to \(t \rightarrow \infty,\) labeling the initial and asymptotic values of \(x_{\mathrm{p}}\) on the plot. Explain your reasoning. (c) How long will it take to concentrate the solution to \(15 \% \mathrm{H}_{3} \mathrm{PO}_{4} ?\)

An electrical coil is used to heat \(20.0 \mathrm{kg}\) of water in a closed well-insulated vessel. The water is initially at \(25^{\circ} \mathrm{C}\) and 1 atm. The coil delivers a steady \(3.50 \mathrm{kW}\) of power to the vessel and its contents. (a) Write a differential energy balance on the water, assuming that \(97 \%\) of the energy delivered by the coil goes into heating the water. What happens to the other \(3 \% ?\) (b) Integrate the equation of Part (a) to derive an expression for the water temperature as a function of time. (c) How long will it take for the water to reach the normal boiling point? Will it boil at this temperature? Explain your answer.

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