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Chapter 5: Problem 6

A slurry contains crystals of copper sulfate pentahydrate \(\left[\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), \text { specific gravity }=2.3\right]\) suspended in an aqueous copper sulfate solution (liquid SG \(=1.2\) ). A sensitive transducer is used to measure the pressure difference, \(\Delta P(\mathrm{Pa}),\) between two points in the sample container separated by a vertical distance of \(h\) meters. The reading is in turn used to determine the mass fraction of crystals in the slurry, \(x_{\mathrm{c}}(\mathrm{kg}\) crystals/kg slurry). (a) Derive an expression for the transducer reading, \(\Delta P(\mathrm{Pa}),\) in terms of the overall slurry density, \(\rho_{\mathrm{s}}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) assuming that the equation used to calculate the pressure head in Chapter 3 \(\left(P=P_{0}+\rho g h\right)\) is valid for this two-phase system. (b) Validate the following expression relating the overall slurry density to the liquid and solid crystal densities \(\left(\rho_{1} \text { and } \rho_{c}\right)\) and the mass fraction of crystals in the slurry: $$\frac{1}{\rho_{\mathrm{si}}}=\frac{x_{\mathrm{c}}}{\rho_{\mathrm{c}}}+\frac{\left(1-x_{\mathrm{c}}\right)}{\rho_{1}}$$ (c) Suppose \(175 \mathrm{kg}\) of the slurry is placed in the sample container with \(h=0.200 \mathrm{m}\) and a transducer reading \(\Delta P=2775\) Pa is obtained. Calculate \((\mathrm{i}) \rho_{\mathrm{s}_{\mathrm{s}},(\mathrm{ii})} x_{\mathrm{c}},\) (iii) the total slurry volume, (iv) the mass of crystals in the slurry, (v) the mass of anhydrous copper sulfate (CuSO \(_{4}\) without the water of hydration) in the crystals, (vi) the mass of liquid solution, and (vii) the volume of liquid solution. (d) Prepare a spreadsheet to generate a calibration curve of \(x_{c}\) versus \(\Delta P\) for this device. Take as inputs \(\rho_{\mathrm{c}}\left(\mathrm{kg} / \mathrm{m}^{3}\right), \rho_{1}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(h(\mathrm{m}),\) and calculate \(\Delta P(\mathrm{Pa})\) for \(x_{\mathrm{c}}=0.0,0.05,0.10, \ldots, 0.60\) Run the program for the parameter values in this problem \(\left(\rho_{\mathrm{c}}=2300, \rho_{1}=1200, \text { and } h=0.200\right)\) Then plot \(x_{c}\) versus \(\Delta P\) (have the spreadsheet program do it, if possible), and verify that the value of \(x_{c}\) corresponding to \(\Delta P=2775\) Pa on the calibration curve corresponds to the value calculated in Part (c). (e) Derive the expression in Part (b). Take a basis of \(1 \mathrm{kg}\) of slurry \(\left[x_{\mathrm{c}}(\mathrm{kg}), V_{c}\left(\mathrm{m}^{3}\right)\right.\) crystals, \(\left.\left(1-x_{\mathrm{c}}\right)(\mathrm{kg}), V_{l}\left(\mathrm{m}^{3}\right) \text { liquid }\right],\) and use the fact that the volumes of the crystals and liquid are additive.

Short Answer

Expert verified
The expression for the pressure difference is \( \Delta P = \rho_s g h \). The overall slurry density formula is valid: \(1/\rho_s =x_c/\rho_c + (1-x_c)/rho_1\). The values of the overall density of the slurry \(\rho_s\), the mass fraction of crystals \(x_c\), the total volume of slurry, the mass and the volume of crystals, and the mass of liquid solution can be calculated from the given data and the derived/validated formulas. A calibration curve of \(x_c\) versus \(\Delta P\) can be generated as described.

Step by step solution

01

Derive expression for pressure difference

Considering the equation \(P=P_{0}+\rho g h\), replacing \(P_{0}\) with \(0\) (assuming the reference point pressure is negligible or zero), \(\rho\) will be replaced with the slurry density \(\rho_s\), \(h\) is given in the problem, and \(g\) is the gravitational constant \(9.807 m/s^2\). We get: \[ \Delta P = \rho_s g h \]
02

Validate the overall slurry density equation

The given equation states that the inverse of the density of slurry \(\rho_s\) equals the mass fraction of crystals \(x_c\) divided by the crystal density \(\rho_c\), added to the mass fraction of liquid \((1- x_c)\) divided by the liquid density \(\rho_1\). This formula is derived from the definition of density as mass per volume unit. It is assuming that the mass divided by density (which is the volume) of each phase is additive (i.e., the volume of the slurry equals the volume of the crystals plus the volume of the liquid). Therefore, it appears to be valid for this two phase system, given that no chemical reaction or other process that could alter the volume is in place.
03

Calculate \(\rho_s\) and \(x_c\)

Given that the transducer reading \(\Delta P = 2775\ Pa\) and height \(h=0.200 m\), we can substitute these values into the formula derived in step 1 to get the slurry density \(\rho_s\). Then we substitute \(\rho_s\) and the given values for \(\rho_c\), \(\rho_1\) into the formula validated in step 2 to get the mass fraction \(x_c\) of crystals in the slurry.
04

Calculate slurry and crystal mass and volumes, and solution mass

Using the overall density and the total slurry mass, we can calculate the total slurry volume. As for the mass of crystals, it equals the total mass of slurry multiplied by \(x_c\). The volume of the crystals can also be calculated using their mass and density. Consequently, we can calculate the mass of the liquid solution by subtracting the mass of the crystals from the total slurry mass.
05

Generate calibration curve

Following the method described in the guide, create the calibration curve using a spreadsheet. The \(x_c\) values will range from 0.0 to 0.6, incrementing by 0.5, and \(\Delta P\) will be calculated using the formula derived in step 1. The spreadsheet software could be used to plot \(x_c\) versus \(\Delta P\), and the plotted point of \(\Delta P = 2775\) Pa will correspond to the \(x_c\) value calculated in step 3.
06

Derive expression for overall slurry density

The derivation of the expression validated in step 2 is based on the definitions of mass (weight) fraction and density (mass/volume), and the fact that the volumes are additive in this two phase system. Assuming a basis of 1 kg of slurry, it consists of \(x_c kg\) crystals and \((1-x_c) kg\) liquid. Consequently, the volume of the slurry is obtained by adding the crystal volume (mass divided by density) to the liquid volume.

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Key Concepts

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

Slurry Density Calculation
Understanding the density of a slurry is crucial for various industrial processes. Slurry density, represented by \(\rho_s\), is a measure of the mass of the slurry per unit volume. The key to calculating the slurry's density involves the consideration of the solid particles, in this case, copper sulfate pentahydrate crystals, suspended in a liquid medium.

To determine slurry density, one must account for the mass and volume of both the solid and liquid components. The formula \[\Delta P = \rho_s g h\] relates the pressure difference (\(\Delta P\)) measured by a transducer to the slurry density (\(\rho_s\)), where \(g\) is the acceleration due to gravity, and \(h\) is the vertical distance between two measurement points.

If students find the concept of slurry density challenging, one way to improve comprehension is through a visual aid, such as a diagram that represents the components of the slurry (solid particles and liquid matrix) and how their respective densities contribute to the overall slurry density. Additionally, posing real-world scenarios or conducting simple experiments that involve measuring densities could further solidify their understanding.
Mass Fraction Determination
The mass fraction represents the proportion of a component's mass to the total mass of the mixture. For a slurry, it's essentially the concentration of solid particles in the mixture. In mathematical terms, the mass fraction of crystals in the slurry is designated as \(x_{c}\).

To find the mass fraction of the crystals, the equation \[\frac{1}{\rho_{\text{{si}}}} = \frac{x_{c}}{\rho_{c}} + \frac{(1 - x_{c})}{\rho_{1}}\] is used. This equation stems from the principle that the total volume of the slurry is the sum of the volumes of the individual components. The mass fraction is obtained by manipulating this relationship and integrating the given densities of the solids and liquids.

For simplification, it could be beneficial for students to practice calculating mass fractions with mixtures of known quantities in the classroom. This hands-on approach, along with step-by-step guided problem solving, can help students grasp how to determine the mass fraction of a particular substance within a heterogeneous mixture.
Pressure Difference Measurement
The measurement of pressure difference, often denoted as \(\Delta P\), plays a critical role in determining the characteristics of a slurry. This measurement can tell us about the flow and density of the slurry, which are essential for understanding how it will behave under different conditions.

The pressure difference is obtained using a transducer that measures the force exerted by the slurry over an area at two distinct points separated by a height \(h\). The calculated pressure difference directly corresponds to the slurry's density when applied to the formula \[\Delta P = \rho_s g h\].

In educational settings, an effective exercise improvement advice could be running a demonstration using a simple fluid column to measure the pressure difference. This hands-on activity, coupled with explanatory scenarios and examples, can assist students in comprehending how pressure differences are measured and used in practical applications such as determining slurry density.

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Most popular questions from this chapter

Terephthalic acid (TPA), a raw material in the manufacture of polyester fiber, film, and soft drink bottles, is synthesized from \(p\) -xylene (PX) in the process shown below. A fresh feed of pure liquid \(\mathrm{PX}\) combines with a recycle stream containing \(\mathrm{PX}\) and a solution (S) of a catalyst (a cobalt salt) in a solvent (methanol). The combined stream, which contains \(S\) and \(P X\) in a 3: 1 mass ratio, is fed to a reactor in which \(90 \%\) of the \(\mathrm{PX}\) is converted to TPA. A stream of air at \(25^{\circ} \mathrm{C}\) and 6.0 atm absolute is also fed to the reactor. The air bubbles through the liquid and the reaction given above takes place under the influence of the catalyst. A liquid stream containing unreacted \(\mathrm{PX}\), dissolved TPA, and all the S that entered the reactor goes to a separator in which solid TPA crystals are formed and filtered out of the solution. The filtrate, which contains all the \(S\) and \(P X\) leaving the reactor, is the recycle stream. A gas stream containing unreacted oxygen, nitrogen, and the water formed in the reaction leaves the reactor at \(105^{\circ} \mathrm{C}\) and 5.5 atm absolute and goes through a condenser in which essentially all the water is condensed. The uncondensed gas contains 4.0 mole \(\%\) O. (a) Taking \(100 \mathrm{kmol}\) TPA produced/h as a basis of calculation, draw and label a flowchart for the process. (b) What is the required fresh feed rate (kmol PX/h)? (c) What are the volumetric flow rates \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the air fed to the reactor, the gas leaving the reactor, and the liquid water leaving the condenser? Assume ideal-gas behavior for the two gas streams. (d) What is the mass flow rate ( \(\mathrm{kg} / \mathrm{h}\) ) of the recycle stream? (e) Briefly explain in your own words the functions of the oxygen, nitrogen, catalyst, and solvent in the process. (f) In the actual process, the liquid condensate stream contains both water and PX. Speculate on what might be done with the latter stream to improve the economics of the process. [Hint: Note that PX is expensive, and recall what is said about oil (hydrocarbons) and water.]

A nitrogen rotameter is calibrated by feeding \(\mathrm{N}_{2}\) from a compressor through a pressure regulator, a needle valve, the rotameter, and a dry test meter, a device that measures the total volume of gas that passes through it. A water manometer is used to measure the gas pressure at the rotameter outlet. A flow rate is set using the needle valve, the rotameter reading, \(\phi\), is noted, and the change in the dry gas meter reading \((\Delta V)\) for a measured running time \((\Delta t)\) is recorded. The following calibration data are taken on a day when the temperature is \(23^{\circ} \mathrm{C}\) and barometric pressure is \(763 \mathrm{mm} \mathrm{Hg} .\) $$\begin{array}{rrr} \hline \phi & \Delta t(\min ) & \Delta V(\mathrm{L}) \\ \hline 5.0 & 10.0 & 1.50 \\ 9.0 & 10.0 & 2.90 \\ 12.0 & 5.0 & 2.00 \\ \hline \end{array}$$ (a) Prepare a calibration chart of \(\phi\) versus \(\dot{V}_{\text {sid }}\), the flow rate in standard \(\mathrm{cm}^{3} / \mathrm{min}\) equivalent to the actual flow rate at the measurement conditions. (b) Suppose the rotameter-valve combination is to be used to set the flow rate to 0.010 mol \(\mathrm{N}_{2} / \mathrm{min}\). What rotameter reading must be maintained by adjusting the valve?

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.)

An adult takes about 12 breaths per minute, inhaling roughly \(500 \mathrm{mL}\) of air with each breath. The molar compositions of the inspired and expired gases are as follows: $$\begin{array}{lcc} \hline \text { Species } & \text { Inspired Gas (\%) } & \text { Expired Gas (\%) } \\ \hline \mathrm{O}_{2} & 20.6 & 15.1 \\ \mathrm{CO}_{2} & 0.0 & 3.7 \\ \mathrm{N}_{2} & 77.4 & 75.0 \\ \mathrm{H}_{2} \mathrm{O} & 2.0 & 6.2 \\ \hline \end{array}$$ The inspired gas is at \(24^{\circ} \mathrm{C}\) and 1 atm, and the expired gas is at body temperature and pressure \(\left(37^{\circ} \mathrm{C}\right.\) and 1 atm). Nitrogen is not transported into or out of the blood in the lungs, so that \(\left(\mathrm{N}_{2}\right)_{\text {in }}=\left(\mathrm{N}_{2}\right)_{\text {out }}\) (a) Calculate the masses of \(\mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. (b) Calculate the volume of air exhaled per milliliter inhaled. (c) At what rate (g/min) is this individual losing weight by merely breathing? (d) The rate at which oxygen is transferred from the air in the lungs to the blood is roughly proportional to \(\left[\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{air}}-\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\right],\) where \(\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\) is a quantity related to the concentration of oxygen in the blood. Compared to regions where atmospheric pressure is 14.7 psia, what effect does the atmospheric pressure in Denver, which is approximately 12.1 psi, have on the transport rate and breathing rate? How does the body adjust to address this condition?

The van der Waals equation of state (Equation \(5.3-7\) ) is to be used to estimate the specific molar volume \(\hat{V}(\mathrm{L} / \mathrm{mol})\) of air at specified values of \(T(\mathrm{K})\) and \(P(\mathrm{atm}) .\) The van der Waals constants for air are \(a=1.33 \mathrm{atm} \cdot \mathrm{L}^{2} / \mathrm{mol}^{2}\) and \(b=0.0366 \mathrm{L} / \mathrm{mol}\) (a) Show why the van der Waals equation is classified as a cubic equation of state by expressing it in the form $$f(\hat{V})=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+c_{1} \hat{V}+c_{0}=0$$ where the coefficients \(c_{3}, c_{2}, c_{1},\) and \(c_{0}\) involve \(P, R, T, a,\) and \(b .\) Calculate the values of these coefficients for air at \(223 \mathrm{K}\) and 50.0 atm. (Include the units when giving the values.) (b) What would the value of \(\hat{V}\) be if the ideal-gas equation of state were used for the calculation? Use this value as an initial estimate of \(\tilde{V}\) for air at \(223 \mathrm{K}\) and 50.0 atm and solve the van der Waals equation using Goal Seek or Solver in Excel. What percentage error results from the use of the ideal-gas equation of state, taking the van der Waals estimate to be correct? (c) Set up a spreadsheet to carry out the calculations of Part (b) for air at \(223 \mathrm{K}\) and several pressures. The spreadsheet should appear as follows: The polynomial expression for \(\hat{V}\left(f=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+\cdots\right)\) should be entered in the \(f(V)\) column, and the value in the \(V\) column should be determined using Goal Seek or Solver in Excel.
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