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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

The product gas from a coal gasification plant consists of 60.0 mole \(\%\) CO and the balance \(\mathrm{H}_{2}\); it leaves the plant at \(150^{\circ} \mathrm{C}\) and 135 bar absolute. The gas expands through a turbine, and the outlet gas from the turbine is fed to a boiler furnace at \(100^{\circ} \mathrm{C}\) and 1 atm at a rate of \(425 \mathrm{m}^{3} / \mathrm{min}\). Estimate the inlet flow rate to the turbine in \(\mathrm{ft}^{3} / \mathrm{min},\) using Kay's rule. What percentage error would result from the use of the ideal-gas equation of state at the turbine inlet?

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of \(37.289 \mathrm{g}\). The same container filled with carbon dioxide at STP has a mass of 37.440 g. When filled with an unknown gas at STP, the container mass is \(37.062 \mathrm{g}\). Calculate the molecular weight of the unknown gas, and then state its probable identity.

The quantity of sulfuric acid used globally places it among the most plentiful of all commodity chemicals. In the modern chemical industry, synthesis of most sulfuric acid utilizes elemental sulfur as a feedstock. However, an alternative and historically important source of sulfuric acid was the conversion of an ore containing iron pyrites (FeS_) to sulfur oxides by roasting (burning) the ore with air. The following reactions occurred in an oven: $$\begin{array}{c} 2 \mathrm{FeS}_{2}(\mathrm{s})+\frac{11}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+4 \mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{3}(\mathrm{g}) \end{array}$$ The gases leaving the oven were fed to a catalytic converter in which most of the remaining \(\mathrm{SO}_{2}\) produced was oxidized to \(\mathrm{SO}_{3}\). Finally, the gas leaving the converter was sent to an absorption column where the \(S O_{3}\) was taken up by water to produce sulfuric acid \(\left(H_{2} S O_{4}\right)\) (a) The ore fed to the oven was 90.0 wt\% \(\mathrm{FeS}_{2}\), and the remaining material may be considered inert. Dry air was fed to the oven in \(30.0 \%\) excess of the amount required to oxidize all of the sulfur in the ore to \(S O_{3}\). Eighty-five percent of the \(\mathrm{FeS}_{2}\) was oxidized, and \(60 \%\) of the \(\mathrm{SO}_{2}\) produced was oxidized to \(S O_{3}\). Leaving the roaster were (i) a gas stream containing \(S O_{2}, S O_{3}, O_{2},\) and \(N_{2}\) and (ii) a solid stream containing unconverted pyrites, ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right),\) and the inert material. Calculate the required feed rate of air in standard cubic meters per \(100 \mathrm{kg}\) of ore fed to the process. Also determine the molar composition and volume (SCM/100 kg ore) of the gas leaving the oven. (b) The gas leaving the oven entered the catalytic converter, which operated at 1.0 atm. Reaction (2) proceeded to equilibrium, at which point the component partial pressures are related by the expression $$K_{\mathrm{P}}(T)=\frac{p_{\mathrm{SO}_{3}}}{p_{\mathrm{SO}_{3}} p_{\mathrm{O}_{2}}^{0.5}}$$ The gases were first heated to \(600^{\circ} \mathrm{C}\) to accelerate the rate of reaction, and then cooled to \(400^{\circ} \mathrm{C}\) to enhance \(S O_{2}\) conversion. The equilibrium constant \(K_{\mathrm{P}}\) at these two temperatures is 9.53 atm \(^{0.5}\) and 397 atm \(^{0.5}\), respectively. Calculate the equilibrium fractional conversions of \(S O_{2}\) at these two temperatures. (c) Estimate the production rate of sulfuric acid in \(\mathrm{kg} / \mathrm{kg}\) ore if all of the \(\mathrm{SO}_{3}\) leaving the converter was transformed to sulfuric acid. What would this value be if all the sulfur in the ore had been converted?

A stream of hot dry nitrogen flows through a process unit that contains liquid acetone. A substantial portion of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the recovery unit at \(205^{\circ} \mathrm{C}\) and 1.1 bar and enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at \(10^{\circ} \mathrm{C}\) and 40 bar. The partial pressure of acetone in the feed to the condenser is 0.100 bar, and that in the effluent gas from the condenser is 0.379 bar. Assume ideal-gas behavior. (a) Calculate for a basis of \(1 \mathrm{m}^{3}\) of gas fed to the condenser the mass of acetone condensed ( \(\mathrm{kg}\) ) and the volume of gas leaving the condenser \(\left(\mathrm{m}^{3}\right)\) (b) Suppose the volumetric flow rate of the gas leaving the condenser is \(20.0 \mathrm{m}^{3} / \mathrm{h}\). Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.

A gas consists of 20.0 mole \(\% \mathrm{CH}_{4}, 30.0 \% \mathrm{C}_{2} \mathrm{H}_{6},\) and \(50.0 \% \mathrm{C}_{2} \mathrm{H}_{4} .\) Ten kilograms of this gas is to be compressed to a pressure of 200 bar at \(90^{\circ} \mathrm{C}\). Using Kay's rule, estimate the final volume of the gas.
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