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Chapter 14: Problem 15

A total of \(86,000 \mathrm{gal} /\) day of an aqueous solution of 3,000 \(\mathrm{ppm}\) of \(\mathrm{NaCl}\) is to be desalinized to \(400 \mathrm{ppm}\) by electrodialysis, with a \(40 \%\) conversion. The process will be conducted in four stages, with three stacks of 150 cell pairs in each stage. The fractional desalinization will be the same in each stage and the expected current efficiency is \(90 \%\). The applied voltage for the first stage is \(220 \mathrm{~V}\). Each cell pair has an area of \(1,160 \mathrm{~cm}^{2}\). Calculate the current density in \(\mathrm{mA} / \mathrm{cm}^{2}\), the current in \(\mathrm{A}\), and the power requirement in \(\mathrm{kW}\) for the first stage. Reference: Mason, E.A., and T.A. Kirkham, C.E.P. Symp.Ser., 55(24), 173-189 (1959).

Short Answer

Expert verified
Current density: 3.19 mA/cm虏, Current: 1,664 A, Power: 366.08 kW.

Step by step solution

01

Determine Required Desalinization

To find the fractional desalinization per stage, we first need to find the total desalinization required. The initial concentration is 3,000 ppm and needs to be reduced to 400 ppm, resulting in a desalinization requirement of 2,600 ppm.Given the conversion is 40%, the total water processed that actually undergoes desalinization is 40% of 86,000 gal/day. Thus, the conversion will take the concentration to 400 ppm after the last of the four stages. We need to calculate the fraction removed per stage using:\[ F = \left( \frac{1-\text{final ppm}}{\text{initial ppm}} \right)^{1/\text{number of stages}} \]Plug in values:\[ F = \left( \frac{1 - 400}{3000} \right)^{1/4} \approx \left( 0.87 \right)^{1/4} \approx 0.472 \]
02

Calculate Initial Flow Treated Per Stage

The treated flow in each stage is determined by multiplying the conversion rate by the total flow:\[ \text{Treated Flow Volume} = \text{Conversion rate} \times \text{Total flow} = 0.4 \times 86,000 \text{ gal/day} \approx 34,400 \text{ gal/day} \]
03

Calculate the Mass of NaCl in Feed

Convert the treated flow to L/day (1 gal = 3.785 L):\[ \text{Treated Flow Volume} = 34,400 \text{ gal/day} \times 3.785 \text{ L/gal} \approx 130,124 \text{ L/day} \]Now, calculate mass of NaCl at 3,000 ppm:\[ \text{Mass of NaCl} = 130,124 \times \frac{3,000}{1,000,000} \approx 390.372 \text{ kg/day} \]
04

Calculate Required Current Using Faraday's Law

Use Faraday's law to determine the necessary current (A). We'll use the molecular weight of NaCl (58.44 g/mol) and the Faraday constant (96,485 C/mol-e).Convert kg/day to mol/day:\[ \text{Mol/day NaCl} = \frac{390,372 \, g/day}{58.44 \, g/mol} \approx 6,678 \text{ mol NaCl/day} \]With a current efficiency of 90%, the current needed:\[ I = \frac{6,678 \times 2 \times 96,485}{0.9 \times 86400} \approx 1,664 \text{ A} \] (where 86400 is the number of seconds per day).
05

Calculate Cell Area

Calculate the total area per stage. Each stack has 3 stacks and 150 cell pairs:\[ \text{Total area} = 3 \times 150 \times 1,160 \text{ cm}^2 = 522,000 \text{ cm}^2 \]
06

Determine Current Density

Calculate the current density by dividing the total current by the total area:\[ \text{Current Density} = \frac{1,664 \text{ A}}{522,000 \text{ cm}^2} \times 1000 = 3.19 \text{ mA/cm}^2 \]
07

Calculate Power Requirement

Calculate the power requirement using the current and applied voltage:\[ \text{Power} = \text{Voltage} \times \text{Current} = 220 \text{ V} \times 1,664 \text{ A} = 366,080 \text{ W} \approx 366.08 \text{ kW} \]

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Key Concepts

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

Desalination Process
Desalination is the process of removing salts and other impurities from water to make it suitable for consumption or specific industrial uses. In this exercise, desalination is achieved through electrodialysis, a technique using electrical currents to drive ions through selective membranes, thereby separating them from the water.
Electrodialysis involves the passage of an electric current across a series of compartments, separated by ion-selective membranes, which allow certain ions to pass and block others. This helps to decrease the salinity of the water by enabling the transfer of ions from the saline side to a more purified side.
  • The application covered in the solution deals with reducing NaCl concentration from 3,000 ppm to 400 ppm over four stages.
  • Each stage contributes to the gradual reduction in salt concentration.
The mention of "40% conversion" implies that only a portion of the feed water is actually treated during each stage, gradually reaching the desired concentration. Each stage has a set fraction removed, calculated to ensure a consistent reduction per stage. This allows for efficient resource use and effective desalination over multiple stages.
Current Density Calculation
Current density is an important parameter in electrodialysis as it reflects the efficiency with which the electrical current is used to drive ions across the membranes. It is calculated as the current divided by the surface area of the electrodes.
In this exercise, we find the total current required for the electrodialysis from Faraday's laws involving the mass of NaCl to be desalted, taking into account the molecular weight and efficiency factors.
  • The entire cell area across the three stacks is also considered, where each stack contains 150 cell pairs with a given area per pair.
  • This allows us to determine the surface capable of facilitating the ion transfer needed for desalination.
The current density offers insight into how effectively each unit of cell area is used, given as milliamperes per square centimeter. A higher current density can indicate efficient ion transport but may also increase energy consumption and wear on the membranes. Here, the calculated current density is a balance of these factors, optimizing the process.
Power Requirement Calculation
The power requirement in electrodialysis refers to the energy needed to maintain the voltage across the cell stacks to continue moving ions against their concentration gradients. In this process, understanding the power requirement is vital since it influences both operational cost and design efficiency.
The power is calculated by taking the product of the applied voltage and the calculated current. For effective energy use, these values must align with the system鈥檚 capabilities and desired desalination outcomes.
  • Voltage applied in the exercise is specific: 220 volts in the first stage.
  • The computed total current of 1,664 amperes contributes to the total power calculation.
The result is expressed in kilowatts (kW), giving a clear sense of the electrical power consumed per unit time. By tallying these within the context of multiple stages, operators can manage and anticipate energy demands throughout the desalination process. This helps in designing systems that are both energy efficient and aligned with environmental and economic goals.

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Most popular questions from this chapter

A reverse-osmosis process is to be designed to handle a feed flow rate of \(100 \mathrm{gal} / \mathrm{min}\). Three designs have been proposed, differing in the \% recovery of potable water from the feed: Design 1: A single stage consisting of four units in parallel to obtain a \(50 \%\) recovery Design 2: Two stages in series with respect to the retentate (four units in parallel followed by two units in parallel) Design 3: Three stages in series with respect to the retentate (four units in parallel followed by two units in parallel followed by a single unit) Draw the three designs and determine the percent recovery of potable water for Designs 2 and \(3 .\)

An aqueous process stream of \(100 \mathrm{gal} / \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) contains \(8 \mathrm{wt} \% \mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(6 \mathrm{wt} \%\) of a high-molecular-weight substance (A). This stream is processed in a continuous, countercurrent-flow dialyzer using a pure water sweep of the same flow rate. The membrane is a microporous cellophane with pore volume \(=50 \%\), wet thickness \(=0.0051 \mathrm{~cm}\), tortuosity \(=4.1\), and pore diameter \(=\) \(31 \AA\). The molecules to be separated have the following properties: \(\begin{array}{lcc} & \mathrm{Na}_{2} \mathrm{SO}_{4} & \mathbf{A} \\ \text { Molecular weight } & 142 & 1,000 \\ \text { Molecular diameter, } \AA & 5.5 & 15.0 \\ \text { Diffusivity, } \mathrm{cm}^{2} / \mathrm{s} \times 10^{5} & 0.77 & 0.25\end{array}\) Calculate the membrane area in \(\mathrm{m}^{2}\) for only a \(10 \%\) transfer of \(\mathrm{A}\) through the membrane, assuming no transfer of water. What is the percent recovery of the \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in the diffusate? Use log-mean concentration driving forces and assume that the mass-transfer resistances on each side of the membrane are each \(25 \%\) of the total mass- transfer resistances for \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{A}\).

The production of paper involves a pulping step to break down wood chips into cellulose and lignin. In the Kraft process, an aqueous, pulping-feed solution, known as white liquor, is used that consists of dissolved inorganic chemicals such as sodium sulfide and sodium hydroxide. Following removal of the pulp (primarily cellulose), a solution known as weak (Kraft) black liquor (KBL) is left, which is regenerated to recover white liquor for recycle. In the conventional process, a typical \(15 \mathrm{wt} \%\) (dissolved solids) KBL is concentrated to 45 to \(70 \mathrm{wt} \%\) by multieffect evaporation. It has been suggested that reverse osmosis might be used to perform an initial concentration to perhaps \(25 \mathrm{wt} \%\). Higher concentrations may not be feasible because of the very high osmotic pressure, which at \(180^{\circ} \mathrm{F}\) and \(25 \mathrm{wt} \%\) solids is estimated to be \(1,700 \mathrm{psia}\). The osmotic pressure for other conditions can be scaled with (14-68) using wt\% instead of molality. A two-stage RO process, shown in Figure \(14.36\), has been proposed to carry out this initial concentration for a feed rate of \(1,000 \mathrm{lb} / \mathrm{h}\) at \(180^{\circ} \mathrm{F}\). A feed pressure of \(1,756 \mathrm{psia}\) is used for the first stage to yield a permeate of \(0.4 \mathrm{wt} \%\) solids. The feed pressure to the second stage is 518 psia to produce water of \(300 \mathrm{ppm}\) dissolved solids and a retentate of \(2.6 \mathrm{wt} \%\) solids. Permeate-side pressure for both stages is \(15 \mathrm{psia}\). Equation (14-69) can be used to estimate membrane area, where the permeance for water can be taken as \(0.0134 \mathrm{lb} / \mathrm{ft}^{2}-\mathrm{hr}-\mathrm{psi}\) in conjunction with an arithmeticmean osmotic pressure for plug flow on the feed side. Complete the material balance for the process and estimate the required membrane areas for each stage. Reference: Gottschlich, D.E., and D.L. Roberts. Final Report DE91004710, SRI International, Menlo Park, CA, Sept. 28, 1990 .

Two mechanisms for the transport of gas components through a porous membrane that are not discussed in Section \(14.3\) or illustrated in Figure 14.6 are (1) partial condensation in the pores by some components of the gas mixture to the exclusion of other components and subsequent transport of the condensed molecules through the pore, and (2) selective adsorption on pore surfaces of certain components of the gas mixture and subsequent surface diffusion across the pores. In particular, Rao and Sircar [48] have found that the latter mechanism provides a potentially attractive means for separating hydrocarbons from hydrogen for low-pressure gas streams. In porous-carbon membranes with continuous pores 4-15 \(\AA\) in diameter, little pore void space is available for the Knudsen diffusion of hydrogen when the hydrocarbons are selectively adsorbed. Typically, the membranes are not more than \(5 \mu \mathrm{m}\) in thickness. Measurements at \(295.1 \mathrm{~K}\) of permeabilities for five pure components and a mixture of the five components are as follows: $$ \begin{array}{lccc} & {\text { Permeability, barrer }} & \\ \text { Component } & \begin{array}{c} \text { As a } \\ \text { Pure Gas } \end{array} & \begin{array}{c} \text { In the } \\ \text { Mixture } \end{array} & \begin{array}{c} \text { mol\% in the } \\ \text { Mixture } \end{array} \\ \mathrm{H}_{2} & 130 & 1.2 & 41.0 \\ \mathrm{CH}_{4} & 660 & 1.3 & 20.2 \\ \mathrm{C}_{2} \mathrm{H}_{6} & 850 & 7.7 & 9.5 \\ \mathrm{C}_{3} \mathrm{H}_{8} & 290 & 25.4 & 9.4 \\ n \mathrm{C}_{4} \mathrm{H}_{10} & 155 & 112.3 & 19.9 \\ & & & 100.0 \end{array} $$ A refinery waste gas mixture of the preceding composition is to be processed through such a porous-carbon membrane. If the pressure of the gas is \(1.2 \mathrm{~atm}\) and an inert sweep gas is used on the permeate side such that partial pressures of feed gas components on that side are close to zero, determine the permeate composition on a sweep-gas-free basis when the composition on the upstreampressure side of the membrane is that of the feed gas. Explain why the component permeabilities differ so drastically between experiments with the pure gas and the gas mixture.

Explain, as completely as you can, how membrane separations differ from: (a) Absorption and stripping (b) Distillation (c) Liquid-liquid extraction (d) Extractive distillation
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