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Chapter 4: Problem 23

A stream of humid air containing 1.50 mole \(\% \mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) and the balance dry air is to be humidified to a water content of 10.0 mole\% \(\mathrm{H}_{2} \mathrm{O}\). For this purpose, liquid water is fed through a flowmeter and evaporated into the air stream. The flowmeter reading, \(R\), is \(95 .\) The only available calibration data for the flowmeter are two points scribbled on a sheet of paper, indicating that readings \(R=15\) and \(R=50\) correspond to flow rates \(\dot{V}=40.0 \mathrm{ft}^{3} / \mathrm{h}\) and \(\dot{V}=96.9 \mathrm{ft}^{3} / \mathrm{h},\) respectively. (a) Assuming that the process is working as intended, draw and label the flowchart, do the degree-offreedom analysis, and estimate the molar flow rate (lb-mole/h) of the humidified (outlet) air if (i) the volumetric flow rate is a linear function of \(R\) and (ii) the reading \(R\) is a linear function of \(\dot{V}^{0.5}\) (b) Suppose the outlet air is analyzed and found to contain only \(7 \%\) water instead of the desired \(10 \%\) List as many possible reasons as you can think of for the discrepancy, concentrating on assumptions made in the calculation of Part (a) that might be violated in the real process.

Short Answer

Expert verified
The estimated molar flow rate of the humidified (outlet) air is about 0.79 lb-mole/h if the volumetric flow rate is a linear function of R, or about 1.81 lb-mole/h if the reading R is a linear function of \( \sqrt{V} \). The discrepancy between the expected and actual mole percentage of water can be due to a range of factors, such as inaccuracies in the flowmeter, temperature or pressure variations, lack of homogeneity in the air-water mixture, leaks in the system, incomplete evaporation of water, or mechanical malfunctions of the system.

Step by step solution

01

Calculating Volumetric Flow Rate

The problem provides two points on the flowmeter, (15, 40) and (50, 96.9), and states that the volumetric flow rate is either a linear function of R or R is a linear function of \( \sqrt{V} \). Using the formula y = mx + b, where m is the slope and b is the intercept, and the given points, we can calculate the volumetric flow rate corresponding to R = 95.
02

Linear Function of R

If the volumetric flow rate is a linear function of R, then the equation would be \( \dot{V} = mR + b \). Given the two points (15, 40) and (50, 96.9), the slope \( m \) can be calculated as \( \frac{96.9 - 40}{50 - 15} \approx 1.89 ft^{3}/h \). Similarly, the intercept b can be found by substituting the calculated value of m and the given point into the equation to get \( 40 - 1.89(15) \approx -7.35 ft^{3}/h \). The volumetric flow rate at R = 95 can then be calculated as \( 1.89(95) -7.35 \approx 172.65 ft^{3}/h \).
03

Linear Function of \( \sqrt{V} \)

If R is a linear function of \( \sqrt{V} \), then we have \( R = m \sqrt{V} + b \). The same points are used to solve for the new slope and intercept. In this case the slope m equals \( \frac{50 -15}{\sqrt{96.9} - \sqrt{40}} \approx 4.99 \), and b equals \( 15 - m \sqrt{40} \approx -4.17 \). The volumetric flow rate at R = 95 can be calculated by rearranging this equation to \( V = (\frac{R - b}{m})^{2} \) which gives \( V = (\frac{95 +4.17}{4.99})^{2} \approx 396.44 ft^{3}/h \).
04

Molar Flow Rate of Humidified Air

The molar flow rate of the humidified air can be calculated by using the ideal gas law, \( PV = nRT \), where P is pressure, V is volume, n is the number of moles, R is the gas constant and T is temperature. Since the molar flow rate is \(\dot{n} = \dot{V} / RT\), we can take an approximate room temperature of 298.15 K and a constant pressure of 1 atm. We therefore find that \( \dot{n} = \frac{172.65 ft^{3}/h}{0.7302 ft^{3}/(lb-mole K)*298.15 K} \approx 0.79 lb-mole/h \) for the linear function of R, or \( \dot{n} = \frac{396.44 ft^{3}/h}{0.7302 ft^{3}/(lb-mole K)*298.15 K} \approx 1.81 lb-mole/h \) for the linear function of \( \sqrt{V} \).
05

Reasons of Discrepancy

The discrepancy between the expected and actual mole percentage of water might be due to inaccuracies in the flowmeter, temperature or pressure variations, or lack of homogeneity in the air-water mixture. Other reasons might include leaks in the system, incomplete evaporation of water, or mechanical malfunctions of the system itself.

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Key Concepts

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

Degree-of-Freedom Analysis
In chemical process analysis, the degree-of-freedom (DOF) analysis is a critical step to determine if a process system is solvable or if additional information is needed. DOF analysis involves evaluating all the known and unknown variables within a set of equations, helping in decision-making about the number of independent equations required to solve for unknowns.
In our example, the process involves humidifying air, where we start by identifying the various input and output streams, known parameters such as flow rates or mole percentages, and the system boundaries. We then list all the governing equations, such as material balances for each component.
The degree of freedom is determined using the formula: DOF = Number of Unknowns - Number of Independent Equations.
This result indicates whether sufficient data exists for solving the system. A DOF of zero means that the system is exactly solvable, positive DOF signifies underdetermination (need more info), and negative indicates overdetermination (extra restrictions). DOF analysis serves as an essential diagnostic tool to ensure process viability.
Flow Measurement Calibration
Flow measurement calibration ensures that the readings from a flowmeter accurately reflect the actual volumetric flow rate. This is essential for maintaining accurate process control, particularly when monitoring or adjusting the flow of materials through a system.
In the provided example, a flowmeter measures the flow of liquid water being evaporated to humidify an air stream. Calibration data, involving known reference points of readings (R) and corresponding flow rates (\( \dot{V} \)), allow us to establish a relationship model. Whether the flow meter's response is linear with respect to the reading or involves some transformation such as a square root function is determined using data points such as (15, 40) and (50, 96.9).
This assessment allows us to select the correct mathematical function to model the true flow rate, ensuring that the flowmeter can be used reliably for the humidification process. Proper calibration facilitates precise procedure adjustments and maintains product quality.
Ideal Gas Law
The ideal gas law is an equation of state for a hypothetical ideal gas. It is a simplified model that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas using the equation: \[ PV = nRT \] where R is the ideal gas constant.
In our chemical process example, the ideal gas law helps calculate the molar flow rate of the humidified air. Assuming the conditions are close to ideal, such as approximate room temperature and atmospheric pressure, this law facilitates understanding of how the change in water content in air affects the system.
By knowing the volumetric flow rate (\( \dot{V} \)) and assuming a constant pressure and temperature, we can solve for the molar flow rate of the exiting humidified air, which is essential for balancing and controlling the humidification process. This forms the basis to ensure the desired mixing and conversion in practical applications.
Evaporative Humidification
Evaporative humidification is a process where water is added to the air, thereby increasing the air's humidity. This process is particularly useful in climatic control applications and industries requiring specific humidity levels for operations.
In the humidification process described, air initially containing a lower concentration of water (1.50 mole %) is passed over evaporating water to reach a desired higher humidity (10.0 mole %). This method of humidification is efficient and typically involves passing the air over a wet surface or directly adding water vapor.
Ensuring the correct final humidity involves controlling the rate of water evaporation, which can be influenced by factors like temperature, pressure, and flow rates. Such humidification procedures are critical in sectors like food processing, textile manufacturing, and HVAC systems where precise environmental conditions are essential.

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Most popular questions from this chapter

Ethyl acetate (A) undergoes a reaction with sodium hydroxide (B) to form sodium acetate and ethyl alcohol: The reaction is carried out at steady state in a series of stirred-tank reactors. The output from the ith reactor is the input to the \((i+1)\) st reactor. The volumetric flow rate between the reactors is constant at \(\dot{V}(\mathrm{L} / \mathrm{min}),\) and the volume of each tank is \(V(\mathrm{L})\) The concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) in the feed to the first tank are \(C_{\mathrm{A} 0}\) and \(C_{\mathrm{B} 0}(\mathrm{mol} / \mathrm{L})\). The tanks are stirred sufficiently for their contents to be uniform throughout, so that \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) in a tank equal \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) in the stream leaving that tank. The rate of reaction is given by the expression where \(k[\mathrm{L} /(\mathrm{mol} \cdot \mathrm{min})]\) is the reaction rate constant. (a) Write a material balance on \(A\) in the \(i\) th tank, and show that it yields where \(\tau=V / \dot{V}\) is the mean residence time in each tank. Then write a balance on \(\mathrm{B}\) in the ith tank and subtract the two balances, using the result to prove that (b) Use the equations derived in Part (a) to prove that and from this relation derive an equation of the form where \(\alpha, \beta,\) and \(\gamma\) are functions of \(k, C_{\mathrm{A} 0}, C_{\mathrm{B} 0}, C_{\mathrm{A}, i-1},\) and \(\tau .\) Then write the solution of this equation for \(C_{\mathrm{A} i}\) (c) Write a spreadsheet or computer program to calculate \(N\), the number of tanks needed to achieve a fractional conversion \(x_{\mathrm{A} N} \geq x_{\mathrm{Af}}\) at the outlet of the final reactor. Your program should implement the following procedure: (i) Take as input values of \(k, \dot{V}, V, C_{\mathrm{A} 0}(\mathrm{mol} / \mathrm{L}), C_{\mathrm{B} 0}(\mathrm{mol} / \mathrm{L}),\) and \(x_{\mathrm{Af}}\) (ii) Use the equation for \(C_{A i}\) derived in Part ( \(b\) ) to calculate \(C_{\mathrm{Al}}\); then calculate the corresponding fractional conversion \(x_{\mathrm{A} 1}\) (iii) Repeat the procedure to calculate \(C_{\mathrm{A} 2}\) and \(x_{\mathrm{A} 2},\) then \(C_{\mathrm{A} 3}\) and \(x_{\mathrm{A} 3},\) continuing until \(x_{\mathrm{A} i} \geq x_{\mathrm{Af}}\) Test the program supposing that the reaction is to be carried out at a temperature at which \(k=36.2 \mathrm{L} /(\mathrm{mol} \cdot \mathrm{min}),\) and that the other process variables have the following values: Use the program to calculate the required number of tanks and the final fractional conversion for the following values of the desired minimum final fractional conversion, \(x_{\mathrm{Af}}: 0.50,0.80,0.90,0.95\) 0.99, 0.999. Briefly describe the nature of the relationship between \(N\) and \(x_{\mathrm{Af}}\) and what probably happens to the process cost as the required final fractional conversion approaches \(1.0 .\) Hint: If you write a spreadsheet, it might appear in part as follows: (d) Suppose a 95\% conversion is desired. Use your program to determine how the required number of tanks varies as you increase (i) the rate constant, \(k ;\) (ii) the throughput, \(\dot{V} ;\) and (iii) the individual reactor volume, \(V\). Then briefly explain why the results you obtain make sense physically.

A paint mixture containing \(25.0 \%\) of a pigment and the balance binders (which help the pigment stick to the surface) and solvents (which ensure that the paint stays in liquid form) sells for 18.00 dollar/kg, and a mixture containing 12.0\% sells for 10.00 dollar /kg. (a) If a paint retailer produces a blend containing \(17.0 \%\) pigment, for how much (S/kg) should it be sold to yield a 10\% profit? (b) Paint manufacturers have begun to market "low VOC" paint as a more environmentally friendly product. What are VOCs? List some ways in which paint products can be altered to lower the VOC content.

Mammalian cells can be cultured for a variety of purposes, including synthesis of vaccines. They must be maintained in growth media containing all of the components required for proper cellular function to ensure their survival and propagation. Traditionally, growth media were prepared by blending a powder, such as Dulbecco's Modified Eagle Medium (DMEM) with sterile deionized water. DMEM contains glucose, buffering agents, proteins, and amino acids. Using a sterile (i.e., bacterial-, fungal-,and yeast-free) growth medium ensures proper cell growth, but sometimes the water (or powder) can become contaminated, requiring the addition of antibiotics to eliminate undesired contaminants. The culture medium is supplemented with fetal bovine serum (FBS) that contains additional growth factors required by the cells. Suppose an aqueous stream (SG = 0.90) contaminated with bacteria is split, with 75\% being fed to a mixing unit to dissolve a powdered mixture of DMEM contaminated with the same bacteria found in the water. The ratio of impure feed water to powder entering the mixer is 4.4:1. The stream leaving the mixer (containing DMEM, water, and bacteria) is combined with the remaining 25\% of the aqueous stream and fed to a filtration unit to remove all of the bacteria that have contaminated the system, a total of \(20.0 \mathrm{kg}\). Once the bacteria have been removed, the sterile medium is combined with FBS and the antibiotic cocktail PSG (Penicillin-Streptomycin-L-Glutamine) in a shaking unit to generate 5000 L of growth medium (SG = 1.2). The final composition of the growth medium is 66.0 wt\% H_O, 11.0\% FBS, 8.0\% PSG, and the balance DMEM. (a) Draw and label the process flowchart. (b) Do a degree-of-freedom analysis around each piece of equipment (mixer, filter, and shaker), the splitter, the mixing point, and the overall system. Based on the analysis, identify which system or piece of equipment should be the starting point for further calculations. (c) Calculate all of the unknown process variables. (d) Determine a value for (i) the mass ratio of sterile growth medium product to feed water and (ii) the mass ratio of bacteria in the water to bacteria in the powder. (e) Suggest two reasons why the bacteria should be removed from the system.

In the production of soybean oil, dried and flaked soybeans are brought into contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil. (a) Draw a flowchart of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract). (b) The soybeans contain 18.5 wt\% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to \(2.0 \mathrm{kg}\) hexane per \(\mathrm{kg}\) beans. The residual solids leaving the extraction unit contain 35.0 wt\% hexane, all of the non-oil solids that entered with the beans, and \(1.0 \%\) of the oil that entered with the beans. For a feed rate of \(1000 \mathrm{kg} / \mathrm{h}\) of dried flaked soybeans, calculate the mass flow rates of the extract and residual solids, and the composition of the extract. (c) The product soybean oil must now be separated from the extract. Sketch a flowchart with two units, the extraction unit from Parts (a) and (b) and the unit separating soybean oil from hexane. Propose a use for the recovered hexane.

Inside a distillation column (see Problem 4.8), a downward-flowing liquid and an upward-flowing vapor maintain contact with each other. For reasons we will discuss in greater detail in Chapter \(6,\) the vapor stream becomes increasingly rich in the more volatile components of the mixture as it moves up the column, and the liquid stream is enriched in the less volatile components as it moves down. The vapor leaving the top of the column goes to a condenser. A portion of the condensate is taken off as a product (the overhead product), and the remainder (the reflux) is returned to the top of the column to begin its downward journey as the liquid stream. The condensation process can be represented as shown below: A distillation column is being used to separate a liquid mixture of ethanol (more volatile) and water (less volatile). A vapor mixture containing 89.0 mole \(\%\) ethanol and the balance water enters the overhead condenser at a rate of \(100 \mathrm{lb}\) -mole/h. The liquid condensate has a density of \(49.01 \mathrm{b}_{\mathrm{m}} / \mathrm{ft}^{3},\) and the reflux ratio is \(3 \mathrm{lb}_{\mathrm{m}}\) reflux/lb \(_{\mathrm{m}}\) overhead product. When the system is operating at steady state, the tank collecting the condensate is half full of liquid and the mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) is 10.0 minutes. Determine the overhead product volumetric flow rate (ft \(^{3}\) /min) and the condenser tank volume (gal).
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