91Ó°ÊÓ

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.

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.