91Ó°ÊÓ

The final step of a manufacturing process in which a protective coating is applied to the inner surface of a circular tube involves passage of dry, atmosphere air through the tube to remove a residual liquid associated with the process. Consider a coated 5-m-long tube with an inner diameter of \(50 \mathrm{~mm}\). The tube is maintained at a temperature of \(300 \mathrm{~K}\), and the residual liquid exists as a thin film whose corresponding vapor pressure is \(15 \mathrm{~mm}\) Hg. The molecular weight and diffusion coefficient of the vapor are \(\Lambda_{\alpha}=70 \mathrm{~kg} / \mathrm{kmol}\) and \(D_{A B}=10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. Air enters the tube at a mean velocity of \(0.5 \mathrm{~m} / \mathrm{s}\) and a temperature of \(300 \mathrm{~K}\). (a) Estimate the partial pressure and mass density of vapor in the air exiting the tube. (b) What is the rate of liquid removal from the tube in \(\mathrm{kg} / \mathrm{s}\) ?

Step by step solution

01

Calculate Partial Pressure of Vapor Exiting the Tube

The given vapor pressure of the residual liquid is \(15 \mathrm{~mm}\) Hg. Let's convert it to \(Pa\) since we will use the Ideal Gas Law later: \(P = 15 \mathrm{~mm\ Hg} \times \dfrac{101325 \mathrm{~Pa}}{760 \mathrm{~mm\ Hg}} = 2005.197 \mathrm{~Pa}\) The partial pressure of the vapor exiting the tube, \(P_\alpha\), is equal to the vapor pressure, \(P\): \(P_\alpha = P = 2005.197 \mathrm{~Pa}\)

02

Calculate Mass Density of Vapor Exiting the Tube

Now, let's calculate the mass density of the vapor in the air exiting the tube using the Ideal Gas Law: \(\rho_\alpha = \frac{P_\alpha \Lambda_\alpha}{R T}\) where \(P_\alpha = 2005.197 \mathrm{~Pa}\) (partial pressure of vapor) \(\Lambda_\alpha = 70 \mathrm{~kg/kmol}\) (molecular weight of vapor) \(R = 8314.46 \mathrm{~J/(kmol\ K)}\) (universal gas constant) \(T = 300 \mathrm{~K}\) (temperature of the tube) \(\rho_\alpha = \frac{2005.197 \mathrm{~Pa} \times 70 \mathrm{~kg/kmol}}{8314.46 \mathrm{~J/(kmol\ K)} \times 300 \mathrm{~K}} = 0.56269 \mathrm{~kg/m^3}\)

03

Calculate Flow Rate of Air Entering the Tube

Find the flow rate of air entering the tube, \(Q\). Flow rate is the product of cross-sectional area, \(A\), and air mean velocity, \(V\): \(Q = A \times V\) Cross-sectional area, $A = \dfrac{\pi d^2}{4}\) where \(d = 50 \times 10^{-3} \mathrm{~m}\) (inner diameter of the tube) \(A = \dfrac{\pi (50 \times 10^{-3} \mathrm{~m})^2}{4} = 1.9635 \times 10^{-3} \mathrm{~m^2}\) Now, let's calculate the flow rate using the air mean velocity: \(Q = 1.9635 \times 10^{-3} \mathrm{~m^2} \times 0.5 \mathrm{~m/s} = 9.8174 \times 10^{-4} \mathrm{~m^3/s}\)

04

Calculate Rate of Liquid Removal from Tube

Now we will calculate the rate of liquid removal, \(r\), from the tube by multiplying the mass density of the vapor, \(\rho_\alpha\), and the flow rate, \(Q\): \(r = \rho_\alpha \times Q\) Substituting the values from Step 2 and Step 3: \(r = 0.56269 \mathrm{~kg/m^3} \times 9.8174 \times 10^{-4} \mathrm{~m^3/s}\) \(r = 5.525 \times 10^{-4} \mathrm{~kg/s}\) So, the rate of liquid removal from the tube is approximately \(5.525 \times 10^{-4} \mathrm{~kg/s}\).

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Key Concepts

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

Partial Pressure

Understanding partial pressure is key when working with gases. In our exercise, the partial pressure of the vapor is a crucial measurement. It represents the pressure that this vapor alone would exert if it occupied the entire volume. We started with the vapor pressure, given in mm Hg, and converted it to Pascals (Pa):

• The vapor pressure is 15 mm Hg.
• Conversion to Pascals involves the formula: \(P = 15\, \text{mm Hg} \times \frac{101325\, \text{Pa}}{760\, \text{mm Hg}} = 2005.197\, \text{Pa}\).
• This becomes the partial pressure of the vapor exiting the tube.

Knowing partial pressure helps predict how the vapor behaves, interacts with other gases, and contributes to the overall pressure in a system.
This is fundamental in processes like diffusion and chemical reactions.

Ideal Gas Law

The Ideal Gas Law is a powerful tool in predicting the behavior of gases under different conditions. It's defined by the equation \(PV = nRT\), where:

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the universal gas constant,
• \(T\) is the temperature.

In this exercise, we rearranged the formula to find the mass density \(\rho_\alpha\) of the vapor:

• \(\rho_\alpha = \frac{P_\alpha \Lambda_\alpha}{RT}\)
• Substituting the values, we determined the mass density to be \(0.56269\, \text{kg/m}^3\).

The Ideal Gas Law helps understand the gas behavior in various conditions, like changes in pressure or temperature, which is crucial for designing chemical processes and safety measures.

Diffusion Coefficient

The diffusion coefficient is a measure that describes how quickly molecules spread. It's used to model the rate at which substances mix, driven by concentration gradients. Here, the diffusion coefficient for the vapor is given as \(D_{AB} = 10^{-5} \text{ m}^2/\text{s}\).

• This coefficient indicates the ease with which the vapor molecules move through the air inside the tube.
• In practical terms, it affects how quickly the vapor mixes with air and moves through the tube.

The diffusion coefficient is essential in understanding mass transfer processes, especially in systems involving gases and liquids.
By knowing this value, engineers and scientists can model and predict how different gases will interact over time, helping in designing efficient manufacturing processes.