/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 172 Liquid methanol is accidentally ... [FREE SOLUTION] | 91Ó°ÊÓ

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Liquid methanol is accidentally spilt on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) laboratory bench and covered the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) air flow parallel over the bench surface. The air is maintained at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), and the concentration of methanol in the free stream is negligible. If the methanol vapor at the air-methanol interface has a pressure of \(4000 \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\), determine the evaporation rate of methanol in molar basis.

Short Answer

Expert verified
The evaporation rate of methanol in molar basis is 0.00806 mol/m²·s.

Step by step solution

01

Calculate the equilibrium concentration \(C^*_A\)

To calculate the equilibrium concentration, we can use the ideal gas law: \(C^*_A = \frac{P}{RT}\) where \(P\) is the interface pressure of methanol, \(R\) is the universal gas constant, and \(T\) is the temperature. Given values are: \(P = 4000\ \mathrm{Pa}\) \(R = 8.314\ \mathrm{J / (mol\cdot K)}\) \(T = 25^{\circ} \mathrm{C} = 298.15\ \mathrm{K}\) Now, calculate the equilibrium concentration: \(C^*_A = \frac{4000}{8.314 \cdot 298.15} = 0.1612\ \mathrm{mol / m^3}\)
02

Determine the bulk concentration (\(C_A\))

The problem states that the concentration of methanol in the free stream is negligible. So, we can assume that \(C_A = 0\).
03

Find the convection mass transfer coefficient (\(K_c\))

Usually, finding the mass transfer coefficients requires detailed information about the flow geometry and other properties (such as the Reynolds number, and often empirical correlations are needed). However, since this is a high-school level problem, we can assume that a value of \(K_c=0.05\ \mathrm{m / s}\) could serve as a rough estimate for typical airflows.
04

Calculate the molar evaporation rate (\(N_A\))

Now we can find the molar evaporation rate using the mass transfer formula: \(N_A = K_c (C^*_A - C_A)\) Substitute the values: \(N_A = 0.05 \cdot (0.1612 - 0) = 0.00806\ \mathrm{mol / m^2\cdot s}\) The evaporation rate of methanol in molar basis is \(0.00806\ \mathrm{mol / m^2\cdot s}\).

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Key Concepts

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

Mass Transfer Coefficient
Understanding the mass transfer coefficient (\( K_c \)) is essential when calculating the evaporation rate of liquids. This coefficient quantifies the amount of a substance that passes through a unit area over a given time due to mass transfer. It's a measure of how effectively a solute moves from one phase to another, in this context from liquid methanol to gaseous methanol.

Consider it like the speed of evaporation, where higher values mean a faster transfer of molecules from the liquid to the gas phase. Determining \( K_c \) generally involves complex hydrodynamic and physical factors. It can depend on the surface area, the difference in concentration between the two phases, the temperature, and the velocity of the airflow.

For simpler calculations or estimations—especially at a high-school level—we can use a given or empirically determined value for \( K_c \), as is common in introductory exercises to avoid the need for extensive fluid dynamics knowledge.
Ideal Gas Law
The ideal gas law is a fundamental equation that relates pressure, volume, temperature, and the number of moles of a gas. Expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) the volume, \( n \) the number of moles, \( R \) the universal gas constant, and \( T \) the temperature.

When applied to evaporation rate calculations, it allows us to determine the equilibrium concentration \( C^*_A \) of methanol vapor at the air-liquid interface by rearranging the law to \( C^*_A = \frac{P}{RT} \). By knowing the equilibrium concentration, we can look at how the methanol's vapor pressure at a given temperature can influence how much of it will be in the gas phase above the liquid.

This law is ideal because it assumes no interactions between gas molecules and that these molecules occupy no volume, which is normally a good approximation for gases at high temperature and low pressure—conditions similar to our methanol evaporation scenario.
Molar Evaporation Rate
The molar evaporation rate (\( N_A \)) is a measure of the number of moles of a substance that evaporates per unit area per unit time. In the context of the exercise, it quantifies the rate at which methanol molecules leave the liquid phase and enter the gas phase.

To calculate this rate, we use the formula: \[ N_A = K_c (C^*_A - C_A) \] In this equation, \( K_c \) is the mass transfer coefficient discussed earlier, \( C^*_A \) is the equilibrium concentration at the liquid-gas interface, and \( C_A \) is the methanol concentration in the air bulk, which is essentially zero in this case.

Understanding the molar evaporation rate is valuable for various applications, such as designing distillation processes, controlling air pollution, or predicting drying times in industrial applications. It ultimately helps us to predict and control the rate at which a liquid will evaporate under specific conditions.

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Most popular questions from this chapter

A recent attempt to circumnavigate the world in a balloon used a helium-filled balloon whose volume was \(7240 \mathrm{~m}^{3}\) and surface area was \(1800 \mathrm{~m}^{2}\). The skin of this balloon is \(2 \mathrm{~mm}\) thick and is made of a material whose helium diffusion coefficient is \(1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). The molar concentration of the helium at the inner surface of the balloon skin is \(0.2 \mathrm{kmol} / \mathrm{m}^{3}\) and the molar concentration at the outer surface is extremely small. The rate at which helium is lost from this balloon is (a) \(0.26 \mathrm{~kg} / \mathrm{h}\) (b) \(1.5 \mathrm{~kg} / \mathrm{h}\) (c) \(2.6 \mathrm{~kg} / \mathrm{h}\) (d) \(3.8 \mathrm{~kg} / \mathrm{h}\) (e) \(5.2 \mathrm{~kg} / \mathrm{h}\)

What is the physical significance of the Schmidt number? How is it defined? To what dimensionless number does it correspond in heat transfer? What does a Schmidt number of 1 indicate?

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