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Chapter 12: Problem 44

A quantity of \(0.8 \mathrm{~kg}\) of moist air initially at \(27^{\circ} \mathrm{C}, 1\) bar, \(52 \%\) relative humidity is compressed isothermally to \(3.5\) bar. If condensation occurs, determine the amount of water condensed, in \(\mathrm{kg}\). If there is no condensation, determine the final relative humidity.

Short Answer

Expert verified
The amount of water condensed is approximately 0.0092 kg.

Step by step solution

01

- Identify given values

Initial mass of moist air: \(0.8 \,\mathrm{kg}\)Initial temperature \(T_1: 27^{\circ} \,\mathrm{C} = 300.15 \,\mathrm{K}\)Initial pressure \(P_1: 1 \,\mathrm{bar}\)Initial relative humidity \(\phi_1: 52\%\)
02

- Use the psychrometric chart or equations

Using the psychrometric chart or appropriate equations, find the initial specific humidity (\(\omega_1\)) and saturation pressure of water vapor at \(27^{\circ} \,\mathrm{C}\).
03

- Calculate the partial pressure of water vapor

Use the relative humidity to find the initial partial pressure of water vapor: \ \(P_{v1} = \phi_1 \cdot P_{sat}\); \ Here, \(P_{v1} = 0.52 \cdot 3.564 \,\mathrm{kPa} = 1.85328 \,\mathrm{kPa}\)
04

- Determine the new total pressure

New total pressure after isothermal compression: \(P_2 = 3.5 \,\mathrm{bar} = 350 \,\mathrm{kPa}\)
05

- Calculate the specific humidity before condensation

Specific humidity before condensation: \(\omega_1 = 0.622 \cdot \frac{P_{v1}}{P_1 - P_{v1}} = 0.622 \cdot \frac{1.85328}{100 - 1.85328}\)
06

- Calculate the new water vapor partial pressure

Assuming all water vapor exists before condensation, new partial pressure: \ \(P_{v2} = P_{v1} \cdot \frac{P_2}{P_1} = 1.85328 \cdot \frac{350}{100} = 6.48648 \,\mathrm{kPa}\)
07

Check if condensation occurs

Check if \(P_{v2} > P_{sat,2}\) at 27\(^{\circ} \,\mathrm{C}\). In this case, \(P_{sat2} = 3.564\,\mathrm{kPa}\). Since \(P_{v2} \) is greater than \(P_{sat2} \), condensation occurs.
08

Calculate the amount of water condensed

Amount of water vapor in the final state: \( P_{v2}= P_{sat,2} = 3.564 \mathrm{kPa} \). New specific humidity: \(\omega_2 = 0.622 \cdot \frac{3.564}{350-3.564} \approx 0.00645\)Initial mass of water vapor: \(0.8\mathrm{kg} \cdot \omega_1 \approx 0.01435\mathrm{kg} \)Final mass of water vapor: \(0.8\mathrm{kg} \cdot \omega_2 \approx 0.00516\mathrm{kg}\)
09

Final calculation of condensed water

Mass of water condensed: \(0.01435\mathrm{kg} - 0.00516\mathrm{kg} \approx 0.00919\mathrm{kg} \)

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Key Concepts

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

Moist Air
Moist air is simply air that contains water vapor. It's a mixture of dry air and water vapor. The properties of moist air vary depending on its temperature, pressure, and the amount of water vapor in it. Understanding moist air is crucial in thermodynamics because it affects processes like heating, cooling, and phase changes. In the context of the given problem, we're dealing with the air at an initial temperature of 27掳C, a pressure of 1 bar, and a relative humidity of 52%.
Isothermal Compression
Isothermal compression means the process of compressing the air while keeping the temperature constant. For gases, this type of compression affects pressure and volume but keeps the temperature unchanged. This concept is essential for the problem because the problem specifies compressing moist air from an initial pressure of 1 bar to a final pressure of 3.5 bar, all while maintaining the initial temperature of 27掳C. Isothermal processes typically follow the equation \( PV = nRT \), where the temperature does not change, so work and heat interactions are directly proportional to changes in pressure and volume.
Relative Humidity
Relative humidity (RH) measures the amount of water vapor in the air relative to the maximum amount the air can hold at a given temperature. It鈥檚 expressed as a percentage. In the exercise, the initial relative humidity is 52%. It's calculated by the formula \( \text{RH} = \frac{P_v}{P_{sat}} \times 100\text{\%} \), where \( P_v \) is the partial pressure of water vapor, and \( P_{sat} \) is the saturation pressure of water vapor at a given temperature. As air is compressed, its ability to hold water vapor changes, which may lead to condensation if the air becomes saturated (i.e., RH reaches 100%).
Psychrometric Chart
The psychrometric chart is a graphical representation of the physical and thermal properties of moist air. Engineers and scientists use it to solve problems involving air conditioning, heating, and moisture control. The chart includes information about dry bulb temperature, wet bulb temperature, relative humidity, specific humidity, and enthalpy. In the given problem, we use the chart or related equations to find the initial specific humidity and saturation pressure of water vapor at 27掳C. The values obtained from the chart help in determining whether condensation will occur during the isothermal compression process.
Partial Pressure
Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. In our moist air exercise, it鈥檚 important to determine the partial pressure of water vapor to understand how much water vapor the air contains. This is done using the equation: \( P_v = \text{RH} \times P_{sat} \). In our scenario, the initial partial pressure of the water vapor was calculated to be 1.85328 kPa. After isothermal compression, the new partial pressure of the water vapor was recalculated and compared against the saturation pressure to determine if condensation would occur. The final partial pressure is greater than the saturation pressure, confirming condensation.

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Most popular questions from this chapter

A can of soft drink at a temperature of \(4^{\circ} \mathrm{C}\) is taken from a refrigerator into a room where the temperature is \(21^{\circ} \mathrm{C}\) and the relative humidity is \(70 \%\). Explain why beads of moisture form on the can's outer surface. Provide supporting calculations.

Two rigid, insulated tanks are interconnected by a valve. Initially \(0.527 \mathrm{kmol}\) of nitrogen at \(2.5\) bar and \(260 \mathrm{~K}\) fills one tank. The other tank contains \(0.14 \mathrm{kmol}\) of oxygen at 1 bar and \(310 \mathrm{~K}\). The valve is opened and the gases are allowed to mix until a final equilibrium state is attained. During this process, there are no heat or work interactions between the tank contents and the surroundings. Determine (a) the final temperature of the mixture, in \(\mathrm{K}\), (b) the final pressure of the mixture, in bar. (c) the amount of entropy produced in the mixing process, in \(\mathrm{kJ} / \mathrm{K}\).

A vessel whose volume is \(0.8 \mathrm{~m}^{3}\) initially contains dry air at \(0.3 \mathrm{MPa}\) and \(30^{\circ} \mathrm{C}\). Water is added to the vessel until the air is saturated at \(30^{\circ} \mathrm{C}\). Determine the (a) mass of water added, in \(\mathrm{kg}\). (b) final pressure in the vessel, in bar.

An air-handling system is being designed for a \(12 \mathrm{~m} \times\) \(12 \mathrm{~m} \times 2.5 \mathrm{~m}\) biological research facility that houses 3000 laboratory mice. The indoor conditions must be maintained at \(24^{\circ} \mathrm{C}, 60 \%\) relative humidity when the outdoor air conditions are \(32^{\circ} \mathrm{C}, 70 \%\) relative humidity. Develop a preliminary design of an air-conditioning and distribution system that satisfies National Institute of Health (NIH) standards for animal facilities. Assume a biological safety level of one (BSL-1), and that two thirds of the floor space is devoted to animal care. Since an interruption in ventilation or air conditioning could place the laboratory animals under stress and compromise the research under way in the facility, account for redundancy in your design.

Answer the following questions involving a mixture of two gases: (a) When would the analysis of the mixture in terms of mass fractions be identical to the analysis in terms of mole fractions? (b) When would the apparent molecular weight of the mixture equal the average of the molecular weights of the two gases?
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