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Chapter 12: Problem 45

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.

Short Answer

Expert verified
(a) Mass of water added: 0.0205 kg, (b) Final pressure: 3.04246 bar.

Step by step solution

01

Identify Given Data

Volume of the vessel, \(V = 0.8 \, \mathrm{m}^{3}\). Initial pressure, \(P_i = 0.3 \, \mathrm{MPa}\). Initial temperature, \(T = 30^{\circ} \, \mathrm{C}\). Final temperature, \(T_f = 30^{\circ} \, \mathrm{C}\), same as initial because the air is saturated at this temperature.
02

Use the Ideal Gas Law for Initial Condition

The number of moles of air initially present can be found using the ideal gas law:\[ P_i V = nRT \] Here, \(R\) is the specific gas constant for dry air, \(R = 0.287 \, \mathrm{kJ/kg \cdot K}\). Convert temperature to Kelvin: \(T = 30 + 273 = 303 \, \mathrm{K}\). Substituting, we get:\[ n = \frac{P_i V}{RT} = \frac{0.3 \, \mathrm{MPa} \times 0.8 \, \mathrm{m}^3}{0.287 \, \mathrm{kJ/kg \cdot K} \times 303 \, \mathrm{K}} \]
03

Calculate Mass of Air

The mass of air can be calculated from the number of moles:\[ m_{\text{air}} = nM_{\text{air}} = \frac{P_i V}{RT} M \] Where \(M\) is the molar mass of dry air (approximately 29 g/mol or 0.029 kg/mol). Substitute the values to get mass of air.
04

Find Saturation Pressure of Water

Use standard tables to find the saturation pressure of water at \(30\,^{\circ} \mathrm{C}\). From the tables, \(P_{sat} = 4.246\, \mathrm{kPa}\).
05

Calculate Final Pressure in Vessel

Because the air is saturated at the final state, the final pressure in the vessel will include the partial pressure of the air (which remains the same) plus the saturation pressure of water. The final pressure is:\[ P_f = P_{\text{air}} + P_{sat} \] Here, \(P_{\text{air}}\) is the partial pressure of the air, which remains as the initial air pressure (since we assume the volume remains constant). So, \(P_{\text{air}} = P_i\). Therefore, \(P_f = 0.3 \, \mathrm{MPa} + 4.246 \, \mathrm{kPa}\).
06

Convert Final Pressure to Bar

Convert \(P_f\) to bar:\[ P_f = 0.304246 \, \mathrm{MPa} = 3.04246 \, \mathrm{bar}\]
07

Determine the Mass of Water Added

The mass of water added corresponds to the mass of water vapor needed to saturate the air at \(30\,^{\circ} \mathrm{C}\). Using the ideal gas law for the water vapor:\[ m_{\text{water}} = \frac{P_{sat}V}{RT} M_{\text{water}} \] Where \(M_{\text{water}} = 0.018 \, \mathrm{kg/mol}\). Substitute the values to get the mass of water.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics, represented as:
\( PV = nRT \)
Here, \(P\) represents pressure, \(V\) is volume, \(n\) stands for the number of moles, \(R\) denotes the specific gas constant, and \(T\) indicates temperature.
In this exercise, we use the ideal gas law to calculate the initial number of moles of dry air. We also use it to determine the mass of water vapor needed at the final state.
Important points to remember about the ideal gas law:
  • It's an approximation that works best at high temperatures and low pressures.
  • The gas constant \(R\) varies depending on the gas; for dry air, \(R = 0.287 \, \text{kJ/kg}\text{路}\text{K}\).
  • Always use absolute temperature in Kelvin.
Saturation Pressure
Saturation pressure (\(P_{sat}\)) is the pressure at which a gas (like water vapor) is in equilibrium with its liquid phase at a given temperature.
At this point, the rate at which the liquid evaporates equals the rate at which the vapor condenses. To determine if air is saturated, saturation pressure tables or charts are typically used.
For example, at \(30^{\text{C}}\), the saturation pressure of water is \(4.246 \, \text{kPa}\).
Key points to note:
  • Saturation pressure is temperature-dependent.
  • As temperature increases, saturation pressure also increases.
  • At the saturation point, air cannot hold more water vapor, leading to condensation if more water is added.
Molar Mass
Molar mass (\(M\)) represents the mass of one mole of a substance and is expressed in grams per mole (g/mol) or kilograms per mole (kg/mol).
For dry air, the molar mass is approximately 29 g/mol (0.029 kg/mol), and for water, it is 18 g/mol (0.018 kg/mol).
Knowing the molar mass allows us to convert between the number of moles and the mass of a substance using the formula:
\( m = nM \)
In this exercise, we use it to find the mass of air and the mass of water added.
Some significant points include:
  • For accurate thermodynamic calculations, always use the correct molar mass value.
  • Molar mass ties directly into the ideal gas law for calculating species' masses.
Dry Air Properties
Dry air is a mix of gases excluding water vapor. It primarily consists of nitrogen (~78%), oxygen (~21%), and small amounts of other gases.
Key properties of dry air used in thermodynamic problems include:
  • Specific gas constant \(R = 0.287 \, \text{kJ/kg}\text{路}\text{K}\).
  • Molar mass approximately 29 g/mol (0.029 kg/mol).

In this exercise, dry air鈥檚 properties help determine the initial number of moles and final conditions in the vessel. Remember:
  • Dry air assumptions simplify calculations by ignoring water vapor content.
  • Changes in temperature and pressure affect dry air鈥檚 behavior based on the ideal gas law.
Thermodynamic Calculations
Thermodynamic calculations involve applying laws and equations to determine states and properties of systems.
Here鈥檚 a summary of the steps processed in this exercise:
  • Initialize with the ideal gas law to find the initial moles of air.
  • Determine dry air mass using its molar mass.
  • Find saturation pressure of water at given temperature from standard tables.
  • Calculate final pressure as the sum of partial pressures of air and water vapor.
  • Use the ideal gas law again to find the mass of water added.

Remember:
  • Always keep track of units and consistently use Kelvin for temperature.
  • Understanding properties like saturation pressure helps predict system behavior.
  • Thermodynamic tables are crucial for accurate values like saturation pressures.

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Most popular questions from this chapter

Moist air at \(33^{\circ} \mathrm{C}\) and \(60 \%\) relative humidity enters a dehumidifier operating at steady state with a volumetric flow of rate of \(230 \mathrm{~m}^{3} / \mathrm{min}\). The moist air passes over a cooling coil and water vapor condenses. Condensate exits the dehumidifier saturated at \(12^{\circ} \mathrm{C}\). Saturated moist air exits in a separate stream at the same temperature. There is no significant loss of energy by heat transfer to the surroundings and pressure remains constant at 1 bar. Determine (a) the mass flow rate of the dry air, in \(\mathrm{kg} / \mathrm{min}\). (b) the rate at which water is condensed, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of dry air flowing through the control volume. (c) the required refrigerating capacity, in tons.

On entering a dwelling maintained at \(20^{\circ} \mathrm{C}\) from the outdoors where the temperature is \(10^{\circ} \mathrm{C}\), a person's eyeglasses are observed not to become fogged. A humidity gauge indicates that the relative humidity in the dwelling is \(55 \%\). Can this reading be correct? Provide supporting calculations.

Natural gas at \(28^{\circ} \mathrm{C}, 2\) bar enters a furnace with the following molar analysis: \(50 \%\) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right), 25 \%\) ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right), 25 \%\) methane \(\left(\mathrm{CH}_{4}\right)\). Determine (a) the analysis in terms of mass fractions. (b) the partial pressure of each component, in bar. (c) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), for a volumetric flow rate of \(30 \mathrm{~m}^{3} / \mathrm{s}\).

A gas mixture at \(1600 \mathrm{~K}\) with the molar analysis \(20 \%\) \(\mathrm{CO}_{2}, 30 \% \mathrm{H}_{2} \mathrm{O}, 50 \% \mathrm{~N}_{2}\) enters a waste-heat boiler operating at steady state, and exits the boiler at \(700 \mathrm{~K}\). A separate stream of saturated liquid water enters at 35 bar and exits as saturated vapor with a negligible pressure drop. Ignoring stray heat transfer and kinetic and potential energy changes, determine the mass flow rate of the exiting saturated vapor, in \(\mathrm{kg}\) per kmol of gas mixture.

A device is being designed to separate into components a natural gas consisting of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) in which the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{6}\), denoted by \(y\), may vary from \(0.05\) to \(0.50\). The device will receive natural gas at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) with a volumetric flow rate of \(100 \mathrm{~m}^{3} / \mathrm{s}\). Separate streams of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) will exit, each at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Heat transfer between the device and its surroundings occurs at \(20^{\circ} \mathrm{C}\). Ignoring kinetic and potential energy effects, plot versus \(y\) the minimum theoretical work input required at steady state, in \(\mathrm{kW}\).
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