Problem Statement
A cylinder of 0.6 m\(^3\) in volume contains air at 50°C and 0.3 N/mm\(^2\) absolute pressure. The air is compressed to 0.3 m\(^3\). Find (i) pressure inside the cylinder assuming isothermal process and (ii) pressure and temperature assuming adiabatic process. Take \(k = 1.4\).
Given Data
- Initial volume, \(V_1 = 0.6 \, \text{m}^3\)
- Initial temperature, \(t_1 = 50^\circ\text{C}\)
- Initial pressure, \(P_1 = 0.3 \, \text{N/mm}^2 = 0.3 \times 10^6 \, \text{N/m}^2 = 30 \times 10^4 \, \text{N/m}^2\)
- Final volume, \(V_2 = 0.3 \, \text{m}^3\)
- Adiabatic index, \(k = 1.4\)
Convert initial temperature to Kelvin: \(T_1 = 50 + 273 = 323 \, \text{K}\)
Solution
(i) Isothermal Process
For an isothermal process, the product of pressure and volume remains constant (Boyle's Law):
1. Apply Isothermal Process Equation
Where \(P_2\) is the final pressure.
$$ P_2 = \frac{P_1 V_1}{V_2} $$ $$ P_2 = \frac{30 \times 10^4 \, \text{N/m}^2 \times 0.6 \, \text{m}^3}{0.3 \, \text{m}^3} $$ $$ P_2 = 30 \times 10^4 \times 2 \, \text{N/m}^2 $$ $$ P_2 = 60 \times 10^4 \, \text{N/m}^2 = 0.6 \times 10^6 \, \text{N/m}^2 $$ $$ P_2 = 0.6 \, \text{N/mm}^2 $$Pressure inside the cylinder, \(P_2 = 0.6 \, \text{N/mm}^2\).
(ii) Adiabatic Process
For an adiabatic process, the relationship between pressure and volume is given by \(P V^k = \text{Constant}\).
1. Apply Adiabatic Process Equation for Pressure
Where \(P_2\) is the final pressure.
$$ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^k $$ $$ P_2 = 30 \times 10^4 \, \text{N/m}^2 \times \left( \frac{0.6 \, \text{m}^3}{0.3 \, \text{m}^3} \right)^{1.4} $$ $$ P_2 = 30 \times 10^4 \times (2)^{1.4} $$ $$ P_2 = 30 \times 10^4 \times 2.639 $$ $$ P_2 = 79.17 \times 10^4 \, \text{N/m}^2 = 0.7917 \times 10^6 \, \text{N/m}^2 $$ $$ P_2 \approx 0.7917 \, \text{N/mm}^2 $$2. Apply Adiabatic Process Equation for Temperature
For an adiabatic process, the relationship between temperature and volume is given by \(T V^{k-1} = \text{Constant}\).
Where \(T_2\) is the final temperature.
$$ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{k-1} $$ $$ T_2 = 323 \, \text{K} \times \left( \frac{0.6 \, \text{m}^3}{0.3 \, \text{m}^3} \right)^{1.4-1} $$ $$ T_2 = 323 \, \text{K} \times (2)^{0.4} $$ $$ T_2 = 323 \, \text{K} \times 1.3195 $$ $$ T_2 \approx 426.2 \, \text{K} $$Convert final temperature to Celsius:
$$ t_2 = 426.2 - 273 = 153.2^\circ\text{C} $$Pressure inside the cylinder, \(P_2 \approx 0.7917 \, \text{N/mm}^2\).
Temperature inside the cylinder, \(t_2 \approx 153.2^\circ\text{C}\).
Explanation
1. Isothermal Process:
An isothermal process occurs at a constant temperature. According to Boyle's Law, for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional. This means as volume decreases, pressure increases proportionally. In this case, since the volume is halved, the pressure doubles.
2. Adiabatic Process:
An adiabatic process occurs without any heat transfer into or out of the system. During compression, work is done on the gas, which increases its internal energy, leading to an increase in both pressure and temperature. The relationship \(P V^k = \text{Constant}\) and \(T V^{k-1} = \text{Constant}\) are derived from the first law of thermodynamics for ideal gases under adiabatic conditions, where \(k\) is the adiabatic index (ratio of specific heats).
3. Comparison of Processes:
For the same compression ratio, the final pressure and temperature in an adiabatic process are higher than in an isothermal process. This is because in an isothermal process, heat is allowed to escape to maintain constant temperature, whereas in an adiabatic process, the energy added through compression is retained within the system, leading to a temperature rise.
Physical Meaning
1. Compression Work:
In both processes, work is done on the air to compress it. This work manifests as an increase in pressure. In an adiabatic compression, some of this work also goes into increasing the internal energy of the gas, which is observed as a rise in temperature.
2. Ideal Gas Behavior:
These calculations assume ideal gas behavior, where intermolecular forces are negligible and gas particles occupy negligible volume. While real gases deviate from ideal behavior, especially at high pressures and low temperatures, this assumption provides a good approximation for many engineering applications.
3. Engineering Applications:
Understanding isothermal and adiabatic processes is critical in the design and analysis of various thermodynamic systems, such as internal combustion engines, refrigerators, and compressors. For example, in an engine cylinder, compression is often approximated as an adiabatic process due to its rapid nature, leading to significant temperature increases that aid in combustion.
4. Significance of Adiabatic Index (k):
The adiabatic index \(k\) (also known as the heat capacity ratio or isentropic expansion factor) is a property of the gas that reflects how its specific heat capacities at constant pressure and constant volume are related. It dictates the steepness of the pressure-volume and temperature-volume curves during adiabatic processes and is crucial for accurate calculations in such systems.
