91Ó°ÊÓ

The ideal gas law is a crucial concept in thermodynamics and is particularly essential in solving compressor problems like the one in our exercise. This law connects the pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of an ideal gas using the equation: \[PV = nRT\] Here,

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

For our specific problem, we use a version of this equation tailored for air, replacing \(n\) with the specific gas constant \(R\) (for air, \(R = 0.287 \text{ kJ/kg.K}\)): \[v_1 = \frac{RT_1}{p_1}\] This formula helps us calculate the specific volume \(v_1\) at the inlet. By knowing the temperature \(T_1\) and pressure \(p_1\) at the inlet, we can determine \(v_1\). This understanding lays the foundation for finding other parameters like the exit temperature and mass flow rate.

A polytropic process is a type of thermodynamic process that follows the relation: \[p v^n = \text{constant}\] Here, \(n\) is the polytropic index which can vary depending on the specific process conditions. In our exercise, the polytropic index \(n\) is given as 1.34. This means the relationship between pressure and volume remains constant through the transformation: \[p_1 v_1^{1.34} = p_2 v_2^{1.34}\] Knowing this, we can solve for the specific volume at the exit (\( v_2 \)): \[v_2 = v_1 \left(\frac{p_1}{p_2}\right)^{1/1.34}\] By substituting the known values for \(p_1\), \(v_1\), and \(p_2\), we find \(v_2\). This calculation is critical to determine the new state of the gas as it exits the compressor.

The specific volume of a gas is the volume per unit mass, often denoted as \(v\). In the context of our compressor problem, we first calculate the specific volume at the inlet using the ideal gas law: \[v_1 = \frac{RT_1}{p_1}\] Let's break it down step by step:

• The temperature at the inlet \(T_1 = 293 \text{ K}\)
• The pressure at the inlet \(p_1 = 100 \text{ kPa}\)
• The specific gas constant for air \(R = 0.287 \text{ kJ/kg.K}\)

Substituting these values into our equation gives: \[v_1 = \frac{0.287 \times 293}{100} = 0.8411 \text{ m}^3/\text{kg}\] Next, we use the polytropic process relationship to determine the specific volume at the exit (\(v_2\)): \[v_2 = 0.8411 \left(\frac{100}{1000}\right)^{1/1.34} = 0.2986 \text{ m}^3/\text{kg}\] This allows us to understand the change in specific volume as air undergoes compression.

Mass flow rate is a critical parameter in thermodynamic problems involving fluid flow. It represents the amount of mass passing through a section per unit time, denoted as \( \dot{m} \). In our exercise, to find the mass flow rate at the inlet: \[\dot{m} = \frac{\dot{V}}{v_1}\] Here,

• \( \dot{V} \) is the volumetric flow rate (at the inlet) given as 0.25 \( \text{m}^3/\text{s} \)
• \( v_1 \) is the specific volume at the inlet we previously calculated as 0.8411 \( \text{m}^3/\text{kg} \).

Putting these values into the equation, we get: \[\dot{m} = \frac{0.25}{0.8411} = 0.2972 \text{ kg/s}\] Finally, we calculate the exit area and pipediameter using the mass flow rate and the exit velocity: \[A_2 = \frac{\dot{m} v_2}{v_2} = \frac{0.2972 \times 0.2986}{210} = 4.227 \times 10^{-4} \text{ m}^2\] Converting this area to diameter: \[d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 4.227 \times 10^{-4}}{\pi}} = 0.0232 \text{ m} \times 1000 = 23.2 \text{ mm}\] The calculation of mass flow rate and exit pipe dimensions are vital for ensuring the compressor operates efficiently.