Question

Air enters a compressor operating at steady state with a pressure of 1 bar, a temperature of \(20^{\circ} \mathrm{C}\), and a volumetric flow rate of \(0.25 \mathrm{~m}^{3} / \mathrm{s}\). The air velocity in the exit pipe is \(210 \mathrm{~m} / \mathrm{s}\) and the exit pressure is \(1 \mathrm{MPa}\). If each unit mass of air passing from inlet to exit undergoes a process described by \(p v^{1.34}=\) constant, determine the exit temperature, in \({ }^{\circ} \mathrm{C}\).

Short answer

368.31 \( ^{\circ} \text{C} \)

Step-by-step solution

Identify given values

The problem gives the following values: - Inlet pressure, \( p_1 = 1 \text{ bar} \)- Inlet temperature, \( T_1 = 20^{\text{C}} = 293.15 \, \text{K} \)- Inlet volumetric flow rate, \( \dot{V}_1 = 0.25 \text{ m}^3\text{/s} \)- Exit velocity, \( v_2 = 210 \text{ m/s} \)- Exit pressure, \( p_2 = 1 \text{ MPa} = 1000 \text{ kPa} \)- Polytropic process exponent, \( n = 1.34 \)- Relation equation: \( p v^{1.34}=k \)

Express the inlet specific volume

Using the ideal gas law, the specific volume at the inlet can be expressed as \[ v_1 = \frac{RT_1}{p_1} \]where - R (specific gas constant for air) = 0.287 \text{ kJ/kg⋅K}- \( p_1 \) = 1 bar = 100 kPa- \( T_1 \) = 293.15 K Thus, \[ v_1 = \frac{0.287 \times 293.15}{100} = 0.841\text{m}^3/\text{kg} \]

Use the given polytropic process relation

Knowing the process is described by \( p v^{1.34} = k \), and given the inlet condition \( k = p_1 v_1^{1.34} \), we can write \[ k = 100 \times (0.841)^{1.34} \]Solving for k, \[ k = 76.96 \text{kPa⋅m}^{1.34/\text{kg}} \]

Calculate the exit specific volume

Rearranging the polytropic relation for exit conditions (i.e., \( p_2 v_2^{1.34} = k \)) gives \[ v_2 = \left(\frac{k}{p_2}\right)^{\frac{1}{1.34}}\]Substituting the known values, \[ v_2 = \left(\frac{76.96}{1000}\right)^{\frac{1}{1.34}} = 0.184 \text{m}^3/\text{kg} \]

Calculate mass flow rate using inlet conditions

The mass flow rate at the inlet can be determined by \[ \dot{m} = \frac{\dot{V}_1}{v_1} \] Substituting known values, \[ \dot{m} = \frac{0.25}{0.841} = 0.297 \text{ kg/s} \]

Determine the exit temperature

Use the ideal gas law again to find the exit temperature: \[ T_2 = \frac{p_2 v_2}{R} \] Substituting in the known values, \[ T_2 = \frac{1000 \times 0.184}{0.287} = 641.46 \text{ K} \] Converting to Celsius, \[ T_2 = 641.46 - 273.15 = 368.31 ^{\circ} \text{C} \]

Key concepts

Polytropic Processes

A polytropic process is a type of thermodynamic process that typically occurs in various industrial devices like compressors and turbines. It’s characterized by the equation:
\[ pv^n = k \]
Here, p stands for pressure, v is specific volume, n is the polytropic index, and k is a constant. This equation shows how pressure and specific volume are related during the process.
Different values of n describe different processes:

• n = 1 corresponds to an isothermal process (constant temperature).
• n = γ (where \( \gamma = \frac{C_p}{C_v} \) ) describes an adiabatic process (no heat exchange).

In this exercise, the process follows a polytropic index of 1.34, which is typical for many real-world applications. Understanding this equation helps us predict how gas properties will change during compression or expansion.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that relates pressure, volume, and temperature of an ideal gas. It is represented as:
\[ PV = nRT \]
Here:

• P - Pressure
• V - Volume
• n - Number of moles
• R - Universal gas constant
• T - Temperature

For calculations involving specific volume (volume per unit mass), the equation takes the form:
\[ v = \frac{RT}{P} \]
Where, v is the specific volume, R is the specific gas constant, and P and T are the given pressure and temperature respectively. This law is a key tool in determining properties like specific volume and temperature at different stages in thermodynamic cycles, making it crucial for solving the exercise.

Mass Flow Rate Calculation

The mass flow rate is a measure of the mass of a substance passing through a given surface per unit time. It’s essential for analyzing systems involving fluid flow, such as compressors. For incompressible or ideal gas conditions, it is calculated as:
\[ \dot{m} = \frac{\dot{V}_1}{v_1} \]
Where:

• \dot{m} - Mass flow rate
• \dot{V}_1 - Volumetric flow rate
• v_1 - Specific volume at the inlet

Knowing the mass flow rate helps in further calculations and ensures energy balance across the system. In this exercise, we determine it to understand the rate at which air is moving through the compressor under given conditions.

Specific Volume

Specific volume is an intrinsic property of a substance, defined as the volume occupied by a unit mass. It is the reciprocal of density and plays a vital role in thermodynamic calculations. For ideal gases, it can be calculated using the Ideal Gas Law:
\[ v = \frac{RT}{P} \]
Here, v is the specific volume, R is the specific gas constant, and P and T are the given pressure and temperature.
For the inlet conditions in our problem, we used this formula to find the specific volume of air, which was essential for subsequent steps such as calculating the mass flow rate and applying the polytropic process equation.

Understanding specific volume helps visualize how the same mass of a gas takes up different volumes under varying pressure and temperature conditions.

Exit Temperature Calculation

One of the final steps in analyzing a thermodynamic process like in this exercise is determining the exit temperature of the gas. For this purpose, we use the Ideal Gas Law rearranged to solve for temperature:
\[ T_2 = \frac{p_2v_2}{R}\]
Here:

• p_2 - Exit pressure
• v_2 - Exit specific volume
• R - Specific gas constant

Substituting in the known values provides the exit temperature in Kelvin, which can be further converted to Celsius if needed.
This calculation ensures we understand the thermal effects of the compression process, crucial for designing efficient compressor systems and predicting performance under various operating conditions.