Question

Nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(3.8 \mathrm{~atm}\) and \(170^{\circ} \mathrm{C}\) enters an insulated turbine operating at steady state and expands to \(1 \mathrm{~atm}\). If the isentropic turbine efficiency is \(83.2 \%\), determine the temperature at the turbine exit, in \({ }^{\circ} \mathrm{C}\), using the ideal gas model for the nitrogen and ignoring kinetic and potential energy changes.

Short answer

The temperature at the turbine exit is approximately 148.05°C.

Step-by-step solution

Initial Conditions and Definitions

Identify the initial conditions and define key terms. Initial pressure, \( P_1 = 3.8 \mathrm{~atm} \), initial temperature, \( T_1 = 170^{\circ} \mathrm{~C} \). Also note that the final pressure \( P_2 = 1 \mathrm{~atm} \). Efficiency of the turbine, \( \eta = 83.2\% = 0.832 \). Assume nitrogen behaves as an ideal gas.

Convert Initial Temperature to Kelvin

Convert the initial temperature from Celsius to Kelvin using the relation: \( T(K) = T(^{\circ}C) + 273.15 \). Therefore, \( T_1 = 170 + 273.15 = 443.15 \mathrm{~K} \).

Use Isentropic Relations for Ideal Gas

Using the isentropic relation for an ideal gas: \( \frac{T_2s}{T_1} = \left( \frac{P_2}{P_1} \right)^\frac{\gamma - 1}{\gamma} \), where \( \gamma = 1.4 \) for nitrogen. Calculate isentropic exit temperature \( T_{2s} \): \[ T_{2s} = 443.15 \times \left( \frac{1}{3.8} \right)^\frac{1.4 - 1}{1.4} = 309.8 \mathrm{~K} \]

Calculate Actual Exit Temperature

Use the isentropic turbine efficiency formula: \[ \eta = \frac{T_1 - T_2}{T_1 - T_{2s}} \] Rearrange to find actual exit temperature \( T_2 \): \[ T_2 = T_1 - \eta \times (T_1 - T_{2s}) = 443.15 - 0.832 \times (443.15 - 309.8) = 421.2 \mathrm{~K} \]

Convert Exit Temperature back to Celsius

Convert the exit temperature from Kelvin back to Celsius: \( T_{2C} = T_2 - 273.15 \). Therefore, \[ T_{2C} = 421.2 - 273.15 = 148.05^{\circ} \mathrm{~C} \].

Key concepts

Ideal Gas Model

The **ideal gas model** is a simplified way to understand the behavior of many gases under various conditions. Under this model, gases are assumed to follow the Ideal Gas Law: \[ PV = nRT \] Where:

• P = pressure
• V = volume
• n = number of moles
• R = universal gas constant (8.314 J/(mol·K))
• T = temperature in Kelvin

Nitrogen ( \( \text{N}_2 \)) is often treated as an ideal gas under many practical conditions. This model makes many thermodynamic calculations more straightforward, such as in turbine efficiency problems.

Thermodynamic Processes

In this context, we are dealing with processes occurring in an **isentropic turbine**. An isentropic process is both adiabatic (no heat transfer) and reversible, maintaining constant entropy. In real-world applications, the process isn't perfectly isentropic due to inefficiencies. Therefore, we use the formula for **isentropic turbine efficiency** to estimate real outcomes:\[ \text{Efficiency} ( \( \text{η} \)) = \frac{T_1 - T_2}{T_1 - T_{2s}} \]Here,

• \( T_1 \) = Initial temperature
• \( T_2 \) = Actual exit temperature
• \( T_{2s} \) = Isentropic exit temperature

Combining these definitions helps calculate thermodynamic processes more accurately.

Temperature Conversion

Understanding **temperature conversion** is crucial for thermodynamic calculations. Most engineering problems, like the one at hand, require conversion between Celsius (°C) and Kelvin (K). The conversion formula is straightforward: \[ T(K) = T(^{\text{°}}C) + 273.15 \] This step ensures compatibility with other thermodynamic formulas that rely on the Kelvin scale. In the exercise, we converted the initial temperature from 170°C to Kelvin: \[ 170^{\text{°}}C + 273.15 = 443.15 K \] Later, we used this converted temperature to find the turbine's exit temperature and then converted it back to Celsius for the final answer.

Pressure Relationships

Pressure plays a vital role in thermodynamic calculations. The concept of **pressure relationships** helps us understand how pressure influences other properties like temperature in fluids. In an isentropic process for an ideal gas, the relationship between pressure and temperature is given by: \[ \frac{T_2s}{T_1} = \left( \frac{P_2}{P_1} \right)\ ^ {( \frac{γ - 1}{γ} )} \] Where \( \text{γ} \) is the specific heat ratio (Cp/Cv). For nitrogen, \( \text{γ} = 1.4 \). This relationship allows us to find the isentropic exit temperature ( \( T_{2s} \)) when pressures \( P_1 \) and \( P_2 \) are known. Accurately accounting for pressure changes is essential for correctly evaluating turbine performance.