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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

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

Step by step solution

01

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.
02

Convert Initial Temperature to Kelvin

Convert the initial temperature from Celsius to Kelvin using the relation: \( T(K) = T(^{\circ}C) + 273.15 \). Therefore, \( T_1 = 170 + 273.15 = 443.15 \mathrm{~K} \).
03

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 \( T_{2s} \): \[ T_{2s} = 443.15 \times \left( \frac{1}{3.8} \right)^\frac{1.4 - 1}{1.4} = 309.8 \mathrm{~K} \]
04

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 \( T_2 \): \[ T_2 = T_1 - \eta \times (T_1 - T_{2s}) = 443.15 - 0.832 \times (443.15 - 309.8) = 421.2 \mathrm{~K} \]
05

Convert Exit Temperature back to Celsius

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

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Key Concepts

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

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.

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Most popular questions from this chapter

Methane gas \(\left(\mathrm{CH}_{4}\right)\) enters a compressor at \(298 \mathrm{~K}, 1\) bar and exits at 2 bar and temperature \(T\). Employing the ideal gas model, determine \(T\), in \(\mathrm{K}\), if there is no change in specific entropy from inlet to exit.

Reducing irreversibilities within a system can improve its thermodynamic performance, but steps taken in this direction are usually constrained by other considerations. What are some of these?

An inventor claims to have conceived of a second lawchallenging heat engine. (See H. Apsden, "The Electronic Heat Engine," Proceedings 27th International Energy Conversion Engineering Conference, 4.357-4.363, 1992. Also see U.S. Patent No. 5,101,632.) By artfully using mirrors the heat engine would "efficiently convert abundant environmental heat energy at the ambient temperature to electricity." Write a paper explaining the principles of operation of the device. Does this invention actually challenge the second law of thermodynamics? Does it have commercial promise? Discuss,

Air is compressed in an axial-flow compressor operating at steady state from \(27^{\circ} \mathrm{C}, 1\) bar to a pressure of \(2.1\) bar. The work input required is \(94.6 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing through the compressor. Heat transfer from the compressor occurs at the rate of \(14 \mathrm{~kJ}\) per \(\mathrm{kg}\) at a location on the compressor's surface where the temperature is \(40^{\circ} \mathrm{C}\). Kinetic and potential energy changes can be ignored. Determine (a) the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\). (b) the rate at which entropy is produced within the compressor, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of air flowing.

A system undergoing a thermodynamic cycle receives \(Q_{\mathrm{H}}\) at temperature \(T_{\mathrm{H}}^{\prime}\) and discharges \(Q_{\mathrm{C}}\) at temperature \(T_{\mathrm{C}}^{\prime}\). There are no other heat transfers. (a) Show that the net work developed per cycle is given by $$ W_{\text {cycle }}=Q_{\mathrm{H}}\left(1-\frac{T_{\mathrm{C}}^{\prime}}{T_{\mathrm{H}}^{\prime}}\right)-T_{\mathrm{C}}^{\prime} \sigma $$ where \(\sigma\) is the amount of entropy produced per cycle owing to irreversibilities within the system. (b) If the heat transfers \(Q_{\mathrm{H}}\) and \(Q_{\mathrm{C}}\) are with hot and cold reservoirs, respectively, what is the relationship of \(T_{\mathrm{H}}^{\prime}\) to the temperature of the hot reservoir \(T_{\mathrm{H}}\) and the relationship of \(T_{\mathrm{C}}^{\prime}\) to the temperature of the cold reservoir \(T_{\mathrm{C}}\) ? (c) Obtain an expression for \(W_{\text {cycle }}\) if there are (i) no internal irreversibilities, (ii) no internal or external irreversibilities.

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