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Chapter 6: Problem 105

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掳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路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} ( \( \text{畏} \)) = \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 (掳C) and Kelvin (K). The conversion formula is straightforward: \[ T(K) = T(^{\text{掳}}C) + 273.15 \] This step ensures compatibility with other thermodynamic formulas that rely on the Kelvin scale. In the exercise, we converted the initial temperature from 170掳C to Kelvin: \[ 170^{\text{掳}}C + 273.15 = 443.15 K \] 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: \[ \frac{T_2s}{T_1} = \left( \frac{P_2}{P_1} \right)\ ^ {( \frac{纬 - 1}{纬} )} \] Where \( \text{纬} \) is the specific heat ratio (Cp/Cv). For nitrogen, \( \text{纬} = 1.4 \). 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

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.

Answer the following true or false. If false, explain why. A process that violates the second law of thermodynamics violates the first law of thermodynamics. (b) When a net amount of work is done on a closed system undergoing an internally reversible process, a net heat transfer of energy from the system also occurs. (c) One corollary of the second law of thermodynamics states that the change in entropy of a closed system must be greater than zero or equal to zero. (d) A closed system can experience an increase in entropy only when irreversibilities are present within the system during the process. (e) Entropy is produced in every internally reversible process of a closed system. (f) In an adiabatic and internally reversible process of a closed system, the entropy remains constant. (g) The energy of an isolated system must remain constant, but the entropy can only decrease.

Complete the following involving reversible and irreversible cycles: (a) Reversible and irreversible power cycles each discharge energy \(Q_{\mathrm{C}}\) to a cold reservoir at temperature \(T_{\mathrm{C}}\) and receive energy \(Q_{\mathrm{H}}\) from hot reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{\mathrm{H}}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{H}}^{\prime}>T_{\mathrm{H}}\). (b) Reversible and irreversible refrigeration cycles each discharge energy \(Q_{\mathrm{H}}\) to a hot reservoir at temperature \(T_{\mathrm{H}}\) and receive energy \(Q_{C}\) from cold reservoirs at temperatures \(T_{C}\). and \(T_{C}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{C}}^{\prime}>T_{\mathrm{C}}\). (c) Reversible and irreversible heat pump cycles each receive energy \(Q_{\mathrm{C}}\) from a cold reservoir at temperature \(T_{\mathrm{C}}\) and discharge energy \(Q_{\mathrm{H}}\) to hot reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{\mathrm{H}}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{H}}^{\prime}

One-half kilogram of propane initially at 4 bar, \(30^{\circ} \mathrm{C}\) undergoes a process to 14 bar, \(100^{\circ} \mathrm{C}\) while being rapidly compressed in a piston-cylinder assembly. Heat transfer with the surroundings at \(20^{\circ} \mathrm{C}\) occurs through a thin wall. The net work is measured as \(-72.5 \mathrm{~kJ}\). Kinetic and potential energy effects can be ignored. Determine whether it is possible for the work measurement to be correct.

A closed system undergoes a process in which work is done on the system and the heat transfer \(Q\) occurs only at temperature \(T_{\mathrm{b}}\). For each case, determine whether the entropy change of the system is positive, negative, zero, or indeterminate. (a) internally reversible process, \(Q>0\). (b) internally reversible process, \(Q=0\). (c) internally reversible process, \(Q<0\). (d) internal irreversibilities present, \(Q>0\). (e) internal irreversibilities present, \(Q=0\). (f) internal irreversibilities present, \(Q<0\).
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