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Chapter 11: Problem 90

Consider an air-standard jet engine cycle operating in a \(280-\mathrm{K}, 100\) -kPa environment. The compressor requires a shaft power input of \(4000 \mathrm{kW}\) Air enters the turbine state 3 at \(1600 \mathrm{K}\) and 2 \(\mathrm{MPa}\), at the rate of \(9 \mathrm{kg} / \mathrm{s}\), and the isentropic efficiency of the turbine is \(85 \%\). Determine the pressure and temperature entering the nozzle.

Short Answer

Expert verified
To find out the pressure and temperature at the nozzle of the jet engine, compute the actual temperature leaving the turbine after the compression process. Usage of thermodynamic relationships linked with isentropic efficiency then enables you to calculate the required variables: pressure and temperature entering the nozzle.

Step by step solution

01

Calculate the Isentropic Turbine Output

Given the isentropic efficiency of the turbine as \(85\%\) and the temperature at which air enters the turbine (\(T_3 = 1600K\)), Calculate the temperature at the exit of the turbine (\(T_4s\)) for an isentropic process using the correct thermodynamics relationship. The formula to use is: \[T_4s=T_3-\eta_t(T_3-T_{in})\] where \(\eta_t\) is the isentropic efficiency and \(T_{in}\) is the temperature at the turbine intake.
02

Calculate the Actual Turbine Output

After calculating the isentropic turbine output, we need to calculate the actual turbine output (\(T_4\)). The temperature of the air leaving the turbine in a real process (\(T_4\)) is different from the isentropic process, and can be calculated with the formula: \[T_4=T_3-\frac{(T_3-T_{in})}{\eta_t}\]
03

Find the Pressure and Temperature Entering the Nozzle

Using the ideal gas law and the enthalpy method, the pressure (\(P_5\)) and temperature (\(T_5\)) at the nozzle intake are determined. The equations needed to solve this step are: \[P_5=P_4\cdot\left(\frac{T_5}{T_4}\right)^{\frac{\gamma}{\gamma-1}}\] and \[T_5=T_4+\frac{c_p(T_{turbine} - T_{compressor})}{c_p+\frac{R}{\gamma-1}}\] where \(c_p\) is the specific heat at constant pressure, \(\gamma\) is the heat capacity ratio and \(R\) is the gas constant for air. Apply the formulas and you will have the answers.

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Key Concepts

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

Isentropic Process
An isentropic process is an idealized thermodynamic process that is both adiabatic and reversible. In simple terms, it means no heat is transferred to or from the fluid, and the process occurs without increasing entropy.
A common example is the expansion or compression of gases within engines, refrigerators, or turbines.
  • Understanding the Concept: The term "isentropic" originates from "iso," meaning equal, and "entropy," meaning a measure of disorder. So isentropic means there is no change in entropy.
  • Equations Involved: For an isentropic process in an ideal gas, the state change can be described by the formula: \[T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma -1}{\gamma}} \] Where \(T\) denotes temperature, \(P\) is pressure, and \(\gamma\) is the heat capacity ratio.
These relationships help simplify the evaluation of changes in the gas properties through turbines or compressors. They are essential for understanding how energy transfers within engines and other thermodynamic systems.
Turbine Efficiency
Turbine efficiency is a key performance metric that indicates how well a turbine converts thermal energy into mechanical work.
It is expressed as a percentage, comparing the actual performance of the turbine to the ideal isentropic performance.
  • Importance: A turbine with high efficiency minimizes losses and operates closer to the theoretical output.
  • Calculation: The isentropic efficiency \(\eta_t\) of a turbine is calculated using the formula: \[ \eta_t = \frac{\text{Actual Work Output}}{\text{Isentropic Work Output}} \] This equation highlights how close the actual process is to the ideal, showing energy losses due to friction, heat, and other factors.
  • Thermal Efficiency: High-efficiency indicates better performance, meaning fewer losses and more useful work can be extracted from the energy source.
The efficiency of a turbine directly affects the overall efficiency of the engine or system it powers.
Ideal Gas Law
The ideal gas law is a fundamental equation that relates the four variables of a gas: pressure, volume, temperature, and the amount of gas.
It is a cornerstone of thermodynamics and helps estimate how gases will behave under different conditions.
  • The Law: The ideal gas law is represented as: \[ PV = nRT \] Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount of gas in moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
  • Simplified Assumptions: This formula assumes that the gas particles move randomly and do not interact with each other, which is reasonably accurate for many gases under not-too-extreme conditions.
The ideal gas law is used extensively in engineering to determine the properties and behavior of gases in engines, turbines, and other devices.
Enthalpy Method
The enthalpy method is used to analyze energy transfers in thermodynamic processes, especially where changes in heat content are involved.
It's useful for calculating the heat change during physical or chemical processes.
  • Definition: Enthalpy \(H\) is a measure of the total energy of a thermodynamic system and is defined as: \[ H = U + PV \] Where \(U\) is the internal energy of the system, \(P\) is pressure, and \(V\) is volume.
  • Application in Thermodynamics: In processes where pressure is constant, the change in enthalpy \(\Delta H\) equals the heat added or removed from the system: \[ \Delta H = Q \]
  • Use in the Exercise: In the context of our exercise, the enthalpy method is applied to determine the conditions entering the nozzle by calculating changes in heat content, accounting for the specific heat and temperature differences within the engine cycle.
This approach is fundamental to designing and evaluating thermal systems where heat transfer is prominent.

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Most popular questions from this chapter

Consider a gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each compressor stage and each turbine stage is 8 to \(1 .\) The pressure at the entrance to the first compressor is \(100 \mathrm{kPa}\), the temperature entering each compressor is \(20^{\circ} \mathrm{C}\) and the temperature entering each turbine is \(1100^{\circ} \mathrm{C} .\) A regenerator is also incorporated into the cycle and it has an efficiency of \(70 \%\). Determine the compressor work, the turbine work, and the thermal efficiency of the cycle.

The refrigerant \(R-22\) is used as the working fluid in a conventional heat pump cycle. Saturated vapor enters the compressor of this unit at \(50 \mathrm{F} ;\) its exit temperature from the compressor is measured and found to be 185 F. If the compressor exit is 300 psia, what is the isentropic efficiency of the compressor and the coefficient of performance of the heat pump?

A gasoline engine has a volumetric compression ratio of 10 and before compression has air at \(290 \mathrm{K}, 85 \mathrm{kPa},\) in the cylinder. The combustion peak pressure is 6000 kPa. Assume cold air properties. What is the highest temperature in the cycle? Find the temperature at the beginning of the exhaust (heat rejection) and the overall cycle efficiency.

The air conditioner in a car uses \(R-134 a\), and the compressor power input is \(1.5 \mathrm{kW}\), bringing the R-134a from 201.7 kPa to 1200 kPa by compression. The cold space is a heat exchanger that cools \(30^{\circ} \mathrm{C}\) atmospheric air from the outside down to \(10^{\circ} \mathrm{C}\) and blows it into the car. What is the mass flow rate of the \(\mathrm{R}-134 \mathrm{a}\), and what is the low-temperature heat-transfer rate? What is the mass flow rate of air at \(10^{\circ} \mathrm{C}\) ?

Mention two benefits of a reheat cycle.
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