Question

Hot combustion gases enter the nozzle of a turbojet engine at \(230 \mathrm{kPa}, 627^{\circ} \mathrm{C},\) and \(60 \mathrm{m} / \mathrm{s}\) and exit at \(70 \mathrm{kPa}\) and \(450^{\circ} \mathrm{C}\). Assuming the nozzle to be adiabatic and the surroundings to be at \(20^{\circ} \mathrm{C}\), determine \((a)\) the exit velocity and (b) the decrease in the exergy of the gases. Take \(k=1.3\) and \(c_{p}=1.15 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) for the combustion gases.

Short answer

Question: Calculate (a) the exit velocity, and (b) the decrease in exergy of the gases for a turbojet engine nozzle with the following initial conditions: initial pressure = 230 kPa, initial temperature = 627°C, initial velocity = 60 m/s, exit pressure = 70 kPa, exit temperature = 450°C, surrounding temperature = 20°C, k (the ratio of specific heats) = 1.3, and specific heat at constant pressure (cp) = 1.15 kJ/kg⋅°C. The nozzle is assumed to be adiabatic. Answer: (a) The exit velocity is approximately 115.98 m/s. (b) The decrease in exergy of the gases is approximately 60,440.92 J/kg.

Step-by-step solution

Write down the given data and assumptions

First, let's list the given data: 1. Initial pressure (P1) = 230 kPa 2. Initial temperature (T1) = 627°C 3. Initial velocity (V1) = 60 m/s 4. Exit pressure (P2) = 70 kPa 5. Exit temperature (T2) = 450°C 6. Surrounding temperature (Ts) = 20°C 7. k (the ratio of specific heats) = 1.3 8. Specific heat at constant pressure (cp) = 1.15 kJ/kg⋅°C We are also given that the nozzle is adiabatic (no heat transfer).

Calculate the temperature ratio (T2/T1)

First, we need to convert the temperatures from Celsius to Kelvin: T1 = 627°C + 273.15 = 900.15 K T2 = 450°C + 273.15 = 723.15 K Ts = 20°C + 273.15 = 293.15 K Now, calculate the temperature ratio: T2/T1 = 723.15 K / 900.15 K ≈ 0.803

Calculate the velocity ratio (V2/V1) using energy equation

For an adiabatic nozzle, the energy equation for a steady-flow process can be written as: V2/V1 = sqrt((2*cp*(T1-T2))/(V1^2)) From the given data, cp = 1.15 kJ/kg⋅°C = 1,150 J/kg⋅K, and ΔT = T1-T2 = 900.15 K - 723.15 K = 177 K. Therefore, V2/V1 = sqrt((2*1,150*177)/(60^2)) ≈ 1.933

Calculate the exit velocity (V2)

Using the velocity ratio calculated in Step 3, we can find the exit velocity (V2) as: V2 = (V2/V1) * V1 = 1.933 * 60 ≈ 115.98 m/s The exit velocity (a) is approximately 115.98 m/s.

Calculate the decrease in exergy using exergy balance equation

To find the decrease in exergy of the gases, we will use the exergy balance equation for a control volume enclosing the nozzle: ΔEx = m*(Ψ2 - Ψ1) where Ψ is the specific exergy and m is the mass flow rate. Since we're interested in the decrease in exergy only, we'll use the change in specific exergy (ΔΨ = Ψ2 - Ψ1), which can be written as: ΔΨ = cp * (T1 - T2) - T0 * cp * ln(T2/T1) Substituting the given values and calculated values: ΔΨ ≈ (1,150 * (177)) - (293.15 * 1,150 * ln(0.803)) ≈ 60,440.92 J/kg The decrease in exergy of the gases (b) is approximately 60,440.92 J/kg.

Key concepts

Understanding the Adiabatic Nozzle

The adiabatic nozzle is a vital component in various engineering applications, especially in propulsion systems like jet engines. In its essence, an adiabatic nozzle is a device that ensures there is no heat exchange with the surrounding environment as a fluid flows through it. This is described thermodynamically as an 'adiabatic process'.

An adiabatic nozzle typically accelerates a gas or fluid without adding or removing heat, which raises its velocity. Since there’s no heat transfer, any change in the fluid’s velocity comes from the conversion of the fluid’s internal energy (enthalpy) into kinetic energy, adhering to the principles of the conservation of energy.

Application in Jet Engines

In the context of a jet engine, the hot combustion gases expand through the adiabatic nozzle. This expansion results in an increase in the gas velocity, which is critical for producing thrust. The effectiveness of this process is governed by the nozzle's design and the thermodynamic properties of the gas.

To provide an easy-to-follow approach to solving related problems, let's consider a step-by-step illustration using actual values. For gases entering at 627°C and 230 kPa, then exiting at 450°C and 70 kPa in a turbojet engine, we calculate both the temperature ratio and the velocity ratio. By applying the ideal gas law and conservation of mass and energy principles, we can predict the gas exit velocity, a crucial factor for thrust in jet engines.

The Exergy Balance Equation Explained

Exergy is a measure of the maximum useful work possible during a process that brings a system into equilibrium with its surroundings. It is a useful concept in thermodynamics and is a metric for the 'quality' of the energy.

The exergy balance equation, similar to an energy balance, takes into account both the physical work and the heat transfer while considering the environment's ability to absorb energy without changes in entropy. The exergy balance equation for a control volume can be simply put in the form:
\[ \text{Exergy change} = \text{Mass flow rate} \times (\text{Exergy exit} - \text{Exergy inlet}) \]
For an adiabatic process in which there's no heat transfer across the boundaries of the control volume, such as in an adiabatic nozzle, the exergy change is primarily due to the changes in the thermal and mechanical properties of the fluid or gas flowing through the system.

Solving for Decrease in Exergy

In our example problem, we used the exergy balance equation to determine the decrease in the exergy of the combustion gases after passing through an adiabatic nozzle. This involved calculating the specific exergy change and then applying the known mass flow rate. These calculations are paramount to assess the performance of the nozzle and the efficiency of the system.

Specific Heats Ratio in Thermodynamics

In the field of thermodynamics, the specific heats ratio, also known as the isentropic expansion factor (denoted by k or γ), represents the ratio of the specific heat at constant pressure (cp) to the specific heat at constant volume (cv). For a particular substance, this ratio is a measure of its resistance to changes in temperature during expansion or compression.

The value of k is crucial in predicting how a gas will behave under certain thermodynamic processes, such as when passing through an adiabatic nozzle:
\[ k = \frac{c_p}{c_v} \]
The ratio of specific heats plays a key role in formulations such as the energy equation for ideal gases, which comes into play when finding the exit velocity of gases in an adiabatic nozzle. A higher k corresponds to a higher speed of sound in the gas, which relates directly to the gas's ability to do work when it expands.

Role in Jet Engine Efficiency

For our sample exercise, knowing that k=1.3 for the combustion gases allows us to determine that the gas behaves relatively closely to an ideal gas during expansion. This helps us calculate the significant increase in exit velocity, a telltale of the engine’s efficiency. Using the specific heats ratio helps us simplify the calculations and offers insights into the thermodynamic properties of the combustion gases.