Question

A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as the compressor. Air enters the diffuser with a volumetric flow rate of \(83.7 \mathrm{~m}^{3} / \mathrm{s}\) at \(40 \mathrm{kPa}, 240 \mathrm{~K}\), and a velocity of \(180 \mathrm{~m} / \mathrm{s}\), and decelerates essentially to zero velocity. The compressor pressure ratio is 10 and the compressor has an isentropic efficiency of \(85 \%\). The turbine inlet temperature is \(1140 \mathrm{~K}\), and its isentropic efficiency is \(85 \%\). The turbine exit pressure is \(50 \mathrm{kPa}\). Flow through the diffuser and nozzle is isentropic. Using an air-standard analysis, determine (a) the power delivered to the propeller, in MW. (b) the velocity at the nozzle exit, in \(\mathrm{m} / \mathrm{s}\). Neglect kinetic energy except at the diffuser inlet and the nozzle exit.

Short answer

The power delivered to the propeller is found by calculating the difference in work between the turbine and compressor. The velocity at the nozzle exit is determined using the conservation of energy.

Step-by-step solution

Calculate the Diffuser Exit Conditions

First, calculate the conditions at the exit of the diffuser. Given that the air enters the diffuser with a volumetric flow rate of 83.7 m\textsuperscript{3}/s, at 40 kPa, 240 K, and a velocity of 180 m/s, decelerating to zero velocity, we assume isentropic flow. From the ideal gas law, calculate the mass flow rate \(\rho = \frac{P}{RT}\) and \(\text{mass flow rate} = \rho \times \text{volumetric flow rate}\).

Calculate Compressor Exit Conditions

Using the pressure ratio of 10 and the isentropic efficiency of 85%, apply the isentropic relations and efficiency formulas to find the temperature and pressure at the compressor exit. \(\text{T}_2s = \text{T}_1 \times \bigg( \frac{P_2}{P_1} \bigg)^{\frac{\text{\textgamma}-1}{\text{\textgamma}}}\), where \(\text{\textgamma}\) is the specific heat ratio of air (1.4 for diatomic gases). Then, use the isentropic efficiency to find the actual compressor exit temperature. \(\text{T}_2 = T_1 + \frac{T_{2s} - T_1}{\text{Isentropic Efficiency}}\)

Turbine Inlet and Exit Conditions

Given the turbine inlet temperature of 1140 K and an isentropic efficiency of 85%, apply the isentropic turbine relations and efficiency formulas to find the temperature and pressure at the turbine exit. Since the turbine exit pressure is given (50 kPa), use the isentropic relations \(\text{T}_{4s} = \text{T}_3 \times \bigg( \frac{P_4}{P_3} \bigg)^{\frac{\text{\textgamma}-1}{\text{\textgamma}}}\) and the turbine efficiency: \(\text{T}_4 = \text{T}_3 - \text{\texteta}_{\text{isentropic}}(\text{T}_3 - \text{T}_{4s})\).

Calculate Work Output of the Turbine and Compressor

Use the temperatures at the exit of the turbine and compressor to calculate the work needed by the compressor and the work produced by the turbine. The specific work is given by \(\text{w} = \text{Cp}(\text{T}_\text{in} - \text{T}_\text{out})\), where \(Cp\) is the specific heat at constant pressure (approx. 1005 J/kg·K for air). Calculate the total work by multiplying specific work by the mass flow rate.

Calculate Power Delivered to the Propeller

The power delivered to the propeller is the difference between the turbine work and the compressor work. Convert the work to power: Power (MW) = \( \frac{\text{Total work (kW)}}{1000} \).

Calculate Velocity at the Nozzle Exit

For the nozzle exit velocity, apply the conservation of energy principle. Using the initial and final states: \(\text{T}_5 = \text{T}_4 \times \bigg( \frac{P_5}{P_4} \bigg)^{\frac{\text{\textgamma}-1}{\text{\textgamma}}}\) since the flow through the nozzle is isentropic. Use the formula for velocity: \(v_\text{exit} = \bigg(2C_p(T_4 - T_5) \bigg)^{0.5}\).

Key concepts

isentropic efficiency

Isentropic efficiency is a measure of the effectiveness of a thermodynamic process compared to an idealized process. Think of it as a comparison of the real-world performance of components like compressors and turbines to their ideal, most efficient performance. For a compressor, isentropic efficiency is defined as the ratio of the work input required for an ideal isentropic process to the actual work input. This can be represented as:\[ \eta_c = \frac{T_{2s} - T_1}{T_2 - T_1} \]
Where \( T_{2s} \) is the temperature after an ideal isentropic compression, \( T_2 \) is the actual temperature after compression, and \( T_1 \) is the initial temperature. For turbines, the isentropic efficiency relates to power output rather than input: \[ \eta_t = \frac{T_3 - T_{4s}}{T_3 - T_4} \]
This equation compares the actual temperature drop across the turbine (\( T_3 - T_4 \)) to the ideal temperature drop (\( T_3 - T_{4s} \)). High values of isentropic efficiency mean the device is performing close to the ideal case. Therefore, understanding and calculating isentropic efficiency helps in evaluating and improving the performance of thermodynamic systems.

ideal gas law

The ideal gas law is fundamental to thermodynamics and describes the behavior of an ideal gas in terms of pressure (\( P \)), volume (\( V \)), and temperature (\( T \)). The law is written as \[ PV = nRT \]
Where \( n \) represents the number of moles of the gas, and \( R \) is the ideal gas constant (8.314 J/mol·K). In turboprop engine analysis, you use the ideal gas law to find properties like density (\( \rho \)) and mass flow rate. For instance, density can be calculated from \( \rho = \frac{P}{RT} \). Once you know the density, you can find the mass flow rate as \( \text{mass flow rate} = \rho \times \text{volumetric flow rate} \). This relationship is vital in determining parameters at various engine stages, like the diffuser and compressor exit conditions. Remember, the ideal gas law applies best to gases where intermolecular forces and volumes are negligible, often valid in high-temperature and low-pressure conditions.

compressor pressure ratio

The compressor pressure ratio is a crucial factor in turboprop engine performance. It is defined as the ratio of the pressure after compression (\( P_2 \)) to the pressure before compression (\( P_1 \)): \[ \text{Pressure Ratio} = \frac{P_2}{P_1} \]. This ratio indicates how much the air pressure increases as it passes through the compressor. A higher pressure ratio generally means better engine efficiency and performance, as it implies a significant increase in air density, leading to more air mass for combustion. To calculate the temperature change associated with this pressure rise, you use isentropic relations: \[ T_{2s} = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \]
Where \( T_{2s} \) is the ideal isentropic temperature after compression, and \( \gamma \) is the specific heat ratio (approximately 1.4 for air). Then, using the isentropic efficiency, you find the actual exit temperature: \[ T_2 = T_1 + \frac{T_{2s} - T_1}{\eta} \]. Understanding the compressor pressure ratio allows for better design and optimization of turbo-machinery components.

diffuser exit conditions

The diffuser in a turboprop engine slows down the incoming air, reducing its velocity and increasing its pressure, preparing it for the compressor. The conditions at the diffuser exit are considered crucial as they determine the efficiency of the subsequent compression process. In isentropic flow, the air decelerates without any losses, making calculations simpler. For instance, given the initial conditions in the original exercise—air entering the diffuser at 40 kPa, 240 K, and a velocity of 180 m/s—isentropic relations can be used to find the state at the exit. Using the ideal gas law, first calculate the density as: \( \rho = \frac{P}{RT} \) with: \( P = 40 \text{kPa} \) and \( T = 240 \text{K} \). Given the volumetric flow rate (83.7 m³/s), the mass flow rate is: \( \text{mass flow rate} = \rho \times \text{volumetric flow rate} \). The uniform deceleration to zero velocity transforms kinetic energy into pressure energy, affecting the subsequent stages. Thus, understanding diffuser exit conditions is essential for accurate and efficient engine design.