Question

Air at \(22 \mathrm{kPa}, 220 \mathrm{~K}\), and \(250 \mathrm{~m} / \mathrm{s}\) enters a turbojet engine in flight at an altitude of \(10,000 \mathrm{~m}\). The pressure ratio across the compressor is \(12 .\) The turbine inlet temperature is \(1400 \mathrm{~K}\), and the pressure at the nozzle exit is \(22 \mathrm{kPa.}\) The diffuser and nozzle processes are isentropic, the compressor and turbine have isentropic efficiencies of 85 and \(88 \%\), respectively, and there is no pressure drop for flow through the combustor. On the basis of an air-standard analysis, determine (a) the pressures and temperatures at each principal state, in \(\mathrm{kPa}\) and \(\mathrm{K}\), respectively. (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

Pressures and temperatures at principal states: - (34.5 kPa, 251.1 K), (414 kPa, 552.1 K), (414 kPa, 1400 K), (22 kPa, 814.7 K).- Nozzle exit velocity: 912.1 m/s.

Step-by-step solution

- Identify given values

Given: - Inlet air pressure: \(P_1 = 22 \text{ kPa}\)- Inlet air temperature: \(T_1 = 220 \text{ K}\)- Inlet air velocity: \(v_1 = 250 \text{ m/s}\)- Altitude: \(10,000 \text{ m}\)- Pressure ratio across the compressor: \(r_p = 12\)- Turbine inlet temperature: \(T_3 = 1400 \text{ K}\)- Exit nozzle pressure: \(P_5 = 22 \text{ kPa}\)- Isentropic efficiency of compressor: \(\eta_c = 0.85\)- Isentropic efficiency of turbine: \(\eta_t = 0.88\)- No pressure drop in the combustor.

- Diffuser Process (State 1 to 2)

For an isentropic diffuser process:Calculate the stagnation or total temperature \(T_{01}\): \[ T_{01} = T_1 + \frac{v_1^2}{2c_p} \approx 220 + \frac{250^2}{2\cdot1005} = 251.1 \text{ K}\] Since the diffuser is isentropic: \[ T_2 = T_{01} = 251.1 \text{ K}\] Pressure \(P_2\) can be calculated using isentropic relations: \[ \frac{P_2}{P_1} = \left( \frac{T_{01}}{T_1} \right)^{\frac{\gamma}{\gamma-1}} \Rightarrow P_2 = P_1 \left( \frac{251.1}{220} \right)^{\frac{1.4}{0.4}} \approx 34.5 \text{ kPa}\]

- Compressor Process (State 2 to 3)

The compressor increases the pressure from \(P_2\) to \(P_3\): \[ P_3 = P_2 \cdot r_p = 34.5 \cdot 12 = 414 \text{ kPa}\] For isentropic efficiency: \[ T_{3s} = T_2 \cdot r_p^{\frac{\gamma-1}{\gamma}} = 251.1 \cdot 12^{\frac{0.4}{1.4}} \approx 512 \text{ K}\]Actual temperature \(T_3\) with efficiency: \[ T_3 = T_2 + \frac{T_{3s} - T_2}{\eta_c} = 251.1 + \frac{512 - 251.1}{0.85} \approx 552.1 \text{ K}\]

- Combustor Process (State 3 to 4)

Combustion raises the temperature to \(T_4\): \[T_4 = 1400 \text{ K}\] For pressure: \[P_4 = P_3 = 414 \text{ kPa}\] (No pressure drop)

- Turbine Process (State 4 to 5)

For the turbine with an isentropic efficiency:Calculate the exit pressure from turbine: \[P_5 = 22 \text{ kPa} (Nozzle exit pressure)\]Calculate isentropic temperature drop: \[ s_P_4 = P_4 \Rightarrow T_{5s} = T_4 \cdot \left( \frac{P_5}{P_4} \right)^{\frac{\gamma-1}{\gamma}} = 1400 \cdot \left( \frac{22}{414} \right)^{0.2857} \approx 733.5 \text{ K}\]Actual temperature \(T_5\) using efficiency: \[ T_5 = T_4 - \eta_t \cdot (T_4 - T_{5s}) = 1400 - 0.88 \cdot (1400 - 733.5) \approx 814.7 \text{ K}\]

- Nozzle Process (State 5 to 6)

Since the nozzle process is isentropic and given \( P_6 = 22 \text{ kPa}\): Inlet stagnation temperature \(T_{05} = T_4 = 1400 \text{ K} \) The velocity at the nozzle exit given by: \[ v_6 = \sqrt{2c_p(T_5 - T_6)} = \sqrt{2 \cdot 1005 \cdot (814.7 - 298.17)} \approx 912.1 \text{ m/s}\]

Key concepts

Isentropic Processes

Isentropic processes are thermodynamic processes where entropy remains constant. In simpler terms, these processes are reversible and adiabatic (no heat transfer).
For example, in our turbojet engine problem, both the diffuser and the nozzle are assumed to undergo isentropic processes. This means the efficiency of these processes is ideal, and there are no losses due to friction or other factors.
When solving for properties in such processes, we use isentropic relations. For instance, when the air passes through the diffuser, the inlet pressure and temperature change, but the process is isentropic. We calculate the new states using formulas that relate the pressure and temperature ratios to the stagnation properties. This gives us:
\[ P_2 = P_1 \left(\frac{T_{01}}{T_1}\right)^{\frac{\gamma}{\gamma-1}} \] \[ T_2 = T_{01} \] Such relations make it possible to evaluate changes in thermodynamic properties without losses due to irreversibilities. In real applications, while isentropic assumptions simplify the analysis, actual processes often fall short of this ideal due to factors like friction and turbulence.

Compressor Efficiency

Compressor efficiency is crucial in thermodynamic cycles like those in turbojet engines. Efficiency here describes how effectively the compressor increases the pressure of the air while minimizing energy losses.
In the given problem, the isentropic efficiency (\eta_c) of the compressor is 85%. This efficiency is defined as the ratio of the work required for an ideal (isentropic) compression to the work required for the actual process. The relationship can be expressed as:
\[ \eta_c = \frac{T_{3s} - T_2}{T_3 - T_2} \] where \ T_{3s} is the temperature after isentropic compression and \ T_3 is the actual temperature after compression.
To find the actual temperature, the formula is re-arranged to:
\[ T_3 = T_2 + \frac{T_{3s} - T_2}{\eta_c} \] This formula helps in determining the temperature rise through the compressor, accounting for the efficiency. Thus, a lower efficiency means a higher final temperature than ideal, indicating more energy losses.

Turbine Efficiency

Turbine efficiency, similar to compressor efficiency, measures how effectively the turbine extracts work from the high-pressure, high-temperature air. It also shows the losses due to irreversibilities in the expansion process.
The given turbine efficiency (\eta_t) is 88%. This efficiency is defined as the ratio of the actual work produced by the turbine to the work produced by an ideal isentropic process. It can be expressed as:
\[ \eta_t = \frac{T_4 - T_{5}}{T_4 - T_{5s}} \] where \ T_{4} is the turbine's inlet temperature, \ T_{5} is the actual exit temperature and \ T_{5s} is the isentropic exit temperature.
Solving for the actual exit temperature from the turbine, we use:
\[ T_5 = T_4 - \eta_t \cdot (T_4 - T_{5s}) \] This calculation helps in understanding how much energy is effectively converted to work versus how much is lost due to non-ideal effects. A turbine with higher efficiency will output temperatures closer to the isentropic value, indicating minimal energy losses.

Nozzle Exit Velocity

Nozzle exit velocity is a critical parameter in propulsion systems, including turbojet engines. This velocity indicates how fast the air exits the nozzle, contributing to the thrust generated by the engine.
In our problem, the nozzle process is assumed isentropic, making it an ideal expansion. The exit velocity (\ v_6) depends on the difference between the inlet and exit temperatures of the air passing through the nozzle. The formula used is:
\[ v_6 = \sqrt{2c_p(T_5 - T_6)} \] where \ T_5 is the temperature before the nozzle, \ c_p is the specific heat at constant pressure, and \ T_6 is the temperature at the nozzle exit.
By calculating the temperature difference and applying this formula, we find the nozzle exit velocity. In practice, this high-velocity exhaust contributes to the engine's thrust, effectively propelling the aircraft forward. Understanding and optimizing this parameter is key to designing efficient propulsion systems.