Question

\(10.34\) Air at 2 bar, \(380 \mathrm{~K}\) is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchanger where it is cooled at constant pressure to \(320 \mathrm{~K}\) through heat transfer with the ambient. It then expands adiabatically to \(0.95\) bar through a turbine and is discharged into the cabin. The turbine has an isentropic efficiency of \(75 \% .\) If the mass flow rate of the air is \(1.0 \mathrm{~kg} / \mathrm{s}\), determine (a) the power developed by the turbine, in \(\mathrm{kW}\). (b) the rate of heat transfer from the air to the ambient, in \(\mathrm{kW}\).

Short answer

(a) 75 kW, (b) 60.3 kW

Step-by-step solution

- Identify given information

The problem provides the following details:1. Initial pressure: 2 bar2. Initial temperature: 380 K3. Final temperature after cooling: 320 K4. Final pressure after expansion: 0.95 bar5. Isentropic efficiency of the turbine: 75% or 0.756. Mass flow rate of air: 1.0 kg/s

- Determine specific heats (assumed constant)

Assume specific heat capacities for air as constants: - Specific heat at constant pressure, \( c_p = 1.005 \ \text{kJ/kg.K} \)- Specific heat at constant volume, \( c_v = 0.718 \ \text{kJ/kg.K} \)

- Calculate the change in enthalpy during cooling

The cooling occurs at constant pressure. Use the formula:\[ \Delta h = c_p (T_2 - T_1) \]Substitute the values:\[ \Delta h = 1.005 \times (320 - 380) \ = -60.3 \ \text{kJ/kg} \]

- Calculate the change in enthalpy during isentropic expansion

Use the relation for isentropic process:\[ T_3s = T_2 \left( \frac{P_3}{P_2} \right)^{ \frac{k-1}{k} } \]where \( k = c_p / c_v \ = 1.4 \). Substitute the values:\[ T_3s = 320 \left( \frac{0.95}{2} \right)^{ \frac{0.4}{1.4} } \ = 320 \times 0.682 = 218.24 \ \text{K} \]

- Calculate the actual exit temperature from the turbine

Use the turbine efficiency to find the actual exit temperature:\[ \eta_t = \frac{T_2 - T_3}{T_2 - T_{3s}} \]Rearrange to find \( T_3 \):\[ T_3 = T_2 - \eta_t (T_2 - T_{3s}) \]Substitute the values:\[ T_3 = 320 - 0.75 \times (320 - 218.24) \ = 245.32 \ \text{K} \]

- Determine power developed by the turbine

Power developed by the turbine is given by:\[ W_t = \dot{m} (c_p (T_2 - T_3)) \]Substitute the values:\[ W_t = 1.0 \times 1.005 \times (320 - 245.32) \ = 75 \ \text{kW} \]

- Calculate the rate of heat transfer

Heat transfer rate is given by:\[ Q = \dot{m} c_p (T_1 - T_2) \]Substitute the values:\[ Q = 1.0 \times 1.005 \times (380 - 320) \ = 60.3 \ \text{kW} \]

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics and jet engine mechanics. It involves the movement of thermal energy from one place to another due to temperature differences. In the given problem, the air extracted from the jet engine compressor is initially at 380 K. This air is cooled to 320 K through a heat exchanger before it goes through further thermodynamic processes. The equation used for calculating the rate of heat transfer in a constant pressure process is: \[Q = \dot{m} c_p (T_1 - T_2)\]Where:

• \(\dot{m}\): mass flow rate of air (1.0 kg/s)


• \(c_p\): specific heat at constant pressure (1.005 kJ/kg.K)


• \(T_1\): initial temperature (380 K)


• \(T_2\): final temperature after cooling (320 K)

This simplifies to:\[Q = 1.0 \times 1.005 \times (380 - 320) = 60.3 \ \text{kW}\]This shows the rate at which heat is transferred from the air to the ambient, which is 60.3 kW.

Adiabatic Expansion

Adiabatic expansion is a process where a gas expands without exchanging heat with its surroundings. In this exercise, the air is allowed to expand through the turbine from 320 K to a lower temperature. During this adiabatic expansion, the pressure drops from 2 bar to 0.95 bar, which results in a temperature decrease. The isentropic process formula used to calculate the exit temperature if the process was ideal (isentropic) is:\[T_{3s} = T_2 \left( \frac{P_3}{P_2} \right)^{ \frac{k-1}{k} }\]Where:

• \(T_2\): temperature after cooling (320 K)


• \(P_3\): final pressure (0.95 bar)


• \(P_2\): initial pressure (2 bar)


• \(k\): ratio of specific heats (\(c_p / c_v = 1.4\))

By substituting the values:\[T_{3s} = 320 \left( \frac{0.95}{2} \right)^{ \frac{0.4}{1.4} } = 320 \times 0.682 = 218.24 \ \text{K}\]This value represents the temperature if the process were perfectly adiabatic and isentropic.

Isentropic Efficiency

Isentropic efficiency is a measure of how close a real process comes to the ideal isentropic process. It is defined as the ratio of the actual work output of a device like a turbine to the work output if the process was isentropic. In this problem, the isentropic efficiency of the turbine is given as 75%. The formula used to find the actual exit temperature from the turbine considering isentropic efficiency is:\[\eta_t = \frac{T_2 - T_3}{T_2 - T_{3s}}\]Rearranging for \(T_3\) (actual exit temperature), we have:\[T_3 = T_2 - \eta_t (T_2 - T_{3s})\]By substituting the given values:\[T_3 = 320 - 0.75 \times (320 - 218.24) = 245.32 \ \text{K}\]This gives the actual temperature after the expansion through a real turbine.

Specific Heat Capacities

Specific heat capacities are essential in calculating thermal energy changes in substances. For air, the specific heat at constant pressure \(c_p\) and constant volume \(c_v\) are crucial metrics. In this exercise, we assume the specific heat capacities for air as constants since air behaves nearly ideally over the temperature range considered. The values used are:

• \(c_p = 1.005 \ \text{kJ/kg.K}\)


• \(c_v = 0.718 \ \text{kJ/kg.K}\)

These values help us determine heat transfer and work done during the processes. The ratio \(k = \frac{c_p}{c_v} = 1.4\) defines the adiabatic index, which is used in calculating temperature changes in adiabatic processes.

Turbine Power Calculation

Calculating the power developed by the turbine is one of the final goals of the problem. The power generated by the turbine is given by:\[W_t = \dot{m} (c_p (T_2 - T_3))\]Where:

• \(\dot{m}\): mass flow rate of air (1.0 kg/s)


• \(c_p\): specific heat at constant pressure (1.005 kJ/kg.K)


• \(T_2\): temperature after cooling (320 K)


• \(T_3\): actual exit temperature (245.32 K)

Substituting the values, we arrive at:\[W_t = 1.0 \times 1.005 \times (320 - 245.32) = 75 \ \text{kW}\]This means that the turbine develops 75 kW of power.