Question

In a gas turbine operating at steady state, air enters the compressor with a mass flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) at \(0.95\) bar and \(22^{\circ} \mathrm{C}\) and exits at \(5.7\) bar. The air then passes through a heat exchanger before entering the turbine at \(1100 \mathrm{~K}, 5.7\) bar. Air exits the turbine at \(0.95\) bar. The compressor and turbine operate adiabatically and kinetic and potential energy effects can be ignored. Determine the net power developed by the plant, in \(\mathrm{kW}\), if (a) the compressor and turbine operate without internal irreversibilities. (b) the compressor and turbine isentropic efficiencies are 82 and \(85 \%\), respectively.

Short answer

Without irreversibilities, net power is 1286.63 kW. With given efficiencies, net power is 677.77 kW.

Step-by-step solution

Define the given parameters and assumptions

Identify the given parameters and assumptions in the problem statement. \( \dot{m}=5\, \text{kg/s} \), \( P_1= 0.95\, \text{bar} \), \( T_1=295 \text{K}\) (since \(22^\circ \text{C} + 273 = 295 \text{K}\)), \(P_2 = 5.7\, \text{bar} \), \( T_3 = 1100 \text{K} \), \( P_3 = P_2 = 5.7 \text{bar} \), \( P_4 = 0.95 \text{bar} \). Assume adiabatic processes and negligible kinetic and potential energy effects.

Calculate the isentropic work of the compressor

Use the isentropic relations for an ideal gas: \[ \left( \frac{T_2}{T_1} \right) = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \] where \( \gamma=1.4 \). Calculate the temperature at the compressor outlet: \[ T_2 = 295 \left( \frac{5.7}{0.95} \right)^{\frac{0.4}{1.4}} = 518.62 \text{K} \]. Then use this to find the compressor work: \[ W_c = \dot{m} C_p (T_2 - T_1) \] where \( C_p = 1.005 \text{ kJ/kg.K} \). Thus, \( W_c = 5 \times 1.005 \times (518.62 - 295) = 1125.45 \text{ kW} \).

Calculate the isentropic work of the turbine

Again use the isentropic relations: \[ \left( \frac{T_4}{T_3} \right) = \left( \frac{P_4}{P_3} \right)^{\frac{\gamma-1}{\gamma}} \]. Calculate the temperature at the turbine outlet: \[ T_4 = 1100 \left( \frac{0.95}{5.7} \right)^{\frac{0.4}{1.4}} = 619.87 \text{K} \]. Then use this to find the turbine work: \[ W_t = \dot{m} C_p (T_3 - T_4) \]. Thus, \( W_t = 5 \times 1.005 \times (1100 - 619.87) = 2412.08 \text{ kW} \).

Determine net work assuming no internal irreversibilities

Subtract the compressor work from the turbine work: \[ W_{net} = W_t - W_c = 2412.08 - 1125.45 = 1286.63 \text{ kW} \].

Calculate net work with given efficiencies

Given compressor efficiency \(\eta_c = 82\%\) and turbine efficiency \(\eta_t = 85\%\). Adjust the works: \( W_c' = W_c / \eta_c = 1125.45 / 0.82 = 1372.50 \text{ kW} \), \( W_t' = W_t \times \eta_t = 2412.08 \times 0.85 = 2050.27 \text{ kW} \). Then find revised net work: \[ W_{net}' = W_t' - W_c' = 2050.27 - 1372.50 = 677.77 \text{ kW} \].

Key concepts

Isentropic Process

An isentropic process is one where entropy remains constant. This ideal process is used often in thermodynamics. For a gas turbine, the isentropic process helps in calculating the maximum potential work output.
In the given problem, we use the isentropic relations to find the temperatures after compression and expansion. This helps us determine how effective the turbine and compressor are at energy conversion.
For example: \ \ \[ \left( \frac{T_2}{T_1} \right) = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \ \], where \ \( \gamma=1.4 \). It shows the direct relation between pressure and temperature under ideal gas conditions.

Adiabatic Process

In an adiabatic process, there is no heat transfer into or out of the system. This means all the energy change in the system comes from work. For gas turbines, assuming adiabatic processes simplifies the calculations, focusing only on the work done by the compressor and turbine.
In this exercise, both the compressor and turbine are assumed to be adiabatic. Thus, their performance directly translates to how effectively they perform work.
Example calculation:
The temperature change in the compressor and turbine, assuming adiabatic operation, allows one to calculate the work done using the specific heat (\ \( C_p \)).

Thermodynamics Efficiency

Thermodynamic efficiency measures how well a system converts heat into work. In real-world applications, inefficiencies always exist due to various losses. For gas turbines, efficiencies are crucial in understanding real performance.
In the problem, the ideal (isentropic) efficiencies are given (82% for the compressor and 85% for the turbine). These are applied to adjust the theoretical (ideal) work calculations:
\ \[ \text{Compressor:} \; W_c' = \frac{W_c}{\eta_c} \ \] \ \[ \text{Turbine:} \; W_t' = W_t \times \eta_t \ \] These calculations help in updating the net power output considering real-world inefficiencies.

Energy Conversion

Energy conversion in gas turbines involves converting chemical energy into mechanical work. This process happens through compression, combustion, and expansion. The efficiency of this conversion is significant for power output.
By following steady-state assumptions, potential and kinetic energy impacts are neglected, focusing solely on the work from pressure and temperature changes. The overall efficiency shows how effectively the turbine converts input energy into mechanical energy.

Steady-State Operation

In a steady-state operation, conditions within the system remain constant over time. All the parameters, like pressure and temperature, remain unchanged during this state. This simplifies the calculations as there is no accumulation of mass or energy within the system.
In the given gas turbine problem, steady-state assumptions allow focusing on the inlet and outlet conditions to derive power calculations. It ensures that the mass flow rate and thermodynamic properties are consistent along the turbine and compressor stages.
Steady-state operation makes the complex nature of these systems manageable in practical scenarios.