Question

A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of \(7 .\) The minimum and maximum temperatures in the cycle are 310 and 1150 K. Assuming an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator, determine \((a)\) the air temperature at the turbine exit, \((b)\) the net work output, and \((c)\) the thermal efficiency.

Short answer

The air temperature at the turbine exit (T5) is 670.52 K. b) What is the net work output for this Brayton cycle? The net work output for this Brayton cycle is 87.89 kJ/kg. c) What is the thermal efficiency of the Brayton cycle? The thermal efficiency of the Brayton cycle is 18.26%.

Step-by-step solution

Calculate the temperature and pressure at each state point

Given: Pressure ratio (rp) = 7 Minimum temperature (T1) = 310 K Maximum temperature (T3) = 1150 K Calculate the temperature T2s, assuming an isentropic process for the compressor: $$ T_2s = T_1 \times (rp)^{(\gamma - 1)/\gamma}$$ where γ (gamma) = 1.4 for air. $$ T_2s = 310 \times (7)^{(1.4 - 1)/1.4} = 615.6\,\text{K}$$ Now we need to account for the isentropic efficiency of the compressor, which is given as 75%. The actual temperature at point 2, T2, can be determined using the relationship: $$ \frac{T_2 - T_1}{T_{2s} - T_1} = \eta_c$$ $$ T_2 = T_1 + \eta_c (T_{2s} - T_1)$$ $$ T_2 = 310 + 0.75 \times (615.6 - 310) = 539.7\,\text{K}$$ Now, we can calculate the temperature at state point 4 (T4s) assuming an isentropic process for the turbine: $$ T_4s = T_3 / (rp)^{(\gamma - 1)/\gamma}$$ $$ T_4s = 1150 / (7)^{(1.4 - 1)/1.4} = 725.4\,\text{K}$$ Similarly, consider the isentropic efficiency of the turbine (82%), and find the actual temperature at point 4, T4. Use the relationship: $$ \frac{T_3 - T_4}{T_3 - T_{4s}} = \eta_t$$ $$ T_4 = T_3 - \eta_t (T_3 - T_{4s})$$ $$ T_4 = 1150 - 0.82 \times (1150 - 725.4) = 831.88\,\text{K}$$

Apply regenerator effectiveness

The regenerator effectiveness is given as 65%. We can find the temperature at state point 5 (T5) using the relationship: $$ T_5 = T_4 - \epsilon (T_4 - T_2)$$ $$ T_5 = 831.88 - 0.65 \times (831.88 - 539.7) = 670.52\,\text{K}$$ Now, we can answer part (a) of the problem: The air temperature at the turbine exit (T5) is 670.52 K.

Calculate work input/output and net work output

Now we need to find the specific work input to the compressor (Wc) and the specific work output from the turbine (Wt). Use the specific heat at constant pressure (Cp), which is 1.005 kJ/kgK for air. $$ W_c = C_p (T_2 - T_1) = 1.005 \times (539.7 - 310) = 231.03\,\text{kJ/kg}$$ $$ W_t = C_p (T_3 - T_4) = 1.005 \times (1150 - 831.88) = 318.92\,\text{kJ/kg}$$ Now, we can calculate the net work output (Wnet): $$ W_{net} = W_t - W_c = 318.92 - 231.03 = 87.89\,\text{kJ/kg}$$ Now, we can answer part (b) of the problem: The net work output is 87.89 kJ/kg.

Calculate the thermal efficiency

Finally, we need to calculate the thermal efficiency. Use the relationship: $$ \eta_{th} = \frac{W_{net}}{Q_{in}}$$ The heat input (Qin) can be calculated as: $$ Q_{in} = C_p (T_3 - T_5) = 1.005 \times (1150 - 670.52) = 481.262\,\text{kJ/kg}$$ Now, we can calculate the thermal efficiency: $$ \eta_{th} = \frac{87.89}{481.262} = 0.1826 = 18.26\%$$ Now, we can answer part (c) of the problem: The thermal efficiency of the Brayton cycle is 18.26%.

Key concepts

Isentropic Efficiency

Isentropic efficiency is a measure of the ideal versus the actual performance of devices like compressors and turbines. It is expressed as the ratio of the work output from an isentropic (ideal) process to the work output from the actual process under the same beginning and end conditions.

In the context of the Brayton cycle with regeneration, the isentropic efficiency of the compressor (\( \text{represented as} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: η_c\) and the turbine (\(η_t\) can provide insight into the energy losses and effectiveness of the system in converting thermal energy to mechanical work.

A lower isentropic efficiency indicates higher entropy production during the process, signifying increased irreversibility and energy loss. The Brayton cycle example in the exercise assumes 75% efficiency for the compressor and 82% for the turbine, reflecting real-world deviations from the ideal behavior.

Thermal Efficiency

Thermal efficiency is the key performance indicator of a thermal cycle, defined as the ratio of net work done by the system to the heat energy input into the system. High thermal efficiency is desired as it implies more mechanical work or electrical energy is generated for a given amount of heat energy supplied.

The Brayton cycle with regeneration, as illustrated in this exercise, shows us how heat from the exiting gas is used to preheat the air entering the turbine. This regeneration approach captures waste heat, improving the overall thermal efficiency of the system.

In this exercise, the thermal efficiency is calculated by comparing the net work output to the energy input (\(Q_{in}\) from the combustion process, resulting in an 18.26% efficiency. Improving the regenerator effectiveness, as well as the isentropic efficiencies of the compressor and turbine, would further increase this figure.

Regenerator Effectiveness

The regenerator is a crucial component in the Brayton cycle with regeneration, acting as a heat exchanger that recovers heat from the hot exhaust gases to preheat the compressed air before entering the combustion chamber. Regenerator effectiveness is a measure of how well the regenerator transfers heat from the hot fluid side to the cold fluid side relative to the maximum possible heat transfer.

It is defined as the ratio of actual heat transfer to the maximum possible heat transfer, typically expressed as a percentage. The effectiveness of 65% in the exercise indicates that the regenerator recaptures 65% of the heat that would otherwise be lost from the cycle.

A highly effective regenerator can significantly boost cycle efficiency by reducing fuel consumption and, consequently, operating costs. Challenges in improving regenerator effectiveness include balancing heat transfer capabilities with pressure drop limitations and the physical size and cost of the regenerator.