13468-9-146P AID: 1825 | 20/12/2014

Show theregenerative Brayton cycle with air as the working fluid, on diagram as in Figure (1).

For the given regenerative Brayton cycle with air as the working fluid, the temperature, pressure, specific enthalpy, specific entropy as a function of temperature alone, specific entropy, and relative pressure at state are , ,, , and respectively.

Use ideal-gas properties of air table, to find properties of air at 310K, 650K and 1400 K .

Find the pressure ratio for the regenerative Brayton cycle.

Substitute 900 kPafor , and 100 kPa for .

Express the pressure ratio relation for the process 3-4.

Substitute for , and 450.5 for .

Use ideal-gas properties of air, to find specific enthalpy of air at 50.06. Use interpolation method to find .

Express the efficiency of the turbine .

Rearrange and find . Substitute for , for , and 0.90 for .

Find the heat added due to regeneration .

Here, the effectiveness of the regenerator is .

Substitute 0.80 for , for , and for .

Find the net work output of the regenerative Brayton cycle.

Here, the work output by the turbine is , and the work input to the compressor is .

Substitute for , for , 659.84 for , and 310.24 for .

Express the heat input to the regenerative Brayton cycle as

Substitute for , for , and for to find .

Express the heat rejected by the regenerative Brayton cycle .

Substitute for , and for to find .

Find the specific enthalpy at state 6 , using the heat rejection relation.

Substitute 310.24 for , and 397.80 for .

Express the specific enthalpy relation for the regenerator.

Rearrange and substitute 659.84 for , 900.75 for , and 708.04 for to find .

Use ideal-gas properties of air table, to find and of air at 310 K , and at 1400 K .

Use ideal-gas properties of air table, to find of air at .

Use ideal-gas properties of air table, to find of air at .

Use interpolation method to find .

Use ideal-gas properties of air table, to find of air at .

Use interpolation method to find .

Use ideal-gas properties of air table, to find of air at .

Use interpolation method to find .

Find the exergy destruction associated with the process 1-2 of the given Brayton

cycle .

Here, the temperature of the surroundings is , and the gas constant of air is R.

Substitute 300 K for , for (at ), for (at ), for R, and 9 for .

Hence, the exergy destruction associated with process 1-2 of the given Brayton cycle is .

Find the exergy destruction associated with the process 3-4 of the given Brayton cycle .

Substitute 295 K for , for (at ), for (at ), for R, and for .

Hence, the exergy destruction associated with process 3-4 of the given Brayton cycle is .

Find the exergy destruction associated with the regeneration process of the given Brayton cycle .

Substitute 300 K for , for (at ), for (at ), for (at ), and for (at ).

Hence, the exergy destruction associated with regeneration process of the given Brayton cycle is .

Find the exergy destruction associated with the process 5-3 of the given Brayton cycle .

Here, the temperature of the heat source is .

Substitute 295 K for , for (at ), for (at ), for R, 1260 K for , and for . For the process 5-3, pressure remains constant, hence substitute 0 for .

Hence, the exergy destruction associated with process 5-3 of the given Brayton cycle is .

Find the exergy destruction associated with the process 6-1 of the given Brayton cycle .

Here, the temperature of the sink is .

Substitute 295 K for , for (at ), for (at ), for R, 300 K for , and for . For the process 6-1, pressure remains constant, hence substitute 0 for .

Hence, the exergy destruction associated with process 6-1 of the given Brayton cycle is .

Find the stream exergy at the exit of the regenerator (state 6) .

(1)

Here, the specific enthalpy of the surroundings is .

Find the change entropy for the exit of the regenerator .

Here, entropy of air at the surroundings as a function of temperature alone is , and the pressure of air at the surroundings is .

Substitute for (at ), and for (at ). Also, at the exit of the regenerator, pressure remains constant, hence substitute 0 for .

Substitute for , for , 300 K for , and for in Equation (1) to find .

Thus, the exergy of the exhaust gases at the exit of the regenerator is .