(3) Determine the net work output and the
thermal efficiency of the ideal Brayton cycle with intercooling, reheating and regeneration
The Ts diagram of the ideal Brayton cycle with intercooling, reheating, and regeneration is shown on the left.
Under the coldairstandard assumption, heat input is the sum of the
enthalpy differences in the constantpressure heat addition process and the reheating process. Heat output equals
the sum of the enthalpy differences in the constantpressure heat rejection process and the intercooling process.
q_{in} = (h_{6}  h_{5}) + (h_{8} 
h_{7})
= c_{P}(T_{6}
T_{5}) + c_{P}(T_{8} T_{7})
q_{out} = (h_{10} 
h_{1}) + (h_{2}  h_{3})
= c_{P}(T_{10} T_{1}) + c_{P}(T_{2}
T_{3})
The temperature at each state can be determined by the same way as in (1) and (2).
T_{1} = 300 K (given)
P_{1} = 100 kPa (given)
For optimum operation, equal pressure ratios are maintained across each
state.
The total pressure ratio is given as r_{P} = 8
P_{2} = 2.828P_{1}= 282.8 kPa
12: isentropic compression:
T_{2} = T_{1}(P_{2}/P_{1})^{(k1)/k } =
403.8 K
23: constant pressure intercooling. For optimum operation, air is cooled
to a temperature same as the temperature at the compressor inlet.
Hence,
T_{3} = T_{1} = 300 K
P_{3} = P_{2 }= 282.8
kPa
34: isentropic compression:
P_{4} = 2.828P_{3 }=
800 kPa
T_{4} = T_{3}(P_{4}/P_{3})^{(k1)/k } =
403.8 K
Temperature at state 6 reaches the maximum value. The temperature is
given as
T_{6} = 1,300 K
456: constant pressure process:
P_{6} = P_{4} = 800 kPa
67: isentropic expansion:
P_{7} = P_{6} /2.828 = 282.8 kPa
T_{7} = T_{6}(P_{7}/P_{6})^{(k1)/k } =
965.9 K
78: constant pressure reheating process:
P_{8} = P_{7} = 282.8 kPa
For optimum operation, air is heated to the same temperature at each
inlet of the turbine. Hence,
T_{8} = T_{6} =
1,300 K
9101: constant pressure process:
P_{9} = P_{1} = 100 kPa
89: isentropic expansion:
T_{9} = T_{8}(P_{9}/P_{8})^{(k1)/k } = 965.9 K
45 and 910: constant pressure regeneration:
