(3) Determine the net work output and the
thermal efficiency of the ideal Brayton cycle with intercooling, reheating
and regeneration
The T-s diagram of the ideal Brayton cycle with intercooling, reheating,
and regeneration is shown
on the left.
Under the cold-air-standard assumption, heat input is the sum of the
enthalpy differences in the constant-pressure heat addition process and
the reheating process.
Heat output equals
the sum of the enthalpy differences in the constant-pressure 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
1-2: isentropic compression:
T_{2} = T_{1}(P_{2}/P_{1})^{(k-1)/k } =
403.8 K
2-3: 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
3-4: isentropic compression:
P_{4} = 2.828P_{3 }=
800 kPa
T_{4} = T_{3}(P_{4}/P_{3})^{(k-1)/k } =
403.8 K
Temperature at state 6 reaches the maximum value. The temperature is
given as
T_{6} = 1,300 K
4-5-6: constant pressure process:
P_{6} = P_{4} = 800 kPa
6-7: isentropic expansion:
P_{7} = P_{6} /2.828 = 282.8 kPa
T_{7} = T_{6}(P_{7}/P_{6})^{(k-1)/k } =
965.9 K
7-8: 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
9-10-1: constant pressure process:
P_{9} = P_{1} = 100 kPa
8-9: isentropic expansion:
T_{9} = T_{8}(P_{9}/P_{8})^{(k-1)/k } =
965.9 K
4-5 and 9-10: constant pressure regeneration:
The regeneration has an
effectiveness ε = 0.8
P_{5} = P_{4} = 800 kPa
P_{10} = P_{9} = 100 kPa
The energy balance in the regenerator is:
h_{5} - h_{4} = h_{9} - h_{10}
c_{P}(T_{5} - T_{4}) = c_{P}(T_{9} -
T_{10})
Under cold-air-standard assumption, the temperature at state 10 is
T_{10} = T_{9} - (T_{5} - T_{4}) = 516.2
K
Substitute the temperatures into the expression of q_{in}
and q_{out} gives
q_{in} = c_{P}(T_{6}- T_{5}) + c_{P}(T_{8}-
T_{7})
=
1.005( 1300 _{}- 853.5_{}) + 1.005_{}(1300 _{}-
965.9_{})
= 784.5 kJ/kg
q_{out} = c_{P}(T_{10}- T_{1}) + c_{P}(T_{2}-
T_{3})
=
1.005(516.2_{}- 403.8_{}) + 1.005_{}(403.8_{}-
300_{})
=
321.6 kJ/kg
After determining q_{in} and q_{out}, the net work output
and the thermal efficiency
of this cycle can be calculated.
w_{net,out} = q_{in} - q_{out } = 784.5 -
321.6 = 463.0 kJ/kg
η_{th,Brayton} = w_{net,out} /q_{in} =
463.0/784.5 = 59.0% |