Answer:
The turbine consume 34.7% of the turbine power
The cycle has a thermal efficiency of 57.5%
Explanation:
The air enters the compressor at atmospheric temperature and pressure (T0, P0), it is compressed to a new temperature and pressure (T1, P1), then it burns with the fuel at constant pressure, reaching the highest temperature (T2, P2 = P1), finally it expands in the turbine and exits at atmospheric pressure (T3, P3 = P0).
T0 = 300 K
T2 = 1600 K
P0 = P3 = 100 kPa
Since the compression ratio is 14:1
P1 = P2 = 14 * P0 = 14 * 100 kPa = 1400 kPa
The compression and expansion can be considered as adiabatic processes:
P^(1-k)*T^k = constant
For air k = 1.4, then
P0^(-0.4) * T0^1.4 = P1^(-0.4) * T1^1.4
T1^1.4 = (P0^(-0.4) * T0^1.4) / P1^(-0.4)
T1 = ( (P0^(-0.4) * T0^1.4) / P1^(-0.4) )^(1/1.4)
T1 = ( (100000^(-0.4) * 300^1.4) / 1400000^(-0.4) )^(0.71) = 613 K
And for the expansion in the turbine:
P2^(-0.4) * T2^1.4 = P3^(-0.4) * T3^1.4
T3^1.4 = (P2^(-0.4) * T2^1.4) / P3^(-0.4)
T3 = ( (P2^(-0.4) * T2^1.4) / P3^(-0.4) )^(1/1.4)
T3 = ( (1400000^(-0.4) * 1600^1.4) / 100000^(-0.4) )^(0.71) = 723 K
We also need the specific volumes on these points. These are calculated with the gas equation:
p * v = R * T
v = (R * T) / p
R for air is 287 J/(kg*K)
v0 = (287 * 300) / 100000 = 0.86 m^3/kg
v1 = (287 * 613) / 1400000 = 0.12 m^3/kg
v2 = (287 * 1600) / 1400000 = 0.33 m^3/kg
v3 = (287 * 723) / 100000 = 2.07 m^3/kg
Now, the enthalpy for each point is
h = Cv * T + p * v
The Cv for air is 717 J/(kg*K)
Then:
h0 = 717 * 300 + 100000 * 0.86 = 301100 J/kg
h1 = 717 * 613 + 1400000 * 0.12 = 607521 J/kg
h2 = 717 * 1600 + 1400000 * 0.33 = 1609200 J/kg
h3 = 717 * 723 + 100000 * 2.07 = 725391 J/kg
The enthalpic raise in the compressor is
607521 - 301100 = 306421 J/kg
In the combustion chamber:
1609200 - 604521 = 1004679 J/kg
In the turbine:
725391 - 1609200 = -883809 J/kg
The generator receives 100MW.
The turbine produces more than that power, but part is consumed by the compressor.
The fraction consumed by the compressor is:
306421/883809 = 0.347
The thermal efficiency of the cycle is the useful work obtained divided by the energy consumed (in the combustion chamber)
η = (Δht - Δhc)/Δhcomb
η = (883809 - 306421)/1004679 = 0.575
