Question

A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100 MW to an electric generator. The minimum temperature in the cycle is 300 K, and the maximum temperature is 1600 K. The minimum pressure in the cycle is 100 kPa, and the compressor pressure ratio is 14 to 1. Calculate the power output of the turbine. What fraction of the turbine output is required to drive the compressor? What is the thermal efficiency of the cycle?

Answer 1

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