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. The isentropic efficiencies of a compressor and a turbine are 90% and 95%, respectively. 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? Make appropriate assumptions.

Answer 1

To calculate the power output of the turbine, we need to calculate the work done by the turbine by subtracting the enthalpy at the turbine outlet from the enthalpy at the turbine inlet. The ratio of the work done by the compressor to the work done by the turbine gives us the fraction of the turbine output required to drive the compressor. The thermal efficiency of the cycle can be calculated by dividing the power output of the cycle by the heat input to the cycle.

Step-by-step explanation:

First, we need to calculate the power output of the turbine. The isentropic efficiency of the turbine is 95%. The work done by the turbine is given by:

Wt = (H3 - H4)

Where H3 is the enthalpy at the turbine inlet and H4 is the enthalpy at the turbine outlet. The isentropic enthalpy at the turbine inlet can be calculated using the maximum temperature of 1600 K:

H3s = Cp × T3

H3s = (Cp × T1) × (T3/T1)^(γ/(γ-1))

Where Cp is the specific heat capacity at constant pressure, T1 is the minimum temperature, T3 is the maximum temperature, and γ is the ratio of specific heat capacities.

Substituting the values, we get:

H3s = Cp × T1 × (T3/T1)^(γ/(γ-1))

The isentropic efficiency of the turbine can then be used to calculate the actual enthalpy at the turbine inlet:

H3 = H3s × ηt

Now we can calculate the work done by the turbine:

Wt = (H3 - H4)

Next, we need to calculate the work done by the compressor. The isentropic efficiency of the compressor is 90%. The work done by the compressor is given by:

Wc = (H2 - H1)

Where H2 is the enthalpy at the compressor outlet and H1 is the enthalpy at the compressor inlet. The isentropic enthalpy at the compressor outlet can be calculated using the maximum temperature of 1600 K:

H2s = Cp × T2

H2s = (Cp × T1) × (T2/T1)^(γ/(γ-1))

The isentropic efficiency of the compressor can then be used to calculate the actual enthalpy at the compressor outlet:

H2 = H2s × ηc

Finally, the power output of the turbine and the power input to the compressor can be used to calculate the power input to the turbine. The power input to the turbine is given by:

Wturbine = Wt + Wc

The fraction of the turbine output required to drive the compressor is given by:

Fraction = Wc / Wturbine

To calculate the thermal efficiency of the cycle, we need to calculate the heat input to the cycle and the power output of the cycle. The heat input to the cycle can be calculated using the enthalpy difference at the compressor inlet and the turbine outlet:

Qin = (H2 - H4)

The power output of the cycle is already given as 100 MW. The thermal efficiency of the cycle is given by:

Efficiency = Power output / Heat input

Answer 2

a)
`w_(NET)=511.7kJ/kg`

b)
`q_H=965.7kJ/kg\\`

c)
`\eta_(TH)=0.530`

Step-by-step explanation:

Given That:

Minimum temperature
`T_1` = 300 K

Maximum temperature
`T_2` = 1600 K

Compressor pressure ratio
`(P_2)/(P_1)= 14`

k = 1.4

For the compression in the compressor:

`T_2=T_1((P_2)/(P_1))^{(k-1)/(k) } =300(14)^{(1.4-1)/(1.4) }=638.1K`

`w_c=h_2-h_1=C_(po)(T_2-T_1)=1.004(638.1-300)=339.5kJ/kg`

For the expansion in the turbine:

`T_4=T_3((P_3)/(P_4))^{(k-1)/(k) } =1600((1)/(14) )^{(1.4-1)/(1.4) }=851.2K`

`w_t=h_3-h_4=C_(po)(T_3-T_4)=1.004(1600-752.2)=851.2kJ/kg\\w_(NET)=w_t-w_c=851.2-339.5=511.7kJ/kg`

The overall net and cycle efficiency is given by:

`m=(W_(NET))/(w_(NET))=(100000)/(511.7)=195.4kg/s`

`W_t=mw_t=195.4*851.2=166.32MW\\w_c/w_t=339.5/851.2=0.399`

The energy input to the combustor is:

`q_H=C_(po)(T_3-T_2)=1.004(1600-638.1)=965.75kJ/kg`

`\eta_(TH)=w_(TH)/q_H= 511.7/965.7=0.530`