The document describes the ideal and actual Brayton cycle used in gas turbines. It provides the equations for the ideal cycle including work, heat transfer, and efficiency calculations. It then presents two examples of the Brayton cycle with input conditions and uses an EES program to calculate the output parameters and efficiencies.

4. Idealized Brayton cycle
1
2 3
4 1
2 3
4

5. Idealized Brayton cycle

6. The ideal gas turbine cycle equations
𝑊
𝑐 = ℎ1 − ℎ2
𝑊𝑡 = ℎ3 − ℎ4
𝑊𝑛𝑒𝑡 = 𝑊𝑡 − 𝑊
𝑐
𝑄𝑖𝑛 = ℎ3 − ℎ2
𝑄𝑜𝑢𝑡 = ℎ4 − ℎ1
𝜂𝑡ℎ =
𝑊𝑛𝑒𝑡
𝑄𝑖𝑛
𝑃2
𝑃1
=
𝑇2
𝑇1
𝑘
𝑘−1 𝑃3
𝑃4
=
𝑇3
𝑇4
𝑘
𝑘−1

7. Example 1
compressor Turbine
Combustion
chamber
Heat Exchanger
Input 1
Air
Pressure : 0.1 [MPa]
Temperature : 15 [c]
1
2 3
4
output 2
Air
Pressure : 1.0 [MPa]
Temperature : ?
output 3
Air + exhaust gas
Pressure : ?
Temperature : 1100 [c]
output 4
Air + exhaust gas
Pressure : 0.1 [MPa]
Temperature : 15 [c]
𝑄𝑖𝑛 =?
𝑄𝑜𝑢𝑡 =?
𝑊
𝑐 =? 𝑊𝑡 =?
𝑊𝑛𝑒𝑡 =?
𝜂𝑡ℎ =?

8. EES Program for Example 1 p[1] = 0.1
t[1]=288
h[1]=Enthalpy(Air,t=t[1])
s[1]=Entropy(Air,t=t[1],p=p[1])
p[2]=1
p[2]=p[3]
p[1]=p[4]
t[2]=t[1]*((p[2]/p[1])^0.286)
h[2]=Enthalpy(Air,t=t[2])
s[2]=Entropy(Air,t=t[2],p=p[2])
t[3]=1373
h[3]=Enthalpy(Air,t=t[3])
s[3]=Entropy(Air,t=t[3],p=p[3])
t[4]=t[3]*((p[4]/p[3])^0.286)
h[4]=Enthalpy(Air,t=t[4])
s[4]=Entropy(Air,t=t[4],p=p[4])
w_c =h[2]-h[1]
w_t =h[3]-h[4]
Q_in =h[3]-h[2]
Q_out =h[4]-h[1]
w_net =w_t -w_c
eta_th=(w_net /Q_in)

9. Program Result

10. The actual cycle of the Brayton cycle

11. The actual gas turbine cycle equations
𝑊
𝑐 = ℎ1 − ℎ2
𝑊𝑡 = ℎ3 − ℎ4
𝑊𝑛𝑒𝑡 = 𝑊𝑡 − 𝑊
𝑐
𝑄𝑖𝑛 = ℎ3 − ℎ2
𝑄𝑜𝑢𝑡 = ℎ4 − ℎ1
𝜂𝑡ℎ =
𝑊𝑛𝑒𝑡
𝑄𝑖𝑛
𝑇2𝑠
𝑇1
=
𝑃2
𝑃1
𝑘−1
𝑘
𝜂𝐶 =
ℎ2𝑠 − ℎ1
ℎ2 − ℎ1
𝜂𝑡 =
ℎ3 − ℎ4
ℎ3 − ℎ4𝑠
𝑇3
𝑇4𝑠
=
𝑃3
𝑃4
𝑘−1
𝑘

12. Example 2
compressor Turbine
Combustion
chamber
Heat Exchanger
Input 1
Air
Pressure : 0.1 [MPa]
Temperature : 15 [c]
1
2 3
4
output 2
Air
Pressure : 1.0 [MPa]
Temperature : ?
output 3
Air + exhaust gas
Pressure : ?
Temperature : 1100 [c]
output 4
Air + exhaust gas
Pressure : 0.1 [MPa]
Temperature : ?
𝑄𝑖𝑛 =?
𝑄𝑜𝑢𝑡 =?
𝑊
𝑐 =? 𝑊𝑡 =?
𝑊𝑛𝑒𝑡 =?
𝜂𝑡ℎ =?
𝜂𝑐 = 80% 𝜂𝑡 = 85%
𝑃2 − 𝑃3 = 15 [𝑘𝑝𝑎]

13. p[1] = 0.1
t[1]=288
h[1]=Enthalpy(Air,t=t[1])
s[1]=Entropy(Air,t=t[1],p=p[1])
p[2]=1
t_2s=t[1]*((p[2]/p[1])^0.286)
eta_c=0.80
eta_c=(t_2s-t[1])/(t[2]-t[1])
s_sc=s[1]
h_sc=enthalpy(air,p=p[2],s=s_sc)
eta_c=(h_sc-h[1])/(h[2]-h[1])
p[3]=p[2]-0.015
p[4]=p[1]
t[3]=1373
t_4s=t[3]/((p[3]/p[4])^0.286)
s[3]=Entropy(Air,t=t[3],p=p[3])
s_st=s[3]
h_st=enthalpy(air,p=p[4],s=s_st)
eta_t=0.85
eta_t=(t[3]-t[4])/(t[3]-t_4s)
h[3]=Enthalpy(Air,t=t[3])
eta_t=(h[3]-h[4])/(h[3]-h_st)
w_c =h[2]-h[1]
w_t =h[3]-h[4]
Q_in =h[3]-h[2]
Q_out =h[4]-h[1]
w_net =w_t -w_c
eta_th=(w_net /Q_in)
EES Program for Example 2

14. Program Result

15. comparing results

16. References
1- Wikiwand.com / Accessed 10 July 2017
2- Powergen.gepower.com / Accessed 10 July 2017
3- Siemens.com/ Accessed 10 July 2017
4-Van Wylen Thermodynamics book