Answer:
a) [tex]w_{NET}=511.7kJ/kg[/tex]
b) [tex]q_H=965.7kJ/kg\\[/tex]
c) [tex]\eta_{TH}=0.530[/tex]
Explanation:
Given That:
Minimum temperature [tex]T_1[/tex] = 300 K
Maximum temperature [tex]T_2[/tex] = 1600 K
Compressor pressure ratio [tex]\frac{P_2}{P_1}= 14[/tex]
k = 1.4
For the compression in the compressor:
[tex]T_2=T_1(\frac{P_2}{P_1})^{\frac{k-1}{k} } =300(14)^{\frac{1.4-1}{1.4} }=638.1K[/tex]
[tex]w_c=h_2-h_1=C_{po}(T_2-T_1)=1.004(638.1-300)=339.5kJ/kg[/tex]
For the expansion in the turbine:
[tex]T_4=T_3(\frac{P_3}{P_4})^{\frac{k-1}{k} } =1600(\frac{1}{14} )^{\frac{1.4-1}{1.4} }=851.2K[/tex]
[tex]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[/tex]
The overall net and cycle efficiency is given by:
[tex]m=\frac{W_{NET}}{w_{NET}}=\frac{100000}{511.7}=195.4kg/s[/tex]
[tex]W_t=mw_t=195.4*851.2=166.32MW\\w_c/w_t=339.5/851.2=0.399[/tex]
The energy input to the combustor is:
[tex]q_H=C_{po}(T_3-T_2)=1.004(1600-638.1)=965.75kJ/kg[/tex]
[tex]\eta_{TH}=w_{TH}/q_H= 511.7/965.7=0.530[/tex]
