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Solutionir(a) calculate the isentropic temperature at the compressor exit\begin{array}{l}\frac{T_{25}}{T_{1}}=\left[\frac{P_{2}}{P_{1}}\right]^{\frac{k-1}{k}} \\{\left[\frac{T_{25}}{80+273}\right]=(20)^{\frac{1.4-1}{1.4}}, T_{25}=830.802 \mathrm{k}}\end{array}Express the compresser efficiency.\begin{array}{l}\eta_{1}=\frac{h_{25}-h_{1}}{h_{2}-h_{2}} \\0.85=\frac{T_{25}-T_{1}}{T_{2}-T_{1}} \\\left(T_{2}-353\right)=0.85(830.802-353) \\T_{2}=759.13 \cdot \mathrm{K}\end{array}CacMlate the work done by the Compressor\begin{aligned}W_{c} & =C_{p}\left(T_{2}-T_{1}\right) \\& =(1.004)(759.13-353) \\W_{c} & =407.75 \mathrm{~kJ} / \mathrm{kg}\end{aligned}Calculate the isentropic temperature at the turbine exit.\begin{array}{l}\frac{T_{45}}{T_{3}}=\left[\frac{P_{4}}{P_{3}}\right]^{\frac{k-1}{k}} \\{\left[\frac{T_{45}}{1900+273}\right]=\left[\frac{1}{20}\right]^{\frac{1.4-1}{1.4}}} \\T_{45}=923 \mathrm{k}\end{array}Express the relation for turbine efficiency.\begin{array}{r}n_{r}=\frac{W_{\text {T.act }}}{W_{\text {Tiso }}} \\0.88=\frac{W_{\text {T.act }}}{C_{p}\left(T_{3}-T_{\text {Ts }}\right)} \\W_{\text {T.act }}=1104.4 \mathrm{~kJ} / \mathrm{kg}\end{array}(b) Calculate the net work output.\begin{aligned}W_{\text {net }} & =\eta_{P T}\left(W_{\text {T.act }}-W_{c}\right) \\& =(0.9)(1104.4-407.75) \\W_{\text {net }} & =626.985 \mathrm{~kJ} / \mathrm{kg} .\end{aligned}Calculate the net work autput.\begin{aligned}W_{\text {net }} & =W_{T}-W_{c} \\& =(2)\left(W_{\text {Tact }}\right)-W_{c} \\& =1801.05 \mathrm{~kJ} / \mathrm{kg}\end{aligned}The inclusion of reheat ahed of power furbine increase the net work output of gas turbine.please uprote If useful,Thank you ...