Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
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 ...