Question 9.CSGP.137: Carbon dioxide, CO2, enters an adiabatic compressor at 100 k......

Carbon dioxide, CO_2, enters an adiabatic compressor at 100 kPa, 300 K, and exits at 1000 kPa, 520 K. Find the compressor efficiency and the entropy generation for the process.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

C.V. Ideal compressor. We will assume constant heat capacity.
Energy Eq.6.13:      w _{ c }= h _1- h _2 \text {, }

Entropy Eq.9.8:      s _2= s _1: T _{2 s }= T _1\left\lgroup\frac{ P _2}{ P _1} \right\rgroup^{\frac{ k -1}{ k }}=300\left\lgroup\frac{1000}{100} \right\rgroup^{0.2242}=502.7 \,K

w _{ cs }= C _{ p }\left( T _1- T _{2 s }\right)=0.842(300-502.7)=-170.67 \,kJ / kg

C.V. Actual compressor

\begin{aligned}& w _{ cac }= C _{ p }\left( T _1- T _{2 ac }\right)=0.842(300-520)=-185.2 \,kJ / kg \\& \eta_{ c }= w _{ cs } / w _{ cac }=-170.67 /(-185.2)= 0 . 9 2\end{aligned}

Use Eq.8.16 for the change in entropy

\begin{aligned}& s _{\text {gen }}= s _{2 ac }- s _1= C _{ p } \ln \left( T _{2 ac } / T _1\right)- R \ln \left( P _2 / P _1\right) \\& \quad=0.842 \ln (520 / 300)-0.1889 \ln (1000 / 100)= 0 . 0 2 8 \, k J / k g ~ K\end{aligned}

Constant heat capacity is not the best approximation. It would be more accurate to use Table A.8. Entropy change in Eq.8.19 and Table A.8:

s _{ T 2}^{ o }= s _{ T 1}^{ o }+ R \ln \left( P _2 / P _1\right)=4.8631+0.1889 \ln (1000 / 100)=5.29806

Interpolate in A.8 ⇒  T _{2 s }=481 \,K , \quad h _{2 s }=382.807 \,kJ / kg \Rightarrow

\begin{aligned}- w _{ cs }= & 382.807-214.38=168.43 kJ / kg ; \quad- w _{ cac }=422.12-214.38=207.74 \,kJ / kg \\& \eta_{ c }= w _{ cs } / w _{ cac }=-168.43 /(-207.74)= 0 . 8 1 \\& s _{ gen }= s _{2 ac }- s _1=5.3767-4.8631-0.1889 \ln (10)=0.0786\, kJ / kgK\end{aligned}

104

Related Answered Questions