Question

A simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycle are 295 and 1240K. Assuming an isentropic efficiency of 83 percent for the compressor and 87 percent for the turbine. Determine the second law efficiencies of the compressor, the turbine, and the combustion chamber.

Answer 1

$\textbf{Given data}\rightarrow$

P. Pressure ratio (Tp) = ?

$\textup{Compressor inlet temperature}\left ( T_{1} \right )=T_{o}=295\textup{ K}$

Turbine inlet temperature (73) = 1240 K

$\eta _{c}=0.83$

$\eta _{t}=0.87$

$\textbf{Diagram}\rightarrow$

princ ty į poc > is ? 25 Pic T-S diagzam

Solution →

$\textup{Compressor}\rightarrow$

$T_{2s}=T_{1}*\left ( \frac{P_{2}}{P_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

$T_{2s}=295*\left ( 10 \right )^{\frac{1.4-1}{1.4}}$

$T_{2s}=569.555\textup{ K}$

$\eta c=\frac{T_{2s}-T_{1}}{T_{2}-T_{1}}$

$0.83=\frac{569.55-295}{T_{2}-295}$

$T_{2}=625.79\textup{ K}$

$\textup{Exergy destroyed for compressor}$

$x_{c}=T_{o}*\left [ c_{p}*\ln \left ( \frac{T_{2}}{T_{1}} \right )-R\ln \left ( \frac{P_{2}}{P_{1}} \right ) \right ]$

Ic = 295 * 1.005 * In 625.79 295) -0.287 * In (10)

Te = 28.012 kJ/kg

$\eta _{IIc}=1-\frac{x_{c}}{c_{p}*\left ( T_{2}-T_{1} \right )}$

$\eta _{IIc}=1-\frac{28.012}{1.005*\left ( 625.79-295 \right )}$

$\eta _{IIc}=0.9157$

$\textbf{Turbine}\rightarrow$

$T_{3}=T_{4s}*\left ( \frac{P_{3}}{P_{4}} \right )^{\frac{\gamma -1}{\gamma }}$

$1240=T_{4s}*\left ( 10 \right )^{\frac{1.4-1}{1.4}}$

$T_{4s}=642.254\textup{ K}$

13-14 nt = T-T48

$0.87=\frac{1240-T_{4s}}{1240-642.254}$

$T_{4}=719.96\textup{ K}$

$\textup{Exergy destroyed for turbine}$

$x_{t}=T_{o}*\left [ c_{p}*\ln \left ( \frac{T_{4}}{T_{3}} \right )-R\ln \left ( \frac{P_{4}}{P_{3}} \right ) \right ]$

$x_{t}=295*\left [ 1.005*\ln \left ( \frac{719.96}{1240} \right ) -0.287*\ln \left ( \frac{1}{10} \right )\right ]$

$x_{t}=33.764\textup{ kJ/kg}$

$\eta _{IIt}=1-\frac{x_{t}}{c_{p}*\left ( T_{3}-T_{4} \right )-T_{o}*\left [ c_{p}*\ln \left ( \frac{T_{4}}{T_{3}} \right )-R\ln \left ( \frac{P_{4}}{P_{3}} \right ) \right ]}$

$\eta _{IIc}=1-\frac{33.764}{1.005*\left ( 1240-719.96\right )-295*\left [ 1.005*\ln \left ( \frac{719.96}{1240} \right ) -0.287*\ln \left ( \frac{1}{10} \right )\right ]}$

$\eta _{IIc}=0.9309$

$\textbf{Combustion chamber}\rightarrow$

$q=c_{p}*\left ( T_{3}-T_{2} \right )$

$q=1.005*\left ( 1240-625.79 \right )$

$q=617.281\textup{ kJ/kg}$

$\textup{Exergy destroyed for combustion chamber}$

$x_{cc}=T_{o}*c_{p}*\ln \left ( \frac{T_{3}}{T_{2}} \right )$

$x_{cc}=295*1.005*\ln \left ( \frac{1240}{625.79} \right )$

$x_{cc}=202.744\textup{ kJ/kg}$

$\eta _{IICC}=1-\frac{x_{cc}}{q_{in}}$

$\eta _{IICC}=1-\frac{202.744}{617.28}$

$\eta _{IICC}=0.6715$