Question

A Brayton cycle power plant uses a regenerator as well as reheater and intercooler. It has two stages of compression and expansion with compression starting at 103 kPa and 23°C. The pressure ratio across the compressor is 4.5. In each combustion chamber 250 kj/kg of heat is added to the air. The increase in temperature of cold air in the regenerator is 18°C. Assume isentropic process in all processes and specific heat to be constant. The total heat rejected is

• A. 313.86 kJ/kg
• B. 333.86 kJ/kg
• C. 383.65 kJ/kg
• D. 413.45 kJ/kg

Answer 1

The correct option is A 313.86 kJ/kg

$T_2=T_4=T_1(r_p{)}^{\gamma -1/\gamma }=(23+273\left)\right(4.5{)}^{1.4-1/1.4}=454.9K$

$T_5=T_4+18=454.9+18=472.9K$

$q_{in}=c_p(T_6-T_5)$

$T_6=T_5+\frac{q_{in}}{c_p}=472.9+\frac{250}{1.005}=721.66K$


$T_p=T_6{\left(\frac{1}{r_p}\right)}^{\gamma -1/\gamma }=721.66\times {\left(\frac{1}{4.5}\right)}^{0.4/1.4}=469.57K$

$T_8=T_7+\frac{q_{in}}{c_p}=469.99+\frac{250}{1.005}=718.32K$

$T_9=T_8{\left(\frac{1}{r_p}\right)}^{\gamma -1/\gamma }=467.40K$

$T_{10}=T_9-18=467.40-18=449.40K$

Total heat rejected $=c_p(T_{10}-T_1)+c_p(T_2-T_3)$

$=1.005(449.40-296)+1.005(454.9-296)=313.86kJ/kg$