Question #101004

cycle operating system between 307 and 17 degree Celsius the maximum and minimum pressures are 60 2.4 and 1.04. calculate the cycle efficiency and the work ratio. assume air to be the working fluid

Expert's answer


If we deal with Stirling cycle, we can convert and write temperatures:

T1=T4=290 K,T2=T3=580 K.T_1=T_4=290\text{ K},\\ T_2=T_3=580\text{ K}.

Efficiency:


η=1T1T2=1290580=0.5.\eta=1-\frac{T_1}{T_2}=1-\frac{290}{580}=0.5.p2T2=p1T1, p1=T1T2p2=290/58062.4=31.2 bar.\frac{p_2}{T_2}=\frac{p_1}{T_1},\\ \space\\ p_1=\frac{T_1}{T_2}p_2=290/580\cdot62.4=31.2\text{ bar}.

p3T3=p4T4, p3=T3T4p4=580/2901.04=2.08 bar.\frac{p_3}{T_3}=\frac{p_4}{T_4},\\ \space\\ p_3=\frac{T_3}{T_4}p_4=580/290\cdot1.04=2.08\text{ bar}.

Work ratio:


rw=Q41Q23=RT1 ln(p2/p3)RT2 ln(p1/p4)=0.5r_w=\frac{Q_{41}}{Q_{23}}=\frac{RT_1\text{ ln}(p_2/p_3)}{RT_2\text{ ln}(p_1/p_4)}=0.5


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