Question #207064

In an air standard dual cycle, two-thirds of the total heat supply occurs at constant volume. The state at

the beginning of the compression process is 90kPa and 20

OC and the compression ratio is 9. If the total

heat supply is 2100kJ/kg, determine the maximum pressure and temperature and thermal efficiency of

the cycle.


1
Expert's answer
2021-06-15T10:00:36-0400

Solution.

p1=90103Pa;p_1=90\sdot10^{3}Pa;

T1=293K;T_1=293K;

ν1ν2=9;\dfrac{\nu_1}{\nu_2}=9;

Q=2100kJ=2.1106J;Q=2100kJ=2.1\sdot10^6J;

T2T1=(ν1ν2)k1;\dfrac{T_2}{T_1}=(\dfrac{\nu_1}{\nu_2})^{k-1};

T2=T1(ν1ν2)k1;T_2=T_1(\dfrac{\nu_1}{\nu_2})^{k-1};

T2=29391.41=791K;T_2=293\sdot9^{1.4-1}=791K;

p2p1=(ν1ν2)k;\dfrac{p_2}{p_1}=(\dfrac{\nu_1}{\nu_2})^{k};

p2=p1(ν1ν2)k;p_2=p_1(\dfrac{\nu_1}{\nu_2})^{k};

p2=9010391.4=1950103Pa;p_2=90\sdot10^3\sdot9^{1.4}=1950\sdot10^{3}Pa;

Qin=cv(T3T2)    T3=QcV+T2;Q_{in}=c_v(T_3-T_2)\implies T_3=\dfrac{Q}{c_V}+T_2;

T3=1.4106718+791=2740K;T_3=\dfrac{1.4\sdot10^{6}}{718}+791=2740K;

p3=p2(T3T2);p_3=p_2(\dfrac{T_3}{T_2});

p3=1950103(2740791)=6754103Pa;p_3=1950\sdot10^{3}(\dfrac{2740}{791})=6754\sdot10^3Pa;

T4=T3(ν3ν4)k1;T_4=T_3(\dfrac{\nu_3}{\nu_4})^{k-1};

T4=2740(19)0.4=1141K;T_4=2740\sdot(\dfrac{1}{9})^{0.4}=1141K;

Qout=cv(T4T1);Q_{out}=c_v(T_4-T_1);

qout=718(1141293)=608864J;q_{out}=718(1141-293)=608864J;W=QinQout;W=Q_{in}-Q_{out};

W=1.41060.609106=0.791106J;W=1.4\sdot10^6-0.609\sdot10^6=0.791\sdot10^6J;

η=WQin100%;\eta=\dfrac{W}{Q_{in}}\sdot100\%;

η=0.7911061.4106100%=56.5%;\eta=\dfrac{0.791\sdot10^6}{1.4\sdot10^6}\sdot 100\%=56.5\%;

Answer: p3=6754103Pa;T3=2740K;η=56.5%.p_3=6754\sdot10^3Pa;T_3=2740K; \eta=56.5\%.






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