Question

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.

Answer 1

Solution.

$p_1=90\sdot10^{3}Pa;$

$T_1=293K;$

$\dfrac{\nu_1}{\nu_2}=9;$

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

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

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

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

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

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

$p_2=90\sdot10^3\sdot9^{1.4}=1950\sdot10^{3}Pa;$

$Q_{in}=c_v(T_3-T_2)\implies T_3=\dfrac{Q}{c_V}+T_2;$

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

$p_3=p_2(\dfrac{T_3}{T_2});$

$p_3=1950\sdot10^{3}(\dfrac{2740}{791})=6754\sdot10^3Pa;$

$T_4=T_3(\dfrac{\nu_3}{\nu_4})^{k-1};$

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

$Q_{out}=c_v(T_4-T_1);$

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

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

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

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

Answer: $p_3=6754\sdot10^3Pa;T_3=2740K; \eta=56.5\%.$