Answer to Question #207063 in Molecular Physics | Thermodynamics for Sherine Horo

Question

In an air standard Otto cycle with compression ratio of 10, air is compressed from 100kpa and 27

0C,

after which 1500kJ/kg of heat is added to it at constant volume. Compute (i) thermal efficiency of the

cycle, (ii) air temperature at the end of compression, heat addition and expansion processes and (iii) the

mean effective pressure of the cycle.

Answer 1

Solution.

$\dfrac{\nu_1}{\nu_2}=10;$

$p_1=10^5Pa;$

$T_1=300K;$

$Q_{in}=1.5\sdot10^6J;$

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

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

$T_2=300\sdot10^{0.4}=754K;$

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

$T_3=\dfrac{1.5\sdot10^6}{718}+754=2843K;$

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

$T_4=2843(\dfrac{1}{10})^{0.4}=1132K;$

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

$Q_{out}=718(1132-300)=597376J;$

$W=Q_{in}-Q_{out};$

$W=1500000-597376=902624J=0.9\sdot10^6J;$

$\eta=\dfrac{W}{Q_{in}}100\%;$

$\eta=\dfrac{0.9\sdot10^6}{1.5\sdot10^6}100\%=60\%;$

$MEP=\dfrac{W}{\nu_1-\nu_2};$

$\nu_1=\dfrac{RT_1}{P_1};$

$\nu_1=\dfrac{287\sdot300}{10^5}=0.861m^3/kg;$

$\nu_2=\dfrac{\nu_1}{10};$

$\nu_2=\dfrac{0.861}{10}=0.0861m^3/kg;$

$MEP=\dfrac{0.9\sdot10^6}{0.861-0.0861}=1.16\sdot10^6Pa;$

Answer: $i)\eta=60\%;$

$ii)T_2=754K; T_3=2843K; T_4=1132K;$

$iii)MEP=1.16\sdot10^6Pa.$