Answer:
heat rejection [tex]1,28KJ[/tex]
net work [tex]-1,26KJ[/tex]
termal efficency [tex]\eta =0,53[/tex]
mean effective pressure [tex]MEP=367,5Kpa[/tex]
Explanation:
The cycle can be graphicated as the index image
We can determinate the mass of air contained in the system by (assuming that the air behaves as an ideal gas)
[tex]m=\frac{P_{1} V_{1}}{R T_{1}}=\frac{90 Kpa \times 0,004 m^{3}}{0,287\frac{Kpa m^{3}}{Kg K} \times 300,15K}=4,18 \times 10^{-3} Kg[/tex]
The unknown temperatures for the steps 2 and 4 are (where k is the adiabatic constant for the air)
[tex]T_{2}=T_{1}\times(\frac{V_{1}}{V_{2}})^{k-1}=300K\times(\frac{7}{1})^{1,4-1}=653,37K[/tex]
[tex]T_{4}=T_{3}\times(\frac{V_{2}}{V_{1}})^{k-1}=1440K\times(\frac{1}{7})^{1,4-1}=661,19K[/tex]
Application of the first law of thermodynamics to the four steps of the otto cycle processes gives
[tex]W_{1-2}=m \times C_{v}\times (T_{2}-T_{1})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (653,37-300)K=1,06KJ[/tex]
[tex]Q_{2-3}=m \times C_{v}\times (T_{3}-T_{2})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (1440-653,37)K=2,36KJ[/tex]
[tex]W_{3-4}=m \times C_{v}\times (T_{4}-T_{3})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (661,19-1440)K=-2,32KJ[/tex]
[tex]Q_{4-1}=m \times C_{v}\times (T_{1}-T_{4})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (300-661,19)K=-1,08KJ[/tex]
The heat rejection then is
[tex]Q_{rejected}= Q_{2-3}+Q_{4-1}=1,28KJ[/tex]
The net work
[tex]W_{net}= W_{1-2}+ W_{3-4}=-1,26KJ[/tex] (it’s negative because it gets out of the system)
The termal efficency can be calculated by
[tex]\eta =\frac{\left | W_{net} \right |}{Q_{2-3}}=\frac{1,26KJ}{2,36KJ}=0,53[/tex]
The mean effective pressure in the engine is
[tex]MEP=\frac{\left | W_{net} \right |}{V_{1}-V_{2}}=\frac{1,26KJ}{0,004-\frac{0,004}{7}m^{3}}\times \frac{1KPa m^{3}}{KJ}=367,5Kpa[/tex]
