Question

The heat input to an Otto cycle is 1000kJ/kg. The compression ratio is 8 and the pressure and temperature at the beginning of the compression stroke are 100kPa and 15°C. Determine a. The maximum temperature and pressure in the cycle b. The thermal efficiency of the cycle C. The net work output d. The mean effective pressure

Answer 1

a.
`T_3=2027.1 K,P_3=5750.22 KPa`

b.
`\eta =0.564`

c. Work out put = 564 KJ/kg

d.
`P_(mean)=786.61 KPa`

Step-by-step explanation:

Given that

Heat in put = 1000 KJ/kg

Compression ratio,r = 8

`T_1=15 C`

`P_1=100 KPa`

Process 1-2

`(T_2)/(T_1)=r^(\gamma -1)`

`(T_2)/(273+15)=8^(1.4 -1)`

`T_2=661.65 K`

`(P_2)/(P_1)=r^(\gamma)`

`(P_2)/(100)=8^(1.4)`

`P_2=1837.9 KPa`

Process 2-3

We know that for air

`C_v=0.71\ (KJ)/(kg-K)`

`Q=C_v(T_3-T_2)`

`1000=0.71(T_3-661.5)`

`T_3=2027.1 K`

`(P_3)/(P_2)=(T_3)/(T_2)`

`(P_3)/(1837.9)=(2027.1)/(661.5)`

`P_3=5750.22 KPa`

We know that efficiency of otto cycle

`\eta =1-(1)/(r^(\gamma -1))`

`\eta =1-(1)/(8^(1.4-1))`

`\eta =0.564`

`\eta =(Work\ out\ put)/(Heat\ in\ put)`

`0.564 =(Work\ out\ put)/(1000)`

Work out put = 564 KJ/kg

`v_1=(RT_1)/(P_1)`

`v_1=(0.287* 288)/(100)`

`v_1=0.82656\ (m^3)/(kg)`

So

`v_2=(0.82656)/(8)\ (m^3)/(kg)`

`v_2= 0.103(m^3)/(kg)`

`Work\ out\ put\ = P_(mean)* (v_1-v_2)`

`564= P_(mean)* (0.82-0.103)`

`P_(mean)=786.61 KPa`