Question

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature of 1623°C. The initial volume is 0.05 m^3 and the gas expands through a volume ratio of 20 according to the law PV^1.25 = constant. Calculate: a) Work transfer b) Heat transfer

Answer 1

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Step-by-step explanation:

Given:

`P_(1)` = 5 Mpa

`T_(1)` = 1623°C

= 1896 K

`V_(1)` = 0.05
`m^(3)`

Also given
`(V_(2))/(V_(1)) = 20`

Therefore,
`V_(2)` = 1
`m^(3)`

R = 0.27 kJ / kg-K

`C_(V)` = 0.8 kJ / kg-K

Also given :
`P_(1)V_(1)^(1.25)=C`

Therefore,
`P_(1)V_(1)^(1.25)` =
`P_(2)V_(2)^(1.25)`

`5* 0.05^(1.25)=P_(2)* 1^(1.25)`

`P_(2)` = 0.1182 MPa

a). Work transfer, δW =
`(P_(1)V_(1)-P_(2)V_(2))/(n-1)`

`\left [(5* 0.05-0.1182* 1)/(1.25-1)  \right ]* 10^(6)`

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

=
`(mR(T_(2)-T_(1)))/(\gamma -1)+ (P_(1)V_(1)-P_(2)V_(2))/(n-1)`

=
`\left [ (\gamma -n)/(\gamma -1) \right ]* \delta W`

=
`\left [ (1.4 -1.25)/(1.4 -1) \right ]* 527.200`

= 197.7 kJ