Question

A four stroke gasoline engine has a compression ratio of 10:1 with 4 cylinders of total displacement 2.3 L. the inlet state is 280 K, 70 kPa and the engine is running at 2100 RPM with the fuel adding 1400 kJ/kg in the combustion process. What is the network in the cycle and how much power is produced?

Answer 1

Net work, W = 842.52 kJ/kg

Power, P = 32.82 kW

Given:

compression ratio, r = 10:1

Total volume displaced,
`V_(d) = 2.3 l = 2.3* 10^(- 3) m^(3)`

`T_(inlet)` = 280 K

`P_(inlet)` = 70 kPa

Speed, N = 2100 rpm = 35 rps

Heat energy added, Q = 1400 kJ/kg

Solution:

Now, calculation of the overall efficiency:

`\eta = 1- r^(1 - k)`

where

k = 1.4

`\eta = 1- 10^(0.4) = 0.6018`

Also,

Net work done, W =
`\eta Q = 0.6018* 1400 = 842.52 kJ/kg`

Now, specific volume is given by:

`P_(inlet)\vartheta = RT_(inlet)`

`70* 10^(3)\vartheta = 0.287* 280`

`\vartheta = 1.148 m^(3)/kg`

Now, mean effective pressure, P' can be calculated as:

`W = P'(\vartheta - \vartheta')`

`(W)/((\vartheta - \vartheta')) = P'`

`(842.52)/(\vartheta(1 - (\vartheta')/(\vartheta))) = P'`

(Since, r =
`(\vartheta)/(\vartheta')`)

`(842.52)/(\vartheta(1 - (1)/(r))) = P' = (842.52)/(1.148(1 - (1)/(10)))`

P' = 815.45 kPa

Now, for power produced, P:

P =
`(1)/(2)P'V_(d) N`

P =
`(1)/(2)* 815.45* 2.3\time 10^(- 3)* 35`

P = 32.82 kW