Question

A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which the temperature is 2050 K. Using a cold airstandard analysis, i.e., constant specific heats at 300 K, find the compression ratio, compression specific work, and the highest pressure in the cycle.

Answer 1

Answer:

4883 kPa

Explanation:

Standard Otto Cycle pics are attached

Combustion process:

$T_{3}$ = 2050 K

$U_{2}$ = $U_{3}$ - $Q_{h}$  

$T_{2}$ = $T_{3}$  - $Q_{h}$  / $C_{vo}$

    = 2050 - 1000 / 0.717

    = 655.3 K

Compression process:

$P_{2}$ = $P_{1}$ ($T_{2}$ / $T_{1}$)^k/(k-1)

    = 90(655.3/290) ^3.5

    = 1561 kPa

CR = v1 / v2 = ($T_{2}$  / $T_{1}$))^1/(k-1)

     = (655.3 / 290) ^2.5

     = 7.67

w2 = u2 - u1 = $C_{vo}$( $T_{2}$  -  $T_{1}$)

      = 0.717(655.3 - 290)

      = 262 kJ / kg

Highest pressure is after the combustion

$P_{3}$ = $P_{2}$$T_{3}$  / $T_{2}$

    = 1561 × 2050 / 655.3

    = 4883 kPa