# sol9 - 1 Ideal Otto Cycle p 1 = 1 bar p 2 = p 3 = p 4 = T 1...

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Unformatted text preview: 1 Ideal Otto Cycle p 1 = 1 bar p 2 = ? p 3 = ? p 4 = ? T 1 = 300 K T 2 = 700 K T 3 = 1400 K T 4 = ? Assumptions 1. Air is modeled as an ideal gas with k = 1 . 4 2. Air-standard Otto cycle analysis - all processes are internally reversible 3. The compression and expansion stages are adiabatic, isentropic. 4. Neglect kinetic and potential energy effects 5. System is closed The first thing we should do is determine the pressure, temperature and specific volume at the end of every process. Starting with the initial state 1: p 2 p 1 = T 2 T 1 k k- 1 p 2 = 19 . 4052 bar We can find the volume at state 1 and state 2 by using the ideal gas law: v 1 = RT 1 p 1 v 2 = RT 2 p 2 v 1 = 0 . 8610 m 3 / kg v 2 = 0 . 1035 m 3 / kg To find state 3, we use the ideal gas law again, knowing that process 2-3 is a constant volume process: p 2 v 2 RT 2 = p 3 v 3 RT 3 p 3 = 38 . 8104 bar The volume at state 3 is the same as the volume at state 2: v 3 = 0 . 1035 m 3 / kg To find state 4, we use the fact that process 3-4 is isentropic, and that the volume at 4 is the same as the...
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sol9 - 1 Ideal Otto Cycle p 1 = 1 bar p 2 = p 3 = p 4 = T 1...

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