hey, /sci/, i have a question about the otto cycle (applied thermodynamics

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Anonymous
2016-01-16 17:31:05 Post No. 7788196
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Anonymous 2016-01-16 17:31:05 Post No. 7788196 [Report]

hey, /sci/, i have a question about the otto cycle (applied thermodynamics exam soon). If it is symplified like pic related, is the work done by any single transformation int(pdv) or -int(vdp)? The total work in both cases is equal (same area on p-v diagram). I know that -vdp is for flows through an open system, and a typical cylinder IS an open system, but at the same time there is moving boundary work and you can't realy use a control volume. Help pls

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Anonymous 2016-01-16 18:04:07 Post No.7788274
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Anonymous 2016-01-16 18:04:07 Post No.7788274 [Report]

>>7788196
pls respond

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Okarin_2903 2016-01-16 18:34:22 Post No.7788378
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Okarin_2903 2016-01-16 18:34:22 Post No.7788378 [Report]

>>7788274
an ideal otto cycle, is asummed to be a closed system, because the energy transfer happens in a fixed mass or control mass. then, by the enthalpy definition, dh=du + pdv + vdp, but the latter terms become zero, because is a closed system. given this, using first law, and having in mind that the expansion and compression are adiabatical processes (q=0), then the work is defined by means of the internal energy changes. hope this helps.

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Anonymous 2016-01-16 18:40:16 Post No.7788393
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Anonymous 2016-01-16 18:40:16 Post No.7788393 [Report]

>>7788378
I always thought this was called the Carnot cycle, but yes this anon is right. Also, the entropy for the system will be 0, which I'm sure you'll learn soon.

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Anonymous 2016-01-16 18:40:24 Post No.7788394
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Anonymous 2016-01-16 18:40:24 Post No.7788394 [Report]

>>7788378
thanks, cleared up my ideas a bit