[Boards: 3 / a / aco / adv / an / asp / b / biz / c / cgl / ck / cm / co / d / diy / e / fa / fit / g / gd / gif / h / hc / his / hm / hr / i / ic / int / jp / k / lgbt / lit / m / mlp / mu / n / news / o / out / p / po / pol / qa / qst / r / r9k / s / s4s / sci / soc / sp / t / tg / toy / trash / trv / tv / u / v / vg / vip /vp / vr / w / wg / wsg / wsr / x / y ] [Search | Home]
hey, /sci/, i have a question about the otto...
If images are not shown try to refresh the page. If you like this website, please disable any AdBlock software!

You are currently reading a thread in /sci/ - Science & Math

File: img97.gif (2 KB, 271x303) Image search: [iqdb] [SauceNao] [Google]
2 KB, 271x303
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
>>
>>7788196
pls respond
>>
>>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.
>>
>>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.
>>
>>7788378
thanks, cleared up my ideas a bit