Cheatography

# Thermodynamics final Cheat Sheet (DRAFT) by Bama

This is a draft cheat sheet. It is a work in progress and is not finished yet.

### Air Standard Otto Cycle

 Process 1–2: An isentropic compre­ssion of the air as the piston moves from bottom dead center to top dead center. Process 2–3: A consta­nt-­volume heat transfer to the air from an external source while the piston is at top dead center. This process is intended to represent the ignition of the fuel–air mixture and the subsequent rapid burning. Process 3–4: An isentropic expansion (power stroke). Process 4–1: Completes the cycle by a consta­nt-­volume process in which heat is rejected from the air while the piston is at bottom dead center.

### Effici­encies

 Power = 1 - Qᴄ / Qʜ Turbine = Wᴀᴄᴛᴜᴀʟ / Wɪᴅᴇᴀʟ Refrig­eration = Qᴄ / Qʜ - Qᴄ Comp or Pump = Wɪᴅᴇᴀʟ / Wᴀᴄᴛᴜᴀʟ Heat = Qʜ / Qʜ - Qᴄ Regene­rator = Qᴀᴄᴛᴜᴀʟ / Qɪᴅᴇᴀʟ

### Ideal Gases

 pv = RT pV = mRT Cᴘ = Cᴠ + R Polytropic Process: k = Cᴘ / Cv Pʀ = P / Pᴄ Tʀ = T / Tᴄ pvn = constant

### Exact Analysis of Isentropic Process for Ideal Air

 Pʀ2 / Pr1 = P1 / P2 Vʀ2 / Vʀ1 = V2 / V1 Pressure ratio = P2/ P1

### Air Standard Diesel Cycle

 The air-st­andard Diesel cycle is an ideal cycle that assumes heat addition occurs during a consta­nt-­pre­ssure process that starts with the piston at top dead center. The cycle consists of four internally reversible processes in series. Process 1-2 is the same as in the Otto cycle: an isentropic compre­ssion. Heat is not transf­erred to the working fluid at constant volume as in the Otto cycle, however. In the Diesel cycle, heat is transf­erred to the working fluid at constant pressure. Process 2–3 also makes up the first part of the power stroke. Process 3-4 is an isentropic expansion and is the remainder of the power stroke. Process 4-1 As in the Otto cycle, the cycle is completed by a consta­nt-­volume process in which heat is rejected from the air while the piston is at bottom dead center. This process replaces the exhaust and intake processes of the actual engine.

### Miscel­laneous

 Tds = du + pdv Tds = du + vdp x = Mᴠᴀᴘᴏʀ / Mᴛᴏᴛᴀʟ S = Sғ + (x)Sғɢ = Sғ(1 -x) + Sɢ Expansion valve: Δh = 0 Condensor: Qᴏᴜᴛ Evapor­ator: Qɪɴ First law: Eɪɴ - Eᴏᴜᴛ = ΔEsʏsᴛᴇᴍ h = u +pv W = VI (elect­rical)