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MEC 309 (8)

# MEC 309 Cheat sheet.docx

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School
Department
Mechanical Engineering
Course
MEC 309
Professor
Wey H Leong
Semester
Winter

Description
Definitions Chapter 6 Adiabatic  ΔQ=0 Isothermal  ΔT=0 The direction of the entropy transfer is the same as direction of heat transfer. For perfect gas undergoing isentropic process  k = cp/cv pv = const. Isentropic adiabatic AND internally reversible (S=Const) Get T,s and u from tables. Internally Reversible  σ = 0 AV Volumetric Flow rate Max efficiency, n occurs for reversible Power Cycle n=k for Isentropic process for perfect gas Perfect gas = ideal gas with constant cp, cv and k Control Volumes that are internally reversible - Assume perfect gas when monatomic gases or small ΔT Power is Wcycle (dot) Assumptions and Idealizations (Schematic Diagram) - CHECK for constant volume, pressure, temp, enthalpy, entropy - The ___ system is a closed system - No change in ΔKE and ΔPE - The system consists of ___ and is modelled as ideal gas - Volume change is the only work mode - Polytropic process described as pv =Cn - 1-D flow-the flow is normal to inlet/exit -all intensive properties uniform - Steady State and Steady Flow - Power/Refrigeration Cycle - Identify hot and cold reservoirs - (Cold) Air Standard Analysis - Adiabatic / Isentropic - Identify (ir)reversible processes Notes Chapter 2  ΔE = ΔU + ΔKE + ΔPE = Q - W Chapter 3  superheated/saturated vapor/liquid 2 2 2 14.7lbf/in = 1atm 1Btu=778ft.lbf 144in =1ft Get v f1 ,g1,uf1 xg1 ,v ,u ,u f2RGg2alf2ceg2 m=V/v p 1 1mRT p 1 =mR2 2 T /T =p2v /p 2 ||1m=p2 24)v1 1 Chapter 4  SS,SF,1-D,KE/PE, air m=V/v v= RT/P m,v,V,(AV),d Notice for KE units of energy At Steady State, Steady Flow: P(1-2)  Q/m = T(s -s ),2W/1 =Q/m-(u -u ) 2 1 2 2 P(2-3)  Q=0, W/m = -(u -u ) 3 2 Nozzle/diffuser  W=0 0=Q/m+(h -h )+.5(v -v 1 /2 Q=PE=01 2 Turbines/Pump/Compressor W=m(h -h )//Q=KE=PE=0 P(3-4)  Q/m = T(s -s ),4W/4 =Q/m-(u -u3) 4 1 2 Heat XQ=0, W=0 Q=m (h 1-h 2)=m ah ah )/aKE=PE=b b1 b2 P(4-1)  Q=0, W/m = -(u -u ) 1 4 Throttling  W=0 h =h //Q=1E=K2=0 Chapter 9 Cycle Assumptions (Brayton pg 514) Chapter 5 - Air in piston-cylinder assembly is closed syst
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