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 XQ=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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