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MAAE 2400 (9)
Lecture

Thermodynamics Lecture Notes Summary

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School
Department
Mechanical and Aerospace Engineering
Course
MAAE 2400
Professor
Edgar Matida
Semester
Fall

Description
MAAE2400 Thermo Exam Review Units & Convention: 2 2 1N=1kg∙m/s 1MPa=1N /mm 1¿0.1MPa ∣¿ −P atm 1J=1N /m K=℃+273.15 PgaugeP ¿ Q>0: Heat transfer to the system W>0: Work done by the system Closed Systems: Fixed quantity of matter – use internal energy (u) ∆E=∆U+∆KE+∆ PE=m u −u + ∙m V −V 1 2 2+m(z −z ) First Law: ( 2 1) 2 ( 2 1 ) 2 1 Open Systems: Region of space through which mass may flow – use enthalpy (h) 2 h+ V1 2 2 V2 h+ 2 First Law: m out +gz 2) m (¿+gz 1)− ∑ ¿ ¿ dE cv ́ ́ dt =Q −cv+ cv ∑ ¿ dE cv For steady-state/steady-flow: dt =0 Process Parameters: MAAE2400 Adiabatic Process: No thermal interaction with surroundings Isothermal Process: Constant temperature Isobaric Process: Constant pressure Isentropic Process: Constant entropy Power Cycle: Creates work from heat transfer (Hot -> Cold) Refrigeration Cycle: Uses work to create heat transfer (Cold -> Hot) QC TH Coefficient of Performance: βR= Ẃ , βmax T −T C H C Heat Pump: Refrigeration Cycle, but keeps something hot instead of cold Q H Coefficient of Performance: βHP ́ W C Heat Transfer Q́ =−kA ∆T Conduction (Fourier): Cond ∆ x , where k=thermalconductivity[W /m∙ K] ́ Radiation: Q rε∙σ∙ A∙(T bodyT surface σ=5.67×10 W/m ∙K 2 4 ε=Emissivity Where, and MAAE2400 Q Ch∙A ∙(T bodyT surface Convection: Phase Quality: mvapour m vapour x= = m liquid vapour m h=h +x(h −h ) I.e. For Enthalpy: f g f Ideal Gas: pV=mRT or pv=RT kJ 8.314kmol∙K R =287 J ∙K R= air kg
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