Thermodynamics Final Cheat Sheet

Property- Macroscopic char. of a system (eg. Mass, vol, energy, pressure, temp.) *Don’t need knowledge of prev. state Steady State – properties don’t change w/time Extensive prop. – can add up Intensive – may vary from place to place, can’t add Equilibrium – balanced, wouldn’t change if isolated at that point in time Work W>0 – work done by system; W<0 – work done on system Types of work: Expansion of gas, expansion, extension of solid bar, stretching of liquid film, shaft work Power Rate of Work done Work due to polarization/magnetization Heat Transfer Q>0 – heat into system; Q<0 – heat from system Adiabatic – Q=0 Heat Transfer Conduction Q̇=−κA T 2−T 1 dT =−κA , κ=therm dx L Radiation Q̇e =εσA ( T 4b −T s4 ) , σ =5.67∗10−8 W m Convection Q̇C =hA ( T b−T f ) , h=heat transfer coe Cycles Δ E cycle =Qcycle −W cycle , Δ E cycle =0 ,W c Power Cycle Deliver a net work transfer of energy to their surroundings during each cycle Q¿ is heat transfer of energy into system from hot body a) Power Qout is heat transfer of energy out of system to b) Heat Pump cold body Thermal EfÏciency η= Refrigeration and Heat Pump Cycles Q¿ is heat transfer of energy into system from Quality- Two Phase, Liquid-Vapor Mixture Works for h, too m vapor m liquid + m vapor ν=x ν g +(1−x)ν f ν−ν f x= ν g−ν f cold body x= Qout is heat transfer of energy out of system to hot body Requires net work input W cycle =Qout −Q¿ Refrigeration Aims to cool, has coefÏcient of performance β= Enthalpy Tc Q¿ Q¿ H=U + pV ↔ h=u+ pν ,β = = W cycle Qout −Q¿ max T H −T c Units are Joules Heat Pump Aims to maintain the temp above that of surroundings, has coeff. Of performance Kelvin Plank Statement It is impossible for any system to operate in a Q Qout T H thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving γ= out = >1 , γ max = T H − energy by heat transfer from a single thermal reservoir W cycle Qout −Q¿ W cycle Q ¿ −Qout T <1 ,η max=1− c = Q¿ Q¿ Th Clausius Statement It is impossible for any system to operate in such a way that the sole result would be an energy transfer by heat from a cooler to a hotter body Clausius Inequality =−σ cycle ∮ ( δQ T ) b Irreversibilities Heat Transfer through a finite temperature difference Unrestrained expansion of a gas/liquid to a lower pressure Spontaneous chemical reaction or mixing Friction (flow or sliding) Current flow through a resistance Inelastic deformation Carnot Cycle 1-2 adiabatic compression 2-3 isothermal expansion 3-4 adiabatic expansion 4-1 isothermal compression Control Volume v i2 v 2e dE =Q̇− Ẇ + ∑ ṁi hi−∑ ṁe h e + ∑ ṁi −∑ ṁ e + ∑ ṁi g zi −∑ ṁe g z e 2 2 dt ❑ Steady State A (Vel) WEDNESDAY: 205 ṁ=∫ ρ (Vel)n dA =ρA (Vel )= dE v ∑ ṁi =∑ ṁe , dt =0 A Device Fluid Type Key Terms pe − p i Ẇ non −flow Possible Eqation Compressor V(gas) Ẇ , Δ h ¿0 ¿0 Ẇ =ṁ ( h1−h2 ) Pump L ¿0 ¿0 Ẇ , Δ h Ẇ =ṁ ( h1−h2 ) , if Q̇=0 Hydroturbine L 0 ¿0 Ẇ , Δ gz Throttling Device Δh ¿0 ¿0 Ẇ , Δ h ¿0 ¿0 Δ KE , Δh ¿0 ¿0 0 ¿ 0 , KE↑ ¿ 0 , KE↓ any Gas turbine V v 12−v 22 , h1 =h2 2 Ẇ =ṁ ( h1−h2 ) 0= (h1−h2 ) + Steam Turbine Nozzles Any Diffusers Heat Exchangers any ˙ Δ h ,W 2 2 v 1 −v 2 2 0= Q̇+ ∑ ṁi hi−∑ṁe h e 0= (h1− h2 ) + ¿0 Carnot Refrigeration Qin Qcycle Carnot Heat Pump h2−h3 h2−h1 γ max = Entropy, Steady State CV 0=∑ Q̇ + ∑ ṁi s i−∑ ṁe s e + σ̇ cv T Isentropic Process of an Ideal Gas, specific heats constant ( ) ( ) ( ) T 2 p2 = T 1 p1 One Inlet, One Outlet Q̇ σ̇ 1 s2−s 1= ∑ j + cv ṁ j T j ṁ T2 v1 = T1 v2 Well at least Kristen says its true, Int. Rev. du =TdS − pdV dh −vdp =Tds Isentropic EfÏciencies Turbines (typically 0.7 to 0.9) ( k−1) /k (k −1 ) p2 v1 = p1 v2 cp k= cv Nozzle (typically > 95%) Ẇ CV Q̇ cv = +h1−h2 ṁ ṁ If Int. Rev. ( Ẇṁ ) 2 =∫ TdS CV IntRev 1 2 η= ν /2 2 (ν /2)S Compressor/Pump (typically 75 to 85 %) ˙ CV / ṁ) S h 2 S−h 1 (−W = ηC = −Ẇ CV / ṁ h2−h1 Rankine Cycle η= ( ) Ẇ CV ṁ Rankine Cycle h1−h2−(h 4−h3 ) h −h =1− 2 3 h1−h 4 h1 − h4 Back work ratio – ratio of work input to work output bwr= Ẇ P / ṁ h 4−h 3 = Ẇ t / ṁ h1−h2 s= x s g +(1−x) sf Closed Systems 2 S2−S 1=∫ 1 b Control Volume d S cv Q̇ =∑ + ∑ ṁi s i−∑ ṁ e dt T Polytropic Processes n ( ) ( ) ( Ẇṁ ) Ẇ CV ṁ Ẇ CV ṁ p v =constant −n = ( p v −p v n−1 2 2 1 IntRev p =− p1 v 1 ln 2 , n= p1 IntRev CV =−∫ vdp ( δQT ) +σ Q dS =∑ j + σ̇ dt j Tj k Heat Transfer One inlet, one exit, SS, excluding KE, PE Ẇ CV =h1−h2 ṁ h −h Ẇ CV / ṁ = 1 2 η= (Ẇ CV / ṁ)S h 1−h2 S Entropy =−RT ln IntRev ( ) ( ) p2 ,ide p1 IntRev Rankine Cycle Green Highlighter (area under whole curve): Blue Highlighter (area under bottom): Q̇ ¿ ṁ Red Pen (area enclosed): Q̇out ṁ 1 →2 : Turbine Ẇ t =h 1−h2 ṁ 2 →3: Condensor Q̇out =h2−h3 ṁ Isentropic exp. from sat. vap to condensor pressure Heat transfer from working fluid at const. P., exits as sat. liq. 3 → 4: Pump Ẇ P =h4 −h3 ṁ 4 →1: Boiler Q̇ ¿ =h1−h4 ṁ Isentropic comp. to state for in comp. liq. region Heat transfer at const. P. ( ) Q̇net Ẇ net Ẇ P = = ṁ ṁ ṁ IntRev 4 =−∫ v 3 Irreversibilities Internal – see pump, turbine External – heat transfer from hot combustion gases to working fluid Engines MEP= net work for one cycle displacement volume For 2 engines of equal disp. Vol., the one with hgiher MEP produces more work Air Standard: IG working fluid, CM, HT, no exhaust/intake, int. rev. Cold Air Standard: const. spec. hts. Otto Cycle Otto Cycle W cycle W 34 W 12 =u3 −u4 −( u2−u = − m m m W net u 4−u1 η= =1− Q 23 u3 −u2 1 →2 : 2 →3 : W 12 Q23 =u 2−u1 =u 3−u2 m m Isentropic comp. Const. Vol. HT from ext. source 3 → 4: 4 →1: W 34 Q41 =u3−u 4 =u 4−u 1 m m Compression Ratio r= Const. Vol. Heat rejection *sign convention rules not obeyed here Otto Cycle Diesel Diesel Isentropic exp. For isentropic, cold air: T 2 k −1 cp =r , k= T1 cv 1 η=1− k−1 r Diesel 1 →2 : 2 →3 : W 12 Q23 =u 2−u1 =h3−h2 m m Isentropic comp. W 23 3 =∫ pdv= m 2 W cycle Q 41 u −u m m η= =1− =1− 4 1 Q23 Q23 h3 − h2 m m k r c −1 v 1 η=1− k−1 ,r = 3 k ( r c −1 ) c v 2 r Area enclosed: W cycle = Qcycle ( Const. Pressure. HT from ext. source 3 → 4: 4 →1: W 34 Q41 =u3−u 4 =u 4−u 1 m m Isentropic exp. Blue: W¿ ; m Purple: W out m ; Green: Qout m ; ( ) ( ) () T2 v1 = T1 v2 Q Red: ¿ m Dual T 4 v3 = T3 v4 k−1 =r k−1 = k −1 rc r k−1 Gas Turbine Power Plants Brayton 1 →2 : 2 →3 : W 12 Q23 =u 2−u1 =u 3−u2 m m Isentropic comp. ) Cold-Air Standard Const. Vol. Heat rejection Dual v1 v4 = v2 v3 Const. Vol. Heat addition 3 → 4: 4 →5: W 34 W = p3 (v 4− 45 =u4 −u5 m m Isentropic exp. Q34 =h4 −h3 m Q51 u 5−u1 m =1− η=1− Q23 Q 34 u3−u2 +h 4− h3 + m m Const. P. Heat add 5 →1 : Q51 =u5 −u1 m Const. vol. ht. rejection Brayton 1 →2 : Compressor Ẇ C =h2−h1 ṁ Isentropic exp. from sat. vap to condensor pressure Brayton (ideal) 2 →3 Q̇¿ =h3 −h2 ṁ Heat transfer from working fluid at const. P., exits as sat. liq. Brayton Ẇ T Ẇ C − ṁ ṁ η= = Q̇¿ h3−h2 ṁ Ẇ c / ṁ h2−h 1 bwr= = Ẇ t / ṁ h3−h 4 h1−h2−(h 4−h3 ) Cold-Air 3 → 4: Turbine 4 →1 Ẇ T =h 3−h 4 ṁ Q̇out =h4 −h1 ṁ Isentropic comp. to state for in comp. liq. region Heat transfer at const. P. Refrigeration/Heat Pump 1 → 2 s: isentropic comp. 2 →3 Ht. transfer from refrigerant, exits as liquid 3 → 4: throttling to 2-phase liquid-vapor mix 4 →1: HT. to refrigerant at constant p through evaporator Red: Ẇ C ; ṁ Yellow: Ẇ T ; ṁ Blue: Q̇¿ ; ṁ Green: Q̇out ṁ ( ) T 2=T 1 p2 p1 η=1− (k−1 ) /k ,T 4 =T 3 ( k−1 ) / k ( ) p4 p3 T1 1 =1− ( ) T2 p2 k−1 k p1 ( ) Note Brayton and Rankine the same Refrigeration Q̇¿ =h1−h4 , Qin is refrigeration capacity ṁ Ẇ C =h2−h1 ṁ Q̇ out =h2−h3 , h3 =h4 ṁ h −h Q β= ¿ = 1 4 W cycle h2 −h1