University:
Minnesota State University Moorhead
Course:
PHYS-380 | Thermodynamics
Academic year:

2019
Views:

363

Pages:

3
Author:

Amira Hester

Interpolation x −x a ) y = y a +( y b− y a)( x b− x a Work and Power V2 W e =VI =I R= R W m =Fd W s =T ∅ ´ Ẃ m =Fv W s =Tω 2 Energy 1 2 P 1 2 e mch= + v +gz = ρv + v +gz 2 ρ 2 P 1 2 e mchflow =u+ + v + gz ρ 2 1 2 e mchflow =u+ ρv + v + gz 2 Efficiency Desired Required ηtot =η1+ η2+⋯ η= Energy and Heat Transfer ḿ=ρ V́ = ρAV Qout =−¿ Q¿ =+ ¿ Phase Descriptions V V g → SHV Temperature at Known Pressure T < T sat → SCL T > T sat → SHV Pressure at Known Temperature P> Psat → SCL P< Psat → SHV Quality m X = vap m tot V −V f X= V f −V g Enthalpy h=u+ pv H=U + PV h ≈ hf @T +v f @ t ( Psys + Psat @ t ) Pressure |¿|−P sys Patm =P¿ |¿|=P + P sys atm P¿ Energy Balance ∆ E=∑ E¿ −∑ E out ∆ u+ ∆ KE+∆ PE=Q¿ +W ¿ + ḿ¿ e Ideal Gas Law PV =mRT Pv = RT Conservation of Mass ḿ=ρ V avg A c =ρ V́ = Isobaric (P=constant) V1 V2 = T 1 T2 V́ =V avg A c ρ 1 V 1 A1= ρ 2 V 2 A 2 V 1 A1 V 2 A2 = v1 v2 P1 V 1=P 2 V 2 Isochoric (V=cons) P1 P2 = T1 T 2 Turbines, Compressors, Pumps, and Fans Compressibility Factor Pv RT P Pr= Pcr Pv =ZRT T r= T T cr Using Ideal Gas Law Pr ≪1 IGL T r >2∧Pr ≯1 T r >2∧Pr ≯3 T r ≈ 1∧P r ≈ 1 IGL Comp Chart Property Tables Specific Heats ∆ u=CVavg ∆ T ∆ h =C Pavg ∆ T C p =C v =C h=u+ P ∆ v ∆ h=CVavg ∆ T +v ∆ P Boundary Work W B =Fds= PAds =P ∆ V Polytropic P 2 V 2−P1 V 1 1−n n n P1 V 1 =P2 V 2 W B= Ideal Gas mR ( T 2−T 1 ) where n ≠ 1 1−n Isothermal Ideal Gas where n=1 W B =mRT ln ( Turb :W out =+¿ Comp , P , F :W ¿=−¿ Q́=Ẃ = ḿ ( ∆ h +∆ KE + ∆ PE ) ∆ h=open sys ∆ u=closed sys Conservation of Energy for Steady Flow ( ( ) 1 Q́−Ẃ =m´out h− v 2−gz 2 out 1 2 −ḿ¿ h− v +gz 2 ¿ V 1 A1 V 2 A 2 = ḿ1=m´2= v1 v2 ) Energy Balance for Nozzles and Diffusers Solids and Liquids W B= V́ v One Inlet One Exit Isothermal (T=cons) Z= ∆ U + P ∆ V =Q−W other ∆ H=Q−W other v2 ) v1 First Law for Closed Systems ∆ E =∆ U + ∆ KE + ∆ PE ∆ E=Q¿ +W ¿ −Qout −W out Isochoric (V=constant) ∆ U =Q−W other Isobaric (P=constant) ∆ U =Q− P ∆ V −W other First Law Energy Balance ∆ U =Q− P ∆ V −W other [ Qout=m ( h 1−h2 )+ 1 ( v 21−v22 ) 2 1 2 2 If Q=0 :h2 −h1= ( v 1−v 2 ) 2 IF Q=0∧Ideal Gas : 1 2 2 C p (T 2−T 1) = (V 1 −V 2) 2 ] Heat Engines (Power Plants) W out Q Q =1− out =1− L Q¿ Q¿ QH Q¿ +W ¿=Qout −W out W output =Q¿ −Qout ηth= Refrigerators (Air Conditioners) [ Q QL Q H −1= H −1 = W¿ W ¿ QL QH =W ¿ +Q L CO PR = ] −1 Heat Pump CO P HP= [ QH Q Q =1+ L = 1− L W¿ W¿ QH −1 ] Carnot Cycle (most efficient) Four totally reversible processes 1→ 2 ¿ Isothermal heat addition (T 1 =T 2→ 3 ¿ Isentropic expansion(S2 =S 3) 3 → 4 ¿ Isothermal heat rejection ( T 4 → 1 ¿ Is entropic compression( S 4= W T ηth= net =1− L Q¿ TH Carnot Heat Pump CO P HP= [ ] [ ] QH T = 1− L W¿ TH Carnot Refrigerator T Q CO PR = L = H −1 W¿ T L −1 −1 Approximate Analysis (constant specific heats) A-1/A-2 ∆ s=s 2−s1 ≈C vavg ln ( ) ( ( ) ( T2 T1 +R ln Exact Analysis (variable specific heats) A-17 o o ∆ s=s 2−s1= s2 −s1 −R ln ( ) P2 P1 Single Fluid System h1=h2 If ideal gas :h1=h2 ∧T 1=T 2 Mixing Chamber m1 h1 + m2 h2=m3 h3 m1+ m2 =m3 ∑ m¿ =∑ mout Entropy 2 ∆ s=s 2−s2=∫ 1 ∆ s= dQ T Q To Q ∆ s=s 2−s1= + s gen To ∆ s>0 irreversible ∆ s=0 isentropic ∆ s<0 impossible Change in Entropy T ∆ s=∆ h−v ∆ P Solids and Liquids in contact Q1=−Q2 m1 C 1 ( T 2−T 1 )=m2 C 2 (T 2− T 1 ) where T 2=T 2 Entropy Change of a Pure Substance SM :s = s f + X sfg SCL :s≈ s f @ t Adiabatic + Internally Rev=Isentro Isentropic ≠ Adiabatic + Internally Entropy Change of Ideal Gas W isentropic h2 s−h1 = W actual h2 a−h1 Thermal Efficiency: work out to heat in ηth= W net w net = Q¿ q¿ Three(ish) assumptions: 1) Frictionless 2) Quasi-equilibrium expansion & compression 3) Well insulated pipes 4) Change in KE and PE = 0 Carnot Cycle Four totally reversible processes ∑ ḿ¿ h=∑ m´out h where Q=0 Throttling Valve (Decrease Pressure) v1 T P ∆ s=s 2−s1 ≈C pavg ln 2 −R ln T1 P Guess T 2 , find Cvavg ∨C pavg , calcula Heat Exchanger Systems Q=∑ m´out h−∑ ḿ¿ h Q B → A=m ( h A 2−h A 1 ) Q A → B=m ( h B 2−h B 1 ) Q A → B=−QB → A v2 ηcomp = Isentropic Process ∆ s=0 ; s1=s 2 Ideal Turbine −Ẃ =ḿ∆ h w=h1− h2 Ideal Compressor −Ẃ =ḿ∆ h w=h1− h2 Ideal piston-cylinder ∆ U =Q−W B w b=T ∆ s−∆u Isentropic Ideal Gas ( ) ( ) ( ) K−1 K 2 T2 P = T1 P1 T2 V1 = T1 V2 K−1 P2 V 1 K = P1 V 2 W w= =C p ( T 1−T 2) m Relative Properties (A-17) ( ) ( ) P2 P1 P = r2 Pr 1 ∆ s=0 V2 V1 V = r2 Vr1 ∆ s=0 Reversible Flow/Isentropic Efficiency W rev =v ( P1− P2) −∆ KE−∆ PE Isentropic Turbine Efficiency h −h W actual = 1 2a ηturb = W isentropic h1− h2 s Isentropic Compressor Efficiency 1→ 2 ¿ Isothermal heat addition (T 1 =T 2→ 3 ¿ Isentropic expansion(S2 =S 3) 3 → 4 ¿ Isothermal heat rejection (T 3 = 4 → 1 ¿ Isentropic compression(S 4=S1 W T ηth= net =1− L Q¿ TH Air Standard Assumptions Air as working fluid, closed loop, Ideal Gas All processes are internally reversible 3) Combustion process=heat addition 4) Exhaust process=heat loss 4a) Returns to initial state Cold Air Standard Assumptions 1) Constant pressure specific heats 1a) Evaluated at room temperature (25C or 77F) 1) 2) Reciprocating Engines Top Dead Center (TDC) = Smallest volume Bottom Dead Center (BDC) = Largest volume Compression ratio r= V max V BDC V 1 = = V min V TDC V 2 Cutoff ratio rc= V3 V2 Pressure ratio r p= P2 P1 The Otto Cycle (Ideal Gas Engines) Four internally reversible processes 1→ 2 ¿ Isentropic compression( S1 =S 2 2→ 3 ¿ Constant volume heat add . ( V 2= 3 → 4 ¿ Isentropic expansion( S3 =S 4 ) 4 → 1 ¿Constant volume heat rej .(V 4= *Closed system so use: u* q ¿=u3 −u2=C v ( T 3 −T 2) q out=u4 −u1=C v (T 4−T 1 ) W net q T =1− out =1− 1 Q¿ q¿ T2 1 ηth ,otto =1− K −1 r K−1 T 2 =T 1 r T min=T 1 q T 3 =T 1 r K−1 + ¿ Cv T max=T 3 q ¿=u3 −u2 q out =u4 −u1 w out=u3 −u4 w ¿=u 2−u1 ηth ,otto = T 4− W net q =1− out= 1− T3− Q¿ q¿ 1 ηth ,brayton=1− K −1 rp K ηth ,brayton= K −1 T 2 =T 1 r p K T min=T 1 T 4=T 3 ( r1 ) K−1 K p T max=T 3 The Diesel Cycle (Ideal Diesel Engine) Four internally reversible processes 1→ 2 ¿ Isentropic compression(S1 = 2→ 3 ¿ Constant pres . heat add . (P 3 → 4 ¿ Isentropic expansion( S3 =S 4 → 1 ¿Constant pres .heat rej .(P4 *Closed system so use: u* q ¿=C p( T 3 −T 2 ) q out =C v( T 4 −T 1 ) W q ηth ,diesel = net =1− out Q¿ q¿ r cK −1 K (r c −1) ( r )¿ ηth ,diesel =1− 1 ) K−1 K−1 T 2 =T 1 r T min=T 1 T 3 =T 2 r c T max=T 3 K −1 T 4= T 3 ( rr ) c The Brayton Cycle (Ideal Gas Turbine) Four internally reversible processes 1→ 2 ¿ Isentropic compression∈ pu (S 1=S2 ) 2→ 3 ¿ Constant pressure heat addi ( P2=P3 ) 3 → 4 ¿ Isentropic expansion∈a tur (S 3=S 4) 4 → 1 ¿Constant pressure heat rejec (P4 =P1) q ¿=C p( T 3 −T 2 ) q out =C p (T 4 −T 1 ) Ideal Rankine Cycle (Ideal Vapor Plant) Four internally reversible processes 1→ 2 ¿ Isentropic compression∈ pu (S 1=S2 ) 2→ 3 ¿ Constant pres heat add ∈bo ( P 2= P 3 ) 3 → 4 ¿ Isentropic expansion∈a tur (S 3=S 4) 4 → 1 ¿Constant pres heat rej∈con (P4 =P1) *each component treated as open system: h* pump,∈¿= h2−h1 =v (P2− P 1) W¿ boiler ,∈¿=h 3−h2 q¿ W turb , out=h 3−h4 q cond ,out =h4 −h1 h 4−h 1 W q ηth= net =1− out =1− h3−h2 Q¿ q¿ h2= h1+ w p =h1 + v ( P2− P1 ) W net =m ( wt −w p) =m (h 3−h4 −w p ) Real Rankine Cycle Pumps and turbines have losses W s hs 2−h 1 = W a h2 a − h1 h3−h 4 a W ηt = a = W s h3=h4 s η p= Refrigerators and Heat Pumps *use h* ηcomp = Refrigerator h2 s−h1 h 2a −h1 Cooling QL = W¿ Work QL =m(h 1−h4 ) QH =m(h2 a−h 3) W ¿ =m(h2 a− h1) P2 s=P2 a s 2 s=s1 h −h h2 a = h1 + 2 s 1 h4 = h3 ηc CO PR = Heat Pump Heating QH = Work W¿ QH =m(h2−h 3) QL =m(h 1−h4 ) h4 =h3 P2 s=P2 s 2 s=s1 P3=P2 CO P HP= Reversed Carnot Cycle (Ideal Ref & HP) 1 TH −1 TL 1 CO P HP ,Carnot = TL 1− TH CO PR ,Carnot = ( ) ( ) Ideal Vapor Compression Ref Cycle 1→ 2 ¿ Isentropic comp∈comp (S 1=S2 ) 2→ 3 ¿ Constant pres . heat rej .∈cond ( P2=P3 ) 3 → 4 ¿ Throttling∈expansion device (h3=h4 ) 4 → 1 ¿Constant pres .heat add .∈evap ( P4 =P1) h1−h4 q CO PR = L = W netin h 2−h1 q h −h CO P HP= H = 2 3 W netin h2− h1 q L =h1− h4 q H =h 2−h3 w netin=h 2−h1 h4 =h3 s 2 s=s1

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