上游充通大￥ SHANGHAI JIAO TONG UNIVERSITY a Engineering Thermodynamics I Lecture 47 Chapter 9 Gas Power Systems (section 9.10) Spring,5/15/2018 强 Prof.,Dr.Yonghua HUANG VVAMA http://cc.sjtu.edu.cn/G2S/site/thermo.html SHANG 1日 ERSITY
Engineering Thermodynamics I Lecture 47 Spring, 5/15/2018 Prof., Dr. Yonghua HUANG Chapter 9 Gas Power Systems (section 9.10) http://cc.sjtu.edu.cn/G2S/site/thermo.html
Non-adiabatic compression 2 Assumptions: SSSF A) -△KE=△PE=0 PiV"=P2V2 -0-w+学e-a++g dt 0=Qc-Wc+m(h-h2) →wc-qc=h1-h2 p 2c For an internally reversible,steady flow process: P2 2 cooling wc=Jvdp adiabatic Note:Compressor cooling decreases v, therefore also decreases wc! →Try to exchange heat P1 from the compressor V to environment! 上游充通大率 May15,2018 2 SHANGHA BAO TONG LINIERSITY
May 15, 2018 2 Non ‐adiabatic compression 2 2 CV in out in out in out C C 12 CC12 v v 2 2 dE Q W m h gz m h gz dt 0 Q W mh h w q hh Assumptions: ‐ SSSF ‐ KE = PE = 0 1 2 . W C . Q C CV v 1 2s p p 2 p 1 2c cooling adiabatic For an internally reversible, steady flow process: 2 C 1 w vdp Note: Compressor cooling decreases v, therefore also decreases w C! Try to exchange heat from the compressor to environment! n n 11 2 2 p v pv
Continue Non-Adiabatic Compression Limiting Cases: 1.Adiabatic: S2=S1 (no heat transfer) 2.Isothermal: T2=T(infinite heat transfer) 》 For an adiabatic,reversible process with an ideal gas: 2t22s p---g where k=C 1<n<k pyk=constant Cv For an isothermal process with an ideal gas: (isentropic) n=1 N pv constant (isothermal) 》 Between the limits of isentropic and p---- isothermal compression,we can assume a polytropic process that satisfies: py"constant,where n is the polytropic exponent 上游充通大粤 May15,2018 3 SHANGHAI BAO TONG LINIERSITY
May 15, 2018 3 Continue Non ‐Adiabatic Compression Limiting Cases: 1. Adiabatic: s 2 = s 1 (no heat transfer) 2. Isothermal: T2 = T1 (infinite heat transfer) » For an adiabatic, reversible process with an ideal gas: » For an isothermal process with an ideal gas: » Between the limits of isentropic and isothermal compression, we can assume a polytropic process that satisfies: k p v c pv constant where k c pv constant n pv constant , where n is the polytropic exponent v 1 2s p p 2 p 1 1 < n < k 2t 2 n = k (isentropic) n = 1 (isothermal)
Continue Non-Adiabatic Compression For any internally reversible process:w=-vdp For a polytropic process n1 -n n≠1 For an ideal gas “=nR(c,-T) n n≠1 国 For an isothermal process n=1 Wc=-PiVi In(p2/P)=-piVi In(Vi/V2) n=1 For an ideal gas We=-RTIn(p2/p) n=1 上游充通大率 May15,2018 4 SHANGHA BAO TONG LINIERSITY
May 15, 2018 4 Continue Non ‐Adiabatic Compression For any internally reversible process: For a polytropic process For an isothermal process ܹୡ ൌ െ ݀ݒ න ଶ ଵ C 2 2 11 n w pv pv n 1 n 1 For an ideal gas C 21 n w RT T n 1 n 1 n 1 w p v ln(p / p ) p v ln(v / v ) n 1 C 11 2 1 11 1 2 For an ideal gas w RT ln p / p n 1 C 21 n 1
Two-stage Compression with Intercooling 2 Intercooler / T P2 P2 2s Px Pi 2s T Ti 1 1 2t xt=y S S 上游充通大率 May15,2018 5 SHANGHA BAO TONG LINIERSITY
May 15, 2018 5 Two ‐stage Compression with Intercooling s T 1 xt=y 2 2s x xs 2t T 1 p 2 p x p 1 qI air x Intercooler w c 2 y 1 1 1 2 . W C . Q C CV s T 1 2 2s T 1 p 2 p 1
Continue Two-stage Compression with Intercooling Savings in work 2t2 2s P2 Isentropic compression pvk const Px _Isothermal_ compression Polytropic pv const compression pvh const Pi Limits:px=p2>savings=0 →An optimum exits! px=p1→savings=0 上游文道大学 May15,2018 6 SHANGHA BAO TONG LINIVERSITY
May 15, 2018 6 Continue Two ‐stage Compression with Intercooling Limits: p x = p2 savings = 0 p x = p 1 savings = 0 An optimum exits! v p p 2 2t Isothermal compression pv = const p 1 p x 2 2s Polytropic compression pv n = const 1 y x Isentropic compression pv k = const Savings in work
Continue Two-stage Compression with Intercooling Find intermediate pressure px to minimize work input: ·For an ideal gas: p Savings in work 2t 2s P2 Isentropic compression we-mR(-)mR() pvk =const Isothermal compression Polytropic pv const compression -小{- pva=const Wc- n-1 For an ideal intercooler:T=T (ambient) Py =Px (no pressure drop) 上游充通大率 May15,2018 7 HANGHAI HAO TONG LINIVERSITY
May 15, 2018 7 Continue Two ‐stage Compression with Intercooling Find intermediate pressure p x to minimize work input: • For an ideal gas: C x1 2y 1x 2 y C 1 y n 1 n 1 n n 1x 2 y C 1 y y 1 y x nR nR w TT TT n1 n1 nRT T T nRT w 11 n1T n1T nRT p p nRT w 11 n1 p n1 p For an ideal intercooler: T T (ambient) p p (no pressure drop) v p p2 2t Isothermal compression pv = const p1 px 2 2s Polytropic compression pvn = const 1 y x Isentropic compression pvk = const Savings in work
Continue Two-stage Compression with Intercooling n-l n-1 n-1 n- nRT n Wc= 2 n Minimum occurs at: dwc=0 for fixed Tand P2 dpx n-1 n-l dwc=0= RT n-1 Px n 1+ -1 n P dpx n-1 n Pu p =0 上游充通大率 May15,2018 8 SHANGHAI BAO TONG LINIVERSITY
May 15, 2018 8 Continue Two ‐stage Compression with Intercooling n1 n1 n n 1x 12 C 1 x n1 n1 n n 1x 2 C 1 x C 11 2 x n 1 C 1 x x 1 nRT p nRT p w 11 n1 p n1 p nRT p p w 2 n1 p p dw Minimum occurs at: 0 for fixed T ,p , and p : dp dw nRT p n 1 0 dp n 1 n p n 1 1 1 n n 2 2 2 1x x 0 1 n1 p p p np p
Continue two-stage compression with intercooling n-l n n-1 n p P2 P2 2 P 1\p Px P p n- n- n → P2=Px → p. Px P Px=PiP2 p 上游充通大率 May15,2018 9 SHANGHA BAO TONG LINIVERSITY
May 15, 2018 9 Continue two ‐stage compression with intercooling n1 n n1 n nn nn x 22 2 11 xx 1 1 n n x 22 2 1 1x x 1 1 n n 2 2x x x x1 1 n1 n1 n n 2 x 2x x 1 x1 n 1p 1 n 1p p 0 np p np p p 1p p 0 p pp p p pp p p pp p p p pp p p pp x 12 p p p
Continue two-stage compression with intercooling n-l n-l For dwc=0(minimum work): dpx T n-1 Pt Savings in work 2s P2 P2 n Isentropic Also, compression pvk=const P Ty T Px _Isothermal compression Polytropic pv=const compression Thus,T2=Tx and Ty =T: pva const we-,.-T)-T)=w >At optimal intermediate pressure,equal work in both stages! 上游充通大率 May15,2018 10 SHANGHAI BAO TONG LINIVERSITY
May 15, 2018 10 Continue two ‐stage compression with intercooling n1 n1 n n C xx 2 x 11 x n 1 n 2 2x x y1 2x y1 C,1 x 1 2 y C,2 dw Tp p For 0 (minimum work): dp T p p p TT Also, p TT Thus, T T and T T : nR nR w TT TT w n1 n1 At optimal intermediate pressure, equal work in both stages!
