上游充通大￥ SHANGHAI JIAO TONG UNIVERSITY Engineering Thermodynamics I Lecture 46 Chapter 9 Gas Power Cycles Spring,5/15/2018 Prof.,Dr.Yonghua HUANG 强 LA http://cc.sjtu.edu.cn/G2S/site/thermo.html SHANG 1日 ERSITY
Engineering Thermodynamics I Lecture 46 Spring, 5/15/2018 Prof., Dr. Yonghua HUANG Chapter 9 Gas Power Cycles http://cc.sjtu.edu.cn/G2S/site/thermo.html
Diesel and Dual Cycles Air Standard Diesel Cycle for CI Engines: k p 3 s=const Qin p=const 4 v=const s=const Qout 1 V1=V4 ---。。-1 S1=S2 S3=S4 Define:ir ="cutoff ratio"compression ratio r= V2 上游充通大率 May15,2018 2 SHANGHA BAO TONG LINIERSITY
May 15, 2018 2 Diesel and Dual Cycles Air Standard Diesel Cycle for CI Engines: qout s T 1 2 4 3 s 1=s 2 s 3=s 4 qin qout V p 1 2 4 3 V 1=V 4 qin V 2 V 3 3 1 BDC c 2 2 TDC V V V Define : r " " compression ratio r V V cutoff ratio V
Continue Diesel Cycle qin s=const Diesel Cycle Energy Balances: const ·Assuming: V, V3 ·w=0 only during heat transfer4→1；AKE=APE=0 Air is ideal gas,constant specific heats du=0 Wnet W23 W34-W12=qin -lqoutl . during process2→3，wist0(change in volume) qin-W23=(u3-u2) 今qin=(u3-u2)+W23=(u3-u2)+p23(V3-V2) =h3-h2 dout u4 -u1 上游充通大 May15,2018 3 SHANGHAI BAO TONG LINIERSITY
May 15, 2018 3 Continue Diesel Cycle Diesel Cycle Energy Balances: • Assuming: • w = 0 only during heat transfer 4 1; KE = PE = 0 • Air is ideal gas, constant specific heats wnet = w23 + w34 – |w12| = qin – |qout| • during process 2 3, w is ≠0 (change in volume) qin – w23= (u 3 – u 2) qin = (u 3 – u 2) + w23 = (u 3 – u 2) + p23 (v 3 – v 2) = h 3 – h 2 qout = u 4 – u 1 d 0 u V p 1 2 4 3 V1=V4 qin V2 V3
Continue Diesel Cycle Tt p=const 3 Diesel cycle efficiency: v=const For m-constant:=V3 and r=VL V2 V2 Qout n%-w-99-19-1-u=u=1-cc,-p S1=S2 3S4S Qin Qin Qin h3-h2 c(T3-T2) or1→2是-（) =rk-l For a constant pressure process2-→3：p2=P3→ → V2 V3 T V2 上游充通大率 May15,2018 4 SHANGHAI BAO TONG LINIVERSITY
May 15, 2018 4 Continue Diesel Cycle Diesel cycle efficiency: 3 1 c 2 2 v v For m=constant: r and r v v s T 1 2 4 3 s1=s2 s3=s 4 qin qout net v 4 1 in out out 4 1 th in in in 3 2 p 3 2 w c (T T ) qq q u u 11 1 q q q h h c (T T ) 4 1 3 2 41 1 th 32 2 T T T T 1 (T T ) T 1 1 k(T T ) T k 1 k 1 2 1 k 1 1 2 T v For isentropic process 1 2: r T v 2 3 33 23 c 2 3 22 RT RT T v For a constant pressure process 2 3: p p r v v Tv
Continue Diesel Cycle s=const Continue Diesel cycle efficiency: s=const For isentropic process3-→4： 1V2 V3 Vi=V,V k- Then, 1-1 =1- k。-) (const.k) ≥1 Thus,for a given r:nth.Diesel snth.ot! 上游充通大率 May15,2018 5 SHANGHAI BAO TONG LINIERSITY
May 15, 2018 5 Continue Diesel Cycle Continue Diesel cycle efficiency: k 1 k1 k1 k1 2 k 1 2 3 4 1 1 1 22 k1 k1 4 3 3 3 3 13 k1 k1 k 4 33 33 3 k c 1 22 22 2 For isentropic process 3 4: T V T V V V T TV T V V V V TV T TV VV V r T TV VV V k 1 2 1 1 2 T v T v (const. k) 4 1 3 2 k c k 1 th 2 1 c 1 T T T T 1 T Then, 1 T k 1 1 r 1 1 r k(r 1) V p 1 2 4 3 V1=V4 qin V2 V3 Thus, for a given r : ! th,Diesel th,Otto
Continue Diesel Cycle Continue Diesel cycle efficiency: T v=const 3piesel Qin 2 p=const 4 v=const for a given r:mth.Diesel t.o! I S1=S2 S3=S4 However,Diesel cycle typically operates at higher r! 上游充通大率 May15,2018 6 SHANGHA BAO TONG LINIVERSITY
May 15, 2018 6 Continue Diesel Cycle Continue Diesel cycle efficiency: s T 1 2 4 s 1=s 2 s 3=s 4 qin qout 3Diesel 3Otto th,Diesel th,Otto for a given r : ! However, Diesel cycle typically operates at higher r!
Characteristics of Four Stroke Compression Ignition Spark Ignition Engines Characteristics Compression-Ignition Engine Spark-Ignition Engine Compression Ratio 14-22:1 5-8:1 Ignition Compression Electric Spark Thermal Efficiency 30-60% 25-30% Fuel induction Injector Carburettor (Fuel Injection) Fuel System Fuel Oil/Diesel Gasoline (LP gas) Fire Hazard Less Greater Power Variation Increase in Fuel Increase in Air/Fuel Mixture Air Induction Constant Variable (Throttle Airflow) Air-Fuel Ratio 15-100:1 10-20:1 Relative Fuel Consumption Lower Higher Energy per litre of fuel Higher Lower Manifold Throttle Absent Present Exhaust Gas Temperature 482°C/900F 704°C/1300F Starting Harder Easier Lubricants Heavy duty oils Regular and Premium Oils Speed Range Limited (600-3200 rpm) Wide range (400-6000 rpm) Engine Mass per Horsepower 8kg(17.51b) Average 4 kg (9 1b) Initial Cost 五igh Much Lower Lugging ability (Torque) Excellent ess 上游充通大粤 May15,2018 SHANGHAI BAO TONG LINIVERSITY
May 15, 2018 7 Characteristics of Four Stroke Compression Ignition & Spark Ignition Engines
Example 46.1 for Diesel Cycle Given: ·Ideal Diesel cycle: r=18,re=2,p1=0.1MPa,T1=300K ④Find: ● Pressures,temperatures and specific volumes at all state points Heat input,net work output and thermal efficiency ·MEP Assumption: (1)air-standard assumptions (2)△KE=0,△PE=0 上游充通大 May15,2018 8 HANGHAI HAO TONG LINIVERSITY
May 15, 2018 8 Example 46.1 for Diesel Cycle Given: • Ideal Diesel cycle: r = 18, r c = 2, p 1 = 0.1 MPa, T1 = 300 K Find: • Pressures, temperatures and specific volumes at all state points • Heat input, net work output and thermal efficiency • MEP Assumption: (1) air‐standard assumptions (2) KE=0, PE=0
Continue Example 46.1 Solution: P 3 3 V3-2 Io= 0 S=c p=c S=C =18 U=C P1=0.1 MPa T1=300K U Assumptions: 1.The air in the piston-cylinder assembly is the closed system. 2.The compression and expansion processes are adiabatic. 3.All processes are internally reversible. 4.The air is modeled as an ideal gas. 5.Kinetic and potential energy effects are negligible. 上游充通大￥ May15,2018 9 SHANGHA BAO TONG LINIERSITY
May 15, 2018 9 Continue Example 46.1 Solution: Assumptions: 1. The air in the piston–cylinder assembly is the closed system. 2. The compression and expansion processes are adiabatic. 3. All processes are internally reversible. 4. The air is modeled as an ideal gas. 5. Kinetic and potential energy effects are negligible
Continue Example 46.1 Tab.A-17 ©T1=300K,p1=1atm→u1=214.07k/kg,Vr1=621.2 T=300K ©Process1-2: 621.2 v2二V,n =34.51 18 T2=898.3K h2=930.98k/kg Tab.A-17 Ty=0.) EoS P2=PIT]V: 898.3 300 (18)=5.39MPa ©Process2-3: T3= ,=3=2898.3)=1766K V2 h3=1999.1k/kg Tab.A-17 Vr3=3.97 Process 3-4: (S U3= V2V- 3u4=664.3k/kg Tab.A-17 T4=887.7 Process 4-1: EoS T(0.1 MPa) 887.7K 300K/ =0.3 MPa 上游充通大率 May15,2018 10 SHANGHAI BAO TONG LINIVERSITY
May 15, 2018 10 Continue Example 46.1 T 1 = 300 K, p 1=1atm u 1= 214.07kJ/kg, vr1=621.2 Process 1‐2: Process 2‐3: Process 3‐4: Process 4‐1: Tab. A‐17 S Tab. A‐17 T 2=898.3 K h 2 = 930.98 kJ/kg EoS p Tab. A‐17 h 3=1999.1 kJ/kg vr3 = 3.97 S u 4=664.3 kJ/kg T 4 = 887.7 Tab. A‐17 EoS