Basic Practical Diesel Cycle

Basic Practical diesel cycle define some units 3 kN := 10 ⋅N The textbook Diesel cycle is represented by all heat addition at constant pressure. The Otto cycle which is implemented by the spark ignition internal combustion engine adds all heat at constant volume. We will model a combined or dual (Seiliger) cycle with a portion of the heat added at constant volume, the remainder at constant pressure. Setting some parameters to be defined = 1 will reduce to either the Otto or Diesel cycle. 6 3 kPa := 10 ⋅Pa 3 MPa := 10 Pa kJ := 10 ⋅J 3 kmol := 10 molebar := 0.1MPa This model will use an ideal air standard cycle with air as an ideal gas with constant specific heats and reversible processes to represent the behavior. The gas relationships are useful. air-standard cycles ... 1. air as ideal gas is working fluid throughout cycle - no inlet or exhaust process 2. combustion process replaced by heat transfer process 3. cycle is completed by heat transfer to surroundings 4. all processes internally reversible 5. usually constant specific heat (page 311) basic practical diesel cycle Assumptions for analysis ... 1. reversible cycle with all reversible processes 2. working fluid is air assumed to be a perfect gas with constant specific heats, γ = cp /cv = 1.4 3. mass of air in cylinder remains constant 4. combustion processes are represented by heat transfer from an external source. Constant volume or constant pressure pocesses are done. 5. cycle is completed by cooling heat transfer to the surroundings until the air temperature and pressure return to the initial conditions of the cycle (constant volume process). 1-2 isentropic compression 2-3 constant volume heat addition 3-4 constant pressure heat addition 4-5 isentropic expansion 5-1 constant volume cooling data for plot this is the shape ... p-v Dual Cycle Temperature (K) pressure (bar) T-s Dual Cycle entropy (relative to s1) volume (m^3/kg) next we will put numbers on the plots => themodynamic analysis of dual (Seiliger) cycle 10/24/2006 1 The original notes are sourced from VanWylen and Sonntag. They could be revised to use the form of some of the relationships from Woud, but at considerable effort. Rather what follows is the application of the equations developed in the gas relationships lecture applied to the combined air-standard cycle deriving the relationships summarized in Table 7.3 Analytical prediction of the Selinger process on page 245 of the text. rc , a, b v1 rc = v2 a= p3 b= p2 v4 v3 stage 1-2 isentropic adiabatic compression (expansion) _______________________________ volume ratio known γ γ ⎛ v initial ⎞ ⎛ v1 ⎞ γ rc = = p final = p initial⋅ ⎜ = ⎜ ⎟ = rc ⎟ v final v2 ⎝ vfinal ⎠ ⎝ v2 ⎠ v initial v1 v1 p final = rc v2 p initial = rc ⎛ v initial ⎞ Tfinal = Tinitial⋅ ⎜ ⎟ ⎝ vfinal ⎠ Tfinal γ Tinitial = rc γ−1 ⎛ v1 ⎞ =⎜ ⎟ ⎝ v2 ⎠ γ−1 = rc γ −1 stage 2-3 heat transfer at constant volume .... ________________________________________ p3 p2 =a v1 p initial = R⋅ Tinitial v constant⋅100 p3 =1 v2 p final = R⋅ Tfinal Tfinal v constant⋅100 Tinitial Tfinal =a p2 Tinitial = p final p initial = p3 p2 =a =a stage 3-4 heat transfer at constant pressure .... ________________________________________ b= v4 v3 v final v initial = v4 v3 v initial = R⋅ Tinitial p constant⋅100 p4 =b p3 v final = R⋅ Tfinal T4 p constant⋅100 T4 =1 T3 T3 = Tfinal Tinitial = v final v initial =b =b stage 4-5 isentropic adiabatic compression (expansion) _______________________________ volume ratio known v5 v4 = v5 v3 v5 v3 v1 v3 rc ⋅ = ⋅ = ⋅ = v3 v4 v3 v4 v2 v4 b γ as v5 = v1 and v 2 = v3 γ γ ⎛ v initial ⎞ ⎡ ⎛ v 4 ⎞⎤ 1 ⎞ p final = p initial⋅ ⎜ = ⎢p 4 ⋅ ⎜ ⎟⎥ = p 4 ⋅ ⎛ ⎟ ⎜ r ⎟ = p5 ⎝ vfinal ⎠ ⎣ ⎝ v 5 ⎠⎦ ⎜ c⎟ ⎝ b ⎠ 10/24/2006 2 [W 7.68 & 7.69] γ−1 ⎛ v initial ⎞ Tfinal = Tinitial⋅ ⎜ ⎟ ⎝ vfinal ⎠ v5 = v4 γ−1 ⎛ v4 ⎞ = T4 ⋅ ⎜ ⎟ ⎝ v5 ⎠ ⎛ rc ⎞ =⎜ ⎟ p5 ⎝ b ⎠ rc p4 b γ−1 1 = T4 ⋅ ⎛ ⎜ r c ⎜ ⎝ b ⎞ ⎟ ⎟ ⎠ γ−1 = T5 γ ⎛ rc ⎞ =⎜ ⎟ T5 ⎝ b ⎠ T4 γ−1 N.B. ratios are inconsistent 5/4 ... 4/5 stage 5-1 heat transfer at constant volume .... ________________________________________ v5 v1 =1 p5 p1 v5 v1 R⋅ Tinitial p initial = v constant⋅100 p final = R⋅ Tfinal p initial so ... v constant⋅100 p final = Tinitial Tfinal = p5 p1 T5 = T1 ⎛ p4 = p3 ⎞ p5 p3 p2 1 γ γ ⋅a⋅rc = a⋅ b ⎟= ⋅ ⋅ = p1 γ p4 p4 p2 p1 ⎝ ⎠ ⎛ rc ⎞ ⎜ ⎟ ⎝b⎠ p5 = ⋅⎜ p5 =1 p1 = a⋅ b T5 γ T1 = a⋅ b γ this one is initial/final Now, applying the gas relationships to the calculation of states around the air-standard combined cycle γ := 1.4 constants ... cv := 0.7165 kJ cp := 1.0035 kg⋅ K kJ R := 0.287 kg⋅ K kJ kg⋅ K given ... T1 , v 1 (calc) , s1 , p 1 , rv , rp , rc kJ s1 := 1 kg T1 := 295K p 1 := 1bar rv := 12.5 rp := 1.38 rc := 1.86 v 1 := R ⋅ T1 3 v 1 = 0.847 p1 m kg rv = compression ratio rc in text rp = pressure ratio during constant volume heat addition = a in text rc = cut-off ratio. portion of stroke during which constant pressure heat addition occurs = b in text v1 rv = = rc v2 p3 rp = =a p2 v4 rc = =b v3 1-2 isentropic compression of air γ −1 ⎛ p2 ⎞ =⎜ ⎟ T1 ⎝ p1 ⎠ T2 γ ⎛ v1 ⎞ =⎜ ⎟ ⎝ v2 ⎠ s2 := s1 v 2 := γ −1 v1 rv (7.35) this .. for reversible adiabatic process ⎛ v1 ⎞ T2 := T1 ⋅ ⎜ ⎟ ⎝ v2 ⎠ γ−1 p 2 := 3 v 2 = 0.068 10/24/2006 m kg T2 = 810.188 K p 2 = 34.33 bar 3 R ⋅ T2 v2 ⎛ v1 ⎞ =⎜ ⎟ T1 ⎝ v2 ⎠ T2 γ−1 ⎛ v1 ⎞ p2 = p1⋅ ⎜ ⎟ ⎝ v2 ⎠ γ later we will plot on T-s and p-v so the relationships for intermediate states is shown. Any state value can serve as the plot parameter, but we will use temperature. T1 ≤ T_plot ≤ T2 T2 T1 = ⎛ v1 ⎞ ⎜ ⎟ ⎝ v2 ⎠ s = s1 = s2 = constant γ−1 T2 and ... T1 = γ−1 1 γ γ−1 ⎛ p2 ⎞ ⎜ ⎟ ⎝ p1 ⎠ so ... ⎛ T1 ⎞ v_plot = v 1 ⋅ ⎜ ⎟ T_plot ⎝ γ and ... ⎠ T_plot ⎞ p_plot = p 1 ⋅ ⎛⎜ ⎟ ⎝ T1 ⎠ γ−1 2-3 constant volume heat addition using rp during constant volume portion of heat addition ... v 3 := v 2 p 3 := p 2 ⋅ rp p ⋅ v = R⋅ T (3.2) p 3 = 47.375 bar p1⋅ v1 T1 ⎛ T2 ⎞ ⎛ v2 ⎞ s2 − s1 = cvo⋅ ln⎜ + R ⋅ ln ⎟ ⎜ ⎟ ⎝ T1 ⎠ ⎝ v1 ⎠ = p2⋅ v2 p3 (3.5) T2 (7.21) v2 = v3 need to calculate T3 T3 = p3 T3 := T2 ⋅ p2 p2 T2 ⎛ ⎛ T3 ⎞ ⎞ s3 := ⎜ s2 + cv ⋅ ln⎜ ⎟ ⋅K⎟ ⎝ ⎝ T2 ⎠ ⎠ cvo = constant 3 T3 = 1.118 × 10 K kJ s3 = 1.231 kg for later plotting T2 ≤ T_plot ≤ T3 T_plot ⎞ s_plot = s2 + cv ⋅ ln⎛⎜ ⋅K T2 ⎟ ⎝ ⎠ p-v are end points v = constant (straight lines) although intermediate states would be determined from the state equation ... p_plot = R⋅ T_plot v2 3-4 heat added at constant pressure with rc v 4 := v 3 ⋅ rc p 4 := p 3 T4 := p4⋅ v4 R 3 v 4 = 0.126 m kg p 4 = 47.375 bar ⎛ T2 ⎞ ⎛ p2 ⎞ s2 − s1 = cpo⋅ ln⎜ ⎟ − R⋅ ln⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ for later plotting T3 ≤ T_plot ≤ T4 ⎛ v5 ⎞ =⎜ ⎟ T5 ⎝ v4 ⎠ γ−1 cpo = constant T_plot ⎞ s_plot = s3 + cp ⋅ ln⎛⎜ ⋅K T2 ⎟ ⎝ ⎠ 4-5 isentropic expansion T4 (7.23) 3 T4 = 2.08 × 10 K v 5 := v 1 ⎛ v4 ⎞ T5 := T4 ⋅ ⎜ ⎟ ⎝ v5 ⎠ ⎛ ⎛ T4 ⎞ ⎞ s4 := ⎜ s3 + cp ⋅ ln⎜ ⎟ ⋅K⎟ T3 ⎝ ⎝ ⎠ ⎠ p-v are end points v = constant (straight lines) although intermediate states would be determined from the state equation ... s5 := s4 γ−1 10/24/2006 decreasing ... T5 = 970.553 K p 5 := s = s4 = s5 = constant 4 v_plot = kJ s5 = 1.854 kg as with isentropic compression above using T as the plot parameter ... T4 ≥ T_plot ≥ T5 kJ s4 = 1.854 kg R ⋅ T5 v5 p 5 = 3.29 bar R⋅ T_plot p2 T5 T4 = ⎛ v4 ⎞ ⎜ ⎟ ⎝ v5 ⎠ γ−1 T5 and ... T4 = ⎛ p5 ⎞ ⎜ ⎟ ⎝ p4 ⎠ 5-1 constant volume cooling and for later plotting ... T5 ≥ T_plot ≥ T1 γ −1 1 γ γ γ−1 γ −1 so ... ⎛ T4 ⎞ ⎟ ⎝ T_plot ⎠ v_plot = v 4 ⋅ ⎜ ⎛ ⎛ T1 ⎞ ⎞ s1 := ⎜ s5 + cv ⋅ ln⎜ ⎟ ⋅K⎟ ⎝ ⎝ T5 ⎠ ⎠ T_plot ⎞ s_plot = s5 + cv ⋅ ln⎛⎜ ⎟ ⋅K ⎝ T5 ⎠ ⎛ s1 ⎞ ⎜ ⎟ ⎛ 1 ⎞ ⎜ s2 ⎟ ⎜ 1 ⎟ ⎜s ⎟ ⎜ ⎟ ⎜ 3 ⎟ = ⎜ 1.231 ⎟ kJ ⎜ s4 ⎟ ⎜ 1.854 ⎟ kg ⎜ ⎟ ⎜ 1.854 ⎟ ⎜ s5 ⎟ ⎜ ⎟ ⎜s ⎟ ⎝ 1 ⎠ ⎝ 1⎠ check closure of s1 ⎛ p1 ⎞ ⎜ ⎟ ⎛ 1 ⎞ ⎜ p2 ⎟ ⎜ 34.33 ⎟ ⎜p ⎟ ⎜ ⎟ ⎜ 3 ⎟ = ⎜ 47.375 ⎟ bar ⎜ p4 ⎟ ⎜ 47.375 ⎟ ⎜ ⎟ ⎜ 3.29 ⎟ ⎜ p5 ⎟ ⎜ ⎟ ⎜p ⎟ ⎝ 1 ⎠ ⎝ 1⎠ T_plot ⎞ ⎟ ⎝ T4 ⎠ kJ s1 = 1 kg p-v are end points v = constant (straight lines) although intermediate states would be determined from the state equation ... decreasing ... ⎛ T1 ⎞ ⎜ ⎟ ⎛ 295 ⎞ ⎜ T2 ⎟ ⎜ 810 ⎟ ⎜T ⎟ ⎜ ⎟ ⎜ 3 ⎟ = ⎜ 1118 ⎟ K ⎜ T4 ⎟ ⎜ 2080 ⎟ ⎜ ⎟ ⎜ 971 ⎟ ⎜ T5 ⎟ ⎜ ⎟ ⎜ T ⎟ ⎝ 295 ⎠ ⎝ 1⎠ p_plot = p 4 ⋅ ⎛⎜ and ... p_plot = ⎛ v1 ⎞ ⎜ ⎟ ⎛ 0.847 ⎞ ⎜ v2 ⎟ ⎜ 0.068 ⎟ ⎜v ⎟ ⎜ ⎟ 3 ⎜ 3 ⎟ = ⎜ 0.068 ⎟ m ⎜ v4 ⎟ ⎜ 0.126 ⎟ kg ⎜ ⎟ ⎜ 0.847 ⎟ ⎜ v5 ⎟ ⎜ ⎟ ⎜ v ⎟ ⎝ 0.847 ⎠ ⎝ 1⎠ now for plotting, including the intermediate values ... details in area below, relationships developed above parameterization of T-s, p-v T-s Dual Cycle T3 -s3 shown p-v Dual Cycle p2 v2 shown 60 pressure (bar) Temperature (K) 3000 2000 1000 0 1 1.5 entropy (relative to s1) 10/24/2006 40 20 0 2 0 0.5 volume (m^3/kg) 5 R⋅ T_plot 1 v2 now for calculations of efficiency v1 r = compression ratio rv = v2 v p3 T3 r = pressure ratio during constant volume heat addition rp = at constant volume (ideal gas law pv=RT) = p2 T2 p v4 T4 at constant r = cut-off ratio. portion of stroke during which constant pressure heat addition occurs rc = = v3 T3 c pressure (ideal gas law pv=RT) 1-2 isentropic compression of air γ−1 γ ⎛ p2 ⎞ =⎜ ⎟ T1 ⎝ p1 ⎠ T2 ⎛ v1 ⎞ =⎜ ⎟ ⎝ v2 ⎠ γ−1 ⎛ v1 ⎞ =⎜ ⎟ T1 ⎝ v2 ⎠ T2 this .. for reversible adiabatic process (7.35) γ−1 = rv γ−1 γ ⎛ v1 ⎞ γ p 2 = p 1 ⋅ ⎜ ⎟ = p 1 ⋅ rv ⎝ v2 ⎠ 2-3 constant volume heat addition using rp during constant volume portion of heat addition ... ⎛ T3 ⎞ QH1 = m⋅ cv ⋅ T3 − T2 = m⋅ cv ⋅T2 ⋅ ⎜ − 1⎟ = m⋅ cv ⋅T2 ⋅ rp − 1 ⎝ T2 ⎠ ( ) ( ) 3-4 heat added at constant pressure with rc T3 ⎛ T4 ⎞ QH2 = m⋅ cp ⋅ T4 − T3 = m⋅ cp ⋅T3 ⋅ ⎜ − 1⎟ = m⋅ cp ⋅T3 ⋅ rc − 1 = m⋅ cp ⋅T2 ⋅ ⋅ r − 1 = m⋅γ⋅cv.⋅T2 ⋅rp ⋅ rc − 1 T2 c ⎝ T3 ⎠ ( ) ( ) ( ) ( substituting γ = cp /cv and rp = T3 /T 2 5-1 constant volume cooling ⎞ ⎛ T5 QL = −m⋅cv ⋅ T5 − T1 = −m⋅cv ⋅T1 ⋅ ⎜ − 1⎟ ⎝ T1 ⎠ ( T5 ) p5 p5 p4 p3 p2 = = ⋅ ⋅ ⋅ = T1 p1 p4 p3 p2 p1 γ γ γ ⎛ v4 ⎞ ⎛ v1 ⎞ ⎛ v4 ⎞ γ ⋅1⋅r ⋅ ⎜ ⎟ p ⎜ v ⎟ = ⎜ v ⎟ ⋅rp = rp ⋅rc v5 ⎝ ⎠ ⎝ 2⎠ ⎝ 3⎠ QL = −m⋅cv ⋅T1 ⋅ ⎛ rp ⋅rc − 1⎞ ⎝ ⎠ γ => /\ as v5 = v1 and v 2 = v3 combining these for thermal efficiency of the cycle ... η th = 1 + QL QH =1− ( T1 ⋅ ⎛ rp ⋅rc − 1⎞ ⎝ γ ) ( ⎠ γ ) T2 ⋅ ⎡ rp − 1 + γ ⋅ rp ⋅ rc − 1 ⎤ ⎣ ⎦ rp ⋅rc − 1 =1− rv γ−1 ( γ rv 10/24/2006 γ−1 ( ) ( ) rv = 12.5 rp = 1.38 rp ⋅rc − 1 η th := 1 − ) ⋅ ⎡ rp − 1 + γ⋅rp ⋅ rc − 1 ⎤ ⎣ ⎦ ( ) ⋅ ⎡ rp − 1 + γ⋅rp ⋅ rc − 1 ⎤ ⎣ ⎦ η th = 0.592 6 rc = 1.86 ) and we could calculate the work per cycle ( ) ( ) ( W = QH1 + QH2 + QL = m⋅ cv ⋅ T3 − T2 + m⋅ cp ⋅ T4 − T3 − m⋅ cv ⋅ T5 − T1 m := 1 for per unit mass calculation in mcd ( QH1 := m⋅ cv ⋅ T3 − T2 ) kJ QH1 = 220.59 kg ( ) kJ QH2 = 964.897 kg ( ) kJ QL = −484.034 kg QH2 := m⋅ cp ⋅ T4 − T3 QL := −m⋅cv ⋅ T5 − T1 W := QH1 + QH2 + QL W = 701.453 can also express as ... ( kJ W = 701.453 kg N.B. this specific power is more than double LM 2500 swept_volume W⋅ Imep = W⋅ m V1 − V2 = = work m= V1 − V2 p 1 ⋅V1 R ⋅ T1 V1 − V2 W⋅ = p1 R ⋅ T1 1− V2 Imep := in this example ... W⋅ = unit_time 2-stroke ne = # power strokes/time Sulzer RT-flex96C, Sulzer RTA96C (two stroke) p1 R ⋅ T1 1 rv R ⋅ T1 1 Imep = 9.005 bar n e = engine_rpm time is period of power strokes = 1/freq = 1/(n e =engine_rpm) Wi Pi = = period_power_stroke Wi 1 freq 4-stroke ne = 2*# power strokes/time with ... i = number_of_cylinders s wartsila 64 (four stroke) R ⋅ T1 1− rv consider indicated power, ref: Woud 7.4.2-3 work_per_cycle kg p1 1− power_per_cyl = kg ne Pi = Wi⋅ 2 = 1 = Wi⋅n e n e = engine_rpm k = if ( stroke = 2 , 1 , 2) engine power is ... n e⋅i n e⋅i Pi = Wi⋅ = Imep⋅ VS⋅ k k brake power PB, power at engine drive flange, after mechanical losses in engine see Woud (7.12) and 7.4.1 10/24/2006 ) wartsila 32 (four stroke) p 1 ⋅ V1 V1 W⋅ kW some diesel examples ... Indicated Mean Effective Pressure; Imep work_per_stroke W1 := η th⋅ QH1 + QH2 kW W1 = 701.296 s another parameter describing diesel engines Imep = ) 7 We = effective_work unit_mass = Wi mass ⋅η mechanical n e⋅i n e⋅i Pe = We⋅ = mepe⋅VS⋅ k k Pe = constant⋅ mepe⋅ n e for later discussion mepe = mean_effective_pressure = brake_mean_effective_pressure = BMEP Pe⋅k mepe = = VS⋅n e⋅i Pe ne VS⋅ ⋅i k = power_per_cyl ne VS⋅ k COLT-PIELSTICK PC4.2B DATA Configuration Vee Only Bore 570 mm Stroke 660 mm Engine Version 60 Hz Propulsion Cylinder (nos) 10-12-16-18 Output Range (kW) 12,500-22,500 13,250-23,850 Speed (rpm) 400 400/430 Mean Eff. Pressure (bar) 22.3 22.3/22.0 Mean Piston Speed (m/s) 8.8 8.8/9.5 Output/cyl kW (hp) 1250 (1676) 1250 (1676)/1325 (1777) from: page 16 of Fairbanks Morse medium speed diesel handbook power_per_cyl := 1250kW π 2 VS := ⋅bore ⋅stroke 4 n e := 400 n_stroke := 4 min VS = 5.948 ft 3 mepe := k := if ( n_stroke = 2 , 1 , 2) bore := 570mm power_per_cyl as stated in data above mepe = 22.266 bar ne VS⋅ k stroke := 660mm two special cases (can be calculated above setting r p and rc appropriately ... Otto cycle - spark ignition engine heat added at TDC (constant volume) only extra subscript added for special designation air-standard Otto cycle: spark ignition internal-combustion engine 1-2 isentropic compression of air 2-3(=4) heat added at constant volume (piston momentarily at rest at tdc 4-5 isentropic expansion 5-1 heat rejection at constant volume (piston at crank-end dead center) from air standard dual cycle above η th = 1 + 10/24/2006 QL QH =1− ( T1 ⋅ ⎛ rp ⋅ rc − 1⎞ ⎝ γ ) ( ⎠ γ ) T2 ⋅ ⎡ rp − 1 + γ ⋅ rp ⋅ rc − 1 ⎤ ⎣ ⎦ rp ⋅ rc − 1 =1− rv γ−1 8 ( ) ( ) ⋅ ⎡ rp − 1 + γ⋅rp ⋅ rc − 1 ⎤ ⎣ ⎦ γo rco := 1 rpo⋅rco η th_otto := 1 − γo− 1 rc := 1.0 ) ( ) 1 η th_otto := 1 − rv γo− 1 rvo ⎛ 0.475 ⎞ η th_otto = ⎜ 0.602 ⎟ ⎜ ⎟ ⎝ 0.661 ⎠ removing o designation γ−1 1 η th_otto → 1 − ⋅ ⎡ rpo − 1 + γo ⋅ rpo⋅ rco − 1 ⎤ ⎣ ⎦ rvo ⎛5⎞ rp := 5 rv := ⎜ 10 ⎟ ⎜ ⎟ ⎝ 15 ⎠ ( −1 and ... Diesel cycle ... all heat added at constant pressure air-standard Diesel cycle: compression ignition internal-combustion engine 1-2(=3) isentropic compression of air 3-4 heat added at constant pressure (gas expanding during heat addition) 4-5 isentropic expansion 5-1 heat rejection at constant volume (piston at crank-end dead center) γd rpd := 1 rpd⋅rrcd η th_diesel := 1 − γd− 1 rvd ( ) ( η th_diesel → 1 − ) ⋅ ⎡ rpd − 1 + γd ⋅ rpd⋅ rcd − 1 ⎤ ⎣ ⎦ ⎛ 10 ⎞ rp := 1 rv := ⎜ 15 ⎟ ⎜ ⎟ ⎝ 20 ⎠ rc := 2.5 γd −1 γ rv γd− 1 γ−1 ( ⋅γ⋅ rc − 1 ) −1 ( ⋅ γd ⋅ rcd − 1 rvd rc − 1 η th_diesel := 1 − rcd ) ⎛ 0.506 ⎞ η th_diesel = ⎜ 0.58 ⎟ ⎜ ⎟ ⎝ 0.625 ⎠ Relate rc, a, b to efficiency v1 rv = = rc v2 p3 rp = =a p2 v4 rc = =b v3 reset variables γ rp ⋅rc − 1 η th := 1 − rv γ−1 ( ) ( ) ⋅ ⎡ rp − 1 + γ ⋅ rp ⋅ rc − 1 ⎤ ⎣ ⎦ substitute, var1 = var2, where var1 is to be replaced κ η th substitute , rv = rcc , rp = a, rc = b , γ = κ → 1 − 10/24/2006 a⋅ b − 1 .4 rcc ⋅ ⎡⎣a − 1 + a⋅( b − 1)⋅κ⎦⎤ 9 [W 7.87]