Polytropic Efficiency

Polytropic Efficiency kJ := 1000J consider a two stage compressor with a stage efficiency = 0.9 p 1 := 1bar p 2 := 5bar kJ s1 := 1 kg⋅ K p 3 := 20bar η stage := 0.9 calculations the states resulting are plotted ... Temperature 800 600 400 200 0.95 1 1.05 1.1 1.15 entropy using the gas laws and stage effiiciency = ηstage , after the first stage the states will be kJ s2 = 1.04 kg⋅ K T2 = 491.556 K p 2 = 5 bar after the second stage the states will be T3 = 756.994 K kJ s3 = 1.076 kg⋅ K p 3 = 20 bar now if we calculate efficiency of the compressor η comp = ∆Ts ∆T = T3s − T1 T3s − T1 T3 − T1 T3 − T1 = 0.88 as a check ... T33s − T2 T3 − T2 = 0.9 This effect can be accounted for by using polytropic efficiency - or small stage efficiency. reset variables T, s, p ηc = h 2s − h 1 h2 − h1 = T2s − T1 T2 − T1 = ∆Ts ∆T compressor 11/9/2005 1 T2s − T1 T2 − T1 = 0.9 as the pressure ratio approaches unity ∆Ts ∆T => ds = dh T − = d Ts dT γ= d Ts dT dT s = η pc⋅dT for simple compressible substance dT dp ⋅dp = cpo⋅ − R⋅ T T p ideal gas and definition of c po v isentropic ds = 0 substitute η pc = define ... cpo⋅ dT s = η pc⋅dT cpo cvo := cvo dT s T = R⋅ cpo⋅ η pc⋅ cpo dp p dT T = R⋅ dp  T2  rearrange and integrate ..... p R := cpo − cvo γ R ln  T1  simplify → η pc⋅cpo T2 T1 =  p2  p  1 = use in definition of ηc ηc = cpo⋅η pc ⋅ln  p1  γ ⋅ η pc T2s γ ⋅ η pc  p2  R γ−1 γ−1 raise to exponents (7.7) T⋅ ds = dh − v ⋅ dp T2s − T1 T2 − T1 = T1 T2 T1 γ−1 −1 γ r = −1 γ−1 −1 r γ ⋅ η pc −1 example polytropic efficiency = 0.9; calculate isentropic efficiency for p 2 /p1 = 2, 16, 30; use air as working fluid γ−1 i := 0 .. 2 η pc := 0.9 if T1 were 25 deg C γ := 1.4 2 r :=  16   30   p2  = T1  p1  T2 := T1 ⋅r 11/9/2005 T2 := T1 + ηc = T2s − T1 T2 = T1 + T2 − T1 T2s − T1 η pc −1 γ ⋅ η pc  0.89  η c =  0.856   0.845  −1  371.535  T2 =  718.944   877.733  γ ⋅ η pc whereas T2 calculated using ηpc as ηc  363.449  T2s =  658.369   787.897  (ri) γ −1 γ ⋅ η pc γ γ−1 i T1 := 25 + 273.15 γ−1 T2 η c := (ri)  370.704  T2 =  698.393   842.313  2 T2s − T1 ηc γ−1 T2s := T1 ⋅r γ temperature above is higher indicating more energy required for compressor consistent with lower efficiency another observation ... if we say the two stages have a polytropic efficiency of 0.9 then using ... γ−1  p2  = T1  p1  T2 γ−1 γ ⋅ η pc and ... γ−1 T3 T1 =  p3   p3  = T2  p2  T3 γ−1 γ ⋅ η pc T2 T3 T3 = = ⋅ T1 T2 T1 =>  p2    p1  γ−1 γ ⋅ η pc  p3  ⋅  p2  γ−1 γ ⋅ η pc  p3  =  p1  the polytropic efficiency of the compressor is identical, i.e T 3 is determined from γ ⋅ η pc the polytropic efficiency that is the same as the two stages since polytropic efficiency approaches isentropic efficiency for pressure ratio ~ 1, this is the same as saying that for a compressor with a large number of stages each with pressure ration near 1, the polytropic efficiency of the compressosor is isentropic efficiency of the individual stages   p1  turbine would be similar with exception of inversion of relationship ηt = T3 − T4 η pt = => T3 − T4s dT dT s = dT s dT η pt turbine cpo⋅ dT s T = R⋅ dp p cpo dT dp ⋅ = R⋅ p η pt T => => R⋅η pt dp = ⋅ T cpo p dT  p3  = T4  p4  T3 could do by direct analogy if write ηt = 4=2 T4s − T3 => T4 − T3 and ... T4 T3 = 1  r 3=1 η pt⋅  p2  = T1  p1  T2 γ ⋅ η pc η pc⋅ γ−1 γ γ−1 γ−1 1  1   η pt  γ⋅  morphs to ... T4 T3 γ−1 =  p4    p3  γ same example polytropic efficiency = 0.9; calculate isentropic efficiency for p 3 /p4 = 2, 16, 30; use air as working fluid i := 0 .. 2 η pt := 0.9 if T3 were 700 deg C 11/9/2005 γ ⋅ η pc γ := 1.4 1 1 −   r  i η t := i 2 r :=  16   30  1− T3 := 700 + 273.15 3 η pt⋅ 1 r  i γ−1 γ γ−1 γ  0.909  η t =  0.932   0.938  T4 1 =   T3  r  η pt⋅ γ−1 γ 1 T4 := T3 ⋅   r whereas T4 calculated using ηpt as ηt  798.309  T4s =  440.702   368.252  ηt = η pt⋅ γ−1  814.277  T4 =  477.034   405.834  γ T3 − T4 T3 − T4s ( T4 − T3 = T4 := T3 + η pt⋅ T4s − T3 ( T4 = T3 + η pt⋅ T4s − T3 T4s − T3  815.793  T4 =  493.947   428.742  ) γ−1 ) 1 T4s := T3 ⋅   r γ   above temperature is lower indicating more energy extracted from fluid, consistent with higher efficiency direct approach for calculating T 2 modeling as discrete multiple stages. increasing number_of_stages should make ηc_1 approach ηc (back to compressor for calculations) number_of_stages := 4 j := 0 .. 2 TT 1 r_per_stage := r number_of_stages 0, j := 25 + 273.15 power := T r_per_stage = ( 1.189 2 2.34 ) 2 r =  16   30  γ := 1.4 γ−1 γ temperature after each stage TT 1 , ns = TT 0 , ns (r_per_stage j) + n := 0 .. number_of_stages power ⋅TT 0, j − TT power T2s = T ⋅r TT 1 η c_1 := j  0.892  η c_1 =  0.869   0.862  ηc = η pc TT number_of_stages = 4 0, j TT isentropic continuous model  0.89  η c =  0.856   0.845  ( j) ⋅ r 0, j power n+ 1, j − TT number_of_stages , j  0.891  η c_1 =  0.862   0.852  T2 − T1 := TT n, j T2 = T1 + (r_per_stage j) + T2s − T1 ηc power ⋅TT 4 − TT η pc 0, j 20 stages  0.89   η c_1 = 0.859   0.849  50 stages  0.89   η c_1 = 0.857   0.846  further evidence that for a compressor with a large number of stages each with pressure ration near 1, the polytropic efficiency of the compressosor is isentropic efficiency of the individual stages. this follows through to determine isentropic efficiency for the compressor based on equating polytropic efficiency of small stages to the isentropic efficiency of the (small) stage. 11/9/2005 n, j 0, j − TT 10 stages T2s − T1 n, j