Compressor Instability with Integral Methods Episode 1 Part 6 pps
... xL =1. 0 1 1 1 1 0.3 15 1. 0 0.5004074 319 23 014 2 1 1 2 0.3 15 0.9 0.49829 062 59 819 02 3 1 1 3 0.3 15 0.8 0.49 61 3 212 40 510 77 4 1 2 1 0.3 20 1. 0 0.50 319 2 36 719 266 9 5 1 2 2 0.3 20 0.9 0.4988 464 360 967 71 ... 11 2 1 2 0.5 15 0.9 0.498393 918 0485 01 12 2 1 3 0.5 15 0.8 0.4972 968 67242345 13 2 2 1 0.5 20 1. 0 0.5 015 53 818 57 319 6 14 2...
Ngày tải lên: 07/08/2014, 12:21
... J V p = 10 0 m/s 1 2 3 4 5 6 glass bead d p = 1. 5 mm 10 –5 10 –4 10 –3 10 1 10 –2 10 0 10 50 10 0 250 Fig. 2 . 16 Kinetic energies of abrasive particles; calculated with (2 .14 ). 1 to 3: particle ... al., 19 93) Material Damaged grains (%) Lattice constant ( ˚ A) Cell volume ( ˚ A 3 ) Almandine 5 60 11 .522 (0.0 06) 1, 529 .62 Spessartine – 11 . 61 3 (0.005)...
Ngày tải lên: 07/08/2014, 12:21
... model, 10 8 11 0 numerical methods, 11 0 11 4 results, 11 5 11 8, 11 8 12 0, 12 0 12 1 and inlet distortion propagation in compressor, 57–73 Tangent of flow angle in inlet, 52 Turbomachinery, 1 V ... numerical method equations, 11 4th-order Runge-Kutta Equations, 11 12 results, 13 17 asymptotic behavior results, 17 –20 compressor performance and charac- teristic, 23–...
Ngày tải lên: 07/08/2014, 12:21
Compressor Instability with Integral Methods Episode 1 Part 1 doc
... concrete – 5.0 17 .1 Fusion bonded epoxy (FBE) 6 .1 8.9 >40 Heat shrink sleeve 12 .7 18 .0 27.3 Tape 14 .9 31. 8 28.5 Coal tar urethane 13 .6 16 .5 26. 9 a See Sect. 8.2.3 for surface preparation grades the ... (ISO 12 944-4, 19 98). To define a blast cleaning method completely, the following information is required: Table 1. 1 Blast cleaning methods according to ISO 12 944...
Ngày tải lên: 07/08/2014, 12:21
Compressor Instability with Integral Methods Episode 1 Part 2 ppt
... ALFA(5 ,1) =0.0 ! PRESSURE 1 SI0=SITA1 SITA0=SITA1*PI /18 0. GAMA=TAN(SITA0) AL0=ALFA (1, 1) DAL0=0.0 01 AL1=ALFA (1, 1) 10 CK1=Y (1) *ALFA (1, 1) ! K1 CK0=CK1+ (1- CK1/ALFA (1, 1))*ALFA(3 ,1) ! K0 ****** ... F(4,I)=FY0U(I)/(GAMA*A(3))-A(4)/A(3)*CK2*F (1, I) F(5,I)=(FXU(I)-F (1, I))/(GAMA**2*2) DO 10 1 K =1, 5 10 1 RK (1, K,I)=DX*F(K,I) ! K 11, K12,K13,K14,K15 DO 20...
Ngày tải lên: 07/08/2014, 12:21
Compressor Instability with Integral Methods Episode 1 Part 4 ppt
... 0.2 0 .1 9.0())= Γ Φ 0. 61 7 66 0. 718 73 0. 819 80 0.92087 1. 0 219 4 )( Φ ε (%) -45.0 - 36. 0 -27.0 -18 .0 -9.0 9.0)0( P = Γ Δ 1. 303 86 1. 304 71 1.30595 1. 30793 1. 311 86 )P( Δ ε (%) -33 .60 -33. 56 -33.50 ... 0 .10 000 0 .10 000 0.50000 )10 ( α 0 .10 108 0.09847 0.08992 0.50 867 )( α ε (%) 1. 080 -1. 530 -10 .080 1. 734 )0( 0 α 1. 00000 1. 00000 1. 000...
Ngày tải lên: 07/08/2014, 12:21
Compressor Instability with Integral Methods Episode 1 Part 5 doc
... and Hutchings, 19 94) Particle diameter in Particle velocity Abrasive mass flow Average distance L P /d P μm in m/s rate in g/min in μm 63 –75 70 50 900 ∼ 13 12 5 15 0 52 6 3,200 ∼ 23 212 –250 45 31 4,000 ∼ 17 65 0–750 ... velocity (Lecoffre et al., 19 93) 1. 8 1. 2 relative velocity 0 .6 0 0 200 400 60 0 particle radius in μ m gas particle R m 800 10 00 0 .1 0.5 1 2 4 6 F...
Ngày tải lên: 07/08/2014, 12:21
Compressor Instability with Integral Methods Episode 1 Part 7 ppt
... A (1) =1. 775 460 69 A(2)= - 16 .62 88937 A(3) =11 5.42272 A(4)=-394.29835 A(5) =65 7.747843 A (6) =- 4 16 .67 7009 HMAX=0.003 0 16 7777 Appendix 4.A. Fortran Program: Chebyshev Curve Fitting 10 1 89 x ξ 02 46 810 0.48 0.49 0.50 0. 51 0.52 0.53 0.54 0.55 0. 56 0.57 θ =1 θ=5 θ =10 θ =15 θ=20 θ=25 o o o o o o o o o o o o α(0)=0 .1 ... 0 .15 α =+ ( 810 α −°≤ ≤ ° ) (4.26a)...
Ngày tải lên: 07/08/2014, 12:21
Compressor Instability with Integral Methods Episode 2 Part 1 docx
... 2 5.4 11 .3 28.3 3 13 .7 20.0 41. 4 3.5 17 .3 19 .2 54 .1 4.5 19 .7 19 .7 59 .6 Garnet 2 2 .6 6 .6 23.8 2.5 5 .6 19 .6 17 .2 3 10 .1 20.9 28.8 3.5 10 .6 21. 1 30.8 4.5 21. 3 23.7 54 .1 Glass blast 2.75 7 .6 19 .3 ... coal slag 2 7 .6 30.4 15 .2 3 9.5 32.4 17 .7 5 12 .6 23.0 32.9 Star blast 2 5.5 24.3 13 .6 3 13 .9 38 .1 22.2 5 17 .5 36 29.3 Aluminium oxide 2 6...
Ngày tải lên: 07/08/2014, 12:21
Compressor Instability with Integral Methods Episode 2 Part 2 potx
... PEEK 10 45 30 11 14 20 28 62 43 21 22 14 7 3 30 42 37 40 32 5 24 13 6 40 62 42 21 47 4 10 13 1 50 68 35 14 46 5 16 13 2 60 80 14 7 59 4 22 9 6 70 90 36 6 46 0 16 4 3 80 86 3 1 10 90 90 31 5 46 0 ... (Spangenberg, 19 72) Parameter σ M in N/cm 2 Steel type Armco iron St 42 St 70 200 1, 60 0 10 0 1, 800 200 1, 60 0 C 1 5 31 1 ,64 9 2, 019 C 2 11 .2 3.5 5...
Ngày tải lên: 07/08/2014, 12:21