ADVANCED THERMODYNAMICS ENGINEERING phần 1 ppt

... K 1 , atm 1 R 1 , bar 1 0.555, 1. 013 β s atm 1 bar 1 1. 013 γ k activity coefficient, ˆ α k / ˆ α k id ˆ φ k /φ k Gruneisen constant λ thermal conductivity kW m 1 K 1 BTU ft 1 R 1 0 .16 05 ηFirst ... kJ kg 1 BTU lb 1 0.4299 q c charge R gas constant kJ kg 1 K 1 BTU lb 1 R 1 0.2388 R universal gas constant kJ kmole 1 BTU lbmole 1 0.2388 K 1 R 1 S e...

Ngày tải lên: 08/08/2014, 15:21

... 405.5 12 6.2 304 .15 385.0 T ref (K) 273 .16 233 .14 233 .14 233 .15 64 .14 3 216 .55 233 .15 P ref (bar) 0.00 611 3 1. 0495 0. 512 615 . 717 7 .12 53 5 .17 8 .6 417 h ref (kJ kg 1 ) 0. 01 0.0 0.0 15 0.3 3 01. 45 ... in ln γ 1 = X 2 2 (A 12 X 2 + 2(A 21 –A 12 ) X 1 ), and ln γ 2 = X 1 2 (A 21 X 1 + 2(A 12 –A 21 ) X 2 ). (11 8) As X 1 →0, X 2 1, and ln γ 1...

Ngày tải lên: 08/08/2014, 15:21

... KK + => − − − −−− − −− 1 12 1 12 1 111 112 11 11 111 2 11 0 M . (26) For disturbance of order K +1, the stability criterion D 1 < 0, D 2 > 0, D 3 > 0, D K +1 > 0 results in k +1 inequalities. e. ... the figure, and the extensive 14 6 14 6.5 14 7 14 7.5 14 8 14 8.5 14 9 14 9.5 15 0 15 0.5 15 1 0.04 0.05 0.06 0.07 0.08 0.09 0 .1 0 .11 0 .12 0...

Ngày tải lên: 08/08/2014, 15:21

... m/Masker 1 m [-30 -20 -10 0 10 ] 0 Target 0.25 m/Masker 1 m [-40 -30 -20 -10 0] +14 Target 0 .12 m/Masker 1 m [-40 -30 -20 -10 0] +27 Target 1 m/Masker 0.25 m [-20 -10 0 10 20] -9 Target 1 m/Masker ... 20] -9 Target 1 m/Masker 0 .12 m [-20 -10 0 10 20] -13 Table 1. The range of TMR values tested and normalization shifts for each spatial configuration in Experiments 1...

Ngày tải lên: 19/06/2014, 12:20

... Computer Congress, October 19 80. pp. 32 1- 26. (Boehm, 1 98 1 1 B. W. BOEHM, Software Engineering Economics Prentice Hall, Englewood Cliffs, NJ, 1 98 1 . (Brereton et a1., 19 9 91 P. BRHRETON, D. BUDGEN, ... (No. I , 1 994), pp. 1 82-2 14 . IBoehm, 1 97 61 B. W. BOEHM, 'Esoftware Engineering. '' IEEE Transaaions on Computers (2-25 (December 19 76), pp....

Ngày tải lên: 07/07/2014, 06:20

... which leakage occurs is 1 mm 2 , assuming the density of air to be 1. 1 kg m –3 , the mass flow exiting the deforming balloon equals 10 1 10 –9 1. 1 = 1. 1x10 –8 kg s 1 . C. SUMMARY In this chapter ... yields Figure 13 : Steady flow through a capillary tube. Solution Q 12 – W 12 = = U 2 – U 1 + KE 2 – KE 1 + PE 2 – PE 1 . 0+3000=U 2 –U 1 +((2000÷2)(24.6ms 1 ) 2 –...

Ngày tải lên: 08/08/2014, 15:21

... (0ºC, 0. 611 kPa) = 0 s g – s f = 9 .16 kJ kg 1 K 1 = σ. (C) s(393 K, 10 0 kPa) – s(273 K, 0. 611 ) = 2.02 ln (393÷273) – (8. 314 18 .02)ln (10 0÷0. 611 ) = – 1. 616 kJ kg 1 K 1 , or s(393, 10 0 kPa) ... ( P N 2 ) 2 = 0.6 11 = 6.6 bar. Therefore, at conditions 1 and 2, respectively, s CO 2 (12 00K, 4.4 bar) = s CO 2 0 (12 00 K) – R ln(4.4 1) = 234 .1 – 8. 314 × ln(4.4 1...

Ngày tải lên: 08/08/2014, 15:21

... (0. 21/ 1)) + 10 0 × (0.08 314 × 298/0. 21) = –469 81 kJ kmole 1 . Likewise, φ N 2 ,∞ (T 0 ,p N2 , ∞ ) = 619 0 – 298 × (19 1.5 – 8. 314 × ln (0.79 /1) ) + 10 0 × (0.8 314 × 298/0.79) = 619 0–298 × 19 3.5 + 10 0 ... –0.0 01 × (60 – 0 .1) × 10 0 = – 6 kJ kg 1 . From Eq. (B) h 3 = 19 2 kJ kg 1 (sat liquid at 0 .1) , and h 4 = 19 2 + 6 = 19 8 kJ kg 1 . In the boiler q in = q...

Ngày tải lên: 08/08/2014, 15:21

... s o (T,v) = R ln (1 (b/ v )) – (1/ 2)(a/(bT 3/2 )) ln (1+ (b/ v )) = 8. 314 ×ln (1 (0.0 211 ÷0.245) – (0.5 14 2.64 10 0÷(0.0 211 ×873 1. 5 )) ln (1+ (0.0 211 /0.245)) = 1. 826 kJ kmole 1 K 1 . Then using Eq. ... u(T,v) – u o (T) = –(3/2)(a/(bT 1/ 2 )) ln (1+ (b/v)), i.e., u Res = –(3÷2)× (14 2.64 10 0÷(0.0 211 ×273 .15 1/ 2 )) ln (1+ (0.0 211 ÷3 717 )) = – 0.348 kJ kmole 1 or...

Ngày tải lên: 08/08/2014, 15:21

... pressures (p CO /1) 1 (p O 2 /1) 1/ 2 /(p CO 2 /1) 1 = (6 /1) 1 (1/ 1) 1/ 2 /(2 /1) 1 = 3. The criterion K o (T) = 0.0002 ≥ (p CO /1) 1 (p O 2 /1) 1/ 2 /(p CO 2 /1) 1 , or p CO /1) 1 (p O 2 /1) 1/ 2 /(p CO 2 /1) 1 ... 0.0 21 0.0527 310 .2 334.7 –225887 O 2 9.75 0 0.2057 0. 514 1 205 .1 210 .7 –627 81 N 2 36.66 0 0.7733 1. 933 19 1.6 18 6 .1 –55428 CO 2 4 –3...

Ngày tải lên: 08/08/2014, 15:21