ADVANCED THERMODYNAMICS ENGINEERING phần 9 pdf
ADVANCED THERMODYNAMICS ENGINEERING phần 9 pdf
... 361 .93 0. 097 6 17.1488 0.14088
Nitric Oxide NO 30.01 180 65 0.252 0.058 0.588 1 09. 5 121.4 76.7 460.5 19. 76 0.0 199 1.45357 0.02878
Nitrogen N
2
28.02 126.2 33 .9 0.2 89 0.0 895 0.0 39 63.3 77.4 25.7 199 .2 ... 144.45 0.0 394 6.86 798 0.05682
Sulfur trioxide SO
3
80.06 490 .9 82.07 0.256 0.1271 0.422 290 .0 317 .9 24.6 502.5 192 .24 0.0431 8.5626 0.06216
Sulfuric acid H
2
SO
4
98 .0...
ADVANCED THERMODYNAMICS ENGINEERING phần 3 pdf
... P
3
/P
2
= T
3
/T
2
= 90 0÷574 , i.e., P
3
= 15.68 atm, and s
3
= 2.8 49 – 0.287 ln
(15.68÷1) = 2.0 59 kJ kg
–1
K
–1
. Therefore,
2.0 59 – 1.702 – (2 89/ 1000) – (– 29/ 298 ) = σ so that
σ = 0.165 ... ln(350 ÷ 273) = 1.0 395 kJ K
–1
.
Similarly, at temperatures of 3 49. 5, 3 49, 348.5, , K, S
W
= 1.0336, 1.0276, , kJ K
–1
.
For a 0.5 K temperature drop dS
W
= 1.0336 – 1.0 395 = –0....
Software Engineering (phần 9) pdf
... River. NJ, 199 5.
1 (Bowen and Hinchev, 1 99 5a1 J. P. BOwEN AND M. G. HINCHEY, ifrlkn Commandments Of .j 'e '' '
;
Formal Methods,'' IEEE Computer 28 (April 1 99 5). pp. ... Electric Re- Jonsson, and Overgaard, 199 21 in Sweden, starting in i
search and Development Center in Schenectady, New 1 96 7. Version 1.0 of UM L was published in 199 7. lt ' !
!Yo...
ADVANCED THERMODYNAMICS ENGINEERING phần 1 ppt
... in
–2
0.1450
PE potential energy kJ BTU 0 .94 78
pe specific potential energy
Q heat transfer kJ BTU 0 .94 78
q heat transfer per unit mass kJ kg
–1
BTU lb
–1
0.4 299
q
c
charge
R gas constant kJ kg
–1
... R
–1
0.2388
T temperature °C, K °F, °R (9/ 5)T+32
T temperature °C, K °R 1.8
t time s s
U internal energy kJ BTU 0 .94 78
u specific internal energy kJ kg
–1
BTU lb
–1
0.4 299
u
interna...
ADVANCED THERMODYNAMICS ENGINEERING phần 2 ppsx
... Air Composition
Species Mole % Mass %
Ar 0 .93 4 1.288
CO
2
0.033 0.050
N
2
78.084 75.521
O
2
20 .94 6 23.1 39
Rare gases 0.003 0.002
Molecular Weight: 28 .96 kg kmole
–1
.
2. Proof of the Euler Equation
Assume ... ∂∂∂∂
(92 )
The LHS of Eq. (94 ) is a vector called curl
r
F
. If
rr
∇×F
= 0, then the two are parallel to each
other, i.e., the vector field is irrotational. Assume that
r
F
=...
ADVANCED THERMODYNAMICS ENGINEERING phần 4 potx
... –T
0
) = 4 .97 .
= 6203 – 298 × (205.0 – 8.314 × ln (0.21/1)) + 100 × (0.08314 ×
298 /0.21)
= –4 698 1 kJ kmole
–1
.
Likewise,
φ
N
2
,∞
(T
0
,p
N2
,
∞
)
= 6 190 – 298 × ( 191 .5 – 8.314 × ln (0. 79/ 1)) + ... =1 × 673 – 298 × (1 × ln (673/ 298 )) = 430.2 kJ kg
–1
,
ψ
a,e
= 473 – 298 × 1 × ln (473/ 298 ) = 335.3 kJ kg
–1
,
ψ
w,i
= 4 19 – 298 × 1.31 = 28.6 kJ kg
–1
, and
ψ
w,e
= 2676.1 –...
ADVANCED THERMODYNAMICS ENGINEERING phần 5 potx
... and used in Eq. ( 69) to determine ω.
ii. Empirical Relations
Empirical relations are also available, e.g.,
ω = (ln P
R
sat
– 5 .92 714 + 6. 096 4/T
R,BP
+1.28862 ln T
R,BP
– 0.1 693 5 T
R,BP
)/
(15.2578 ... R–134a,
c
f
= 1.464 kJ kg
–1
K
–1
, and c
p,o
at 298 K = 0.851 kJ kg
–1
K
–1
.
Using the values for P = 690 kPa, R = 0.081 49 kJ kg
–1
K
–1
, c
vo
= 0.7 697 kJ kg
–1
K
–1
,
v
f...
ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt
... 2501.3 233.18 13 89. 0 215.188 524.534 1 69. 59
A
o
30.54 22.54 16.778 17.2 29 29. 75 28.58 29. 15 19 26.765
†
B
o
0.01030 0.1141077 0.2865 0.4757 .025 .00377 –.001573 0.17 594
†
C
o
0.0 130 196 .35 0 0 –154808. ... subscript “o”
µ
N
2
,
.mix
= µ
N
2
( 298 , 2 bar) + RT ln X
N
2
= 298 – 298 ×(ln( 298 ÷273) – 0. 297 ×ln(2÷1)) + 0. 297 × 298 ln(0.6)
A mixture contains 60% N
2
and 40% O
2...
ADVANCED THERMODYNAMICS ENGINEERING phần 7 ppt
... liquid) (cf.
Figure 11), σ = ( 95 30.5 – ( 97 76.2))÷ 593 = 0.414 kJ kmole
-1
K
-1
, while for a change from D
to the metastable vapor state K, σ = ( 95 30.5 – ( 96 38.8)) ÷ 593 = 0.183 k kmole
-1
K
-
.
-4000
-3500
-3000
-2500
-2000
-1500
-1000
-500
0
500
1000
0 ...
¯s
=
1 49. 83 kJ kmole
–1
. The state is represented by point D in the figure, and the extensive
146
146.5
147
147.5
148
1...
ADVANCED THERMODYNAMICS ENGINEERING phần 8 doc
... Hence,
p
O
2
= X
O
2
P = (50– 0.0005) ÷(0.001+(50–0.0005)) = 0 .99 99 P = 0 .99 99 bar.
Therefore,
s
O
2
= 205.03 – 8.314 × ln 0 .99 99 = 205.03 kJ K
–1
kmole
–1
, i.e.,
-6.0E+05
-5.0E+05
-4.0E+05
-3.0E+05
-2.0E+05
-1.0E+05
0.0E+00
0.0 ...
h
HO
2
– T
s
HO
2
= –285830 – 298 × 69. 95 = –306675 kJ kmole
–1
of H
2
O(l) ,
g
CO
o
= –110530 – 298 × 197 .56 = –1 694 03 kJ kmole
–1
of CO,...