Specific-heat Power-law ratio Molecular R, m²/(s² · K) pg, N/m³ µ,N· s/m² Gas weight V, ms exponent n* На 2.016 4124 0.822 9.05 E-6 1.08 E-04 1.41 0.68 Не 4.003 2077 1.63 1.97 E-5 1.18 E-04 1.66 0.67 Н.о 18.02 461 7.35 1.02 E-5 1.36 E-05 1.33 1.15 0.72 0.67 0.79 Ar 39.944 208 16.3 2.24 E-5 1.35 E-05 1.67 Dry air CO2 CO 28.96 287 11.8 1.80 E-5 1.49 E- 05 1.40 44.01 189 17.9 1.48 E-5 8.09 E-06 1.30 28.01 297 11.4 1.82 E-5 1.56 E-05 1.40 0.71 297 N2 O2 28.02 11.4 1.76 E-5 1.51 E-05 1.40 0.67 32.00 260 13.1 2.00 E-5 1.50 E-05 1.40 0.69 NO 30.01 277 12.1 1.90 E-5 1.52 E-05 1.40 0.78 N20 Cl2 CH4 44.02 189 17.9 1.45 E-5 7.93 E- 06 1.31 0.89 70.91 16.04 117 28.9 1.03 E-5 3.49 E- 06 1.34 1.00 518 6.54 1.34 E-5 2.01 E-05 1.32 0.87
Specific-heat Power-law ratio Molecular R, m²/(s² · K) pg, N/m³ µ,N· s/m² Gas weight V, ms exponent n* На 2.016 4124 0.822 9.05 E-6 1.08 E-04 1.41 0.68 Не 4.003 2077 1.63 1.97 E-5 1.18 E-04 1.66 0.67 Н.о 18.02 461 7.35 1.02 E-5 1.36 E-05 1.33 1.15 0.72 0.67 0.79 Ar 39.944 208 16.3 2.24 E-5 1.35 E-05 1.67 Dry air CO2 CO 28.96 287 11.8 1.80 E-5 1.49 E- 05 1.40 44.01 189 17.9 1.48 E-5 8.09 E-06 1.30 28.01 297 11.4 1.82 E-5 1.56 E-05 1.40 0.71 297 N2 O2 28.02 11.4 1.76 E-5 1.51 E-05 1.40 0.67 32.00 260 13.1 2.00 E-5 1.50 E-05 1.40 0.69 NO 30.01 277 12.1 1.90 E-5 1.52 E-05 1.40 0.78 N20 Cl2 CH4 44.02 189 17.9 1.45 E-5 7.93 E- 06 1.31 0.89 70.91 16.04 117 28.9 1.03 E-5 3.49 E- 06 1.34 1.00 518 6.54 1.34 E-5 2.01 E-05 1.32 0.87
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
Air, assumed to be a perfect gas from Table A.4, fl ows
through a long, 2-cm-diameter insulated tube. At section 1,
the pressure is 1.1 MPa and the temperature is 345 K. At
section 2, 67 meters further downstream, the density is
1.34 kg/m 3 , the temperature 298 K, and the Mach number
is 0.90. For one-dimensional fl ow, calculate ( a ) the mass
fl ow; ( b ) p 2 ; (c) V 2 ; and ( d ) the change in entropy between
1 and 2. ( e ) How do you explain the entropy change?
Transcribed Image Text:Specific-heat Power-law
ratio
Molecular
R, m²/(s² · K)
pg, N/m³ µ,N· s/m²
Gas
weight
V, ms
exponent n*
На
2.016
4124
0.822
9.05 E-6
1.08 E-04
1.41
0.68
Не
4.003
2077
1.63
1.97 E-5
1.18 E-04
1.66
0.67
Н.о
18.02
461
7.35
1.02 E-5
1.36 E-05
1.33
1.15
0.72
0.67
0.79
Ar
39.944
208
16.3
2.24 E-5
1.35 E-05
1.67
Dry air
CO2
CO
28.96
287
11.8
1.80 E-5
1.49 E- 05
1.40
44.01
189
17.9
1.48 E-5
8.09 E-06
1.30
28.01
297
11.4
1.82 E-5
1.56 E-05
1.40
0.71
297
N2
O2
28.02
11.4
1.76 E-5
1.51 E-05
1.40
0.67
32.00
260
13.1
2.00 E-5
1.50 E-05
1.40
0.69
NO
30.01
277
12.1
1.90 E-5
1.52 E-05
1.40
0.78
N20
Cl2
CH4
44.02
189
17.9
1.45 E-5
7.93 E- 06
1.31
0.89
70.91
16.04
117
28.9
1.03 E-5
3.49 E- 06
1.34
1.00
518
6.54
1.34 E-5
2.01 E-05
1.32
0.87
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