Helium (Table A.4) enters a 5-cm-diameter pipe at p1 =550 kPa, V1 = 312 m/s, and T1 = 40 °C. The friction factoris 0.025. If the flow is choked, determine (a) the length ofthe duct and (b) the exit pressure. Specific-heat Power-lawratioMolecularR, m²/(s² · K)pg, N/m³ µ,N· s/m²GasweightV, msexponent n*На2.01641240.8229.05 E-61.08 E-041.410.68Не4.00320771.631.97 E-51.18 E-041.660.67Н.о18.024617.351.02 E-51.36 E-051.331.150.720.670.79Ar39.94420816.32.24 E-51.35 E-051.67Dry airCO2CO28.9628711.81.80 E-51.49 E- 051.4044.0118917.91.48 E-58.09 E-061.3028.0129711.41.82 E-51.56 E-051.400.71297N2O228.0211.41.76 E-51.51 E-051.400.6732.0026013.12.00 E-51.50 E-051.400.69NO30.0127712.11.90 E-51.52 E-051.400.78N20Cl2CH444.0218917.91.45 E-57.93 E- 061.310.8970.9116.0411728.91.03 E-53.49 E- 061.341.005186.541.34 E-52.01 E-051.320.87

Given, D = 5 cm = 0.05 m P1 = 550 kPa V1 = 312 m/s T1 = 40°C+273 = 313 friction factor, f = 0.025

Answered: Specific-heat Power-law ratio Molecular…

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

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Question

Helium (Table A.4) enters a 5-cm-diameter pipe at p₁ =
550 kPa, V₁ = 312 m/s, and T₁ = 40 °C. The friction factor
is 0.025. If the flow is choked, determine (a) the length of
the duct and (b) the exit pressure.

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

Expert Solution

Step 1

Given,

D = 5 cm = 0.05 m

P₁ = 550 kPa

V₁ = 312 m/s

T₁ = 40°C+273 = 313

friction factor, f = 0.025

Step 2

From the given table,

For helium, k =1.66 ; R = 2077 J/kg.K

$Ma = V / a a = \sqrt{k \times R \times T} a = \sqrt{1.66 \times 2077 \times 313} a = 1038.82 m / s$

$Ma = \frac{312}{1038.82} Ma = 0.3$

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