Knowing μ for air at 20 ° C from Table 1.4, estimate itsviscosity at 500 ° C by ( a ) the power law and ( b ) the Sutherlandlaw. Also make an estimate from ( c ) Fig. 1.6. Comparewith the accepted value of μ ≈ 3.58 E-5 kg/m ∙ s. RatioRatioFluidkg/(m - s)μμ (H)kg/mm/svlv(Hg)HydrogenAir9.0 E-61.00.0841.05 E-49101.8 E-52.11.201.50 E-5130GasolineWaterEthyl alcoholMercury2.9 E-4336804.22 E-73.71141351.0 E-39981.01 E-68.71.2 E-37891.52 E-6131.5 E-317013,5501.16 E-71.0SAE 30 oil0.2933,000170,0008911,2603.25 E-42,85010,300Glycerin1.51.18 E-3'1 kg/(m - s) = 0.0209 slug/(ft - s); 1 m/s = 10.76 f/s. 10· Liquid-6.4Dense gasTwo-phaseregion252.10Hi=#Criticalpoint30.80.70.50.60.5P, = p/P = 0.20.4Low-density limit100.3of0.20.45 6 7 89100.60.8 14T=2.

For air n = 0.7. Write the expression of the power law to determine the viscosity of air. Comparing…

Answered: Ratio Ratio Fluid kg/(m - s) μμ (H)…

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

Knowing μ for air at 20 ° C from Table 1.4, estimate its
viscosity at 500 ° C by ( a ) the power law and ( b ) the Sutherland
law. Also make an estimate from ( c ) Fig. 1.6. Compare
with the accepted value of μ ≈ 3.58 E-5 kg/m ∙ s.

Ratio
Ratio
Fluid
kg/(m - s)
μμ (H)
kg/m
m/s
vlv(Hg)
Hydrogen
Air
9.0 E-6
1.0
0.084
1.05 E-4
910
1.8 E-5
2.1
1.20
1.50 E-5
130
Gasoline
Water
Ethyl alcohol
Mercury
2.9 E-4
33
680
4.22 E-7
3.7
114
135
1.0 E-3
998
1.01 E-6
8.7
1.2 E-3
789
1.52 E-6
13
1.5 E-3
170
13,550
1.16 E-7
1.0
SAE 30 oil
0.29
33,000
170,000
891
1,260
3.25 E-4
2,850
10,300
Glycerin
1.5
1.18 E-3
'1 kg/(m - s) = 0.0209 slug/(ft - s); 1 m/s = 10.76 f/s.

10
· Liquid-
6.
4
Dense gas
Two-phase
region
25
2.
10
Hi=#
Critical
point
3
0.8
0.7
0.5
0.6
0.5
P, = p/P = 0.2
0.4
Low-density limit
10
0.3
of
0.2
0.4
5 6 7 8910
0.6
0.8 1
4
T=
2.

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