How to find the given biot constants through the interpolation method from the table given. First Eigenvalue, ₁, for One-dimensional TransientConduction with External ConvectionBi=0.010.020.030.040.050.060.070.080.090.100.150.200.250.30.40.50.6hλ₂k0.70.80.91.02.03.04.05.06.07.08.09.010.020.030.040.050.0100.0∞oInfinite Plane Wall0.09980.1400.17320.19870.22170.24250.26150.27910.29560.31110.37790.43280.48010.52180.59320.65330.70510.75060.79100.82740.86031.07691.19251.26461.31381.34961.37661.39781.41491.42891.49611.52021.53251.5400..5552π/2GeometryInfinite Cylinder0.14120.19950.24390.28140.31420.34380.37080.39600.41950.44170.53760.61700.68560.74650.85160.94081.01851.08731.14901.20481.25581.59951.78771.90811.98982.04902.09372.12862.15662.17952.28812.32612.34552.3572.38092.4050Sphere0.17300.24450.29890.34500.38520.42170.45500.48600.51500.54230.66080.75930.84480.92081.05281.16561.26441.35251.43201.50441.57082.02882.28892.45562.57042.65372.71652.76542.80442.83632.98573.03723.06323.07883.1102π A₁0.1039 and A₁Bi =The Biot number for this process = Bi = 0.0109The constants A₁ and A₁ corresponding to this Biot are, from 11-2 tables.The interpolation method used to find the=hLBi = ht1.0018(80W/m².Â°C)(0.015m)(110W/m.Ã°C)

Bi = 0.0109 The Given value of Biot number lies between 0.01 and 0.02 value. I have attached one…

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How to find the given biot constants through the interpolation method from the table given.

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

How to find the given biot constants through the interpolation method from the table given.

First Eigenvalue, ₁, for One-dimensional Transient
Conduction with External Convection
Bi=
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.15
0.20
0.25
0.3
0.4
0.5
0.6
hλ₂
k
0.7
0.8
0.9
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
20.0
30.0
40.0
50.0
100.0
∞o
Infinite Plane Wall
0.0998
0.140
0.1732
0.1987
0.2217
0.2425
0.2615
0.2791
0.2956
0.3111
0.3779
0.4328
0.4801
0.5218
0.5932
0.6533
0.7051
0.7506
0.7910
0.8274
0.8603
1.0769
1.1925
1.2646
1.3138
1.3496
1.3766
1.3978
1.4149
1.4289
1.4961
1.5202
1.5325
1.5400
..5552
π/2
Geometry
Infinite Cylinder
0.1412
0.1995
0.2439
0.2814
0.3142
0.3438
0.3708
0.3960
0.4195
0.4417
0.5376
0.6170
0.6856
0.7465
0.8516
0.9408
1.0185
1.0873
1.1490
1.2048
1.2558
1.5995
1.7877
1.9081
1.9898
2.0490
2.0937
2.1286
2.1566
2.1795
2.2881
2.3261
2.3455
2.3572
.3809
2.4050
Sphere
0.1730
0.2445
0.2989
0.3450
0.3852
0.4217
0.4550
0.4860
0.5150
0.5423
0.6608
0.7593
0.8448
0.9208
1.0528
1.1656
1.2644
1.3525
1.4320
1.5044
1.5708
2.0288
2.2889
2.4556
2.5704
2.6537
2.7165
2.7654
2.8044
2.8363
2.9857
3.0372
3.0632
3.0788
3.1102
π

A₁0.1039 and A₁
Bi =
The Biot number for this process = Bi = 0.0109
The constants A₁ and A₁ corresponding to this Biot are, from 11-2 tables.
The interpolation method used to find the
=
hL
Bi = ht
1.0018
(80W/m².Â°C)(0.015m)
(110W/m.Ã°C)

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