Presented by,Pradeep Kumar Somasundaram
MTH5230
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AbstractThe process of finding the minerals under the earth’s crustusing the earthed electrodes. The current from the batteryconducted through the earth and the field of constantcurrent created on the surface of the earth are mapped. Byusing Linear Partial Differential Equations the potentials aredetermined and with help of Bessel functions and method ofseparation of variable the prospecting is found in differentmedium and found that electrolytic tank measurementsreplace the direct measurements.
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Instrument
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Introduction
Underground minerals, surface potentials
Homogeneous medium satisfies Laplace equation𝛻2 𝑉 = 0 −−−→ (1)
𝜕V
𝜕r|z=0 = 0 −−−→ 2
Considering a point electrode at point A
Potential of the field
V =Iϱ
2πR−−−→ (3)
where,
R is the distance of the potential field from the source point Aϱ is the specific resistance of the mediumI is the intensity of the current
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Potential field Potentials differ for an infinite medium
•---------•--r--•--r--•
A M O N
V M − V N =𝜕V
𝜕r∆r −−−−→ 4
V M − V(N)
∆r≅
𝜕V
𝜕r≅
Iϱ
2πr2−−−−→ 5
where,
r is the distance between the point O to the points M and N.
O is the mid-point of the receiving circuit from the feeding electrode.
I is the current intensity of the feeding circuit which is known value.
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Homogeneous resistance Two layers
homogeneous resistance − ϱ0homogeneous resistance -- ϱ1thickness l
The resistance can be represented as ϱ z = ϱ0 where 0 ≤ z < lϱ1 where l < z
r<<l the impedance will be ϱk = ϱ0 r>>l the impedance will be ϱk = ϱ1 Conditions of continuity
V0 |z=l = V1 |z=l −−−−→ 6
1𝜕V0
ϱ0𝜕r|z=l =
1𝜕V1
ϱ1𝜕r|z=l −−−−→ 7
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Cylindrical symmetry
𝜕2V
𝜕r2+
1
r
𝜕V
𝜕r+
𝜕2V
𝜕z2= 0 −−−−→ 8
e±λzJ0 λr −−−−→ 9
where, J0 is the Bessel function of the zero order
λ is the separation parameter. The solutions will be of
V0 r, z =ϱ0I 1
2π (z2+r2)+ 0
∞(A0e
−λz + B0eλz )J0 λr dλ −−−−→ 10
V1 r, z = 0∞(A1e
−λz + B1eλz )J0 λr dλ −−−−→ 11
Find A0, B0, A1, B1 which are the functions of λ
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Special functionsFor arbitrary r, A0 = B0
For V1 the condition of the bounded nature as z∞; B1 = 0
V1 r, z = 0∞(A1e
−λz )J0 λr dλ
Formula found in the boundary value problem by the equations of special functions
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(z2+r2)= 0
∞J0 λr e−λz dλ
𝑞 =ϱ0I
2π
(12)
(13)
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By substituting the known values
By using the equations (6) and (7)
Derivation
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Solving equations (A) and (B)
Finding the values
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Contd.
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since |k|<1
The equation of V0 can be written as
Assuming z=0 we obtain the distribution of the potential on the earth’s surface by solving the problem using the method of images.
Distribution of the potential
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Change of variables
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The limit of the nth term of the sum will be equal to Kn, from which it follows that
To prove the impedance at infinity
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Conclusion Different conductivity profiles the impedances are also different.
𝜌𝑘 𝑟1 ≠ 𝜌𝑘(𝑟2)
Defects are determined by the presence of cavity under the surface.
The cavity of the surface can be measured by placing a metallic piece between the poles of a magnet and the magnetic field on the surface.
Electrolytic tank.
Replaces effectively the direct measurements of temperature, magnetic and other fields.
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