Coil:
When the length of a component is several times larger than its diameter, then we need Multi-turn coil to produce longitudinal magnetic field in the component.
Circular current pass through the coil, and as per Right-Hand rule, the coil produce Longitudinal Magnetic field. A Longitudinal magnetic field has magnetic line of force that run parallel to the axis of the coil. Any discontinuity that is perpendicular to the magnetic lines of force will show, if the discontinuity is parallel to the line of magnetic line of force, there will be no leakage in magnetic flux and therefore no indication observed.
Right-hand rule:
Right-Hand rule says, if the person grasps a conductor in his right hand with the thumb pointing in the direction of the current, the fingers will point in the direction of the Magnetic field.
In This case, if the Fingers point to the direction of current, the Thumb will point to the direction of Magnetic Field.
Fill Factor:
The Magnetisation with a coil is depend upon the Fill Factor. The fill factor is the ratio of the cross-sectional area of the test piece to the cross-sectional area of the coil.
fill factor = cross-sect. area of test piece / (÷ by)
cross-sect. area of coil
High Fill-Factor:
A high fill factor is where Cross-sectional area of the coil is less than twice the cross-sectional area of the part including hollow portions. in other words, the test piece takes up one half or more of the area of the coil. A High Fill-Factor is greater than 0.5:
Intermediate Fill-Factor:
Intermediate fill factor is where Cross-sectional area of the coil is greater than twice and less than ten times the cross-sectional area of the part being examined. in other words, Intermediate Fill-Factor is greater than 0.1 and less than 0.5:
Intermediate Fill-Factor > 0.1 & < 0.5
Low Fill-Factor:
Low Fill factor is where cross-sectional area of the coil is equal or less than one-tenth of the cross-sectional area of the part being examined. in other words, Low Fill-Factor is less than or equal to 0.1:
How to Calculate Fill-Factor?
In order to calculate Fill-Factor, first we need to calculate the Cross-sectional area of the coil and part being inspected.
Cross-sectional area of Coil:
In order to calculate Cross-sectional area, we use following formula: πr2
Where, π(pi) is special number, roughly equal to 3.14, and r is the radius of coil, to calculate radius of coil, divide the Internal dia of coil by 2.
Measure Internal Dia of coil, In my case its 11" ID of coil. so, radius will be half, that is 11/2 = 5.5inch. Let's calculate cross sectional area of coil.
The Cross-sectional area of coil = πr2
Value of PI (π) = 3.14
radius r = 5.5inch
Using Formula = π(r)2
= π(5.5)2
= π x 30.25
= 94.958 inch2 or
= 95inch2
Cross-sectional area of coil = 95inch2
Above, we have calculated the cross-sectional area of coil. using the same method, we can calculate the cross-sectional area of the pipe. Let's take some example to clear this concept.
Calculate the Fill-Factor for the following items. State whether the fill factor is high or low.
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A solid bar of 6" diameter.
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We are using 9" coil, having radius 9/2 = 4.5".
The area of coil is πr2 = 3.14 x (4.5)2 = 64 inch2
The area of 6" diameter is πr2 = 3.14 x (3)2 = 28 inch2.
The Fill-Factor is ratio of cross-sectional area of Part being inspected to the cross-sectional area of coil. so, we have = 28 ÷ 64 = 0.43:
Ans: 0.43 is greater than 0.1 and less than 0.5 so this is Intermediate Fill-Factor.
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A Pipe of 9" diameter with internal Id of 3"?
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The fill factor is the same as for the bar, because the hollow area (ID) of the pipe are counted as part of its area in this case.
We are using 12" coil, having radius 12/2 = 6".
The area of coil is πr2 = 3.14 x (6)2 = 113 inch2
The area of 9" diameter is πr2 = 3.14 x (4.5)2 = 64 inch2.
The Fill-Factor is ratio of cross-sectional area of Part being inspected to the cross-sectional area of coil. so, we have = 64 ÷ 113 = 0.56:
Ans: 0.56 is greater than 0.5, so this is High Fill-Factor.
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A Pipe of 3" diameter?
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We are using 12" coil, having radius 12/2 = 6".
The area of coil is πr2 = 3.14 x (6)2 = 113 inch2
The area of 3" diameter is πr2 = 3.14 x (1.5)2 = 7 inch2.
The Fill-Factor is ratio of cross-sectional area of Part being inspected to the cross-sectional area of coil. so, we have = 7 ÷ 113 = 0.06:
Ans: 0.06 is less than 0.1, so this is low Fill-Factor.
Length/Diameter Ratio:
The formula for Calculating the current for coil shots contain the ratio L/D. this is the ratio of the object length to the diameter of the object.
The L/D ratio Affects the ease of Magnetisation: The poles of a bar magnet tend to work against the magnetic field in the magnet. to demagnetize it. Consider the North pole of a magnet. The field lines tend to spread out from the pole in all directions. some of the field from the north pole passes back through the body of the magnet toward the south pole, subtracting from the magnet's primary field. The closer the north pole to the south pole of the magnet, the stronger is this self-demagnetizing effect.
Long & thin parts are in fact easier to magnetize in the coil than short and fat parts. because the shorter parts have a stronger self-demagnetization.
L/D Calculation:
Q: A Pipe has a diameter of 9inch and a length of 24 inch, what is the L/D ratio?
Ans: L/D = 24 ÷ 9 = 2.6:
Formulas for ampere-turns in Coil shots:
The following table shows the formulas used to calculate the maximum current for coil shots. We can use these equations only when Length is less than or equal to 18inch , and L/D is in the range 2 to 15.
| Fill-Factor | Formula | Definition |
| High Fill-Factor | NI = K / {(L/D) + 2} | Where:
» N = Number of turns in the coil.
» I = Coil Current (A).
» K = 35,000 (Constant).
» L = Part Length.
» D = Part Diameter.
» NI = Ampere turns.
|
| Intermediate Fill-Factor | NI = (NI)hf (10 - Y) + (NI)lf (Y - 2)/8 |
Where:
» NIhf = Value of NI calculated for high fill-factor coils using High Fill-Factor formula.
» NIlf = Value of NI calculated for Low Fill-Factor coils using Low Fill-Factor formula.
» Y = ratio of the cross-sectional area of the coil to the cross sectional of the part.
|
| Low Fill-Factor | NI = K / (L/D) |
Where:
» N = Number of turns in the coil.
» I = Coil Current (A).
» K = 45,000 (Constant).
» L = Part Length.
» D = Part Diameter.
» NI = Ampere turns.
|
Examples:
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Calculate the Required Current for 3.5" diameter, 16" long Pipe using 1200 turn, 12" coil.
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The L/D ratio = 16/3.5 = 4.5
coil radius = 12/2 = 6inch
Cross-sectional area of coil = πr2 = 3.14 x 36 = 113inch2
Pipe radius = 3.5/2 = 1.75inch
Cross-sectional area of Pipe = πr2 = 3.14 x 3.06 = 9.6inch2
Fill-factor = 9.6/113 = 0.08
Fill-Factor is Low so we use formula:
NI = K/(L/D)
NI = 45,000/4.5
NI = 10,000 (ampere turns)
I = 10,000/N
I = 10,000/1200
I = 8.33 A
Ans: 8.33A current required for this tool.
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Calculate the Required Current for 8" diameter, 18" long Pipe using 1200 turn, 10" coil.
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The L/D ratio = 18/8 = 2.25
coil radius = 10/2 = 5inch
Cross-sectional area of coil = πr2 = 3.14 x 25 = 78.5 inch2
Pipe radius = 8/2 = 4inch
Cross-sectional area of Pipe = πr2 = 3.14 x 16 = 15.24 inch2
Fill-factor = 15.24/78.5 = 0.64
Fill-Factor is high so we use formula:
NI = K/{(L/D) + 2}
NI = 35,000/2.25+2
NI = 35,000/4.25
NI = 8235 (ampere turns)
I = 8235/N
I = 8235/1200
I = 6.86 A
Ans: 6.86A current required for this tool.
NOTE: The current required is higher with the part at the center of the coil. it is common to place the part at the edge of the coil so that a lower current may be used. In addition, it is usually easier to position the part at the base of the coil. Often the part is simply laid on the coil structure itself.