Figure 1-3
Figure 4-5

As of January 2000, my experience in magnetic particle testing spans 16 years, three employers, 15 Level II colleagues and nine individuals claiming ASNT NDT Level III certification. Out of all of this, only my current sub-contracted ASNT NDT Level III has any knowledge of the intermediate fill factor equation.

The most common excuses I have encountered are "I can't make that equation work" and "I don't understand the symbols used" (despite the fact that the equation's symbols are defined in the specification). This simply illustrates the lack of algebraic skills with many Level II and ASNT NDT Level III personnel or perhaps it demonstrates a lack of interest in investigating and using the equation. By using the low or high fill factor equations instead of the intermediate fill factor equation when needed, wrong amperages can result in either hidden or insufficient definition of indications.

The equation looks formidable. However, by using the basic rules of mathematics and reorganizing the equation, it all becomes simple. After repeated use, it becomes a nice way to kill a little more time before beginning the drudgery of repetitive mag processing.

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By using the basic rules of mathematics and reorganizing the equation, it all becomes simple.

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How to Determine if the Intermediate Fill Factor Equation is Necessary
This involves the comparative sizes of the cross sectional area of the coil to be used and the cross sectional area of the part to be magnetized in it (Figure 1).

(1)

If the value of t is between two and 10, then the intermediate fill factor equation is necessary to calculate the required amperages.

 

Calculating the Coil Cross Sectional Area
The cross sectional area is the total area enclosed in the area of the coil. This is calculated by using Equation 2. There are many equations capable of calculating the area enclosed within a circle. The one used here will be Equation 2. In this equation D represents the inside diameter of the coil or outside diameter of a part as required. The value of 0.7854 is a constant derived from the equation used to determine the cross sectional area of a circle. That is,  pr²=p(D/2)²=(p/4)(D²)=(0.7854)(D²).

(2)

We can see that the cross sectional area of a 305 mm (12 in.) diameter coil is 0.073 m² (113 in.²) using this equation. Likewise, the cross sectional area of a 406 mm (16 in.) diameter coil is 0.13 m² (201 in.²) These numbers will not change because the coil diameter is not adjustable (ignoring the use of cables). The cross sectional area of the coil is then divided by the cross sectional area of the part to be tested.

 

Calculating the Cross Sectional Area of the Part to Be Tested
By definition, any part to be longitudinally magnetized must at least be linear in dimension. The usual definition of linear is three times longer than wide. It is the width and height, or diameter, which are used to determine the cross sectional area, not the length. Using Figure 2, it is easy to solve the equation necessary to achieve this: w x h = cross sectional area of the part. So for Figure 2a, the area is 102 x 76 mm = 0.008 m² (4 x 3 in. = 12 in.²) and for Figure 2b, the area is 102 x 102 mm = 0.01 m² (4 x 4 in. = 16 in.²).

In cylindrical parts, Equation 2 is used for the configuration of the part, where D is the largest diameter of the part. Don't confuse this with D_eff if the part is hollow. Using Figure 3 as an example, by plugging the values into Equation 2 we get: 140² x 0.7854 = 0.015 m² (5.5² x 0.7854 = 24 in.²)

At this point we return to Equation 1. With a 406 mm (16 in.) coil and the part from Figure 2a, the equation would be

(3)

Note that for ease of reading, those equations requiring conversions are given in SI in the body of the text and in the English equivalent in the appendix. With a 305 mm (12 in.) coil and using the part from Figure 3, the equation would be

(4)

Consequently, the intermediate fill factor equation would be required only for the part using the 305 mm (12 in.) coil. If the value were below two, then the high fill factor equation would be used. If the calculated value is above 10, the low fill factor equation is used.

 

Intermediate Fill Factor Equation
The equation found in most specifications is from ASTM E 1444:

(5)

The equation itself consists of terms and factors. Terms are either added together or subtracted from each other. Factors are either multiplied together or divided by one another. In this equation (NI)_h is the ampere turns when calculated using the high fill factor equation, (NI)_l is the ampere turns when calculated using the low fill factor equation and t is the ratio of the cross sectional area divided by the part cross sectional area.

What becomes obvious from Equation 5 is that seven or eight other equations will have to be solved in order to use their answers to begin the intermediate fill factor equation. These equations are:

• the coil area
• the part area
• the ratio of coil area to part area (t)
• D_eff if the part is hollow or x if square or rectangular
• the number of parts to be shot at one time (this involves the effective coil length)
• L for L/D, which includes whether more than one part is to be coil shot at a time
• the low fill factor equation
• the high fill factor equation.

Computing the coil area, part area and the ratio of coil and part areas have been covered above. Let's continue with D_eff in the event that the part is hollow.

If the outside diameter (OD) of the part is 109 mm (4.3 in.) and the inside diameter (ID) is 102 mm (4 in.), then the D_eff equation would apply to determine the effective diameter of the part.

(6)

so that, in this case,

(7)

The answer, 38.4 mm (1.6 in.), can then be substituted for D in both the low and high fill factor equations.

If more than one part is going to be shot in the coil at the same time, the total length of the parts must not exceed the effective length of the coil. Some specifications define the coil effective length as the coil radius from the center of the coil out on both sides. Others define it as the coil width plus 152 mm (6 in.) on both sides. No doubt there are other definitions in other specifications.

However the appropriate specification defines it, the effective length must be divided by the length of one part. The quotient of this equation produces the maximum number of parts that can be shot in the coil at one time (rounded down, of course). This number multiplied by the length of the part represents L in the L/D denominator of the ratio for both high and low fill factor equations.

Let's go through this mathematically. If the effective coil length is 152 mm (6 in.) on both sides, measured from the center of the coil, then any coil used would have an effective length of 305 mm (12 in.), even a 406 mm (16 in.) coil. Shooting parts 71 mm (2.8 in.) in length, the equation would be effective coil length/part length = 305/71 = 4.3 mm (12/2.8 = 4.3 in.). Therefore, a maximum of four pieces could be shot at one time.

The high and low fill factor equations should be very familiar to the average Level II, so they will not be covered in detail. Using these formulas, you will calculate the values necessary for use in the intermediate fill factor equation:

low fill factor equation

high fill factor equation

Having calculated the necessary values from the above equations, simply substitute them in the indicated positions of the equation. Solving the equation from this point is a problem for some people. There is a simple answer to this problem. After plugging the numbers into the equation, do the addition last. That is, do all the math on the left side of the addition sign, then do all the math on the right side of the addition sign. Lastly, add both sides together to get the ampere turns necessary. I will give two examples to illustrate.

 

Example 1
Consider a situation where the part configuration is round and hollow, OD = 147 mm (5.8 in.), ID = 104 mm (4.1 in.) and length is 66 mm (2.6 in.). The machine coil is 305 mm (12 in.) in diameter. Specifications define coil effective length as 394 mm (15.5 in.); the coil is 89 mm (3.5 in.) wide plus 152 mm (6 in.) on both sides. The part's effective diameter, then, is:

(10)

Coil area is:

(11)

Part area is:

(12)

Ratio of coil area to part area t is:

(13)

The number of parts per shot is:

(14)

L for L/D_eff is:

(15)

The low fill factor equation is:

(16)

The high fill factor equation is:

(17)

Do all of the above equations one at a time and label them: this type of organization helps to create the clarity necessary to find any problems that may later arise. I suggest recording the equations on the back of the technique sheet, which will not become easily lost. Now it is a matter of plugging the above answers into the intermediate fill factor equation. This will give you:

(18)

Then solve the equation one side at a time and one line at a time.

• (NI) = (7000) (6)/8 (simplifying left side only)
• (NI) = 42 000/8 (multiplying first)
• (NI) = 5250 ampere turns (then dividing)
• + (15 000) (2)/8 (simplifying right side)
• + 30 000/8 (multiply first)
• + 3750 ampere turns (then dividing)
• 5250 + 3750 (then add both sides)
• 9000 ampere turns (at this point your would divide by the number of turns in the coil ± 10 percent).

 

Example 2
The part configuration is square, 86 mm (3.4 in.) wide and high, 127 mm (5 in.) long (Figure 4). The machine coil is 305 mm (12 in.) in diameter. Specifications define the effective length of coil equal to coil diameter, 152 mm (6 in.) on both sides measured from the center of the coil outward. The first task is to calculate the cross sectional area of the part. Using the equation w x h = cross sectional area, we get 86 x 86 = 0.007 m² (3.4 x 3.4 = 12 in.²) As the example part is not round in configuration, the D in the L/D ratio is no longer usable. It must be substituted with another value. As the part is square (or if it were rectangular) it is replaced by the value of x as illustrated in Figure 5. To calculate x, use Equation 19:

(19)

So, for example 2, we get

(20)

As 86² = 7396, we end up calculating the square root of 14 792 and rounding off, getting an answer of 122 mm (5 in.). Consequently, 122 mm (5 in.), the value of x, replaces D in the L/D ratio for calculating high and low fill factors. Our calculation then proceeds in the following order.

Coil area:

(21)

Part area:

(22)

Ratio of coil area to part area t:

(23)

Number of parts per shot:

(24)

L for (L/x):

(25)

x to replace D:

(26)

Low fill factor equation:

(27)

High fill factor equation:

(28)

Hence, we get our variables and can plug them into the intermediate fill factor equation for example 2:

• (NI) = (8750) (10 – 9)/8 + (22500) (9 – 2)/8
• (NI) = (8750) (1)/8 + (22500) (7)/8 (simplifying terms)
• (NI) = (8750/8) + (157500/8) (multiplying)
• (NI) = 1093 + 19 687 (then dividing)
• (NI) = 20 780 ampere turns (adding; at this point you would divide by the number of turns in the coil ±10 percent).

The second example of the intermediate fill factor equation does the equation line by line, one step at a time, both sides at the same time. This too can be done on the back of the technique sheet for future reference.

 

Discussion
What can the final answer tell us? The final answer should be between the low and high fill factor answers. The low fill factor answer was 22 500 and the high fill factor answer was 8750. The intermediate fill factor answer was 20 780. Though closer to 22 500, it is still between the two numerical values. This gives us a hint that the answer should be correct. If the answer were lower than the lowest answer or higher than the highest answer, then something must be wrong.

What doesn't the equation tell us? If the part has more than one diameter, it may need multiple coil shots at various amperages. If one area of the part needs intermediate fill factor amps, the other may need low fill factor amps or high fill factor amps. If it does, the high and low fill factor ampere turns have already been worked out. Just divide them by the number of turns in the coil being used.

 

Acknowledgments
I wish to thank Betty Dunckley, Armando Del Atorre and Paul Dozois of Glendale Community College, Glendale, California, for their assistance.

 

References
ASTM E 1444, Standard Practice for Magnetic Particle Examination, West Conshohocken, Pennsylvania, ASTM, 1994.

 

Appendix
Below are equations from the main body of the feature with English units replacing the SI units. Any equations not using measurements are not reproduced here.

(3)

 

(4)

 

(7)

 

(10)

 

(11)

 

(12)

 

(13)

 

(14)

 

(15)

 

(16)

 

(17)

 

(20)

 

(21)

 

(22)

 

(23)

 

(24)

 

(25)

 

(26)

 

(27)

 

(28)

 

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