Half-Value Layer Calculation The thickness of any given material where 50% of the incident energy has been attenuated is know as the half-value layer layer (HVL). The HVL is inversely inversely proportional to the attenuation attenuation coefficient (m) and the two values are related by -1 the following equation. equation. Since m is normally normally given in units of cm , the HVL is commonly expressed in units of cm.
This relationship comes from the intensity attenuation equation.
It can be seen that if an incident energy of 1 and a transmitted energy is 0.5 (1/2 the incident energy) is plugged into the equation, the thickness (x) multiplied by m must equal 0.693 (since the number 0.693 is the exponent value that give a value of 0.5). X in this case is the half-value layer. Example Calculation What is the HVL for a material with a attenuation coefficient of 0.45/cm?
Radiographic Inspection - Formula Based on Newton's Inverse Square Law Radiographic Inspection - Exposure-Distance Relationship
Where: E1 = Exposure at D1 E2 = Exposure at D2 D1 = Distance 1 from source D2 = Distance 2 from source
When producing radiographs, it is sometimes necessary to change the source-tofilm distance. Since the intensity of the source varies inversely with the square of the distance from the source, the exposure must be adjusted. When the exposure at one distance is known, this information can be used to calculate the new exposure with the equation above. Since exposure is the product of time and amperage, either of these variables can be substituted directly for exposure in the equation. Example 1) An exposure of 560 milliampere – seconds produces an acceptable radiograph at a source-to-film distance of 30 inches. What would the exposure need to be if the source-to-film distance is decreased to 24 inches? Solve the equation for E2, plug in known values, and solve.
Example 2) An exposure time of 1.86 minutes and an amperage of 5.6 mA produces an acceptable radiograph at a source-to-film distance of 30 inches. What would the exposure time need to be to produce a similar radiograph at a
source-to-film distance of 24 inches. Solve for E2 or in this case T 2 since only the exposure time will be adjusted. Then plug in the known values and solve for the new exposure time.
Example 3) An exposure of 5.6 milliamperes with a 30 inch tube to film distance produced a good radiograph. What would the milliamperes need to be, if the tube to film distance is changed to 24 inches? Solve for E2 or in this case C 2 since only the exposure current will be adjusted. Then plug in the known values and solve for the new exposure time.
In radiographic inspection, the radiation spreads out as it travels away from the gamma or X-ray source. Therefore, the intensity of the radiation follows Newton's Inverse Square Law. As shown in the image to the right, this law accounts for the fact that the intensity of radiation becomes weaker as it spreads out from the source since the same about of radiation becomes spread over a larger area. The intensity is inversely proportional to the distance from the source. In industrial radiography, the intensity at one distance is typically known and it is necessary to calculate the intensity at a second distance. Therefore, the equation takes on the form of:
Where: I1 = Intensity 1 at D1 I2 = Intensity 2 at D2 D1 = Distance 1 from source D2 = Distance 2 from source ote: This is the commonly found form of the equation. However, for some it is easier to remember that the intensity time the distance squared at one location is equal to the intensity time the distance squared at another location. The equation in this form is: I1 x d12 = I2 x d22
Example 1) Use Newton's Inverse Square Law to calculate the intensity of a radioactive source at a different distance than the distance it was originally measured. If the intensity of a Iridium 192 source was found to be 62 milliroentgen/hour 100 feet, what is the exposure at a distance of 1 foot.
Where: I1 = Intensity at D
1
I2 = Intensity at D 2 D1 = Distance 1 D2
= Distance 2
Reworking the equation to solve for I2
Substitute in the known values and solve for I 2
Example 2) A source is producing an intensity of 456 R/h at one foot from the source. What would be the distance in feet to the 100, 5, and 2 mR/h boundaries.
Convert Rem per hour to mRem per hour 456R/h x 1000 = 456,000 mR/h Rework the equation to solve for D 2
Plug in the known values and solve
D2= 67.5 feet Using this equation the 100mR/h boundary would be 68 feet, the 5mR/h boundary would be 301.99 feet, and the 2mR/h boundary would be 477.5 feet.
Time-Current Reciprocity Calculations Radiographic exposure is the product of the X-ray system current and the time of the exposure. If it is necessary to change the current or the exposure time while maintaining the overall exposure, the Law of Reciprocity can be used. The equation for this specific case is:
Where: C1 = Current 1 C2 = Current 2 T1 = Time at current 1 T2 = Time at current 2 Example Calculation
An acceptable radiograph is produced using a current of 5mA and an exposure time of 10 minutes. If it is necessary to use a current of 8mA instead of 10mA, what does the new exposure time need to be to produce a similar radiograph? First solve the equation for T 2.
Substitute in the known values and solve.
Radiographic Inspection - Exposure-Density Relationship When it is necessary to adjust the density of a radiography, a simple ratio can be used to estimate the exposure necessary to produce the change. In the straight line portion of the characteristic curves of many films, it can be seen that doubling the exposure will produce a doubling of the film density. Therefore, the following equation can be used to estimate the change in exposure needed to produce a change in the film density. A more accurate calculation can be made using the film characteristic curve and the characteristic curve must be used when one exposure is outside the straight line portion of the curve.
Where: E1= Exposure 1 E2= Exposure 2 FD1= Film density at exposure 1 FD2= Film density at exposure 2 Example Calculation If a exposure of 6.2 mA-minutes produces a film density of 1.5, what exposure will produce a film density of approximately 2.5? Assume that both densities fall on the straightened portion of the film characteristic curve.
Solve the equation for E2, substitute in known values and solve for E 2.
Procedure for Using the Film Characteristic Curve to Adjust the Exposure
1. Locate the measured density (Dm) on the characteristic curve of film being used. 2. Record the relative exposure corresponding to this density. Call this value E m. 3. Record the relative exposure that would produce the target density. Call this value E T 4. Compute the ratio R=EM/ET. This is the amount the actual exposure needs to be adjusted to produce the target density. 5. Compute the adjusted exposure by dividing the actual exposure used to produce the initial radiograph by this ratio (Ea = Ei/R).
Magnification Calculations Sometimes the distance between test specimen and image detector is increased to obtain magnification in the image. Magnification is especially useful when parts being inspected and their details are very small. The farther the test specimen is from the image detector, the greater the magnification achieved. The amount of magnification can be calculated using the following formula. M = (a + b) / a
Where: M = magnification a = distance from source to object b = distance from object to detector
Example Calculation Calculated the geometric magnification when the source to object distance is 80cm and the object to detector distance is 30cm.
M = (a + b) / a M = (80cm + 30cm)/80cm M = 1.375
Geometric Unsharpness Calculations When performing an X-ray inspection, the geometric unsharpness of the inspection setup needs to be taken into consideration, especially when using geometric magnafication. The size of the X-ray tube focal-spot and the magnification factors, namely the source-to-specimen and specimen-to-detector distances, are used to calculate the geometric unsharpness of the inspection setup. The allowable amount of unsharpness is controlled by specification being followed. In general, the allowable amount is 1/100 of the material thickness up to a maximum of 1mm (0.040 inch).
Geometric unsharpness or the size of the penumbra (Ug) shown in the image above can be calculated using the following equation: Ug = f * b/a
Where: f = X-ray generator focal-spot size. a = distance from x-ray source to front surface of material/object b = distance from the material/object to the detector When magnafication is not needed, the test specimen is usually placed as close as possible to the detector and the source is placed some distance from the sample to minimize the penumbra. A greater distance between the source and the object will reduce geometric unsharpness. However, the intensity of the source decreases as the distance increases. Therefore, the source should be placed only as far away as necessary to control the penumbra. If the test object is placed in direct contact with the detector (like is often done in film radiograph) the following formula can be used that takes into account the material thickness instead of the object-to-detector distance. This formula is: Ug = f * t/d
Where: f = X-ray generator focal spot size t = the thickness of the material d = distance from x-ray source to front surface of material/object
Example Calculation Calculate the geometric unsharpness when using an X-ray generator with a 3mm spot size and the test component is 100 cm from the x-ray tube and 50 cm from the detector.
First convert all lengths to like units. Then plug the known information into the equation and solving for Ug. Ug = f * b/a Ug = 0.3cm * 50cm/100cm Ug = 0.15cm
Defect Depth Calculation The depth of a object such as a defect can be determined by taking two exposures with the source shifted in second exposure. As shown in the image, a marker is placed on the surface of the component. Two radiographs are
produced with the alignment between the source and the specimen different for each exposure. Using the following equation, the depth of the defect can be determined by relating the shift of the defect to the shift in the marker in the two radiographs.
Where: D = Defect depth t = thickness of the component SD = Shift of defect in the radiographs SM = Shift of marker in the radiographs Example Calculation When radiographing a component with a thickness of 1.5 inches, a defect is detected. A second radiograph is produced with sample shifted relative to the center of the source cone beam. A penetrameter on the surface of the component is found to shift 0.45 inch between the two radiographs. The defect is found to shift 0.18 inch between the two radiographs. Determine the depth of the defect.
Substitute known values into the equation and solve for defect depth.
Attenuation Calculation The linear attenuation coefficient (m) describes the fraction of a beam of x-rays or gamma rays that is absorbed or scattered per unit thickness of the absorber. m basically accounts for the number of atoms in a cubic cm volume of material and the probability of a photon being scattered or absorbed f rom the nucleolus or an electron of one of these atoms. m is used in the following equation to calculate the intensity of a narrow beam of penetrating radiation after it has traveled a given distance in a material.
Where: Ix =
the intensity of photons transmitted across some distance x
I0 =
the initial intensity of photons
m = x =
the linear attenuation coefficient distance traveled
Using the transmitted intensity equation above, linear attenuation coefficients can be used to make a number of calculations. These include:
the intensity of the energy transmitted through a material when the incident x-ray intensity, the material and the material thickness are known. the intensity of the incident x-ray energy when the transmitted x-ray intensity, material, and material thickness are known. the thickness of the material when the incident and transmitted intensity, and the material are known. and the material can be determined be determined from the value of m when the incident and transmitted intensity, and the material thickness are known.
