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The following question was answered by an expert in the appropriate field:

Q

While calculating dose rate from a pipeline carrying radioactive solution (which I suppose can be approximated to a line source), what should be the length of the source that should be considered in arriving at dose (1) on contact and (2) at a distance from the surface of the source. The question has more relevance when the pipeline runs to several tens of meters.

A

It is not unusual to simulate an extended straight run of pipe carrying radioactive material as a line source for some dose-rate estimations. The line source approximation is often made in cases for which the length of the pipe is much greater than the diameter of the pipe and when the length is also much greater than the distance from the surface of the pipe to the dose point (a factor of 5 is usually a reasonable minimum factor to use in making this judgment). In some instances—depending on the diameter of the pipe, the wall thickness of the pipe, the density of material filling the pipe, the distance of the dose point from the pipe surface, and the energies of the photons—one may have to take into account the attenuation of photons in the material filling the pipe (e.g., aqueous solution) and in the pipe wall. In the simplest case, one may be able to neglect photon attenuation and still obtain a reasonable estimate of dose rate; this is especially true for small pipe diameters and small pipe wall thicknesses.

For the case in which photon attenuation may be neglected, the dose rate, D, at a point, P, at a perpendicular distance, R, from the center of the pipe (pictured below) of length L1+L2 and carrying a uniform concentration of a specific radionuclide can be shown to be given by:

 L1 L2 R • P

D = (SLΓ/R)(tan-1(L1/R) + tan-1(L2/R)) (1)

where SL is the radionuclide activity per unit length of the line source (e.g., in megabecquerels per meter); for a cylindrical pipe of inner radius, I, carrying a radionuclide concentration, C, the SL value would be given by CπI2. The quantity Γ is the point source dose rate constant for the radionuclide, for example in units of sieverts per hour per megabecquerel at one meter (Sv h-1 MBq-1 m2). The angles given by the arctangent values must be expressed in radian measure. For the common case where the dose point is opposite the center of the line, the above reduces to:

D = (2 SLΓ/R) tan-1(L/R) (2)

where L is half the line length (L1 = L2 = L). We can see from equations 1 and 2 that, for a given radionuclide and a fixed concentration, the dose rate will depend on both the length of the line segment and the distance from the line. We can look at a brief example to demonstrate how great the effect of the selected line segment length is for a selected dose point. We will assume that the dose point is opposite the center of the line length (equation 2); since the 2SLΓ will be constant for any distances selected, we will calculate only the values of tan-1(L/R) for our example. I have assumed an arbitrary total pipe length of 20 meters, giving L = 10 meters, and selected representative distances R of 0.10 m, 0.50 m, and 1.0 m. For each assumed value of R, I have calculated values of tan-1(L/R) for values of L of 0.2 m, 0.5 m, 1.0 m, 2.0 m, 5.0 m, and 10.0 m.

R, meters L, meters
0.2 0.5 1.0 2.0 5.0 10.0
0.10 1.107 1.373 1.471 1.521 1.551 1.561
0.50 0.381 0.785 1.107 1.326 1.471 1.521
1.0 0.197 0.464 0.785 1.107 1.373 1.471

For each value of R, the respective value of tan-1 (L/R) at L = 10 m represents the true value for dose rate when multiplied by 2 SLΓ/R. The degree of uncertainty you are willing to accept will determine how long a length of source you should consider when making calculations/estimates using the line source expression. You can see from the above table that the closer the dose point is to the source, the smaller the assumed line segment may be and still yield a particular level of accuracy. Thus, for the 0.10 m point, the value at L = 1.0 m is about 94 percent of the true value; for the 0.5 m point we would have to go to somewhat less than the L = 5.0 m to obtain comparable accuracy, and for the 1.0 m point we would have to assume a length of between 5 and 6 m to obtain the same accuracy. This is because when the dose point is very close to the extended source, the activity in the source closest to the dose point has a much greater influence than does activity that is farther away in the pipe, this effect being associated with the well-known inverse square law. As we move farther away from the source, activity from all portions of the source exhibit a greater influence on the dose rate. In the extreme, if you imagined being 100 m away from the 20 m long source, all portions of the source would contribute about equally to dose rate at the dose point because every piece of the pipe source is about equidistant from the dose point. From the above we might conclude that if the distance from the dose point to the source is small compared to the segment length, L, if we select a segment length that is at least five times the distance, R, from the source to the dose point the calculated result will likely be within about 15 percent of the true value.

Keep in mind that the above discussion assumed no attenuation of photons in the pipe solution or pipe wall. This is often not the case, and the attenuation will alter the result. Often, when the dose point is close to the pipe source, the attenuation effect will reduce the length of pipe that might have to be considered compared to the unattenuated case. This has to do with the fact that the attenuation is an exponential process that depends on the pathlength of photons through the attenuating material; when the dose point is close to the source, photons traveling at oblique angles from more distant parts of the source must travel through more attenuating material than do photons coming from parts of the source closer to the dose point. The significance of those distant photons then becomes greatly diminished.

The above may be more (or possibly less) than you wanted to abide, but your question is a good one that I think deserves the attention.

George Chabot, PhD

Answer posted on 14 May 2014. The information posted on this web page is intended as general reference information only. Specific facts and circumstances may affect the applicability of concepts, materials, and information described herein. The information provided is not a substitute for professional advice and should not be relied upon in the absence of such professional advice. To the best of our knowledge, answers are correct at the time they are posted. Be advised that over time, requirements could change, new data could be made available, and Internet links could change, affecting the correctness of the answers. Answers are the professional opinions of the expert responding to each question; they do not necessarily represent the position of the Health Physics Society.