I am working on a X ray tube project and I want to calculate approximate dose. Do you know any equation ?

I have found one on internet but I think this is wrong:

$$D=g\cdot kV^2\cdot \dfrac{mAs}{d^2}= \left[\dfrac{Sv}{h}\right]$$

So here is $g$-factor a constant and how can I find it (I mean if it is what is it for this equation, if it is not then how can I calculate it, because as well as I know $~g~$-factor depends on the angle of the anode and these things). And here the $~d^2~$ stands for distance but this the distance of what ? The distance between the tube and the object or the distance between cathode (filament) and anode (tungsten) ?

And can I calculate $~mAs (I\cdot t)~$ by $~\dfrac{kV}{R}~$ of the filament ?

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    $\begingroup$ Proportional to voltage squared? What is your source? (Not the x-ray source, I mean where did you get this from?) $\endgroup$ – Pieter Feb 19 at 9:32
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    $\begingroup$ I am trying to understand the expression. I suppose "mA" means milliamps? And "k" is not some constant but also a prefix, so the voltage in kilovolt? Units do not seem to match: the left hand side multiplies by time (in seconds), the right hand side has a dose rate in Sv/h? No, the current of a vacuum tube does not follow Ohm's law. $\endgroup$ – Pieter Feb 19 at 9:44
  • $\begingroup$ @Pieter I found it on researchgate.net, but it seemed me wrong so I wanted to verify it. mAs is miliampere second and kV is kilovolt yes. Do you know any other equation for my purpose sir ? $\endgroup$ – fissile_uranium Feb 19 at 11:06
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    $\begingroup$ @nmasanta It's great that you've converted the formulae to MathJax, but I think you may have changed the meaning slightly. It's hard to tell because the OP hasn't responded to Pieter's requests for clarification. But I think $m$ is supposed to be the milli- prefix, not a separate quantity (eg mass). Also, $mAs (I\cdot t)$ is (probably) supposed to be 2 separate expressions, with the $(I\cdot t)$ being a synonym or explanation of $mAs$. Ah, fissile_uranium just responded as I was writing this comment. ;) $\endgroup$ – PM 2Ring Feb 19 at 11:09
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    $\begingroup$ @fissile_uranium Can you give us a link for the source of that expression from researchgate.net? $\endgroup$ – PM 2Ring Feb 19 at 11:11

This seems to be some engineering-type formula. From a different questions site, and it is difficult to make sense of.

X-ray flux is proportional to current $I$ (which is normally of the order of milli-ampères for x-ray tubes) and inversely proportional to $d^2$ when the distance $d$ is large compared to source size. Dose is also proportional to time $t$ (which can be measured in seconds or hours - unclear what here).

Electrical power is $P = IV$ and when voltage is in kV and the current in milliamps you get the product in watt.

Only a very small part of this power is emitted as x-rays. That fraction is increasing with voltage. A proportional increase would result in that $V^2$ factor, but there is no theoretical justification for proportionality.

So the $g$ in this parametrisation will depend on lots of things: anode material, for example. And angle. And on voltage.

If the left-hand side includes a multiplication with time to compute a time-integrated flux, you would need to multiply with the area of the target to get an absorbed dose. The unit for that is sievert. Dose rate can be expressed in Sv/h, but then you must not multiply with time on the left-hand side.

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