# 1 Tube heating

```AHXR108N
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J. Boger FVCC Physics
1
Tube heating
1
Heat is developed in the tube when very energetic electrons crash into the anode. Only a small fraction
of these collisions result in x-ray production, typically less then 1%. For the purpose of tube heating
calculations, x-ray production can be ignored and we can assume all of the electric energy put into
the system results in tube heat given by Q. We get a measure of this energy by taking the product
of voltage, current and exposure time, Q = P t = V It. Heat generated at the focal spot is given by
equation 1 and is measured in joules.
Q = kV e &times; mAs [J]
(1)
Here kVe is the effective voltage applied to the anode. But we know that voltage applied to the anode
is often a rectified AC voltage. So there is a waveform factor to apply to the kVp to get to this effective
voltage. By including a waveform factor w, we get heat in terms of kVp, the common voltage used in
radiography. For a constant voltage the waveform factor is 1, and for single phase with large ripple in
the rectified voltage, the waveform factor is .707. Simply put, the more ripple found in the rectified
anode voltage, the less effectively energy is transferred.
Q = w &times; kV p &times; mAs [J]
Voltage Supply
Constant V
High Frequency
3φ 12-pulse
3φ 6-pulse
1φ
w
Voltage Ripple
1
0%
1
1%
0.99
4%
0.96
14%
.707
100%
It is most important to understand that the unit for heat energy calculated in equation 2 is joules.
1
Much of this discussion was presented in ’Physical Principles of Medical Imaging’, 2nd edition, P. Sprawls
(2)
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J. Boger FVCC Physics
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Heat Unit [HU]
Single phase equipment was common in the early days of x-ray devices, and single phase has the largest
ripple in its rectified output. To make calculations easier, a new unit for energy was introduced called
the Heat Unit (HU). With this definition, a simplified calculation of Q was introduced.
Q = kV p &times; mAs, [HU] For 1φ equipment
(3)
Unfortunately, this simplification added a complication to calculating heat in more advanced systems
like 3φ or hf systems. We need a basic conversion factor from the standard unit of joules to the radiographic unit of HU. Comparing equations 2 and 3 we find a conversion between the unit [HU] and [J].
Q[J] = w1φ &times; kV p &times; mAs
Q[HU ] = kV p &times; mAs
In order to get the HU equation to match
the joule equation, I must introduce a conversion factor. That
J
. Inverting the conversion factor gives us a way to go from joules
conversion factor is simply w1φ HU
to HU and is given in equation 4.
HU
1 = 1.41
J
(4)
Using this conversion with equation 2 produces an equation finding tube heat Q directly in units of [HU].
For example, the formula for Q in units of [HU] for a 3φ x-ray tube is
Q[J] = w3φ &times; kV p &times; mAs
Q[HU ]3φ = Q[J] &times; 1.41
[HU ]
[J]
Q[HU ]3φ = (0.96 &times; kV p &times; mAs) &times; 1.41
= 1.35 &times; kV p &times; mAs.
(5)
(6)
[HU ]
[J]
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2.1
J. Boger FVCC Physics
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Examples
Single Phase tube
A single phase tube is operated at 90 [kVp] and 300 [mAs]. Compute the thermal energy delivered to
the anode in units of [HU] and [J].
Step 1: Calculate the heat energy in joules, [J].
Q = w1φ &times; kV p &times; mAs
= .707 &times; 90[kV p] &times; 300[mAs]
= 19.1[kJ]
Step 2: Now convert from joules [J] to heat units [HU].
Q = 19.1[kJ] &times; 1.41
[HU ]
[J]
= 27, 000[HU ]
Note that in the case of single phase, we can get to heat units directly, which was the purpose of finding
the special unit (HU) in the first place.
Q = 90[kV p] &times; 300[mAs]
= 27, 000[HU ]
2.2
Three Phase, six pulse tube
A three phase tube is operated at 90 [kVp] and 300 [mAs]. Compute the thermal energy delivered to
the anode in units of [HU] and [J].
Step 1: Calculate the heat energy in joules, [J].
Q = w &times; kV p &times; mAs
= .96 &times; 90[kV p] &times; 300[mAs]
= 25.9[kJ]
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J. Boger FVCC Physics
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Step 2: Now convert from joules [J] to heat units [HU].
Q = 25.9[kJ] &times; 1.41
[HU ]
[J]
= 36, 600[HU ]
2.3
High Frequency tube
A high frequency tube is operated at 90 [kVp] and 300 [mAs]. Compute the thermal energy delivered
to the anode in units of [HU] and [J].
Step 1: Calculate the heat energy in joules, [J].
Q = w &times; kV p &times; mAs
= 1.0 &times; 90[kV p] &times; 300[mAs]
= 27.0[kJ]
Step 2: Now convert from joules [J] to heat units [HU].
Q = 27.0[kJ] &times; 1.41
[HU ]
[J]
= 37, 800[HU ]
3
Heat Unit Formulas for various x-ray tubes
Equation 4 is the real conversion factor to convert Heat Units to Joules, .... period. Still, some sources
try to make life easier by providing a different factor for each different type of x-ray tube. I find this
approach does little more than confuse the subject. You do not need these fudged formulas ( they
are not conversion factors) if you know the waveform factor and you follow the examples given above.
Nevertheless, here are the formulas for each type of x-ray tube. I’ll call them the ’cheat-sheet’ of
calculating Heat units directly.
FVCC Physics, J.Boger, 2013
AHXR108N
J. Boger FVCC Physics
X-ray tube type
w
Multiplication factor
w
High Frequency
1
1.41 = whf
1φ
w
3φ 12-pulse
0.99
1.40 = w3φ12
1φ
w
3φ 6-pulse
0.96
1.35 = w3φ6
1φ
w1φ
1φ
.707
1.00 = w1φ
Q[HU ]hf
Q[HU ]3φ12
Q[HU ]3φ6
Q[HU ]1φ
5
Use
= 1.41 (kV p &times; mAs)
= 1.40 (kV p &times; mAs)
= 1.35 (kV p &times; mAs)
= 1.00 (kV p &times; mAs)
```