Military roentgenology / War Department.

Date:: [1944]

Catalogue details

Licence: Public Domain Mark

Credit: Military roentgenology / War Department. Source: Wellcome Collection.

41/470 page 31

the permissible exposures for half-wave and self- rectification respectively. It will be noted that on full-wave at 80 kvp for 100 ma, the tube may be operated for 5 seconds; whereas, for half-wave the exposure time ts limited to 3 seconds and with self- rectification, to 1¢ second. It should be clear that for the short exposure time on self-rectified circuits as compared with full-wave and half-wave there is a greater strain of the inverse voltage on the tube. This condition becomes more pronounced when the temperature of the target is increased following elec- tron bombardment. As explained in paragraph 44, the valve serves to suppress the inverse voltage, even though the temperature of the target is in- creased. Therefore, in self-rectified circuits, the X-ray tube ts subjected to great electrical and heat loads and hence has a lower capacity than with half or full-wave circuits. To understand the difference between half-wave and full-wave, it must be realized that on half-wave the milliammeter is recording only half the time, since only half of the impulses flow through the circuit. The instrument, however, due to the inertia, records only a steady mean, which in this case is approximately one-third of the peak value attained by the current during any one impulse. The actual current passed through the tube may be as high as 300 ma even though the meter records only 100 ma. For full-wave rectification, since all of the current is rectified and passes through the tube, the moving parts of the milliammeter are activated more frequently. Therefore, the mean current recorded by the milliammeter is nearer the actual peak value, and it has been determined that the peak is approximately one and one-half times the meter reading. Hence, the actual tube current would be 150 ma; whereas, the meter would be reading 100 ma. In the above example, the total power delivered, irrespective of the type of rectification is the same. For half-wave, more electrons for a given meter reading strike the focal spot than for full-wave and, consequently, the temperature of the focal spot approaches its maximum point more rapidly. On full-wave, the current is passed in smaller amounts, but more frequently, (permitting the focal spot to tolerate a greater load and allowing a greater quan- tity of the accumulated heat to be conducted away from the target). The type of rectification, therefore, is an important factor to be considered in connection with maximum energy loads permissible for a single exposure. e. Kilowatt rating. A method of determining the approximate milliamperage capacity of a tube with a given focal spot is based upon the ability of one square millimeter of tungsten to sustarn 200 watts per second or 300 watts for 14 second. This method is designated as the kilowatt rating and is of value, if a rating chart is not available. Some X-ray tubes may be identified as 2, 4, 6, or 10 kw (Philips). The 2 kw tube, for example, has a projected effective focal spot of 1.7 mm. The kilowatt rating and, hence, the permissible kvp and ma-sec. for 1 second may be determined, if the projected effective focal spot (PEFS) is known, by reference to the followmg formula: (PEFS? x 34%) x 200 1000 where PEFS? = the dimensions of the projected focal spot squared, 314 is a constant value, which is used to convert the size of the PEFS into the actual focal spot under bombardment, 200 is the wattage sustained by 1 square millimeter of actual focal spot and 1000 is used to convert watts to kw solving: VE Pie Rp AN 1000 = 2.02 or 2ikw Thus, a tube having a 1.7 mm effective focal spot and a target inclination of 19° may be identified as 2 kw. If the tube is operated at 80 kvp for 1 second, the maximum milliamperage for a 2 kw tube can be determined: Kw rating = Kw rating = fea Tate oi | Where P equals power available to the tube, E is equal to root mean square voltage and I is equal to average current, converting 80 kvp to rms voltage. 80,000 peak volts x .707 = 56,560 volts 3) 2000 7 E 7 565600 .035 or 35 ma Hence, a 2 kw tube, lacking a rating chart may be operated at 80 kvp for 1 second at 35 ma. f. Anode thermal capacity. When exposures are made in rapid succession, heat is stored temporarily in the anode, and then transferred into the surround- ing volume of oil and metal parts. Figure 23a, Is an example of an anode cooling chart. Heat produced in the X-ray tube is expressed in terms of arbitrary units called Heat Units (H. U.) and Is equal to the product of kvp x ma-sec, (based upon Joule’s Law: par. 18). When the heat storage capacity of the anode of an X-ray tube is 75,000 H. U., a series of exposures may be made safely in rapid succession, until the capacity is reached, at which time the anode must be allowed to cool sufficiently, before any other exposures are attempted. g. Thermal storage capacity and heat dissipation rate. In addition to the anode thermal capacity, consideration must be given to the over-all thermal storage capacity and heat dissipation rate. The over- all thermal storage capacity of a tube varies, depend- ing upon the size and rate of heat dissipation of the anode. Various cooling methods are used to cool the anode. (See par. 33e.) The thermal storage capacity of X-ray tubes may vary from 100,000 to 300,000 H. U. for stationary anode tubes to as high as 250,000 H. U. per hour for rotatmg anode tubes. (See fig. 23b.) When the heat storage capacity has been fulfilled, then the rate of thermal dissipation per minute becomes the governing factor in cal- culating the maximum number of exposures that

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