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Radiopharmaceutical Production

Radiopharmaceutical Production. Heat transfer Beam heating Density reduction. STOP.

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Radiopharmaceutical Production

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  1. Radiopharmaceutical Production Heat transfer Beam heating Density reduction STOP

  2. Heat is deposited in the target material as the beam passes through it. This heat deposition has several important consequences in terms of target design. In this section, we will look at how heat can be removed from the target and how the temperatures in the target can affect the target density and other physical characteristics Contents Heat Balance Conduction Radiation Convection Heat transfer in gas targets Heat transfer in liquid targets Heat transfer in solid targets Heat Transfer and Density Reduction STOP

  3. Heat Balance Qin = Qout • In order to have a stable radionuclide production, the heat deposited in the target must be removed. • The beam power in watts is the beam energy loss in MeV times the beam current in microamps • There are three modes of heat transfer • Conduction • Radiation • Convection • Conduction is important in target bodies and foils • Radiation can be important in foils and solid metal targets • Convection is important inside gas and liquid targets

  4. Conduction The first form of heat transfer to be considered is conduction since in many ways it is the easiest to model. Heat applied to one side of a material will be transferred to the other side at a rate based on the thermal conductivity. Heat Source Heat Sink • Qcond = -kA dT/dx • where • Qcond = heat transferred by conduction (watts) • A = cross-sectional area (cm2) • k = thermal conductivity (watts/cm-°C) • dT = temperature differential (°C) • dx = distance differential (cm)

  5. Conduction In the more usual form of the equation, the heat removed is directly proportional to the thermal conductivity times the area times the temperature difference and inversely proportional to the distance the heat must travel through the substance • where • k = coefficient of thermal conductivity (watt/cm-°C) • A = cross-sectional area (cm2) • x = distance (cm) • T1 = temperature of the hotter part (°C) • T2 = temperature of the cooler part (°C)

  6. Conduction Example We will take an example of a target generating 600 watts of power and see how much heat can be removed through two different metal target bodies of the same thickness and size. Target Interior Beam Metal Body Cooling Water Typical Beam 20 MeV and 30μA = 600 watts If we compare aluminum with nickel for the same thickness, area and temperature differential QAl =(2.37watt-cm-1°K-1)(2 cm2)(100°K)/(0.5 cm)=948 watts QNi =(0.91watt-cm-1°K-1)(2 cm2)(100°K)/(0.5 cm)=364 watts It is clear that it would be possible to remove the heat through the aluminum target body, but that the nickel target would continue to heat up

  7. Radiation The next form of heat transfer is radiation. The amount of heat transferred is very dependent on the temperature of the material. The heat transferred is given by the following equation. • Qrad = AσT4 • where • Qrad = heat emitted as radiation (watts) • A = area of the surface (cm2) • σ = Stefan-Boltzman constant • T = absolute temperature (K)

  8. Radiation Heat Loss The table on the right gives the amount of heat that can be transferred at a given temperature. These calculations assume a perfect blackbody. In reality, the amount of heat being transferred must be multiplied by the emissivity which is a number less than one which represents how close the material is to a blackbody radiator. As can be seen, the amount of heat given off by this mechanism is rather low until very high temperatures are reached

  9. Convection • Qconv = hA(T1-T2) • where • Qconv = heat transferred by convection • A = surface area (cm2) • h = film coefficient (watt/cm2-K) • T = temperature (K) Convection is perhaps the most difficult form of heat transfer to model accurately. Although the equation is rather simple, the value of the film coefficient is rather difficult to estimate or measure and depends on a great number of factors.

  10. Convective heat transfer for laminar flow in tubes As an example, we can take the simple case for convective heat transfer for laminar flow of gas inside tubes. The equation describing this situation is given below. The equation look more complicated and the dimensionless quantities of Reynolds number (Re) and the Prandl modulus (Pr) are made up of combinations of more fundamental constants. Evaluation of these quantities and calculations on the heat removal are beyond the scope of this presentation. The key parameter in convective heat transfer is the film coefficient h which is usually tabulated in books for various geometries and materials More information can be found in texts devoted to heat transfer.

  11. Heat transfer by convection A key parameter in the efficiency of heat transfer will be whether is the flow is laminar or turbulent. Turbulent flow is much more efficient at removing heat than laminar flow.

  12. One of the most common problems for the production of radioisotopes in gas targets is density reduction As the beam is put on target, the pressure and temperature increase as is shown in the figure. The pressure rise is correlated with the temperature rise in the target The target comes to equilibrium fairly quickly The problem has been approached in several different ways. Heat Transfer in Gas Targets

  13. In small volume targets, this problem is more serious The small size means that these targets must be run at high pressure where foil rupture is a distinct possibility In a typical gas target, the total length is on the order of 10 to 15 cm In order to produce carbon-11, they must be operated at about 20 atmospheres Since the volume is small, the pressure rise is large Heat Transfer in Gas Targets 12.5 cm

  14. Heat Transfer in Gas Targets • In the section on target physics, we learned that the beam deposits its energy in the target in the form of heat. This heat raises the temperature of the target material which in gas targets causes an increase in the pressure. The pressure with the beam on over the pressure with the bam off is called the pressure rise ratio. • The pressure rise ratio reaches a maximum when there is just enough gas in the target to stop the beam as shown in the figure • As the pressure increases beyond that point, the ratio falls as the heat transfer is improved

  15. Heat Transfer in Gas Targets • The pressure to stop the beam is dependent on both the energy and the beam current • The variable is the total power deposited in the gas by the beam • As shown in the figure, the higher the total power deposited in the target, the higher the pressure needed to stop the beam.

  16. Heat Transfer in Gas Targets Figure taken from E. Hugel PhD thesis • The fraction of heat transferred by convection is high except very close to the wall • The fraction is relatively independent of gas temperature

  17. 1 2 3 4 5 0 Centimeters Heat Transfer in Gas Targets Beam • Density Reduction is a result of target heating • The heating is not uniform in the beam strike area as is shon in this photograph of the light emission from the beam in a gas target. • There is spreading of the beam as well as a non-uniform distribution from the top to the bottom of the target. Figure courtesy of S-J Heselius

  18. Heat Transfer in Gas Targets • There are convection flow patterns set up inside the target • These currents help in the convective heat transfer • The temperature profile as measured in a gas target shows the non-uniform distribution and suggests the heat flow pattern as shown in the figure below.

  19. Beam Heating in Different Gases The pictures below show proton beams stopping in different gases. The different colors are due to the electronic excitation of the gas by the beam. It should also be noted that the beam areas have different shapes depending on the thermal characteristics of the gases. Figure courtesy of O. Solin

  20. The usual method for the production of fluorine-18 used in FDG is from an irradiation of oxygen-18 enriched water with a beam of protons. The water target used for this production needs to be able to withstand high beam currents and long irradiation times of about 2 hours. If the removal of heat is inadequate, there can be boiling in the target which reduces the yield of fluorine-18 The target itself is usually just a depression cut into a block of metal or a small volume between two metal foils. Liquid Targets

  21. There is substantial density reduction in the water during irradiation if the target is boiling and if no provision has been made for this boiling in the target design. This has several consequences The yield is reduced in the target Higher pressures are created Mechanical action may affect surfaces At right are views of water targets with insufficient heat transfer at beam currents of 5 µA (a) and 10 µA (b) Note the increase in the intensity of the boiling at the higher beam current [18O] H2O Cooling Water Beam Boiling in the target

  22. The boiling in the water can be reduced by raising the pressure on the water, effectively increasing the density In the figure, the light emission is a measure of the amount of boiling in the target. It deceases with increasing pressure as the boiling point of the water is raised. Fluorine-18 Fluoride Production

  23. The pressure build up in the target causes a deformation of the front window, increasing the target volume The foil must be supported or thicker The target shown on the bottom has a support grid to maintain the shape of the foil which keeps the volume constant and allows higher pressures to be applied to the foil without rupturing. Cooling Water Target Body Foil Support Fluorine-18 Fluoride Production

  24. If the beam is too tightly focused, the beam will cause the water to boil Use a diffuse beam for water targets Fluorine-18 Fluoride Production Courtesy of Wake Forest

  25. Another Solution: Recirculate the water through the target Fluorine-18 Fluoride Production

  26. Conductive Heat Transfer Radiative Heat Transfer (>500°C) Beam Convective Heat Transfer Backing Plate Target Material Cooling water flow Heat Transfer in Solid Targets The main operational mode of heat transfer is conduction. Heat loss by convection and radiation are usually very small in comparison to loss by conduction. The thinner the layer between the target and the cooling water, the better the heat removal.

  27. Thallium Layer Copper b layer Copper c layer Incident 30 MeV Proton Beam Cooling Water Flow cc Vacuum bb aa b c a Heat transfer in Thallium Target The figure at right represents a practical situation where there are several layers of conduction in the target. The beam passes through the thallium layer and part way into the copper backing plate before losing all its energy The temperature profile for this target is given on the nest slide.

  28. Thallium Layer Copper b layer Copper c layer Cooling Water Tm Ti Tb Temperature Tc Tw Distance Heat transfer in Thallium Target

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