- Email: [email protected]
242mAm fueled nuclear battery
ARTICLE IN PRESS
Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
242m
Am fueled nuclear battery
Yigal Ronen*, Amir Hatav, Nir Hazenshprung Department of Nuclear Engineering, Ben-Gurion University of the Negev, Beer-Sheva 84120, Israel Received 22 October 2003; received in revised form 5 April 2004; accepted 14 April 2004 Available online 17 June 2004
Abstract A nuclear battery based on the direct energy conversion of the fission products is presented. Such energy conversion is possible by using a nuclear reactor with ultra-thin fuel elements of 0.2 mm of 242mAm. The amount of nuclear fuel is 376 g and the dimensions of the battery are 2:4 2:4 2:4 m3 (including the vacuum spacing), with a BeO moderator and Be electrodes. The total power of the reactor is 10.6 MW and the electrical power is 0.652 MW. r 2004 Elsevier B.V. All rights reserved. PACS: 84.60.d; 28.41.Ak; 28.50.Ma Keywords: Nuclear battery;
242m
Am; Special reactor
1. Introduction There is a growing interest [1–3] in nuclear batteries, in particular for outer space applications. The oldest type of nuclear battery was suggested by Moseley in 1913 [4]. His device was composed of two electrodes with a 150 kV potential and a vacuum between the electrodes; b particles from a radium source of 20 mCi produced a current of 1011 A, leading to a power of 1.5 mW. The best sources for the direct-charging or direct-conversion of nuclear batteries [4,5] are fission products, which have very high energy and high ionization. However, in order to use *Corresponding author. Tel.: +972-7646-1372; fax: +9727647-2955. E-mail address: [email protected] (Y. Ronen).
these fission products directly, the fuel elements must be very thin (on the order of microns), so that the fission products can escape from the fuel. However, in order to obtain a critical reactor with high power, conventional nuclear fuels cannot make the reactor critical. The use of 242mAm as a nuclear fuel [6] creates a critical reactor with ultrathin fuel elements. The principal behind direct-charging or directconversion [5] is based on the direct conversion of fission product energy into electrical energy, using a high voltage potential. The kinetic energy of the fission products is thus converted to potential energy and the charges collected in the conductive electrodes create an electrical current. High-power nuclear batteries are important, due to the fact that they have almost no moving parts. As a result, maintenance problems (especially
0168-9002/$ - see front matter r 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.nima.2004.04.238
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Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
important in outer space) are considerably reduced. In this paper, we present a preliminary design of a nuclear battery based on a 242mAm nuclear fuel. The main characteristic of the proposed design is its simplicity, even though, as a result of this, efficiency is reduced.
2. The reactor The reactor is composed of 242mAm as a nuclear fuel with a thickness of 0.2 mm and a moderator of 4 cm of BeO and two 0.5 cm thick electrodes of Be, as presented in Fig. 1. The moderator-to-fuelvolume ratio is Vm =Vf ¼ 250 000: The infinite multiplication factor for this design is [6] kN ¼ 1:8: In order to obtain an actual reactor, we have considered the form of a cube with x–y dimensions of 2:4 2:4 m2 and with 24 electrical units, as presented in Fig. 1. Ignoring the He coolant channels, as well as the vacuum for the buckling calculations, we have an actual cube of 2:4 2:4 2:4 m3 ; with a buckling of 5:14 104 cm2 :
Using the BOXER code system [7], we have calculated two parameters. The first parameter is keff ; the value of keff ¼ 1:477: In order to reduce keff to keff ¼ 1; we need burnable poison. Burnable poison rods should be inserted into the moderator. We also need control rods, which will be inserted into the moderator, as presented in Fig. 2. In this preliminary study, we have not discussed these aspects in any detail. The second parameter that we have calculated with the BOXER code system [7] is the amount of fuel needed for criticality and we found that it is 193 g, compared to 376 g in the present reactor. We need 1.34 g of 242mAm in order to obtain 1 MWD, so the difference between the actual fuel weight and the fuel weight needed for criticality can give us an indication of the upper limit of the energy that can be obtained from this reactor. In this case, it is 136.6 MWD. Increasing the fuel thickness from 0.2 to 0.4 mm will increase the fuel weight to 752 g, which will enable the reactor to produce more than 400 MWD. In order to get more accurate results on the amount of energy that can be obtained, detailed burn-up calculations are needed.
Fig. 1. Battery layout (side).
ARTICLE IN PRESS Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
641
Given all the above, the voltage which will yield the highest efficiency is determined by V¼
Fig. 2. Battery layout (top).
The weight of the Be (electrodes) is 2:54 103 kg and the moderator BeO weight is 16:5 103 kg; so the total weight of a bare reactor is about 19:04 103 kg: We have determined the neutron thermal flux to be 5 1013 neutrons=cm2 s: As a result, the power of one unit will be 441.5 kW and of the entire reactor—10.6 MW (th).
3. The electrical consideration In order to calculate the electrical power and the total efficiency, we made several minor assumptions and did not deal with certain other factors: 1. We did not take leakage currents into account. 2. Nor the effect of space charges, nor 3. the effects of radiation loss (bremsstrahlung) of the fission products. 4. We assumed that all fission products have two average energies and the same electrical charge of 20 (e+), 5. and that the emission of the fission products is isotropic. 6. The ratio of space between the electrodes and 4 the electrode dimensions was taken to be 240 240; which indicates that the electrodes can be considered as infinite.
bT0 q
ð1Þ
where T0 is the value of the kinetic energy of the fission product and q ¼ 20ðeþ Þ: The best value of the reduced voltage b was determined to be 0.445 [5]. The voltage of Eq. (1) was determined by the average energy of the fission products. However, fission products are characterized as being light or heavy. As a result, two average energies should be considered one for the heavy and one for light fission products. The average energy of these two fission product groups is 104.5 and 76.5 MeV. Thus, using Eq. (1) with V ¼ 2:024 MV; we can determine the reduced voltages b1 and b2 of the fission product groups. These values are 0.389 and 0.533. When the relative effects are insignificant as in this case of fission products, the maximum efficiency of the electrical energy to the kinetic energy of the fission products is given by [5] pffiffiffiffi b Zi ¼ i ð1 bi Þ i ¼ 1; 2: ð2Þ 2 The maximum efficiencies for the two fission product groups are Zi ¼ 0:0732 and 0.0719. The relative currents of the fission products, which are overcoming the potential barrier, are given by pffiffiffiffi ð3Þ ji ¼ 1 bi i ¼ 1; 2: The results are j1 ¼ 0:3763 and j2 ¼ 0:2699: However, in order to obtain more realistic values, we need to multiply these relative currents by two factors. The first factor is 0.5, which considers the fact that only half of the fission products are moving in the direction of the anode. The second factor is 0.95, which is the relative ratio of the fission products escaping the fuel [6]. The fission rate density is Sf f: Since there are two fission products per fission, the fission products’ density is 2Sf f ¼ 2:37 1016 ½#=cm2 s: Thus, the number of fission products per cm2 of fuel is: S ¼ 2:37 1016 0:2 104 ¼ 4:74 1011 ½#=cm2 s:
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Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
implemented, using thinner fuel elements in the center of the reactor and thicker fuel elements far from the center. As a result, flux flattening will be obtained in the azimuthal direction. The amount of cooling needed in the hottest channel will be Q’ max 414500 3:875 m ¼ 1:546 kg=s: ð6Þ ¼ ’ ¼ 5193 200 Cp DT
Fig. 3. System layout.
As a result the total current will be JðV Þ ¼
S qðj1 þ j2 Þ0:95 0:5: 2
ð4Þ
Thus, the overall current is JðV Þ ¼ 0:233½mA=cm2 : Therefore, the electrical power of the battery will be Pe ¼ JðV Þ A V n
ð5Þ
where n is the number of units in the battery, and, in this case, it is 24. With an electrical potential of 2.02 MV and an area A of 240 240 cm2 ; the battery power is 0.652 MW. The total reactor power is 10.6 MW (th), yielding a battery efficiency of Z ¼ 0:652=10:6 ¼ 0:0615; which is quite low. Using electrical field suppression grids or using magnetic suppression or Venetian-Blinds [8] might improve the efficiency, but it will complicate the system (Fig. 3). So simplicity versus efficiency should be weighed. In this work, the ‘‘simplicity’’ aspect was discussed.
4. Heat removal The total power of the reactor is 10.6 MW and the electrical power is 0.652 MW. We need to remove 9.948 MW of heat, or 0.4145 MW of heat per one unit. To remove this amount of heat, we have considered helium cooling [9] in the moderator, as demonstrated in Fig. 1 The ratio of maximum power density to the average power density in a rectangular reactor is 3.875. This ratio can be reduced if flux flattening is
We assume a maximum of 200 C difference between the temperature of the helium entering the hottest channel of the reactor, which is in the middle, and leaving it. The Cp of helium is 5193 J=kg C at 1200 K and Q’ represents the average heat power to be removed. Given a channel of 1 cm width, we can calculate the coolant velocity m ’ v¼ ð7Þ rc A where A is the channel cross-section 2:4 0:01 ¼ 0:024 m2 and the density of helium at a pressure of 30 atm is rc ¼ 1:346 kg=m3 at 1200 K. This gives us 1:546 ¼ 47:86 ½m=s: ð8Þ v¼ 1:346 0:024 The average temperature difference in the reactor coolant will be Q’ av: 414500 DT ¼ ¼ 51:6 C: ¼ ð9Þ 5193 1:546 Cp m ’ We have also calculated the temperature distribution for the hottest spot in the reactor. The highest value for the power density is in the middle of the reactor and it is larger by a factor of 3.875 than the average power density. The hottest channel is in the middle of the reactor and the highest temperature is about 24 cm from the edge of the reactor. Assuming a cooling rate of 0:985 kg=cm; we get the highest temperature distribution, as given in Fig. 4. We can see that the highest temperature is at the surface of the Be anode, which is also the fuel temperature. The highest calculated temperature is 1069 C. However, the actual temperature will be lower, since there is axial cooling in the metal beryllium. We see that the fuel temperature of 1069 C is below
ARTICLE IN PRESS Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644 1080
Be Am
Beo
He 1060
1063.27
1069
1040 1040.27 1020
T[C]
1000 980 960 940 920 900 880
892.26 0
0.5
1
1.5
2
2.5
3
Element - width[cm]
Fig. 4. Temperature distribution in the reactor.
the melting point of the fuel, which is 1176 C. If a lower fuel temperature is needed, which is probably the case, we have to reduce the average temperature of the coolant, which will lead to a larger radiator with a larger surface area. Reducing the coolant temperature by 100 C will increase the radiator area by a factor of 1.44. The melting point of the beryllium metal is 1280 C, well below the temperatures in question. We can see that the hottest temperature of the coolant is 892 C, leading to an entrance temperature of 692.3 C and the outlet average temperature of the coolant is 692:3 þ 25:8 ¼ 718:1 Cð991 KÞ: The helium coolant is cooled in the radiator by radiating its heat to the outer areas (Fig. 3). The total area of the radiator is given by A¼
sðT24
Q’ T14 Þ
ð10Þ
where T2 is the average temperature of the coolant and T1 ; the temperature of outer space which is 0 K. The Stefan–Boltzmann constant is: s ¼ 5:67 108 W=m2 K: Thus: A¼
9:928 106 ¼ 181:3 ½m2 : 5:67 108 9914
ð11Þ
Therefore, 181.3 m2 is the lower limit value. The actual area will be larger.
643
Assuming that heat radiates from both sides of the radiator, we need at least a 13:5 13:5 m size radiator. The net electrical power of 0:652 MW does not include a loss of electrical power due to pressure losses in the system. Major pressure losses may occur in the radiator and in the core. In order to calculate pressure losses in a radiator, we need an actual design, but the detailed structure of our radiator cannot yet be determined at this early stage of the design. However, the pressure losses in the reactor were calculated. The power needed to overcome the pressure drop in the core was found to be 38:5 kW: So the total power needed to overcome the total pressure drops in the core and the radiator will be greater by a factor of at least two.
5. Summary In this paper, we present preliminary results of a nuclear battery. This battery is based on the direct energy conversion of the fission products of 242m Am nuclear fuel. The reactor/battery is 2:4 2:4 2:4 m3 (including the vacuum between the anodes, the cathodes and the cooling channels) has a 0.2 mm thickness of 242mAm fuel; the total fuel weight is 376 g. The moderator is composed of 16.5 tons of BeO, and the anodes and cathodes are of Be metal, with a total weight of 2.54 tons. The thermal power of the reactor is 10.6 MW and the electrical power is 0.652 MW, having an efficiency of only 6.15%. A radiator is used in order to cool the reactor and cooling is obtained by heat radiation to the cold outer areas. Such cooling requires a radiator with an area of about 181.3 m2. More accurate calculations are needed in order to obtain a more exact value for the radiator’s area. We also need a compressor for the helium coolant pressure drop in the reactor core and the radiator. As a result, there is a loss of power. Using hydrogen, instead of helium, as a coolant might reduce the power of the compressor by a factor of about five. We do have preliminary results of direct nuclear energy conversion to electricity with minimal
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moving parts. Only the compressor, needed to overcome the coolant pressure losses, has moving parts. If more moving parts are included in the design, more electrical power can be generated by using the helium coolant in a gas turbine connected to a generator to produce electricity in a conventional heat cycle. The upper limit of heat-cycle efficiency is determined by the Carnot cycle as T1 T2 Z¼ ð12Þ T1 where T1 ¼ 991 K and T2 is determined by the area of the radiator. The increase in the radia’ tion area is given by ðT1 =T2 Þ4 ½QðradiatorÞ= ’ QðreactorÞ: In order to increase the electrical power of the reactor beyond the power obtained by direct energy conversion, we need to have a heat cycle which will include a gas turbine and an electric generator. Furthermore, we will have to increase the area of the radiator. Optimization of all these factors will be determined by the goals set. Using a 0.2 mm 242mAm fuel will enable us to get 136.6 MWD of energy and 8.66 MWD of electrical energy. Increasing the fuel thickness to 1 mm, we will have 1880 g of 242mAm, which will give us ca. 1303 MWD of energy or about 80 MWD of electrical energy. With such an amount of fuel, the battery can work at full power for about 123 days without refueling, however, the problems associated with the burnable poison will increase. In order to obtain a better performance of the nuclear battery, several options should be considered. The first is flux flattening, in order to reduce the maximum-to-average power density in the reactor. Reduction of this ratio will lead to a reduction in the velocity of the coolant, leading to lower pressure drops and, as a result, reducing the power
of the compressor. Power flattening in the axial direction can be obtained by using different fuel thicknesses. In the middle of the reactor, we can use a thin fuel element which gradually gets wider toward the ends of the core. The second option that should be considered is a constant refueling mechanism. This, if possible, will lead to an increase in the electrical energy (MWD) of the battery. It will also eliminate the need for burnable poison. These are only preliminary results, to show that this concept, in principal, can work and that all the basic parameters are rational. However, final conclusions must be based on a more accurate analysis of a more detailed design.
References [1] G. Chapline, Y. Matsuda, Fusion Technol. 20 (1991) 719. [2] P.V. Tsvetkov, R.R. Hart, T.A. Parish, Highly efficient power system based on direct fission fragment energy conversion utilizing magnetic collimation, 11th International Conference on Nuclear Engineering, Tokyo, Japan, 2003, ICONE II-36275. [3] S.A. Slutz, D.B. Seidel, R.J. Lipinski, G.E. Rochau, L.C. Brown, Phys. Plasmas 10 (7) (2003) 2983. [4] S.L. Soo, Direct Energy Conversion, Prentice-Hall, Engelwood Cliffs, NJ, 1968. [5] G.H. Miley, Direct Conversion of Nuclear Radiation Energy, American Nuclear Society, La Grange Park, IL, 1970. [6] Y. Ronen, E. Shwageraus, Nucl. Instr. and Meth. A 455 (2000) 442. [7] J.M. Paratte, et al., ELCOS—The PSI Code System For LWR Core Analysis, Paul Scherrer Institute, PSI, Villigen, Switzerland, February 1996. [8] R.W. Moir, W.L. Barr, ‘‘Venetian blind’’ direct energy converter for fusion reactors, Nucl. Fusion 13 (1973) 35. [9] G. Melese, R. Katz, Thermal and Flow Design of Helium Cooled Reactors, American Nuclear Society, La Grange Park, IL, 1984.
Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
242m
Am fueled nuclear battery
Yigal Ronen*, Amir Hatav, Nir Hazenshprung Department of Nuclear Engineering, Ben-Gurion University of the Negev, Beer-Sheva 84120, Israel Received 22 October 2003; received in revised form 5 April 2004; accepted 14 April 2004 Available online 17 June 2004
Abstract A nuclear battery based on the direct energy conversion of the fission products is presented. Such energy conversion is possible by using a nuclear reactor with ultra-thin fuel elements of 0.2 mm of 242mAm. The amount of nuclear fuel is 376 g and the dimensions of the battery are 2:4 2:4 2:4 m3 (including the vacuum spacing), with a BeO moderator and Be electrodes. The total power of the reactor is 10.6 MW and the electrical power is 0.652 MW. r 2004 Elsevier B.V. All rights reserved. PACS: 84.60.d; 28.41.Ak; 28.50.Ma Keywords: Nuclear battery;
242m
Am; Special reactor
1. Introduction There is a growing interest [1–3] in nuclear batteries, in particular for outer space applications. The oldest type of nuclear battery was suggested by Moseley in 1913 [4]. His device was composed of two electrodes with a 150 kV potential and a vacuum between the electrodes; b particles from a radium source of 20 mCi produced a current of 1011 A, leading to a power of 1.5 mW. The best sources for the direct-charging or direct-conversion of nuclear batteries [4,5] are fission products, which have very high energy and high ionization. However, in order to use *Corresponding author. Tel.: +972-7646-1372; fax: +9727647-2955. E-mail address: [email protected] (Y. Ronen).
these fission products directly, the fuel elements must be very thin (on the order of microns), so that the fission products can escape from the fuel. However, in order to obtain a critical reactor with high power, conventional nuclear fuels cannot make the reactor critical. The use of 242mAm as a nuclear fuel [6] creates a critical reactor with ultrathin fuel elements. The principal behind direct-charging or directconversion [5] is based on the direct conversion of fission product energy into electrical energy, using a high voltage potential. The kinetic energy of the fission products is thus converted to potential energy and the charges collected in the conductive electrodes create an electrical current. High-power nuclear batteries are important, due to the fact that they have almost no moving parts. As a result, maintenance problems (especially
0168-9002/$ - see front matter r 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.nima.2004.04.238
ARTICLE IN PRESS 640
Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
important in outer space) are considerably reduced. In this paper, we present a preliminary design of a nuclear battery based on a 242mAm nuclear fuel. The main characteristic of the proposed design is its simplicity, even though, as a result of this, efficiency is reduced.
2. The reactor The reactor is composed of 242mAm as a nuclear fuel with a thickness of 0.2 mm and a moderator of 4 cm of BeO and two 0.5 cm thick electrodes of Be, as presented in Fig. 1. The moderator-to-fuelvolume ratio is Vm =Vf ¼ 250 000: The infinite multiplication factor for this design is [6] kN ¼ 1:8: In order to obtain an actual reactor, we have considered the form of a cube with x–y dimensions of 2:4 2:4 m2 and with 24 electrical units, as presented in Fig. 1. Ignoring the He coolant channels, as well as the vacuum for the buckling calculations, we have an actual cube of 2:4 2:4 2:4 m3 ; with a buckling of 5:14 104 cm2 :
Using the BOXER code system [7], we have calculated two parameters. The first parameter is keff ; the value of keff ¼ 1:477: In order to reduce keff to keff ¼ 1; we need burnable poison. Burnable poison rods should be inserted into the moderator. We also need control rods, which will be inserted into the moderator, as presented in Fig. 2. In this preliminary study, we have not discussed these aspects in any detail. The second parameter that we have calculated with the BOXER code system [7] is the amount of fuel needed for criticality and we found that it is 193 g, compared to 376 g in the present reactor. We need 1.34 g of 242mAm in order to obtain 1 MWD, so the difference between the actual fuel weight and the fuel weight needed for criticality can give us an indication of the upper limit of the energy that can be obtained from this reactor. In this case, it is 136.6 MWD. Increasing the fuel thickness from 0.2 to 0.4 mm will increase the fuel weight to 752 g, which will enable the reactor to produce more than 400 MWD. In order to get more accurate results on the amount of energy that can be obtained, detailed burn-up calculations are needed.
Fig. 1. Battery layout (side).
ARTICLE IN PRESS Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
641
Given all the above, the voltage which will yield the highest efficiency is determined by V¼
Fig. 2. Battery layout (top).
The weight of the Be (electrodes) is 2:54 103 kg and the moderator BeO weight is 16:5 103 kg; so the total weight of a bare reactor is about 19:04 103 kg: We have determined the neutron thermal flux to be 5 1013 neutrons=cm2 s: As a result, the power of one unit will be 441.5 kW and of the entire reactor—10.6 MW (th).
3. The electrical consideration In order to calculate the electrical power and the total efficiency, we made several minor assumptions and did not deal with certain other factors: 1. We did not take leakage currents into account. 2. Nor the effect of space charges, nor 3. the effects of radiation loss (bremsstrahlung) of the fission products. 4. We assumed that all fission products have two average energies and the same electrical charge of 20 (e+), 5. and that the emission of the fission products is isotropic. 6. The ratio of space between the electrodes and 4 the electrode dimensions was taken to be 240 240; which indicates that the electrodes can be considered as infinite.
bT0 q
ð1Þ
where T0 is the value of the kinetic energy of the fission product and q ¼ 20ðeþ Þ: The best value of the reduced voltage b was determined to be 0.445 [5]. The voltage of Eq. (1) was determined by the average energy of the fission products. However, fission products are characterized as being light or heavy. As a result, two average energies should be considered one for the heavy and one for light fission products. The average energy of these two fission product groups is 104.5 and 76.5 MeV. Thus, using Eq. (1) with V ¼ 2:024 MV; we can determine the reduced voltages b1 and b2 of the fission product groups. These values are 0.389 and 0.533. When the relative effects are insignificant as in this case of fission products, the maximum efficiency of the electrical energy to the kinetic energy of the fission products is given by [5] pffiffiffiffi b Zi ¼ i ð1 bi Þ i ¼ 1; 2: ð2Þ 2 The maximum efficiencies for the two fission product groups are Zi ¼ 0:0732 and 0.0719. The relative currents of the fission products, which are overcoming the potential barrier, are given by pffiffiffiffi ð3Þ ji ¼ 1 bi i ¼ 1; 2: The results are j1 ¼ 0:3763 and j2 ¼ 0:2699: However, in order to obtain more realistic values, we need to multiply these relative currents by two factors. The first factor is 0.5, which considers the fact that only half of the fission products are moving in the direction of the anode. The second factor is 0.95, which is the relative ratio of the fission products escaping the fuel [6]. The fission rate density is Sf f: Since there are two fission products per fission, the fission products’ density is 2Sf f ¼ 2:37 1016 ½#=cm2 s: Thus, the number of fission products per cm2 of fuel is: S ¼ 2:37 1016 0:2 104 ¼ 4:74 1011 ½#=cm2 s:
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Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644
implemented, using thinner fuel elements in the center of the reactor and thicker fuel elements far from the center. As a result, flux flattening will be obtained in the azimuthal direction. The amount of cooling needed in the hottest channel will be Q’ max 414500 3:875 m ¼ 1:546 kg=s: ð6Þ ¼ ’ ¼ 5193 200 Cp DT
Fig. 3. System layout.
As a result the total current will be JðV Þ ¼
S qðj1 þ j2 Þ0:95 0:5: 2
ð4Þ
Thus, the overall current is JðV Þ ¼ 0:233½mA=cm2 : Therefore, the electrical power of the battery will be Pe ¼ JðV Þ A V n
ð5Þ
where n is the number of units in the battery, and, in this case, it is 24. With an electrical potential of 2.02 MV and an area A of 240 240 cm2 ; the battery power is 0.652 MW. The total reactor power is 10.6 MW (th), yielding a battery efficiency of Z ¼ 0:652=10:6 ¼ 0:0615; which is quite low. Using electrical field suppression grids or using magnetic suppression or Venetian-Blinds [8] might improve the efficiency, but it will complicate the system (Fig. 3). So simplicity versus efficiency should be weighed. In this work, the ‘‘simplicity’’ aspect was discussed.
4. Heat removal The total power of the reactor is 10.6 MW and the electrical power is 0.652 MW. We need to remove 9.948 MW of heat, or 0.4145 MW of heat per one unit. To remove this amount of heat, we have considered helium cooling [9] in the moderator, as demonstrated in Fig. 1 The ratio of maximum power density to the average power density in a rectangular reactor is 3.875. This ratio can be reduced if flux flattening is
We assume a maximum of 200 C difference between the temperature of the helium entering the hottest channel of the reactor, which is in the middle, and leaving it. The Cp of helium is 5193 J=kg C at 1200 K and Q’ represents the average heat power to be removed. Given a channel of 1 cm width, we can calculate the coolant velocity m ’ v¼ ð7Þ rc A where A is the channel cross-section 2:4 0:01 ¼ 0:024 m2 and the density of helium at a pressure of 30 atm is rc ¼ 1:346 kg=m3 at 1200 K. This gives us 1:546 ¼ 47:86 ½m=s: ð8Þ v¼ 1:346 0:024 The average temperature difference in the reactor coolant will be Q’ av: 414500 DT ¼ ¼ 51:6 C: ¼ ð9Þ 5193 1:546 Cp m ’ We have also calculated the temperature distribution for the hottest spot in the reactor. The highest value for the power density is in the middle of the reactor and it is larger by a factor of 3.875 than the average power density. The hottest channel is in the middle of the reactor and the highest temperature is about 24 cm from the edge of the reactor. Assuming a cooling rate of 0:985 kg=cm; we get the highest temperature distribution, as given in Fig. 4. We can see that the highest temperature is at the surface of the Be anode, which is also the fuel temperature. The highest calculated temperature is 1069 C. However, the actual temperature will be lower, since there is axial cooling in the metal beryllium. We see that the fuel temperature of 1069 C is below
ARTICLE IN PRESS Y. Ronen et al. / Nuclear Instruments and Methods in Physics Research A 531 (2004) 639–644 1080
Be Am
Beo
He 1060
1063.27
1069
1040 1040.27 1020
T[C]
1000 980 960 940 920 900 880
892.26 0
0.5
1
1.5
2
2.5
3
Element - width[cm]
Fig. 4. Temperature distribution in the reactor.
the melting point of the fuel, which is 1176 C. If a lower fuel temperature is needed, which is probably the case, we have to reduce the average temperature of the coolant, which will lead to a larger radiator with a larger surface area. Reducing the coolant temperature by 100 C will increase the radiator area by a factor of 1.44. The melting point of the beryllium metal is 1280 C, well below the temperatures in question. We can see that the hottest temperature of the coolant is 892 C, leading to an entrance temperature of 692.3 C and the outlet average temperature of the coolant is 692:3 þ 25:8 ¼ 718:1 Cð991 KÞ: The helium coolant is cooled in the radiator by radiating its heat to the outer areas (Fig. 3). The total area of the radiator is given by A¼
sðT24
Q’ T14 Þ
ð10Þ
where T2 is the average temperature of the coolant and T1 ; the temperature of outer space which is 0 K. The Stefan–Boltzmann constant is: s ¼ 5:67 108 W=m2 K: Thus: A¼
9:928 106 ¼ 181:3 ½m2 : 5:67 108 9914
ð11Þ
Therefore, 181.3 m2 is the lower limit value. The actual area will be larger.
643
Assuming that heat radiates from both sides of the radiator, we need at least a 13:5 13:5 m size radiator. The net electrical power of 0:652 MW does not include a loss of electrical power due to pressure losses in the system. Major pressure losses may occur in the radiator and in the core. In order to calculate pressure losses in a radiator, we need an actual design, but the detailed structure of our radiator cannot yet be determined at this early stage of the design. However, the pressure losses in the reactor were calculated. The power needed to overcome the pressure drop in the core was found to be 38:5 kW: So the total power needed to overcome the total pressure drops in the core and the radiator will be greater by a factor of at least two.
5. Summary In this paper, we present preliminary results of a nuclear battery. This battery is based on the direct energy conversion of the fission products of 242m Am nuclear fuel. The reactor/battery is 2:4 2:4 2:4 m3 (including the vacuum between the anodes, the cathodes and the cooling channels) has a 0.2 mm thickness of 242mAm fuel; the total fuel weight is 376 g. The moderator is composed of 16.5 tons of BeO, and the anodes and cathodes are of Be metal, with a total weight of 2.54 tons. The thermal power of the reactor is 10.6 MW and the electrical power is 0.652 MW, having an efficiency of only 6.15%. A radiator is used in order to cool the reactor and cooling is obtained by heat radiation to the cold outer areas. Such cooling requires a radiator with an area of about 181.3 m2. More accurate calculations are needed in order to obtain a more exact value for the radiator’s area. We also need a compressor for the helium coolant pressure drop in the reactor core and the radiator. As a result, there is a loss of power. Using hydrogen, instead of helium, as a coolant might reduce the power of the compressor by a factor of about five. We do have preliminary results of direct nuclear energy conversion to electricity with minimal
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moving parts. Only the compressor, needed to overcome the coolant pressure losses, has moving parts. If more moving parts are included in the design, more electrical power can be generated by using the helium coolant in a gas turbine connected to a generator to produce electricity in a conventional heat cycle. The upper limit of heat-cycle efficiency is determined by the Carnot cycle as T1 T2 Z¼ ð12Þ T1 where T1 ¼ 991 K and T2 is determined by the area of the radiator. The increase in the radia’ tion area is given by ðT1 =T2 Þ4 ½QðradiatorÞ= ’ QðreactorÞ: In order to increase the electrical power of the reactor beyond the power obtained by direct energy conversion, we need to have a heat cycle which will include a gas turbine and an electric generator. Furthermore, we will have to increase the area of the radiator. Optimization of all these factors will be determined by the goals set. Using a 0.2 mm 242mAm fuel will enable us to get 136.6 MWD of energy and 8.66 MWD of electrical energy. Increasing the fuel thickness to 1 mm, we will have 1880 g of 242mAm, which will give us ca. 1303 MWD of energy or about 80 MWD of electrical energy. With such an amount of fuel, the battery can work at full power for about 123 days without refueling, however, the problems associated with the burnable poison will increase. In order to obtain a better performance of the nuclear battery, several options should be considered. The first is flux flattening, in order to reduce the maximum-to-average power density in the reactor. Reduction of this ratio will lead to a reduction in the velocity of the coolant, leading to lower pressure drops and, as a result, reducing the power
of the compressor. Power flattening in the axial direction can be obtained by using different fuel thicknesses. In the middle of the reactor, we can use a thin fuel element which gradually gets wider toward the ends of the core. The second option that should be considered is a constant refueling mechanism. This, if possible, will lead to an increase in the electrical energy (MWD) of the battery. It will also eliminate the need for burnable poison. These are only preliminary results, to show that this concept, in principal, can work and that all the basic parameters are rational. However, final conclusions must be based on a more accurate analysis of a more detailed design.
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