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18. Estimate the available DD fusion energy in an 8 ounceglassof water. For how long could this energy provide the energy needsof a house with an average power consumption of 10 kW? Solution: (a) First find the number of deuterium atoms in the glass of water. From Table 1.3, 1 oz = 29.57cm3. Thus in 8 oz of water we have 8 x 29.57 = 236.6 cm3 of water, or with a density of 1 g/cm3, a mass m of 236.6 g. The number of atoms of hydrogen NH is NH = 2NH 0 = 2~
2 AH20 = 2(236.6)(6.022 X 1023) = 1.583 X 1025atoms.
18 From Table A.4, the atomic abundanceof deuterium in elemental hydrogen is 0.015%. Thus, the glass of water contains ND = (0.00015)(1.583 X 1025)= 2.375 X 1021atoms of deuterium. From Example 6.6, we see that each atom of deuterium, when fully fussioned to produce 4He, has a fusion energy potential of ED = 23.82/2 = 11.9 MeV. Thus the DD fusion energy potential in an 8oz glassof water is Etot = ED X ND = 2.826 X 1022 MeV = 4.53 X 109 J. (b) The annual energy consumption in the house is PH = (104 J/s)(365.25 X 24 x 3600) sly = 3.16 x 1011J/y. Thus, the deuterium in the 8oz glass of water could provide the house'senergy needs for Etotl PH = 0.014 Y = 5.24 d.
19. The sun currently is still in its hydrogen burning phase, converting hydrogen into helium through the net reaction of Eq. (6.46). It produces power at the rate of about 4 x 1026 . (a) What is the rate (in kg/s) at which massis being W converted to energy? (b) How many 4He nuclei are produced per second? (c) What is the radiant energy flux (W cm2) incident on the earth? The average distance of the earth from the sun is 1.5 x 1011m. Solution:
The sun's power production is Ps = (4 X 1026J/s)[1.602 X 1013 J/MeYj1 = 2.50 x 1039MeV/s. July 24, 2002 ...
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This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 CADY

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