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18. Estimate the available D-D 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 D-D fusion energy potential in an 8-oz 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 8-oz 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 cm-2) 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 10-13 J/MeYj-1 = 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