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The uncompressed radius of the fuel pellet of Sample Problem 43.05 is 20μm. Suppose that the compressed fuel pellet “burns” with an efficiency of 10%—that is, only 10% of the deuterons and 10% of the tritons participate in the fusion reaction of Eq. 43-15. (a) How much energy is released in each such micro explosion of a pellet? (b) To how much TNT is each such pellet equivalent? The heat of combustion of TNT is 4.6 MJ/kg . (c) If a fusion reactor is constructed on the basis of 100 micro explosions per second, what power would be generated? (Part of this power would be used to operate the lasers.)

Short Answer

Expert verified

a) The energy released in micro-explosionis 227 J .

b) The amount of TNT required is 49.3 mg.

c) The power generated is 22.7 kW.

Step by step solution

01

Describe the expression for energy released in micro-explosion

The mass is given by,

m=ρV=43πr3ρ

Here, p is density, and V is volume.

There are equal numbers of H2andH3present, therefore the expression for the number of H2andH3is given by,

N2H=N2H=NAmM2H+M3H

Each fusion reaction releases Q=17.59 MeV of energy, with 10% efficiency, then the total energy released by the pellet is given by,

E=0.10N2HQ=0.10(NAmM2H+M3H)Q=4(0.10)πr3ρNAQ3(M2H+M3H).....(1)

02

Find the energy released in micro-explosion

(a)

The molar mass M2Hof deuterium atoms is 2.0×10-3kg/mol, and the molar mass m3Hof tritium atoms is 3.0×10-3kg/mol. The uncompressed radius of the fuel pellet is 20μmand its density is ρ=200kg/m3.

Substitute all the known values in equation (1).

E=40.10π20×10-6m3200kg/m36.022×1023mol-117.59MeV32.0×10-3kg/mol+3.0×10-3kg/mol=1.42×1015MeV=1.421×1015Mev1.602×1013J/MeV227J

Therefore, the energy released in micro-explosionis 227 J.

03

Find the amount of TNT

(b)

The expression to calculate the amount of TNT is given by,

m=EETNT......2

Substitute all the known values in equation (2).

m=227J4.6×106J=4.93×10-5kg=49.3mg

Therefore, the amount of TNT required is 49.3 mg.

04

Find the power generated

(c)

The expression to calculate power generated is given by,

P=dNdtE.......3

Substitute all the known values in equation (3).

P=100s-1227J=22700W=22.7KW

Therefore, the power generated is 22.7 kW.

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Most popular questions from this chapter

(a) – (d) Complete the following table, which refers to the generalized fission reaction

U+nX+Y+bn235

In certain stars the carbon cycle is more effective than the proton–proton cycle in generating energy.This carbon cycle is

C12+H113N+γ,Q1=1.95MeV,N1313C+e++v,Q2=1.19,C13+H114N+γ,Q3=7.55,C14+H115O+γ,Q4=7.30,15O15N+e++v,Q5=1.73,C15+H112C+4He,Q6=4.97

(a) Show that this cycle is exactly equivalent in its overall effects to the proton–proton cycle of Fig. 43-11. (b) Verify that the two cycles, as expected, have the same Q value.

Question: A 66 kiloton atomic bomb is fueled with pure U235(Fig. 43-14), 4.0%of which actually undergoes fission. (a) What is the mass of the uranium in the bomb? (It is not 66 kilotons—that is the amount of released energy specified in terms of the mass of TNT required to produce the same amount of energy.) (b) How many primary fission fragments are produced? (c) How many fission neutrons generated are released to the environment? (On average, each fission produces 2.5 neutrons.)

A thermal neutron (with approximately zero kinetic energy) is absorbed by a238Unucleus. How much energy is transferred from mass-energy to the resulting oscillation of the nucleus? Here are some atomic masses and neutron mass.

U237237.048723uU237238.050782uU237239.054287uU237240.056585un1.008664u

Roughly 0.0150% of the mass of ordinary water is due to “heavy water,” in which one of the two hydrogens in anH2O molecule is replaced with deuterium,H2 . How much average fusion power could be obtained if we “burned” all theH2 in 1.00 litre of water in 1.00 day by somehow causing the deuterium to fuse via the reactionH2+H2H3e+n ?

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