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Unformatted text preview: 22.3 12 ENGINEERING OF NUCLEAR REACTORS Fall 1993 ##5## PROBLEM 3 1 l N Ciﬁwh :nﬁ—trﬂux LPG En — {truth rt. ~3§N~l1c¢ whats Fu—t. 1 attire 5: 5“ Elf" "‘5'! I]. Consider a a nular cylindrical fuel pellet ef length L. in the radius, RV: and eutside radius, am It is operating at qr. such that far a given outside surface temperature, ‘l'fu, the inside 5url'ace temperature TV is lust at the fuel melting litnit Tmeh. A fellew engineer claims that if the same vnlume of fuel is arranged as a sphere with an
inside veided tegien of radius R1,: and eperated between the same twn surface temperature limits
i_e.: Ta and Tm. mere newer can he extracted frees the spherical fuel vnlume than than: the. ITjill'tﬂdl'lEEtl fuel pellet. In hath cases the velumetric energy generatien rate is radially censtant. is this claim cerrect? Preve nr dispruue it. Please use the nomenclature of Fig. l Assume ne sintering nccurs. GWEN: The cinedimensional heat cunductlnn equatidn in the radial directinn in spherical teertliuat'es is
i _d kl; E + m 3
g! dr ( :1: q
l‘et a Sp here: V5=4f dall3 A5=4KRE. Cylinder:
Rv=D.ESEI'I'I
L Rtec=l cm
L=l ern Cylindrical Annular
Fuel Pellet Sphere: Rv=ﬂ.25CI'TI Rtss=? [te he determined tram prehlem statement.) it _
Spherical Hellew Fuel Pellet 22 MG OF NUCLEAR RE 3 Solution to Problem 811 1. Method of solution: Determineﬂ = yq'” for two different shapes of
fuel pellet given the following: a.Steady State h. v“ = y, c.llo sintering d.q"’ is radially independent for the cylinder and the sphere.
Thus a one dimensional soution to the energy equation can he found. For a cylinder q’” can be determined directly from Eq 3454 in text:
T
4 rllll if!“ El»; T,“ arr _ 4J1.” err “123 FM W
I. — __,
aim 1_’ aL] _[ at. hf”) L {[1—{nes} J_[[}_33} £n{4}} Tm
Iker J. Rm“; R.Ir J For the Sphere we must first solve for the radius of the sphere given
the volume. Ins: em = than?“ — a3) + a: =[I.39{:io1n
Starting with the oliedirnEnsional heat conduction equaLion in the
radial direction in spherical coordinates:
1 a 1 3T]
—— r k—— + ”’=t'l
1'1 all Elr q‘ Using the following boundary condition since no heat can ﬂow into
the void inside the sphere: y, =sm1ai H a1) = VI: a sun; — ail qNLEHv = _ ______l F: G Since qg” is not a function of r we can multiply through by r. integrate and apply me above boundary condition:
a: 1 at _
_ _ . rr.i_£ =
ar[fkar]+tl {J
 't "J 9*
[Vi{ﬂ '+ q’ r
t. o'lr J a +CI=CII saving for qf’ : {1 MT .41, =J—an5[ Rf] 12.1 I I _ E ansJ 3 Rf R :'—_———'———I——__________ T'u
new: I 11253 LEE5_[ 1 1 ]
‘ 3 new £1.35 e _ new TmI‘I.
qf’= REEL 'de Thus. mmparing the total power generated in each of the pellets, we Can see the fellow engineer was indeed cnrreet. L1”; 9.22]T'”kd1' é”: _ =ne
_._§_::E_u= r Tami2,. 5Q: n “1:” ...
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 Fall '08
 acosta
 Physics, Thermodynamics, Heat

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