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Iil”. Win13 22.312 Homework set #1 Fall 19!]? SnLUImNs: PRﬂELEM 2—1 Due to the diEerent geometry and arrangement of the pellets within the heel pin.
we are faced with two different cases. 1. F llthereactnrs exc r : _. e The [net pellets are cylindrical and are Stacked in the ﬁle] pin. Theret'ere we can use the
relatiens, straightforwardly derived from eqs. 25: (am) 65;: _ ' m
(mg—5 (2) I: Hate that, with these deﬁrdtiens. D“ is the cuter cladding diameter and the heat ﬂux is
evaluated at the outer surface at the cladding. Using the 1talues found in the referenced Tables: AGE.
1132.3 T515 BWR. PWIt
m
595.4 524.5 363.4 mean '54.} "32
Whine2 Homework set #1 Fall 1991‘ 22.312 . sis"? can not he computed as in eq. 1. In fact, by deﬁnition (textbook, pg. 22), <q '“P is
the power generated per unit volume of the fuel material {both ﬁssile and fertile material}. As in a HTGR the ﬁrel particles are niierospheres, the channel is not uniformly loaded with
fuel, due to heth voids leFt among the mierespheres and their siliconer graphite layers. oid we start noting that the spheres, having a total diameter of .T to
compared to the diameter of the channel (15.? mm}._ whiten _ conditions, the spheres can he assumed to he packed as tightly as pessihle. This leads to a hexagonal closed peeked {HEP} conﬁguration, which can he thought of as a cubic face
3' centered {refs} lattice. Fer this lattice the packing factor. deﬁned as:  To account for the v
1.1 mm are small as _ total  void volume ' l. total volume has the value
.1! 2
fr : — E'— E [1:74 to situations like this. where the packing is the result of a Physics studiesI show that
icing lhctor should he relaxed by lﬁ~2ﬂ “In. Using 15% we random disposition, the pee "III1 . ohtain:
f" .163.
With this deﬁnition, the total volume oi'the rr .
a 2 E D; IV where L is the channel‘s height.
d with first. Assuming an inner radius R; of still) the and This volume is not corttpletelig.r ﬁlls d
_ _ storal shell micEness 5 of [15 tinl. Elie liaetion of the spheia'mﬁnemﬁeﬂpied
by fuel is' 4 3
FILE
fr=T3 l —:—R']—3=.335
' EEIRJ+§}J Liar“?
I  H Finally. the ﬁdel‘svglurne in the ehannel is:
.I J:
=_ S 4 :1 Charles Elect, Eleoteetmjv solid state physics: it short course, What. New 1iI’orlr. 196.1. pg. 2?. {Bardeen emanates}. *' 22.3 Is Homework set #1 Fall was (‘4; “"2 ean be calculated using its deﬁnition as ratio of the average heat generation rate to
the volume of the fuel which generates this power Referring to a pin of height fl, we can 4' write: ; w ._E a L ._ EDlLfoF Zﬂefefa '
'.  Using the values «1;: 5=1ST sum and £1.51 5.? mm, the ﬁnal result is; I cq":==1ss_sstwrm‘ . — which is in the range of the values previoule found; ﬁg; :11”; At the surface of the sphere (and thus, at the surface of the cladding}. ﬁg“?! can be
_ obtained from the ratio of the sphere power to the sphere's surface. "t Shore trwcr (gull—Ea
tar—P p r 3  Sphere surface _ 47d}: Using the result above, the sphere power: lsjlﬁ ngnd the heat ﬂux
ﬁq'ﬁe 11.3 it"llil'll'rn2
At the surface of the channel ﬁq ' '3’ can he obtained as in case t: From a thermedynamical point efview. the heat ﬂtut of interest is the one calculated at the
surface of the cooling channels. Nonetheless. not enough utt‘oonati on is found in the
textbook to calculate this quantin ...
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This note was uploaded on 02/02/2012 for the course PHY 4049 taught by Professor Acosta during the Fall '08 term at University of Florida.
 Fall '08
 acosta
 Physics

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