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Unformatted text preview: PROBLEM 4,1 KNOWN: Method of separation of variables ( Section 4.2) for two~dimensionaL steadystate conduction. . A 2 . . . .
HM): Show that negative or zero values of IL , the separation constant, result in solutions which cannot satisfy the boundary conditions.
SCHEMATIC: ASSUMPTIONS; (1) Two—dimensional}, steedynstate conduction, (2) Constant properties. ANALYSIS: From Section 4.2, identification of the separation constant, P? leads to the two ordinary
differential equations, 4.6 and 4.75 having the forms 2 2 i§+12Xm0 d—‘é—Ltzreo {1,2} dx dy
and the temperature distribution is 6 (soy) :2 X (x) Y (y ). (3)
Consider now the situation when k2 2 O. From Eqs. {1), (2}, and {3), ﬁnd that XﬁCgFCZX, Y=C3+C4y and 9(X,y)m(C;+C2X) (C3+C4y). (4)
Evaluate the constants  C1? C2, (33 and C4 — by subsnmtion of the boundary conditions: X202 9 0,31): C}+C20)(C3+C4y}=0 C130 ymO: e X,O)m 0+C2X)(C3+C4O)x0 53:0 XmL: 9{L,O)K(O+C2L)(O+C4y)20 A {32:0 er: 9(X,W)=(0+0~X)(0+C4W)=l OH The last boundary condition leads to an impossibility (0 :ﬁ 1), We therefore conclude that a A2 value
of zero wiii not result in a form of the temperature distribution which will satisfy the boundary conditions. Consider now the situation when R2 < 0. The solutions to Eqs. (1) and (2) will be
X 2 (35311); + (Egg/“x, Y = C7005 ﬂy + Cgsin ﬂty  (5,6)
and Q(X,y) :[C5e' “X +C6e+ﬂxj [C7eos Ay+Cgsin Fty] {7)
L r m P , .
K . C ‘«— " mi KW L” ,7.» w 5‘9: “1:“
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ﬂy
’ ‘P
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1:
w
r, m f‘ , g I r" 3 f p. M, 3,:
~ 4 z ,u. PM. 1 f ”2; ,1 awe, w
L; . ‘ g «a; .‘ x ,r w .,,,,.J“ ,_.".w"._ w _ . 1' SUIEMA'NC: To : 106),ro
' I 93.4%:
T; m 89.5”6 I g r“ ‘ 1 xi“ . ‘
5% 64“. W5: film. ., mg... i. 53“} 75W égwwm mg” {$2, is E245» 5%: “gym
E: j Wgﬁ ﬁfﬁﬂg’g‘w my “7ng Wﬁgﬁ p543?" WE} Efﬁgy w i gﬁéﬁw {aim am WWW? wﬁQEW am “ﬁgmg WW Ewﬁw‘ . m? @ng 4;; Ewgw ”3% ...
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 Fall '09
 varischetti
 Heat Transfer

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