Problem 3.84 of Bergman. Consider one-dimensional conduction in a plane composite wall. The outer surfaces are exposed to a fluid at T

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1 1/10 bergman3-84.xmc Problem 3.84 of Bergman. Consier one-imensional conuction in a plane composite wall. The outer surfaces are expose to a flui at T 5 C an a convection heat transfer coefficient of h1000 W/(m K). The mile wall B experiences uniform heat generation qot B, while there is no generation in walls A an C. The temperatures at the interfaces are 61 C an T 11 C. Instructor: Nam Sun Wang T s.a T at air-a interface T s.c T at C-air interface Since there is no raiation, we can work in C, rather than K. T inf : 5 C h : 1000 W/(m K) : 5 W/(m K) : 50 W/(m K) : 61 C T : 11 C :.03 m L B :.03 m :.0 m x0 x+... x0 is at the air-a bounary. Sign convention: right irection is +. x A : x B : + L B x C : + L B + a) Assuming negligible contact resistance at the interfaces, etermine the volumetric heat generation qot B an the thermal conuctivity. Heat balance at the air-a interface gives heat flux on the A sie k q'' A h ( T inf T s.a ) A 1 L ( T s.a ) A T inf R aira : + q'' A : 10773W/m h R aira Heat balance at the C-air interface gives heat flux on the C sie - sign means towar left. k h ( T s.c T inf ) C 1 L ( T s.c T ) C T T inf R Cair : + : W/m h R Cair Heat transfer rate on the A sie an the C sie equals the heat generation rate in B. q'' ( L B A) qot B A ( q'' A + ) A + qot B : W/m 3 L B Heat balance equation with heat generation qot B in B accumulationin-out+generation T ρ c p t x k T B x y k T + B y 1-imension in x-irection T ρ c p t x k T B + qot B x z k T + B + qot B z

2 /10 bergman3-84.xmc Constant T T ρ c p T + qot t B steay-state 0 T x t B + qot B x Integrate once T qot B x + C 1 q'' A k T qot B B x x A + C 1 x x xa k qot B x A C 1 B Integrate one more time qot B T x + C 1 x + C k B We apply B.C.s to fin 3 unknowns simultaneously: integration constants C 1 & C an. The numerical values of C 1 an C epen on where we place x0. Ha we place x0 at the center of wall B, everything shoul work out just fine, although the values of C 1 an C woul be ifferent. C 1 : C : : 1... provie initial guess Given at A-B interface, xx A q'' A qot B x A + C 1 qot B at A-B interface, xx A x A + C 1 x A + C k B qot B at B-C interface, xx B + L B T x B + C 1 x B + C k B C C : Fin C 1, C, T B ( x) qot B : x k B + C 1 x + C Temperature at the two surfaces T s.a & T s.c (not aske in the problem statement) q'' A h T inf T s.a W/(m K) q'' A T s.a T s.a : T inf T s.a : + q'' A h h T s.c T inf T s.c T T s.c : T inf T s.c : T h T A q'' A q'' A T A ( x) : x + T s.a check: T A ( x A ) x T C T C ( x) : x x C x + T s.c check: Temperature that is uniformly vali in x[0 x C + L B + ] T A x ( x x B ) Tx : x < x A + x A x T B ( x) + x B < x T C ( x B ) 11 T 11 T C x

3 3/10 bergman3-84.xmc b) Plot the temperature istribution, showing its important features. 400 x A x B Temperature (C) Tx T x Position, x (m) c) Consier conitions corresponing to a loss of coolant at the expose surface of material A (h0). Determine an T an plot the temperature istribution throughout the system. Heat balance at the air-a interface gives heat flux on the A sie q'' A T inf T s.a T s.a T const... isothermal in A Heat transfer rate on the C sie alone equals the heat generation rate in B, because there is no heat transfer on the A sie. ( L B A) qot B A : L B qot B W/m Heat balance at the C-air interface gives heat flux on the C sie h T s.c T inf T s.c T T s.c : + T inf C T : + T s.c C h Heat balance equation with heat generation qot B in B T T ρ c p T + qot t B steay-state 0 T x t B + qot B x Integrate once T qot B x + C 1 x Apply B.C. at xx A L T qot B qot B A x x A + C 1 C 1 : x A 781 x xa Check: k T qot B B x B + C 1 x x xb k qot B x B C 1 B qot B x B C Integrate one more time qot B T x + C 1 x + C k B

4 4/10 bergman3-84.xmc Apply B.C. at B-C interface, x B + L B T qot B x B + C 1 x B + C k B qot B C : T + x B C 1 x B k B qot B T B ( x) : x + C 1 x + C k B Temperature in A is constant at : T B ( x A ) T A ( x) : Temperature in C is linear (because of conuction /wo heat generation) h T s.c T inf T s.c T T s.c : T inf check: T s.c : T h T C T C ( x) : ( x x B ) + T x check: Temperature that is vali in x[0 + L B + ] T A x ( x x B ) Tx : x < x A + x A x T B ( x) + x B < x b) Plot the temperature istribution, showing its important features. T C ( x B ) T T C x x A x B Temperature (C) Tx T x Position, x (m)

5 5/10 bergman3-84.xmc Re-o Part a). Alternatively, we can apply heat balance step by step to list all applicable equations in a systematical manner (starting from left of the figure). Likewise for Part c).. Reset back to Part a) : 61 T : 11 T s.a : T s.c : q'' A : : qot B : 10 6 : 1 C 1 : C :... provie initial guess Given LHS of air-a interface q'' A h ( T inf T s.a ) RHS of air-a interface & A wall q'' A T s.a at A-B interface, xx A q'' A qot B x A + C 1 qot B at A-B interface, xx A x A + C 1 x A + C k B qot B at B-C interface, xx B + L B T x B + C 1 x B + C k B at B-C interface, xx B + L B qot B x B + C 1 LHS of C-air interface & C wall T s.c T ( ) RHS of C-air interface h T s.c T inf Solve the above 8 equations for 8 unknowns. T s.a T s.c q'' A qot B C 1 C : Fin T s.a, T s.c, q'' A,, qot B,, C 1, C Note that the overall heat balance aroun wall B is automatically satisfie, because the following equation is not an inepenent new equation. When we work out heat balance for each component, the overall heat balance is automatically satisfie. ( L B A) qot B A q'' A + Check: L B qot B ( q'' A + )

6 6/10 bergman3-84.xmc Re-o Part c). Exactly the same set of equations, except that h0 at the air-a interface an we solve for & T, instea of qot B & as we i in Part a). All equations remain ientical to Part a) -- just copy/paste from Part a). Basically, there are 8 equations that govern this problem, an we can solve any combination of 8 unknowns. Fo example, in a ifferent problem statement,, L B,, etc may not be given, an we may be aske to fin these instea. T s.a : T s.c : q'' A : : : T : C 1 : C :... provie initial guess Given LHS of air-a interface q'' A ( T inf T s.a ) From Part a), we change h to 0 for Part c) RHS of air-a interface & A wall q'' A ( T s.a ) at A-B interface, xx A q'' A qot B x A + C 1 qot B at A-B interface, xx A x A + C 1 x A + C k B qot B at B-C interface, xx B + L B T x B + C 1 x B + C k B at B-C interface, xx B + L B qot B x B + C 1 LHS of C-air interface & C wall T s.c T ( ) RHS of C-air interface h T s.c T inf Solve the above 8 equations for 8 unknowns. T s.a T s.c q'' A T C 1 C : Fin T s.a, T s.c, q'' A,,, T, C 1, C Note that the overall heat balance aroun wall B is automatically satisfie, because the following equation is not an inepenent new equation. When we work out heat balance for each component, the overall heat balance is automatically satisfie. ( L B A) qot B A ( q'' A + ) Check: ( L B ) qot B ( q'' A + )

7 7/10 bergman3-84.xmc Re-o Part a). Brute-force numerical integration of ODEs with "rkfixe". This is a bounary value problem, because not all conitions are given at one single x value. For a more general problem, say qot B qot B (x) constant an/or (T) constant, etc, an analytical solution for T A (x), T B (x), T C (x) may not exist, an we will nee to resort to numerical ODE solving. Below are the ODEs to be solve an the corresponing B.C.s. There are a total of 8 bounary conitions; these correspon to the 8 algebraic equations from the last approach. Three n-orer ODEs require 6 B.C.s to be specifie. Two aitional B.C.s allow the etermination of aitional unknowns: qot B an. If a software package emans the orer of the ODE to match the number of B.C.s, we satisfy this eman by supplying two aitional ODEs: qot B /x0 & /x0 T A ( x) x h T inf T A ( 0) T B ( x) x T C ( x) x h T C x C x T A 0 x T A x A + qot B T B ( x A ) T B ( x B ) T ( T inf ) x T C x C Reset back to Part a) : 61 T : 11 # steps in integration n A : 30 n B : 60 n C : : TAx( T A, T' A ) T' A T'Ax( T A, T' A ) 0 f T' A.1, T' B.1, T' C., qot B,, flag yax( x, ya) xtt'a rkfixe xtt'a TBx T B, T' B x T B x B TAx ya, ya 0 1 T'Ax ya, ya 0 1 T' A.1 reverse( xtt'a) T' B T'Bx( T B, T' B ) ybx( x, yb) xtt'b rkfixe TCx T C, T' C T'Cx T C, T' C qot B x T B x A T A x A x T C x B T C x B, x A, 0, n A, yax TBx yb, yb 0 1 T'Bx yb, yb 0 1 T' B.1 T' C 0 ycx( x, yc), x A, x B, n B, ybx TCx yc, yc 0 1 T equations being implemente T A ( x) x T A ( x A ) T' A ( x A ) guess T A is integrate backwar reverse the rows x T B ( x) + qot B T B ( x A ) T' B ( x A ) guess The points at x A an x B are repea T C ( x) x

8 y (, y ) 8/10 bergman3-84.xmc T'Cx yc, yc 0 1 T xtt'c rkfixe, T' C. x B, x C, n C, ycx Nee to satisfy 5 more B.C.s: h ( T inf T A ( 0) ) f h T 0 ( inf xtt'a 01, ) + xtt'a x T A( 0) 0, f k 1 A T' A.1 + T'.1 A x T A( x A ) x T B( x A ) f xtt'b T nb, 1 T B ( x B ) T f k 3 B xtt'b + k nb, C T' C. x T B( x B ) x T C( x B ) f h xtt'c T 4 ( nc, 1 inf + k ) C xtt'c nc, h ( T C ( x C ) T inf ) x T C( x C ) return f if flag When flag0, return the B.C.s; return stack( xtt'a, xtt'b, xtt'c) otherwise when flag1, return temperature. 5 equations containe in f0 solve for 5 unknwons: 3 initial temperature graients (T' A.1, T' B.1, T' C. ) an parameters (qot B & ) provie initial guess: T' A.1 : T' B.1 : T' C. : qot B : : 1 Given f T' A.1, T' B.1, T' C., qot B,, 0 T' A.1 T' B.1 T' C. qot B : Fin T' A.1, T' B.1, T' C., qot B, After we fin the correct B.C.s (T' A.1, T' B.1, T' C. ) an parameters (qot B & ), we call function f one more time to integrate with these correct values to fin temperature T. xtt' : f ( T' A.1, T' B.1, T' C., qot B,, 1) x : xtt' 0 T : xtt' 1 T' : xtt' Temperature (C) T x A x Position, x (m) x B T T s.a : T T 0 s.c : T last( T) T C ( x B ) T T' C ( x B ) guess

9 9/10 bergman3-84.xmc Re-o Part a). Similar to the last brute-force numerical integration of ODEs with "rkfixe", except that we guess the initial conitions T 0 an T' 0 at x0. : yax x, ya f T 0, T' 0, qot B,, flag xtt'a rkfixe ybx( x, yb) xtt'b rkfixe ycx( x, yc) xtt'c rkfixe ya 1 0 T 0 T' 0, 0, x A, n A, yax yb 1 qot B xtt'a na, 1 xtt'a k na, B yc 1 0 xtt'b nb, 1 xtt'b k nb, C, x A x B,, n B, ybx, x B x C f h T 0 inf xtt'a + k 01, A xtt'a 0, f xtt'a na, 1 1 f xtt'b T nb, 1 f h xtt'c T 3 ( nc, 1 inf + k ) C xtt'c nc, return f if flag return stack( xtt'a, xtt'b, xtt'c) otherwise equations being implemente T A ( x) x T A ( 0) guess T' A ( 0) guess T A is integrate forwar,, n C, ycx T B ( x) + qot B x x T A( x A ) x T B( x A ) The points at x A an x B are repeate T C ( x) x x T B( x B ) x T C( x B ) Nee to satisfy 4 more B.C.s: h ( T inf T A ( 0) ) x T A( 0) T A ( x A ) T T B x B ( T inf ) h T C x C x T C x C When flag0, return the B.C.s; when flag1, return temperature. 4 equations containe in f0 solve for 4 unknwons: initial temperature & graient (T 0, T' 0 ) an parameters (qot B & ) provie initial guess: T 0 : T' 0 : qot B : : 1 Given f T 0, T' 0, qot B,, 0 T 0 T' 0 qot B : Fin T 0, T' 0, qot B, After we fin the correct I.C.s (T 0, T' 0 ) an parameters (qot B & ), we call function f one more time to integrate with these correct values to fin temperature T.

10 10/10 bergman3-84.xmc Temperature (C) xtt' : f ( T 0, T' 0, qot B,, 1) x xtt' 0 : T xtt' 1 : T x A x Position, x (m) x B T T s.a : T T 0 s.c : T last( T) Mathca bug/limitation. Mathca's "Given..Oesolve" can hanle only B.C.s given at ifferent x values, not at 4 ifferent x values (x0, x A, x B, x C ). Thus, below is theoretically vali, but, ue to Mathca bug/limitation, is no goo. combine all three regions into one ( x A x) ( x x B ) qot x, qot B Given : qot B kxk, B : ( x < x A ) ( x x B ) + x A x + x B < x kxk (, B ( x) ) Tx + qot ( x, qot B ( x) ) Tx ( A ) Tx ( B ) T x x qot B( x) x ( x) h ( T inf T0 ) T' ( 0) h T x C T qot B : Oesolve T qot B, x, x C... no goo ( T inf ) T' ( x C )