assignchap3-77-92

assignchap3-77-92 - PROBLEM 3.77 KNOWN: Geometry and...

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Unformatted text preview: PROBLEM 3.77 KNOWN: Geometry and boundary conditions of a nuclear fuel element. FIND: (a) Expression for the temperature distribution in the fuel, (b) Form of temperature distribution for the entire system. SCHEMATIC: Steel "—46 ASSUMPTIONS: (l) One—dimensional heat transfer, (2) Steady-state conditions, (3) Uniform generation, (4) Constant properties, (5) Negligible contact resistance between fuel and cladding. ANALYSIS: (a) The general solution to the heat equation, Eq. 3.39, 2 . £1_T+.S_: (—LSXS+L) (1)12 kf is T=—ix2+Clx+C2. 2kf The insulated wall at x = - (L+b) dictates that the heat flux at x = — L is zero (for an energy balance applied to a control volume about the wall, Em : Eout = 0). Hence E] =—i(—L)+C1:0 01‘ C1=“% dx X- L kf kf T =—ix2 ~£x+Cz Zkf kf The value of TSJ may be determined from the energy conservation requirement that Eg = qcond : qconv, or on a unit area basis. . k q(2L)=?S(TS,1—TS,2)=h(TS!2—Too). Hence, ' 2Lb ' 2L T51=q( )+T52 where T52=q( )+T°,, 9 ks s 1 h ' ' 2L T,,=q<2Lb>+q< as. 1 ks h Continued .. PROBLEM 3.77 (Cont) Hence from Eq. (1), - 2 '2Lb ' 2L 9(L) T(L)-Tsl-q( )+q( LT“ 3 :C2 ’ ks h 2 M which yields C2=Tw+qL fi+3+E i ks h 2 kf Hence, the temperature distribution for (—L S x S +L) is 2 1 h T q x2 qu:c'1L 2": 3 L +10 Zkf kf s 2 kf ‘\ (b) For the temperature distribution shown below, (—L — b) s x s ~L: dT/dx=0, T=Tmax —LSxS+L: IdT/dxl T with T x +L S x S L+bz (dT/dx) is const. PROBLEM 3.92 KNOWN: Long rod experiencing uniform volumetric generation encapsulated by a circular sleeve exposed to convection. FIND: (a) Temperature at the interface between rod and sleeve and on the outer surface, (0) Temperature at center of rod. SCHEMATIC: Sleeve} [(3 = 4 W/m-K I; T 0C T T T hfzé‘g/mi'K ASSUMPTIONS: (l) Onegdimensional radial conduction in rod and sleeve, (2) Steady-state conditions, (3) Uniform volumetric generation in rod, (4) Negligible contact resistance between rod and sleeve. R0 clJ kr': 0.5 W/m-K; Q=24,OOOW/m3 ANALYSIS: (a) Construct a thermal circuit for the sleeve, where q’zEflgen 2 C177: D12 /4 : 24,000 W/rn3 ><7r ><(0.20 m)2 1'4 : 754.0 W/rn , ln(r2/r1) ln (400/200) R5 = —-——— — _ ———— = 2.758x10‘2m- KW 27: kS 27rx4W/m-K l l 2%:3.l83X10’2m-WW ha D2 25 W/m -K><7r><0.400 m Rconv : The rate equation can be written as , T1 —Too 7 T2 —Tm q : I I I Rs + Rconv Rconv Tl = Too + q’(Ré + Réom ) = 27°c+754 vv/n—i(2.758><10‘2 +3.183x10'2)K/w- m=71.s°c < T2 : Tm + q’Rgom, : 27°e+754 W/mx3.183><10'2m4 KIW=51.0°C. < (b) The temperature at the center of the rod is . 2 3 2 r 24000 W/ - 0.100 o a “(3:3)qu +T1=____.M_I‘L(—“”*)+713 C2192 C. < 4k]. 4x05 W/m-K COMMENTS: The thermal resistances due to conduction in the sleeve and convection are comparable. Will increasing the sleeve outer diameter cause the surface temperature T2 to increase or decrease? ...
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assignchap3-77-92 - PROBLEM 3.77 KNOWN: Geometry and...

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