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be answered to without any information about the nature of the T(x) function, using qualitative arguments only. At steady state, heat cannot go from right to left at any location. There is no way out through the left surface because of the adiabatic insulation, so it would accumulate somewhere, contradicting the steady state assumption. Therefore, the temperature must continually decrease from left to right, and Tmax is at x = L. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-76 2-141 Heat is generated in a plane wall. The temperature distribution in the wall is given. The surface temperature, the heat generation rate, the surface heat fluxes, and the relationship between these heat fluxes, the heat generation rate, and the geometry of the wall are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Analysis (a) The variation of temperature is symmetric about x = 0. The surface temperature is C72°=°×\u2212°=\u2212=\u2212== 2242 )m 02.0)(C/m102(C80)()( bLaLTLTTs The plot of temperatures across the wall thickness is given below. -0.02 -0.01 0 0.01 0.02 70 72 74 76 78 80 82 x [m] T [C ] T\u221e h x k gene& 72ºC L 72ºC -L (b) The volumetric rate of heat generation is 35 W/m102×=°×°\u22c5=\u2212\u2212=\u23af\u2192\u23af=+ )C/m102(C) W/m5(2)2(0 24gengen2 2 bkee dx Tdk && (c) The heat fluxes at the two surfaces are [ ] 2 2 W/m4000 W/m4000 \u2212=°×°\u22c5\u2212=\u2212\u2212\u2212=\u2212=\u2212 =°×°\u22c5=\u2212\u2212=\u2212= m) 02.0)(C/m102(C) W/m5(2)(2()( m) 02.0)(C/m102(C) W/m5(2)2()( 24 24 Lbk dx dTkLq bLk dx dTkLq L s L s & & (d) The relationship between these fluxes, the heat generation rate and the geometry of the wall is [ ] [ ] LeLqLq LWHeWHLqLq eALqLq EE ss ss ss gen gen gen genout 2)()( )2()()( )()( &&& &&& &&& && =\u2212+ =\u2212+ =\u2212+ = V Discussion Note that in this relation the absolute values of heat fluxes should be used. Substituting numerical values gives 8000 W/m2 on both sides of the equation, and thus verifying the relationship. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-77 2-142 Steady one-dimensional heat conduction takes place in a long slab. It is to be shown that the heat flux in steady operation is given by \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + += wTT TT W kq * 0 ** ln& . Also, the heat flux is to be calculated for a given set of parameters. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness. Analysis The derivation is given as follows \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + += \u2212=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + + \u2212\u2212=+ \u2212=+ + \u2212=\u2212= \u222b\u222b w w T T WT T TT TT W kq k Wq TT TT W k qTT dx k q TT dT dx dT TT k dx dTkq w w * 0 ** * 0 * * * * 0 ** * * ln ln )0()ln( or )( 0 0 & & & & & The heat flux for the given values is 25 W/m101.42×\u2212=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212×=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + += K)4001000( K)6001000(ln m 0.2 W/m107ln 4 * 0 ** wTT TT W kq& 2-143 A spherical ball in which heat is generated uniformly is exposed to iced-water. The temperatures at the center and at the surface of the ball are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional., and there is thermal symmetry about the center point. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity is given to be k = 45 W/m\u22c5°C. h T\u221e gene& D Analysis The temperatures at the center and at the surface of the ball are determined directly from C86.7°=° ×+°=+= \u221e C). W/m1200(3 m) 12.0)( W/m106.2( C0 3 2 36 gen h re TT os & C225°=° ×+°=+= C) W/m.45(6 m) 12.0)( W/m106.2( C7.86 6 2362 gen 0 k re TT os & PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-78 Fundamentals of Engineering (FE) Exam Problems 2-144 The heat conduction equation in a medium is given in its simplest form as 01 gen =+\u239f\u23a0 \u239e\u239c\u239d \u239b e dr dTrk dr d r & Select the wrong statement below. (a) the medium is of cylindrical shape. (b) the thermal conductivity of the medium is constant. (c) heat transfer through the medium is steady. (d) there is heat generation within the medium. (e) heat conduction through the medium is one-dimensional. Answer (b) thermal conductivity of the medium is constant 2-145 Consider a medium in which the heat conduction equation is given in its simplest form as t T r Tr rr \u2202 \u2202 \u3b1=\u239f\u23a0 \u239e\u239c\u239d \u239b \u2202 \u2202 \u2202 \u2202 11 2 2 (a) Is heat transfer steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the thermal conductivity of the medium constant or variable? (e) Is the medium a plane wall, a cylinder, or a sphere? (f) Is this differential equation for heat conduction linear or nonlinear? Answers: (a) transient, (b) one-dimensional, (c) no, (d) constant, (e) sphere, (f) linear 2-146 An apple of radius R is losing heat steadily and uniformly from its outer surface to the ambient air at temperature T\u221e with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are absolute temperatures). Also, heat is generated within the apple uniformly at a rate of per unit volume. If Tgene& s denotes the outer surface temperature, the boundary condition at the outer surface of the apple can be expressed as (a) )()( 4surr 4 TTTTh dr dTk ss Rr \u2212+\u2212=\u2212 \u221e = \u3b5\u3c3 (b) genss Rr eTTTTh dr dTk &+\u2212+\u2212=\u2212 \u221e = )()( 4surr 4\u3b5\u3c3 (c) )()( 4surr 4 TTTTh dr dTk ss Rr \u2212+\u2212= \u221e = \u3b5\u3c3 (d) genss Rr e R RTTTTh dr dTk & 2 3 4 surr 4 4 3/4)()( \u3c0 \u3c0\u3b5\u3c3 +\u2212+\u2212= \u221e = (e) None of them Answer: )()( 4surr 4 TTTTh dr dTk ss Rr \u2212+\u2212=\u2212 \u221e = \u3b5\u3c3 Note: Heat generation in the medium has no effect on boundary conditions. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-79 2-147 A furnace of spherical shape is losing heat steadily and uniformly from its outer surface of radius R to the ambient air at temperature T\u221e with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are absolute temperatures). If To denotes the outer surface temperature, the boundary condition at the outer surface of the furnace can be expressed as (a) )()( 4surr 4 TTTTh dr dTk oo Rr \u2212+\u2212=\u2212 \u221e = \u3b5\u3c3 (b) )()( 4surr4 TTTThdr dTk oo Rr \u2212\u2212\u2212=\u2212 \u221e = \u3b5\u3c3 (c) )()( 4surr 4 TTTTh dr dTk oo Rr \u2212+\u2212= \u221e = \u3b5\u3c3 (d) )()( 4surr4 TTTThdr dTk oo Rr \u2212\u2212\u2212= \u221e = \u3b5\u3c3 (e) )()()4( 4surr 42 TTTTh dr dTRk oo Rr \u2212+\u2212= \u221e = \u3b5\u3c3\u3c0 Answer (a) )()( 4surr 4 TTTTh dr dTk oo Rr \u2212+\u2212=\u2212 \u221e = \u3b5\u3c3 2-148 A plane wall of thickness L is subjected to convection at both surfaces with ambient temperature T\u221e1 and heat transfer coefficient h1 at inner surface, and corresponding T\u221e2 and h2 values at the outer surface. Taking the positive direction of x to be from the inner surface to the outer surface, the correct expression for the convection boundary condition is (a) [ ))0()0( 11 \u221e\u2212= TThdx dTk ] (b) [ ))()( 22 \u221e\u2212= TLThdx LdTk ] (c) [ ))0( 211 \u221e\u221e \u2212=\u2212 TThdx dTk ] (d) [ ]))(