北京科技大学传热学第2章习题答案.pdf

1 2-10 The variation of the absorption of solar energy in a solar pond with depth is given. A relation for the total rate of heat generation in a water layer at the top of the pond is to be determined. Assumptions: Absorption of solar radiation by water is modeled as heat generation. Analysis: The total rate of heat generation in a water layer of surface area A and thickness L at the top of the pond id determined by integration to be () b egA b e gAAdxegdVgG bL L bx bx L xV )1 ( 0 0 = = = ? ? 2-12 We consider a small rectangular element of lengthx, width y, and height1=z, the density of the body is and the specific heat is C. Noting that heat conduction is two-dimensional and assuming no heat generation ,an energy balance on this element during a small time interval t can expressed as = + elementtheof contentenergythe ofchangeofRate yxandyx atsurfacestheat conductionheatofRate yandxatsurface theatconduction heatofRate or t E QQQQ element yyxxyx =+ + + Noting that the volume of the element is 1=yxzyxVelement, the change in the energy content of the element can be expressed as ()() tttttttttelement TTyxCTTmCEEE= + Substituting, t TT yxCQQQQ ttt yyxxyx =+ + + + Dividing by yx gives t TT C y QQ xx QQ y ttt yyy xzx = + + + 11 Taking the thermal conductivity to be constant and noting that the heat transfer surface areas of the element for heat conduction in the x and y directions are 1=yAx and 1=xAy, respectively, and taking the limit as , and0xyt yields Solar energy Solar pond Radiation beam being absorbed 0 L x 2 t T y T x T = + 1 2 2 2 2 Since from the definition of the derivative and Fouriers law of heat conduction, 2 2 0 111 lim x T k x T k xx T zyk xzyx Q zyx QQ zy xxxx x = = = = + 2 2 0 111 lim y T k y T k yy T zxk yzxy Q zxy QQ zx yyyy y = = = = + Here the property Ck/= is the thermal diffusivity of the material. 2-16 The base plate of a household iron is subjected to specified heat flux on the left surface and to specified temperature on the right surface. The mathematical formulation, the variation of temperature in the plate, and the inner surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions: 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is large relative to its thickness, and the thermal conditions on both sides of the plate are uniform . 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. 4 Heat loss through the upper part of the iron is negligible. Properties: The thermal conductivity is given to be20/kW mC= ? . Analysis: (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to be 2 24 /000,50 10160 800 mW m W A Q q base = = ? ? Taking the direction normal to the surface of the wall to be the x direction with x=0 at the left surface, the mathematical formulation of this problem can be expressed as 0 2 2 = dx Td And CTLT mWq dx dT ? ? 85)( /000,50 )0( 2 2 = = (b) Integrating the differential equation twice with respect to x yields 21 1 )(CxCxT C dx dT += = 3 where 12 and CC are arbitrary constants. Applying the boundary conditions give :0=x = ? ? q CqC 11 :Lx = Lq TCLCTCTCLCLT +=+= ? 22122221 )( Substituting 1 C and 2 C into general solution, we variation of temperature is determined to be () ()() ()85006. 02500 85 /20 006. 0/000,50 2 22 += + = + =+= x Cx CmW mxmW T xLqLq Tx q Tx ? ? ? (c) The temperature at x=0 (the inner surface of the plate) is ()CT ? 100850006. 02500)0(=+= Note that the inner surface temperature is higher than the exposed surface temperature, as expected. 2-17 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions: 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Properties: The thermal conductivity is given to be14/kW m C= ? . Analysis: (a) Noting that the 85% of the 300W generated by the strip heater is transferred to the pipe, the heat flux through the outer surface is determined to be ()() 2 2 /1 .169 604. 02 30085. 0 2 mW mm W Lr Q A Q q SS S = = Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r direction, the mathematical formulation of this problem can be expressed as 0= dr dT r dr d and 4 ( ) ( ) S q dr rdT rTTh dr rdT = = 2 1 1) ( (b) Integrating the differential equation once with respect to r gives 1 C dr dT r= Dividing both sides of the equation above by rto bring it to a readily integrable form and then integrating, ( ) 21 1 lnCrCrT r C dr dT += = where 1 Cand 2 C are arbitrary constants. Applying the boundary conditions give : 2 rr = 2 1 2 1 rq Cq r C S S = : 1 rr = () 2 1 11 1 12211 1 1 lnlnln rq hr rTC hr rTCCrCTh r C S = =+= Substituting 1 C and 2 C into the general solution, we variation of temperature is determined to be ( ) ()() ()() += += += += += 61.12ln483. 010 /14 04. 0/1 .169 037. 0/30 /14 ln10 lnlnlnlnln 1 2 2 1 2 11 1 1 11 1 11 r r CmW mmW mCmW CmW r r C rq hrr r TC hr rrTC hr rTrCrT S ? ? ? (c) The inner and outer surface temperature are determined by direct substitution to be Inner surface(): 1 rr = ( )()C r r rT ? 91. 361.120483. 01061.12ln483. 010 1 1 1 =+= += Outer surface (): 2 rr = ( )C r r rT ? 87. 361.12 037. 0 04. 0 ln483. 01061.12ln483. 010 1 2 1 = += += Note that the pipe is essentially isothermal at a temperature of about -3.9C ? . 5 2-19 A long resistance heater wire is subjected to convection at its outer surface. The surface temperature of the wire is to be determined using the applicable relations directly and by solving the applicable differential equation. Assumptions: 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the wire is uniform. Properties: The thermal conductivity is given to be 15.1/kW mC= ? . Analysis: (a) The heat generation per unit volume of the wire is () () 38 22 /10061. 1 6001. 0 2000 mW mm W Lr Q V Q g gen wire gen = ? The surface temperature of the wire is then (Eq.2-68) ()() () C CmW mmW C h gr TTS ? ? ? 409 /1402 001. 0/10061. 1 30 2 2 38 = +=+= (b) The mathematical formulation of this problem can be expressed as 0 1 =+ g dr dT r dr d r and ( ) ( ) =TrTh dr rdT ? ? (convection at the outer surface) ( ) 0 0 = dr dT (thermal symmetry about the centerline) Multiplying both sides of the differential equation by r and integrating gives 1 2 2 C rg dr dT rr g dr dT r dr d += (a) Applying the boundary condition at the center line, B.C. at :0=r ( ) 00 2 0 0 11 =+= CC g dr dT Dividing both sides of Eq.(a) by r to bring it to a readily integrable form and integrating, ( ) 2 2 42 Cr g rTr g dr dT += (b) Applying the boundary condition at 0 rr =, B.C. at 0 rr =: 2 22 4242 ? ? ? ? r g h gr TCTCr g h rg += += 6 Substituting this 2 C relation into Eq.(b)and rearranging give ( )() h gr rr g TrT 24 2 2 += ? ? which is the temperature distribution in the wire as a function of r. Then the temperature of the wire at the surface () ? rr =is determined by substituting the known quantities to be ( )() ()() () C CmW mmW C h rg T h rg rr g TrT ? ? ? ? 409 /1402 001. 0/10061. 1 30 224 2 38 22 = +=+=+= Note that both approaches give the same result. 2-21 Heat is generated in a large plane wall whose one side is insulated while the other side is maintained at a specified temperature. The mathematical formulation, the variation of temperature in the wall, and the temperature of the insulated surface are to be determined for steady one-dimensional heat transfer. 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, and there is thermal symmetry about the center plane. 3 Thermal conductivity is constant. 4 Heat generation varies with location in the x direction. Properties: The thermal conductivity is given to be30/kW m C= ? . Analysis: (a) Noting that heat transfer is steady and one-dimensional in x direction, the mathematical formulation of this problem can be expressed as ( ) 0 2 2 =+ xg dx Td where L x egg 5 . 0 = ? and 36 /108mWg= ? and ( ) 0 0 = dx dT (insulated surface at x=0) ( )CTLT ? 30 2 = (specified surface temperature) (b)Rearranging the differential equation and integrating, 1 5 . 0 1 5 . 0 5 . 0 2 2 2 /5 . 0 Ce Lg dx dT C L eg dx dT e g dx Td L x L x L x +=+ = ? Integrating one more time, ( )( ) 21 5 . 0 2 21 5 . 0 4 /5 . 0 2 CxCe Lg xTCxC L eLg xT L x L x +=+ = ? (1) Applying the boundary conditions: B.C. at x=0: ( ) Lg CC Lg Ce Lg dx dT l ? =+=+= 22 0 20 111 05 . 0 7 B.C at x=L: ( ) 2 5 . 0 2 2221 5 . 0 2 2 244Lg e Lg TCCLCe Lg TLT l L ? +=+= Substituting the 1 C and 2 C relations into Eq.(1) and rearranging give ( )() + += Lxee Lg TxT L x /24 5 . 0 5 . 0 2 2 ? which is the desired solution for the temperature distribution in the wall as a function of x. The temperature at the insulate surface (x=0) is determined by substituting the known quantities to be ( )()() ()() () ()() 2 0.50 2 2 63 0.5 0420/ 8 10/0.05 304120 30/ 314 g L TTeeL W mm Ce W mC C =+ =+ = ? ? ? ? Therefore, there is a temperature difference of almost 300C ? between the two sides of the plate. 2-24 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions: 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is no heat generation. Properties: The thermal conductivity is given to be( )()1k TkT=+ ? . Analysis: The average thermal conductivity of the medium in this case is simply the conductivity value at the average temperature since the thermal conductivity varies linearly with temperature, and is determined from Eq.2-74 to be () () () 21 41 1 2 500350 (25/) 18.7 10 2 34.2/ aveave TT kk Tk K W m KK W m K + =+ + =+ = ? Then the rate of heat conduction through the plate can be determined from Eq.2-70 to be ()() () 12 500350 34.2/1.50.630,780 0.15 ave KTT QkAW m KmmW Lm = Discussion We would obtain the same result if we substituted the given ( )k Trelation into the second part of Eq.2-70, and performed the indicated integration. 8 2-28 The surface and interface temperature of a resistance wire covered with a plastic layer are to be determined. Assumptions: 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since this two-layer heat transfer problem possesses symmetry about the center line and involves no change in the axial direction, and thus( )rTT =. 3 Thermal conductivities are constant. 4 Heat generation in the wire is uniform. Properties: It is given that15/ wire kW mC= ? and 1.2/ plastic kW mC= ? . Analysis: Letting l T denote the unknown interface temperature, the mathematical formulation of the heat transfer problem in the wire can be expressed as 1 0 ddTg r r drdrk += with ( ) l TrT= 1 and ( ) 0 0 = dr dT Multiplying both sides of the differential equation by r, rearranging, and integrating give 1 2 2 C rg dr dT rr g dr dT r dr d += (a) Applying the boundary condition at the center ()0=r gives B.C. at r=0: ( ) 11 0 000 2 dTg CC drk = += Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, ( ) 2 2 24 dTgg rT rrC drkk = = + (b) Applying the other boundary condition at 1 rr =, B.C. at 1 rr =: 22 1221 44 ll gg TrCCTr kk = +=+ Substituting this 2 C relation into Eq. (b) and rearranging give ( ) () 22 1 4 wirel wire g TrTrr k =+ (c) The mathematical formulation of heat transfer problem in the plastic can be expressed as 9 0= dr dT r dr d with ( ) l TrT= 1 and ( ) ( ) =TrTh dr rdT 2 2 The solution of the differential equation is determined by integration t