Fundamentals of heat and mass transfer frank p incropera david p dewitt solution manual ch3 (51 100)

PROBLEM 3.51 KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipe and ice layer formation on the inner surface FIND: Ice layer thickness δ SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermal resistance, (3) negligible ice/wall contact resistance, (4) Constant k PROPERTIES: Table A.3, Ice (T = 265 K): k ≈ 1.94 W/m⋅K ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it follows that, for a unit length of pipe, q′conv = q′cond ( ) h i ( 2π r1 ) T∞,i − Ts,i = Ts,i − Ts,o ln ( r2 r1 ) 2π k Dividing both sides of the equation by r2, ln ( r2 r1 ) ( r2 r1 ) = Ts,i − Ts,o k 1.94 W m ⋅ K 15$ C × = × = 0.097 $ h i r2 T∞,i − Ts,i C 2000 W m ⋅ K (0.05 m ) ( ) The equation is satisfied by r2/r1 = 1.114, in which case r1 = 0.050 m/1.114 = 0.045 m, and the ice layer thickness is δ = r2 − r1 = 0.005 m = mm < COMMENTS: With no flow, hi → 0, in which case r1 → and complete blockage could occur The pipe should be insulated PROBLEM 3.52 KNOWN: Inner surface temperature of insulation blanket comprised of two semi-cylindrical shells of different materials Ambient air conditions FIND: (a) Equivalent thermal circuit, (b) Total heat loss and material outer surface temperatures SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction, (3) Infinite contact resistance between materials, (4) Constant properties ANALYSIS: (a) The thermal circuit is, R ′conv,A = R ′conv,B = / π r2 h R ′cond ( A ) = R ′cond ( B ) = ln ( r2 / r1 ) < π kA ln ( r2 / ri ) π kB The conduction resistances follow from Section 3.3.1 and Eq 3.28 Each resistance is larger by a factor of than the result of Eq 3.28 due to the reduced area (b) Evaluating the thermal resistances and the heat rate ( q′=q′A + q′B ) , ( R ′conv = π × 0.1m × 25 W/m ⋅ K R ′cond ( A ) = q′= q′= ln ( 0.1m/0.05m ) π × W/m ⋅ K Ts,1 − T∞ R ′cond ( A ) + R ′conv + ) −1 = 0.1273 m ⋅ K/W = 0.1103 m ⋅ K/W R ′cond ( B ) = R ′cond ( A ) = 0.8825 m ⋅ K/W Ts,1 − T∞ R ′cond( B) + R ′conv (500 − 300 ) K (500 − 300 ) K + = (842 + 198 ) W/m=1040 W/m (0.1103+0.1273) m ⋅ K/W (0.8825+0.1273) m ⋅ K/W < Hence, the temperatures are W m⋅K × 0.1103 = 407K m W W m⋅K Ts,2( B) = Ts,1 − q′BR ′cond ( B ) = 500K − 198 × 0.8825 = 325K m W Ts,2( A ) = Ts,1 − q′A R ′cond ( A ) = 500K − 842 ( < < ) COMMENTS: The total heat loss can also be computed from q′= Ts,1 − T∞ / R equiv , −1 −1 −1   ′ ′ ′ ′ where R equiv =  R cond ( A ) + R conv,A + ( R cond(B) + R conv,B )  = 0.1923 m ⋅ K/W   Hence q′= (500 − 300 ) K/0.1923 m ⋅ K/W=1040 W/m ( ) PROBLEM 3.53 KNOWN: Surface temperature of a circular rod coated with bakelite and adjoining fluid conditions FIND: (a) Critical insulation radius, (b) Heat transfer per unit length for bare rod and for insulation at critical radius, (c) Insulation thickness needed for 25% heat rate reduction SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3) Constant properties, (4) Negligible radiation and contact resistance PROPERTIES: Table A-3, Bakelite (300K): k = 1.4 W/m⋅K ANALYSIS: (a) From Example 3.4, the critical radius is k 1.4 W/m ⋅ K rcr = = = 0.01m h 140 W/m ⋅ K < (b) For the bare rod, q′=h (π Di ) ( Ti − T∞ ) q′=140 W m2 ⋅ K (π × 0.01m ) ( 200 − 25)$ C=770 W/m < For the critical insulation thickness, ( 200 − 25) C Ti − T∞ q′= = ln ( rcr / ri ) ln (0.01m/0.005m ) 1 + + 2π rcr h 2π k 2π × 1.4 W/m ⋅ K 2π × (0.01m ) × 140 W/m ⋅ K $ q′= 175$C = 909 W/m (0.1137+0.0788) m ⋅ K/W < (c) The insulation thickness needed to reduce the heat rate to 577 W/m is obtained from ( 200 − 25) C Ti − T∞ W q′= = = 577 ln ( r/ri ) ln ( r/0.005m ) m 1 + + 2π rh 2π k 2π ( r )140 W/m ⋅ K 2π × 1.4 W/m ⋅ K $ From a trial-and-error solution, find r ≈ 0.06 m The desired insulation thickness is then δ = ( r − ri ) ≈ (0.06 − 0.005 ) m=55 mm < PROBLEM 3.54 KNOWN: Geometry of an oil storage tank Temperature of stored oil and environmental conditions FIND: Heater power required to maintain a prescribed inner surface temperature SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial direction, (3) Constant properties, (4) Negligible radiation PROPERTIES: Table A-3, Pyrex (300K): k = 1.4 W/m⋅K ANALYSIS: The rate at which heat must be supplied is equal to the loss through the cylindrical and hemispherical sections Hence, q=qcyl + 2q hemi = qcyl + qspher or, from Eqs 3.28 and 3.36, q= q= Ts,i − T∞ ln ( Do / Di ) + 2π Lk π Do Lh + Ts,i − T∞  1  − +   2π k  Di Do  π Do2h ( 400 − 300 ) K ln 1.04 + 2π ( 2m )1.4 W/m ⋅ K π (1.04m ) 2m 10 W/m ⋅ K ( + ( 400 − 300 ) K ) 1 (1 − 0.962 ) m-1 + 2π (1.4 W/m ⋅ K ) π (1.04m ) 10 W/m ⋅ K 100K 100K q= + -3 -3 -3 2.23 ×10 K/W + 15.30 × 10 K/W 4.32 ×10 K/W + 29.43 ×10-3 q = 5705W + 2963W = 8668W < PROBLEM 3.55 KNOWN: Diameter of a spherical container used to store liquid oxygen and properties of insulating material Environmental conditions FIND: (a) Reduction in evaporative oxygen loss associated with a prescribed insulation thickness, (b) Effect of insulation thickness on evaporation rate SCHEMATIC: ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Negligible conduction resistance of container wall and contact resistance between wall and insulation, (3) Container wall at boiling point of liquid oxygen ANALYSIS: (a) Applying an energy balance to a control surface about the insulation, E in − E out = 0, it follows that q conv + q rad = q cond = q Hence, T∞ − Ts,2 Tsur − Ts,2 Ts,2 − Ts,1 + = =q R t,conv R t,rad R t,cond ( where R t,conv = 4π r22 h ) −1 ( , R t,rad = 4π r22 h r ( (1) ) −1 1.9, the radiation coefficient is h r = εσ Ts,2 + Tsur , R t,cond = (1 4π k )[(1 r1 ) − (1 r2 )] , and, from Eq 2 + Tsur ) (Ts,2 ) With t = 10 mm (r2 = 260 mm), ε = 0.2 and T∞ = Tsur = 298 K, an iterative solution of the energy balance equation yields Ts,2 ≈ 297.7 K, where Rt,conv = 0.118 K/W, Rt,rad = 0.982 K/W and Rt,cond = 76.5 K/W With the insulation, it follows that the heat gain is qw ≈ 2.72 W Without the insulation, the heat gain is q wo = T∞ − Ts,1 Tsur − Ts,1 + R t,conv R t,rad where, with r2 = r1, Ts,1 = 90 K, Rt,conv = 0.127 K/W and Rt,rad = 3.14 K/W Hence, qwo = 1702 W  = q/hfg, the percent reduction in evaporated oxygen is With the oxygen mass evaporation rate given by m % Re duction = Hence, % Re duction =   m wo − m w  m wo q − qw × 100% = wo × 100% q wo (1702 − 2.7 ) W 1702 W × 100% = 99.8% < Continued PROBLEM 3.55 (Cont.)  = (b) Using Equation (1) to compute Ts,2 and q as a function of r2, the corresponding evaporation rate, m  with r2 are plotted as follows q/hfg, may be determined Variations of q and m 10000 0.01 Evaporation rate, mdot(kg/s) Heat gain, q(W) 1000 100 10 0.1 0.001 0.0001 1E-5 1E-6 0.25 0.26 0.27 0.28 0.29 Outer radius of insulation, r2(m) 0.3 0.25 0.26 0.27 0.28 0.29 0.3 Outer radius of insulation, r2(m) Because of its extremely low thermal conductivity, significant benefits are associated with using even a  are achieved with r2 = 0.26 thin layer of insulation Nearly three-order magnitude reductions in q and m -3  decrease from values of 1702 W and 8×10 kg/s at r2 = 0.25 m to 0.627 m With increasing r2, q and m W and 2.9×10-6 kg/s at r2 = 0.30 m COMMENTS: Laminated metallic-foil/glass-mat insulations are extremely effective and corresponding conduction resistances are typically much larger than those normally associated with surface convection and radiation PROBLEM 3.56 KNOWN: Sphere of radius ri, covered with insulation whose outer surface is exposed to a convection process FIND: Critical insulation radius, rcr SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (spherical) conduction, (3) Constant properties, (4) Negligible radiation at surface ANALYSIS: The heat rate follows from the thermal circuit shown in the schematic, q= ( Ti − T∞ ) / R tot where R tot = R t,conv + R t,cond and R t,conv = 1 = hAs 4π hr (3.9) R t,cond =  1  −  4π k  ri r  (3.36) If q is a maximum or minimum, we need to find the condition for which d R tot = dr It follows that d   1   1 1 − + = + − =0     dr  4π k  ri r  4π hr   4π k r 2π h r3  giving k h The second derivative, evaluated at r = rcr, is d  dR tot  1 1 = − + dr  dr  2π k r3 2π h r  r=r rcr = cr =  −  ( 2k/h )3  2π k +  1  3 =  −1 +  > 2π h 2k/h  ( 2k/h ) 2π k  2 Hence, it follows no optimum Rtot exists We refer to this condition as the critical insulation radius See Example 3.4 which considers this situation for a cylindrical system PROBLEM 3.57 KNOWN: Thickness of hollow aluminum sphere and insulation layer Heat rate and inner surface temperature Ambient air temperature and convection coefficient FIND: Thermal conductivity of insulation SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation exchange at outer surface PROPERTIES: Table A-1, Aluminum (523K): k ≈ 230 W/m⋅K ANALYSIS: From the thermal circuit, T −T T1 − T∞ q= ∞ = 1/ 1/r 1/ r R tot − + r2 − 1/ r3 + 4π k A1 4π k I h4π r32 q= ( 250 − 20 )$ C 1/0.15 − 1/ 0.18 1/ 0.18 − 1/ 0.30 K   + + 2W 4π k I  4π ( 230 ) 30 0.3 π ( )( )   = 80 W or 3.84 × 10−4 + 0.177 230 + 0.029 = = 2.875 kI 80 Solving for the unknown thermal conductivity, find kI = 0.062 W/m⋅K COMMENTS: The dominant contribution to the total thermal resistance is made by the insulation Hence uncertainties in knowledge of h or kA1 have a negligible effect on the accuracy of the kI measurement < PROBLEM 3.58 KNOWN: Dimensions of spherical, stainless steel liquid oxygen (LOX) storage container Boiling point and latent heat of fusion of LOX Environmental temperature FIND: Thermal isolation system which maintains boil-off below kg/day SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible thermal resistances associated with internal and external convection, conduction in the container wall, and contact between wall and insulation, (3) Negligible radiation at exterior surface, (4) Constant insulation thermal conductivity PROPERTIES: Table A.1, 304 Stainless steel (T = 100 K): ks = 9.2 W/m⋅K; Table A.3, Reflective, aluminum foil-glass paper insulation (T = 150 K): ki = 0.000017 W/m⋅K ANALYSIS: The heat gain associated with a loss of kg/day is  q = mh fg = 1kg day 86, 400 s day (2.13 ×105 J kg ) = 2.47 W ( ) With an overall temperature difference of T∞ − Tbp = 150 K, the corresponding total thermal resistance is ∆T 150 K R tot = = = 60.7 K W q 2.47 W Since the conduction resistance of the steel wall is R t,cond,s = 1 1  1  − = 2.4 × 10−3 K W  − =   4π k s  r1 r2  4π (9.2 W m ⋅ K )  0.35 m 0.40 m  it is clear that exclusive reliance must be placed on the insulation and that a special insulation of very low thermal conductivity should be selected The best choice is a highly reflective foil/glass matted insulation which was developed for cryogenic applications It follows that R t,cond,i = 60.7 K W = 1  1 1 −   − =  4π k i  r2 r3  4π (0.000017 W m ⋅ K )  0.40 m r3  which yields r3 = 0.4021 m The minimum insulation thickness is therefore δ = (r3 - r2) = 2.1 mm COMMENTS: The heat loss could be reduced well below the maximum allowable by adding more insulation Also, in view of weight restrictions associated with launching space vehicles, consideration should be given to fabricating the LOX container from a lighter material PROBLEM 3.59 KNOWN: Diameter and surface temperature of a spherical cryoprobe Temperature of surrounding tissue and effective convection coefficient at interface between frozen and normal tissue FIND: Thickness of frozen tissue layer SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible contact resistance between probe and frozen tissue, (3) Constant properties ANALYSIS: Performing an energy balance for a control surface about the phase front, it follows that q conv − q cond = Hence, ( )(T∞ − Ts,2 ) = [(1 r1T)s,2− (1−rT2s,1)] 4π k h 4π r22 r22 [(1 r1 ) − (1 r2 )] = ( ) h (T∞ − Ts,2 ) k Ts,2 − Ts,1  r2   r2   k ( Ts,2 − Ts,1 ) 1.5 W m ⋅ K  30  =       − 1 =  r1   r1   hr1 ( T∞ − Ts,2 ) 50 W m ⋅ K ( 0.0015 m )  37  ( )  r2   r2       − 1 = 16.2  r1   r1   ( r2 r1 ) = 4.56 It follows that r2 = 6.84 mm and the thickness of the frozen tissue is δ = r2 − r1 = 5.34 mm < PROBLEM 3.91 (Cont.) < T1 = 931K + 25 K − 18 K = 938 K (b) The temperature distributions may be obtained by using the IHT model for one-dimensional, steadystate conduction in a hollow tube For the fuel element ( q > 0), an adiabatic surface condition is prescribed at r1, while heat transfer from the outer surface at r2 to the coolant is governed by the thermal resistance R ′′tot,2 = 2π r2 R ′tot = 2π(0.011 m)0.0185 m⋅K/W = 0.00128 m2⋅K/W For the graphite ( q = 2500 2500 2100 2100 Temperature, T(K) Temperature, T(K) 0), the value of T2 obtained from the foregoing solution is prescribed as an inner boundary condition at r2, while a convection condition is prescribed at the outer surface (r3) For × 108 ≤ q ≤ × 108 W/m3, the following distributions are obtained 1700 1300 900 500 0.008 0.009 0.01 Radial location in fuel, r(m) qdot = 5E8 qdot = 3E8 qdot = 1E8 0.011 1700 1300 900 500 0.011 0.012 0.013 0.014 Radial location in graphite, r(m) qdot = 5E8 qdot = 3E8 qdot = 1E8 The comparatively large value of kt yields small temperature variations across the fuel element, while the small value of kg results in large temperature variations across the graphite Operation at q = × 108 W/m3 is clearly unacceptable, since the melting points of thorium and graphite are exceeded and approached, respectively To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above q = × 108 W/m3 COMMENTS: A contact resistance at the thorium/graphite interface would increase temperatures in the fuel element, thereby reducing the maximum allowable value of q 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, (b) Temperature at center of rod SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction in rod and sleeve, (2) Steady-state conditions, (3) Uniform volumetric generation in rod, (4) Negligible contact resistance between rod and sleeve ANALYSIS: (a) Construct a thermal circuit for the sleeve, where q′=E ′gen = q π D12 / = 24, 000 W/m3 × π × ( 0.20 m ) / = 754.0 W/m R s′ = ln ( r2 / r1 ) 2π k s R conv = = ln ( 400/200 ) 2π × W/m ⋅ K = 2.758 × 10−2 m ⋅ K/W 1 = = 3.183 × 10−2 m ⋅ K/W hπ D 25 W/m ⋅ K × π × 0.400 m The rate equation can be written as q′= T1 − T∞ T −T = ∞ R ′s + R ′conv R ′conv ) ( T1 = T∞ + q′ ( R ′s + R ′conv ) = 27$ C+754 W/m 2.758 × 10-2 + 3.183 × 10−2 K/W ⋅ m=71.8$C T2 = T∞ + q′R ′conv = 27$ C+754 W/m × 3.183 × 10-2m ⋅ K/W=51.0$C < < (b) The temperature at the center of the rod is T ( ) = To =  12 qr 4k r 24, 000 W/m3 ( 0.100 m ) + T1 = × 0.5 W/m ⋅ K + 71.8$ C=192$C < 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? PROBLEM 3.93 KNOWN: Radius, thermal conductivity, heat generation and convection conditions associated with a solid sphere FIND: Temperature distribution SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Uniform heat generation ANALYSIS: Integrating the appropriate form of the heat diffusion equation, d  dT    kr dr  + q=0 r dr  r2  dT qr =− + C1 dr 3k T (r ) = − or  dT qr C =− + dr 3k r  C1 qr − + C2 6k r The boundary conditions are: −k  d  dT  qr r = − dr  dr  k dT  =0 dr  r=0 hence C1 = 0, and dT  = h T ( ro ) − T∞  dr  r o Substituting into the second boundary condition (r = ro), find  qr  o  o2  o qr  o2 qr qr = h + C2 − T∞  + + T∞ C2 = 3h 6k  6k  The temperature distribution has the form  q 2 qr T (r ) = ro − r + o + T∞ 6k 3h COMMENTS: To verify the above result, obtain T(ro) = Ts,  qr Ts = o + T∞ 3h Applying energy balance to the control volume about the sphere, ( ) 4  q  π ro3  = h4π ro2 ( Ts − T∞ ) 3  find Ts = o qr + T∞ 3h < PROBLEM 3.94 KNOWN: Radial distribution of heat dissipation of a spherical container of radioactive wastes Surface convection conditions FIND: Radial temperature distribution SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across container wall ANALYSIS: The appropriate form of the heat equation is r2 q o d  dT  q r =− =− dr  dr  k k   2  1 −  r     ro     q  r r  dT  + C1 =− o − dr k  5ro2    q  r r  C1 − + C2 T=− o − k  20ro2  r   From the boundary conditions, r2 Hence dT/dr |r=0 = and − kdT/dr |r=ro = h  T ( ro ) − T∞  it follows that C1 = and  q  r r   r  r q o  o − o  = h  − o  o − o  + C2 − T∞   k  20   3 5 C2 = 2roq o 7q o ro2 + + T∞ 15h 60k 4  o2   r  2ro q o qr  r   Hence T ( r ) = T∞ + + −   +   15h k  60  ro  20  ro     COMMENTS: Applying the above result at ro yields < Ts = T ( ro ) = T∞ + ( 2ro q o /15h ) The same result may be obtained by applying an energy balance to a control surface about the container, where E g = qconv The maximum temperature exists at r = PROBLEM 3.95 KNOWN: Dimensions and thermal conductivity of a spherical container Thermal conductivity and volumetric energy generation within the container Outer convection conditions FIND: (a) Outer surface temperature, (b) Container inner surface temperature, (c) Temperature distribution within and center temperature of the wastes, (d) Feasibility of operating at twice the energy generation rate SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional radial conduction ANALYSIS: (a) For a control volume which includes the container, conservation of energy yields  −q E g − E out = , or qV conv = Hence ( ) ( q ( ) π ri3 = h4π ro2 Ts,o − T∞ ) and with q = 10 W/m , Ts,o = T∞ +  qr i 3hro2 = 25$ C + 105 W m ( 0.5 m ) 3000 W m ⋅ K ( 0.6 m ) 2 < = 36.6$ C (b) Performing a surface energy balance at the outer surface, E in − E out = or q cond − q conv = Hence 4π k ss Ts,i − Ts,o = h4π ro2 Ts,o − T∞ (1 ri ) − (1 ro ) ( ) ( ) ( ) 1000 W m ⋅ K  ro  $ Ts,i = Ts,o + −  ro ( Ts,o − T∞ ) = 36.6 C + ( 0.2 ) 0.6 m 11.6$ C = 129.4$ C  k ss  ri 15 W m ⋅ K  h < (c) The heat equation in spherical coordinates is d  dT   k rw  r =  + qr dr  dr  Solving, r2 dT =−  qr + C1 dr 3k rw Applying the boundary conditions, dT and =0 dr r = C1 = and and T (r ) = −  qr C − + C2 6k rw r T ( ri ) = Ts,i  C2 = Ts,i + qr i 6k rw Continued PROBLEM 3.95 (Cont.) Hence ri2 − r ) ( 6k rw q T ( r ) = Ts,i + < At r = 0, T ( ) = Ts,i +  qr i 6k rw = 129.4$ C + 105 W m3 ( 0.5 m ) ( 20 W m ⋅ K ) < = 337.7$ C (d) The feasibility assessment may be performed by using the IHT model for one-dimensional, steadystate conduction in a solid sphere, with the surface boundary condition prescribed in terms of the total thermal resistance ( ) ri2 R ′′tot,i = 4π ri R tot = R ′′cnd,i + R ′′cnv,i = [(1 ri ) − (1 ro )] +  ri 2 k ss   h  ro  where, for ro = 0.6 m and h = 1000 W/m2⋅K, R ′′cnd,i = 5.56 × 10-3 m2⋅K/W, R ′′cnv,i = 6.94 × 10-4 m2⋅K/W, Center temperature, T(0) (C) and R ′′tot,i = 6.25 × 10-3 m2⋅K/W Results for the center temperature are shown below 675 625 575 525 475 2000 4000 6000 8000 10000 Convection coefficient, h(W/m^2.K) ro = 0.54 m ro = 0.60 m Clearly, even with ro = 0.54 m = ro,min and h = 10,000 W/m2⋅K (a practical upper limit), T(0) > 475°C and the desired condition can not be met The corresponding resistances are R ′′cnd,i = 2.47 × 10-3 m2⋅K/W, R ′′cnv,i = 8.57 × 10-5 m2⋅K/W, and R ′′tot,i = 2.56 × 10-3 m2⋅K/W The conduction resistance remains dominant, and the effect of reducing R ′′cnv,i by increasing h is small The proposed extension is not feasible COMMENTS: A value of q = 1.79 × 105 W/m3 would allow for operation at T(0) = 475°C with ro = 0.54 m and h = 10,000 W/m2⋅K PROBLEM 3.96 KNOWN: Carton of apples, modeled as 80-mm diameter spheres, ventilated with air at 5°C and experiencing internal volumetric heat generation at a rate of 4000 J/kg⋅day FIND: (a) The apple center and surface temperatures when the convection coefficient is 7.5 W/m2⋅K, and (b) Compute and plot the apple temperatures as a function of air velocity, V, for the range 0.1 ≤ V ≤ m/s, when the convection coefficient has the form h = C1V0.425, where C1 = 10.1 W/m2⋅K⋅(m/s)0.425 SCHEMATIC: ASSUMPTIONS: (1) Apples can be modeled as spheres, (2) Each apple experiences flow of ventilation air at T‡ = 5°C, (3) One-dimensional radial conduction, (4) Constant properties and (5) Uniform heat generation ANALYSIS: (a) From Eq C.24, the temperature distribution in a solid sphere (apple) with uniform generation is T(r) = 2  qr o r2   −  + Ts 6k  r   o  (1) To determine Ts, perform an energy balance on the apple as shown in the sketch above, with volume V = Sro3 , − q cv + q ∀ = E in − E out + E g = ( )(Ts − T∞ ) + q (4 3π ro3 ) = −7.5 W m ⋅ K ( 4π × 0.0402 m )(Ts − C ) + 38.9 W m3 ( π × 0.0403 m3 ) = − h 4π ro2  (2) $ where the volumetric generation rate is q = 4000 J kg ⋅ day q = 4000 J kg ⋅ day × 840 kg m3 × (1day 24 hr ) × (1hr 3600 s ) q = 38.9 W m3 and solving for Ts, find < Ts = 5.14$ C From Eq (1), at r = 0, with Ts, find T(0) = 38.9 W m3 × 0.0402 m × 0.5 W m ⋅ K + 5.14$ C = 0.12$ C + 5.14$ C = 5.26$ C < Continued PROBLEM 3.96 (Cont.) (b) With the convection coefficient depending upon velocity, h = C1V 0.425 with C1 = 10.1 W/m2⋅K⋅(m/s)0.425, and using the energy balance of Eq (2), calculate and plot Ts as a function of ventilation air velocity V With very low velocities, the center temperature is nearly 0.5°C higher than the air From our earlier calculation we know that T(0) - Ts = 0.12°C and is independent of V Center temperature, T(0) (C) 5.4 5.3 5.2 0.2 0.4 0.6 0.8 Ventilation air velocity, V (m/s) COMMENTS: (1) While the temperature within the apple is nearly isothermal, the center temperature will track the ventilation air temperature which will increase as it passes through stacks of cartons (2) The IHT Workspace used to determine Ts for the base condition and generate the above plot is shown below // The temperature distribution, Eq (1), T_r = qdot * ro^2 / (4 * k) * ( 1- r^2/ro^2 ) + Ts // Energy balance on the apple, Eq (2) - qcv + qdot * Vol = Vol = / * pi * ro ^3 // Convection rate equation: qcv = h* As * ( Ts - Tinf ) As = * pi * ro^2 // Generation rate: qdot = qdotm * (1/24) * (1/3600) * rho // Assigned variables: ro = 0.080 k = 0.5 qdotm = 4000 rho = 840 r=0 h = 7.5 //h = C1 * V^0.425 //C1 = 10.1 //V = 0.5 Tinf = // Generation rate, W/m^3; Conversions: days/h and h/sec // Radius of apple, m // Thermal conductivity, W/m.K // Generation rate, J/kg.K // Specific heat, J/kg.K // Center, m; location for T(0) // Convection coefficient, W/m^2.K; base case, V = 0.5 m/s // Correlation // Air velocity, m/s; range 0.1 to m/s // Air temperature, C PROBLEM 3.97 KNOWN: Plane wall, long cylinder and sphere, each with characteristic length a, thermal  conductivity k and uniform volumetric energy generation rate q FIND: (a) On the same graph, plot the dimensionless temperature, [ T ( x or r ) − T ( a ) ]/[ q a /2k], vs the dimensionless characteristic length, x/a or r/a, for each shape; (b) Which shape has the smallest temperature difference between the center and the surface? Explain this behavior by comparing the ratio of the volume-to-surface area; and (c) Which shape would be preferred for use as a nuclear fuel element? Explain why? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties and (4) Uniform volumetric generation ANALYSIS: (a) For each of the shapes, with T(a) = Ts, the dimensionless temperature distributions can be written by inspection from results in Appendix C.3 T ( x ) − Ts x Plane wall, Eq C.22 = 1−   a  / 2k qa Long cylinder, Eq C.23 Sphere, Eq C.24 1 r  1 −     / 2k   a   qa T ( r ) − Ts   r 2  = 1 −     / 2k   a   qa T ( r ) − Ts = The dimensionless temperature distributions using the foregoing expressions are shown in the graph below Dimensionless temperature distribution (T_x,r-Ts) / (qdot*a^2/2*k) 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 Dimensionless length, x/a or r/a Plane wall, 2a Long cylinder, a Sphere, a Continued … PROBLEM 3.97 (Cont.) (b) The sphere shape has the smallest temperature difference between the center and surface, T(0) – T(a) The ratio of volume-to-surface-area, ∀/As, for each of the shapes is Plane wall ∀ a (1× 1) = =a As (1×1) Long cylinder ∀ π a ×1 a = = As 2π a ×1 Sphere ∀ 4π a / a = = As 4π a The smaller the ∀/As ratio, the smaller the temperature difference, T(0) – T(a) (c) The sphere would be the preferred element shape since, for a given ∀/As ratio, which controls the generation and transfer rates, the sphere will operate at the lowest temperature PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge FIND: Expression relating incident flux to temperature difference between center and edge of gauge SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5) Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contact resistance between foil and heat sink ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr, q r + q′′i ( 2π rdr ) = q r+dr , q r = − k ( 2π rt ) dT , dr q r+dr = q r + dq r dr dr Rearranging, find that q′′i ( 2π rdr ) = d  dT ( −k2π rt )  dr  dr  dr  d  dT  q′′ r = − i r   dr  dr  kt Integrating, r dT q′′r = − i + C1 dr 2kt and q′′r T ( r ) = − i + C1lnr+C2 4kt With dT/dr|r=0 =0, C1 = and with T(r = R) = T(R), q′′R T (R ) = − i + C2 4kt or q′′R C2 = T ( R ) + i 4kt Hence, the temperature distribution is T (r ) = ( ) qi′′ R − r + T ( R ) 4kt Applying this result at r = 0, it follows that q′′i = 4kt 4kt  T ( ) − T ( R ) = ∆T R2 R2 COMMENTS: This technique allows for determination of a radiation flux from measurement of a temperature difference It becomes inaccurate if emission from the foil becomes significant < PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge FIND: Expression relating incident flux to temperature difference between center and edge of gauge SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5) Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contact resistance between foil and heat sink ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr, q r + q′′i ( 2π rdr ) = q r+dr , q r = − k ( 2π rt ) dT , dr q r+dr = q r + dq r dr dr Rearranging, find that q′′i ( 2π rdr ) = d  dT ( −k2π rt )  dr  dr  dr  d  dT  q′′ r = − i r   dr  dr  kt Integrating, r dT q′′r = − i + C1 dr 2kt and q′′r T ( r ) = − i + C1lnr+C2 4kt With dT/dr|r=0 =0, C1 = and with T(r = R) = T(R), q′′R T (R ) = − i + C2 4kt or q′′R C2 = T ( R ) + i 4kt Hence, the temperature distribution is T (r ) = ( ) qi′′ R − r + T ( R ) 4kt Applying this result at r = 0, it follows that q′′i = 4kt 4kt  T ( ) − T ( R ) = ∆T R2 R2 COMMENTS: This technique allows for determination of a radiation flux from measurement of a temperature difference It becomes inaccurate if emission from the foil becomes significant < PROBLEM 3.99 KNOWN: Net radiative flux to absorber plate FIND: (a) Maximum absorber plate temperature, (b) Rate of energy collected per tube SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (x) conduction along absorber plate, (3) Uniform radiation absorption at plate surface, (4) Negligible losses by conduction through insulation, (5) Negligible losses by convection at absorber plate surface, (6) Temperature of absorber plate at x = is approximately that of the water PROPERTIES: Table A-1, Aluminum alloy (2024-T6): k ≈ 180 W/m⋅K ANALYSIS: The absorber plate acts as an extended surface (a conduction-radiation system), and a differential equation which governs its temperature distribution may be obtained by applying Eq.1.11a to a differential control volume For a unit length of tube q′x + q′′rad ( dx ) − q′x+dx = With q′x+dx = q′x + and q′x = − kt dq′x dx dx dT dx it follows that, q′′rad − d 2T dx d  dT  − kt =0  dx  dx  q′′ + rad = kt Integrating twice it follows that, the general solution for the temperature distribution has the form, q′′ T ( x ) = − rad x + C1x+C2 2kt Continued … PROBLEM 3.99 (Cont.) The boundary conditions are: T ( ) = Tw dT  =0 dx  x=L/2 C2 = Tw q′′ L C1 = rad 2kt Hence, q′′ T ( x ) = rad x ( L − x ) + Tw 2kt The maximum absorber plate temperature, which is at x = L/2, is therefore q′′ L2 Tmax = T ( L/2 ) = rad + Tw 8kt The rate of energy collection per tube may be obtained by applying Fourier’s law at x = That is, energy is transferred to the tubes via conduction through the absorber plate Hence,   dT  q′=2  − k t  dx  x=0   where the factor of two arises due to heat transfer from both sides of the tube Hence, q′= − Lq′′rad 800 W 0.2m ) ( m + 60$ C W   180  ( 0.006m ) m ⋅ K  Hence Tmax = or Tmax = 63.7$ C and q′ = −0.2m × 800 W/m or q′ = −160 W/m < < COMMENTS: Convection losses in the typical flat plate collector, which is not evacuated, would reduce the value of q ′ PROBLEM 3.100 KNOWN: Surface conditions and thickness of a solar collector absorber plate Temperature of working fluid FIND: (a) Differential equation which governs plate temperature distribution, (b) Form of the temperature distribution SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Adiabatic bottom surface, (4) Uniform radiation flux and convection coefficient at top, (5) Temperature of absorber plate at x = corresponds to that of working fluid ANALYSIS: (a) Performing an energy balance on the differential control volume, q′x + dq′rad = q′x+dx + dq′conv where q′x+dx = q′x + ( dq′x / dx ) dx dq′rad = q′′rad ⋅ dx dq′conv = h ( T − T∞ ) ⋅ dx Hence, q′′rad dx= ( dq′x / dx ) dx+h ( T − T∞ ) dx From Fourier’s law, the conduction heat rate per unit width is q′′ d 2T h q′x = − k t dT/dx T − T∞ ) + rad = − ( dx kT < kt (b) Defining θ = T − T∞ , d 2T/dx = d 2θ / dx and the differential equation becomes, d 2θ dx − q′′ h θ + rad = kt kt It is a second-order, differential equation with constant coefficients and a source term, and its general solution is of the form θ = C1e+λ x + C2e-λ x + S/λ 1/ λ = ( h/kt ) , S=q′′rad / kt where Appropriate boundary conditions are: θ (0 ) = To − T∞ ≡ θ , dθ /dx) x=L = θ o = C1 + C2 + S/λ Hence, dθ /dx) x=L = C1 λ e+λ L − C2 λ e-λ L = )( ( C1 = θ − S/λ / + e2λ L Hence, ( θ = θ − S/λ ) ) C2 = C1 e2λ L ( )( C2 = θ − S/λ / + e-2λ L  eλ x e -λ x  +   + S/λ L -2 L λ λ 1+e  1+e ) <