4-10 Ball bearings leaving the oven at a uniform temperature of 900 ℃ are exposed to air for a while before they are dropped into the water for quenching. The time they can stand in the air before their temperature falls below 850 ℃ is to be determined.Assumptions 1 The bearings are spherical in shape with a radius of 0r = 0.6 cm. 2 The thermal properties of the bearings are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi ﹤0.1 so that the lumped system analysis is applicable (this assumption will be verified).Properties The thermal conductivity, density, and specific heat of the bearings are given to be k =15.1 W/m ⋅°C,3/8085m kg =ρ, and ⋅=kg kJ C p /480.0°C.Analysis The characteristic length of the steel ball bearings and Biot number arem m D D D A V L c 002.06012.066/23=====ππ0167.0)/1.15()002.0)(/125(2=°⋅°⋅==C m W m C m W khL Bi c <0.1 Therefore, the lumped system analysis is applicable. Then the allowable time is determined to be132016.0)002.0)(/480)(/8085(/125−=°⋅°⋅===s m C kg J m kg C m W L C h V C hA b c p p ρρ s t e e T T T t T t s bt t 7.33090030850)()016.0(1=→=−−→=−−−−−∞∞ The result indicates that the ball bearing can stay in the air 3.7s before being dropped into the water.4-11 Large brass plates are heated in an oven. The surface temperature of the plates leaving the oven is to be determined.Assumptions 1center plane. 3 The thermal properties of the plate are constant. 4 transfer coefficient is constant and uniform over the entire surface. 5Fourier number is τ﹥transient temperature charts) are applicable (this assumption will be verified).Properties The properties of brass at room temperature are given to be C m W k °⋅=/110, s m /109.3326−×=αAnalysis The Biot number for this process is0109.0)/110()015.0)(/80(2=°⋅°⋅==C m W m C m W k hL BiThe constants 1λ and 1A corresponding to this Biot number are, from Table 4-1,Furnace, 700°C1λ=0.1039 and 1A =1.0018The Fourier number is4.90)015.0(min)/60min 10)(/109.33(2262=××==−m s s m Lat τ﹥0.2Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then thetemperature at the surface of the plates becomes376.0)1039.0cos()0018.1()/cos(),(),()4.90()1039.0(111221===−−=−−∞∞e L L e A T T T t x T t L wallλθτλC t L T t L T °=→=−−2.446),(376.070025700),(4-12 A long cylindrical wood log is exposed to hot gases in a fireplace. The time for the ignition of the wood is to be determined.Assumptions 1 Heat conduction in the wood is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the wood are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ﹥0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).Properties The properties of wood are given to be k =0.17 W/m·℃,s m /1028.127−×=α. Analysis The Biot number is20(13.6/)(0.05)4(0.17/)hr W m C m Bi kW m C ⋅°===⋅° The constants1λand 1A corresponding to this Biot number are, from Table 4-1,1λ=1.9081 and 1A =1.46980J is determined from Table 4-2 corresponding to the constant 1λ,00.275J =The Fourier number is determined to be221(1.9081)010100(,)420500(/)(1.4698)(0.275)0.24910500t T r t T A e J r r eT T λττλτ−−∞∞−−=→=→=−− which is > 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used, with the understanding that the error involved will be under 2 percent. Then the length of time before the log ignites is22072(0.249)(0.05)486381.05min (1.2810/)r m t s m s τα−====×4-14 An orange is exposed to very cold ambient air. It is to determine whether the orange will freeze in 4 h in subfreezing temperatures. Assumptions 1 The orange is spherical in shape with a diameter of 8 cm. 2 Heat conduction in the orange is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the orange are constant, and are those of water. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ﹥0.2 so that the one-term approximate solutions (orthe transient temperature charts) are applicable (this assumption will be verified).Properties The properties of the orange are approximated by those of water at the average temperature of about 5℃, k =0.571 W/m.℃ and s m C k p /10136.0)42051000/(571.0/26−×=×==ρα (Table A-9).Analysis The Biot number is0.1051.1)/571.0()04.0)(/15(20≈=°⋅°⋅==C m W m C m W k hr Bi The constants1λand 1A corresponding to this Biot number are, from Table 4-1,1λ=1.5708 and 1A =1.2732The Fourier number is22.1)04.0()/36004)(/10136.0(2262=××==−m h s h s m L tατ﹥0.2 Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then thetemperature at the surface of the oranges becomes0399.05708.1)5708.1sin()2732.1(/)/sin(),(),()22.1()5708.1(001001100221===−−=−−∞∞rad e r r r r e A T T T t r T t r t sphλλθτλC t r T t r T °−=→=−−−−2.5),(0399.0)6(15)6(),(00which is less than 0°C. Therefore, the oranges will freeze.4-15 An area is subjected to cold air for a 10-h period. The soil temperatures at distances 0, 10, 20, and 50 cm from the earth’s surface are to be determined.Assumptions 1 The temperature in the soil is affected by the thermalconditions at one surface only, and thus the soil can be considered tobe a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the soil are constant.Properties The thermal properties of the soil are given to beC m W k °⋅=/9.0 and s m /106.125−×=α.Orange T i = 15°C Ambient air–15°CWinds,–10°CSoil10°CAnalysis The one-dimensional transient temperature distribution in the ground can be determined from⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛+−⎟⎠⎞⎜⎝⎛=−−∞k t h t x erfc k t h k hx t x erfc T T T t x T t t αααα2exp 2),(22 Where7.33/9.0)360010)(/106.1()/40(252=°⋅××°⋅=−Cm W s s m C m W k t h α 11387.332222==⎟⎟⎠⎞⎜⎜⎝⎛=k t h k t h αα Then we conclude that the last term in the temperature distribution relation above must be zero regardless of x despite the exponential term tending to infinity since (1) erfc(ξ)→0 forξ﹥4 (see Table 9-3) and (2)The term has to remain less than 1 to have physically meaningful solutions. That is,03.3321138exp 2exp 22≅⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛+=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛+t x erfc k hx k t h t x erfc k t h k hx αααα Therefore, the temperature distribution relation simplifies to⎟⎠⎞⎜⎝⎛−+=→⎟⎠⎞⎜⎝⎛=−−∞∞t x erfc T T T t x T t x erfc T T T t x T t t t t αα2)(),(2),(Then the temperatures at 0, 10, 20, and 50 cm depth from the ground surface becomex = 0:C T T T T erfc T T T t erfc T T T h T t t t t t t °−==×−+=−+=⎟⎠⎞⎜⎝⎛−+=∞∞∞∞101)()0()(20)()10,0(αx = 0.1m:⎟⎟⎠⎞⎜⎜⎝⎛××−−+=−)/360010)(/106.1(21.0)1010(10)10,1.0(25h s h s m m erfch m T C erfc °−=×−=−=5.89257.02010)066.0(2010 x = 0.2m:⎟⎟⎠⎞⎜⎜⎝⎛××−−+=−)/360010)(/106.1(22.0)1010(10)10,2.0(25h s h s m m erfch m T C erfc °−=×−=−=0.78519.02010)132.0(2010 x = 0.5m:⎟⎟⎠⎞⎜⎜⎝⎛××−−+=−)/360010)(/106.1(25.0)1010(10)10,5.0(25h s h s m m erfch m TC erfc °−=×−=−=8.26418.02010)329.0(20104-19 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each geometry in 10, 20, and 60 min are to be determined.Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in all x -, y , and z - directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x - and radial r - directions. 3 The thermal properties of the granite are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ﹥0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).Properties The thermal properties of the granite are given to be k=2.5 W/m ⋅°C and s m /1015.126−×=α.Analysis:Cubic block : This cubic block can physically be formed by the intersection of three infinite plane walls of thickness 2L =5 cm.After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are4.0)/5.2()025.0)(/40(2=°⋅°⋅==C m W m C m W k hL Bi → 1λ=0.5932 and 1A =1.0580104.1)025.0(min)/60min 10)(/1015.1(2262=××==−m s s m L tατ﹥0.2To determine the center temperature, the product solution can be written as[]3),0(),0,0,0(wall block t t θθ=()3121),0,0,0(τλ−∞∞=−−e A T T T t T t{}369.0)0580.1(50020500),0,0,0(3)104.1()5932.0(2==−−−et T →T (0,0,0,t) = 322.9℃After 20 minutes208.2)025.0(min)/60min 20)(/1015.1(2262=××==−m s s m L tατ﹥0.2{}115.0)0580.1(50020500),0,0,0(3)208.2()5932.0(2==−−−e t T →T (0,0,0,t) = 444.8℃After 60 minutes624.6)025.0(min)/60min 60)(/1015.1(2262=××==−m s s m L tατ﹥0.2 5Hot gases, 500°C{}00109.0)0580.1(50020500),0,0,0(3)624.6()5932.0(2==−−−e t T →T (0,0,0,t) = 499.5℃Note that τ﹥0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable.Cylindrical block : This block can physically be formed by the intersection of a long cylinder of radius r 0=2.5cm and a plane wall of thickness 2L =5 cm. After 20 minutes208.2)025.0(min)/60min 20)(/1015.1(2262=××==−m s s m L tατ﹥0.221(2.5/) 2.5(40/)(0.025)k W m C Bi hL W m C m ⋅°===⋅° We can get: Plane wall:(0,)(0,)0.5wall i T t T t T T θ∞∞−==−Cylinder:(0,)(0,)0.24cyl i T t T t T T θ∞∞−==−So,(0,0,)(0,0,)(0,)(0,)0.50.240.12short cyl wall cyl i T t T t t t T T θθθ∞∞−===×=−→T (0,0,0,t) = 442.4℃。
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