传热学课后题答案-戴锅生主编-第二版

传热学（第二版）戴锅生编习题解1－1解21wwttA6.0)220250(15.002.06.63)(221wwttAw/m·C1－4解)(wfttdLh52873)90200(8563.0w由)(ffPttmC52873)15(1018.436004003ft得8.128151018.44003600528733ftC1－9解热阻网络图：tf2tf11h11h21102.010015001.01011121hhrtm2·C/m（1）0202.010015001.010111211hhrtm2·C/w，减少81.7%（2）1012.0100015001.01012trm2·C/w，减少8.2%（3）11002.0100150001.01013trm2·C/w，减少0.2%结论：①对良导热体，导热热阻在总热阻中所占比例很小，一般可以忽略不计。②降低热阻大的那一个分热阻值，才能有效降低总热阻。1－12解tf2tf11h11h2tw1tw2设热量由内壁流向外壁，结果方程无解。重设热量由外壁流向内壁，则可以看出太阳辐射热流方向与对流换热的热流方向相反，传给外壁的总热量为)5(15480)(6008.02222wfwcttth根据串联热路可知)5(15480212)5(1548049.04.03022wwtt，整理得1.483245.132wt47.362wtC93.749.04.03047.3612wwttqw/m2qhttfw11119.28793.730111hqttwfC2－1解法Ⅰ①由付立叶定律推导取厚度为dr的薄壁微元壳体做为研究对象，根据热平衡drdrddrrrrr·0drdr（1）又drdtrr24（2）（2）代入（1）得048222drdtrdrdtr，整理得0222drdtdrdtr或0)(22drrtd②直接由球坐标导热微分方程式推导球坐标导热微分方程：Ctrtrrrtrat2222222sin1sinsin1)(1根据已知条件：0t，0t，0t，0，代入上式得0)(22drrtd微分方程组：221122,;,0)(wwttrrttrrdrrtd微分方程经两次积分得211CrCt以B.D代入通解得21121rrttCww，2121211rrttCwwrrrttrrttt·1121212112212211·11rrrttdrdtww212121122121212121)(1·11··4··41ddttrrrttrdrdtr212111)(2ddttww解法Ⅱdrdtrr24分离变量得drrdt24Crt4（1）B.D：1rr，1wtt（2）2rr，2wtt（3）（2）代入（1）得114rtCw1144rtrtw（4）（3）代入（4）得112244rtrtww整理得2121212111)(2114)(ddttrrtt或24rdrdt212124rrttrdrdtww121212112114ddrrttww122111)(2ddttww2－3解微分方程：02222ytxtB.D：x＝0，00xxt，x＝a，)(faxaxtthxt；y＝0，00tty，y＝b，bybyhtyt2－5解：设q＝600w/m2221131wwttq3.12.060060130011.0112122qttww＝0.2104m＝210.4mm∵q≤600w/m2∴2≥210.4mmtw3tw111tw2222－9解忽略蒸汽管壁的导热热阻43.0065.0201ddm47.002.0212ddm12320121ln21ln21ddttddttBwwAwwl32211201lnln·18040030180·43.047.0ln3.043.0ln2.0·lnln21321201＝0.5519w/m·C未包材料B时34683.043.0ln5519.02140400ln210121ddttAwwlw/mtw3tw112Alnd1d0tw212Blnd2d12－19已知：1＝250mm，1＝0.28＋0.000233tmw/m·C，2＝0.0466＋0.000213tmw/m·C，3＝250mm，3＝0.7w/m·C，tw1＝1000C，tw4＝50C，q＝759.8w/m2，tw2＝592.7C。求：2。解①设tw2＝700C，则850210007002211wwmtttC4781.0850000233.028.011121wwttq7.6028.7594781.025.010001112qttwwC重设tw2＝602C，则tm1＝801C，4666.0801000233.028.015938.7594781.025.010002wtC取tw2＝593C②3343wwttq4.3218.7597.025.0503343qttwwC2.45724.3215932322wwmtttC144.02.457000213.00466.020515.08.7594.321593144.03222qttwwm＝51.5mm2－24解2.681065.422122.03904AhPm1/m38514407550gerTTC)()]([0mHchXHmch)0625.02.68()]0625.0(2.68[385chXch（ch4.2625＝35.5007）)]0625.0(2.68[5.35385XchK)]0625.0(2.68[85.10XchKXchT)]0625.0(2.68[85.101140)0625.02.68(th)385(2.681065.422)(th40mHAm＝－268.5w[th(mH)＝th4.2625＝0.9996]2－28解设空气温度的真实值为tf52222010827.2)008.001.0(4)(4nddAm203142.001.00dPm57.2310827.22.5803142.01.295AhPm1/m57.13)14.057.23()(chmHch)(10mHchHfgHHttt100，ffttt5000，代入上式得57.1350)(1)50(100ffftmHchtt130757.12ft104ftC测量误差：4100104C，41004*%改用紫铜管后398w/m·C015.910827.239803142.01.295AhPm908.1)14.0015.9()(chHmch908.150100fftt，8.140908.0ft，155ftC55100155C改用紫铜管后，测量误差增加为55C，由)(0mHchH可以看出，随↑，m↓，ch（mH）↓，则必然增加。2－30解032.02004.0)06.009.0(2)(212rrHHCm092.0032.006.012CCHrrm4121028.1)06.0092.0(004.0)(rrACVm2533.106.0092.012rrC3578.0032.01028.1502523423CVHrAh查图2－23，得88.0f)30120()06.0092.0(225))((·2·)(220212200fCfttrrhtthA＝68.76W51.6076.6888.00fW3－8解5.225lcm＝0.025m510382.89502700215Pcam2/s0407.0215025.0350hlBi57.241iB23.40025.030010382.82520laF①查图3－6得2.00m，7430)30250(2.02.00fmttC②又4.05.20.1lx查图3－7得99.0m00099.02.0··mm6.43)30250(99.02.06.73fttC③70010822.2220025.0950270022clQJ/m2220210664.623.400407.0·FBi查图3－8得76.00QQ7010145.276.0QQJ/m2（本题也可直接由集总参数法求解）3－10解①333.1215.0170171hRBi0991.0388153811500ffmmtttt查图3－9得95.12Ra21936094.0018.0075.095.195.122haRs②由1Rr，333.11iB，查图3－10得7.0m9.9138777.07.0fmmttC③)(fwRrtthrt539)389.91(17170)(fwRrtthrtC/m3－4解由Ctaddt2∵02t（物体内温度均匀一致，与坐标无关）∴Cddt由题意：VhAVPCVhACVPddconstCVPCVhAdd为一阶线性非齐次常微分方程，用常数变动法求解。通解CVhAeC)(（ref数学手册P625）式中CehACVCVPCdeCVPCCVhACVhA·)(CehAPCVhA∴CehAPeCVhACVhA以0，0代入上式，得ChAP0，hAPC0hAPehAPeCVhACVhA0CVhACVhAehAPe103－5解335105102.3106.1AVlm，1.010192.22601054.1143hlBiv∴可以用集总参数法求解。设空气温度为20ftC000fttC，51820538fttC5.1096102.34.11403hAP45310072.6106.14208940102.34.11CVhA1/sCVhACVh