传热学课后答案(完结版).pdf
《传热学课后答案(完结版).pdf》由会员分享,可在线阅读,更多相关《传热学课后答案(完结版).pdf(65页珍藏版)》请在冰点文库上搜索。
1绪论思考题与习题(89P)答案:
1冰雹落体后溶化所需热量主要是由以下途径得到:
Q与地面的导热量fQ与空气的对流换热热量注:
若直接暴露于阳光下可考虑辐射换热,否则可忽略不计。
2略3略4略5略6夏季:
在维持20的室内,人体通过与空气的对流换热失去热量,但同时又与外界和内墙面通过辐射换热得到热量,最终的总失热量减少。
(TT外内)冬季:
在与夏季相似的条件下,一方面人体通过对流换热失去部分热量,另一方面又与外界和内墙通过辐射换热失去部分热量,最终的总失热量增加。
(TT外内)挂上窗帘布阻断了与外界的辐射换热,减少了人体的失热量。
7热对流不等于对流换热,对流换热=热对流+热传导热对流为基本传热方式,对流换热为非基本传热方式8门窗、墙壁、楼板等等。
以热传导和热对流的方式。
9因内、外两间为真空,故其间无导热和对流传热,热量仅能通过胆壁传到外界,但夹层两侧均镀锌,其间的系统辐射系数降低,故能较长时间地保持热水的温度。
当真空被破坏掉后,1、2两侧将存在对流换热,使其保温性能变得很差。
10tRRA1tRRA2218.331012m11qtconst直线const而为(t)时曲线212.iR1R3R0R1ftq首先通过对流换热使炉子内壁温度升高,炉子内壁通过热传导,使内壁温度生高,内壁与空气夹层通过对流换热继续传递热量,空气夹层与外壁间再通过热传导,这样使热量通过空气夹层。
(空气夹层的厚度对壁炉的保温性能有影响,影响a的大小。
)13已知:
360mm、0.61()WmK118ft2187()WhmK210ft22124()WhmK墙高2.8m,宽3m求:
q、1wt、2wt、解:
1211tqhh18(10)45.9210.361870.611242Wm3111()fwqhtt11137.541817.5787wfqtth222()wfqhtt22237.54109.7124wfqtth45.922.83385.73qAW14.已知:
3Hm、0.2m、2Lm、45()WmK1150wt、2285wt求:
tR、R、q、解:
40.27.407104532tKRWAHL30.24.4441045tR2mKW3232851501030.44.44410tKWqmR3428515010182.37.40710ttKWR15已知:
50idmm、2.5lm、85ft、273()WhmK、25110Wqm求:
iwt、()iwfqhthttiwfqtth511085155730.052.551102006.7iAqdlqW16.已知:
150wt、220wt、241.23.96()WcmK、1200wt求:
1.2q、1.2q、1.2q解:
12441.21.2()()100100wwttqc44227350273203.96()()139.2100100Wm412441.21.2()()100100wwttqc442273200273203.96()()1690.3100100Wm21.21.21.21690.3139.21551.1Wqqqm17已知:
224Am、215000()WhmK、2285()WhmK、145t2500t、2285()WkhmK、1mm、398()WmK求:
k、解:
由于管壁相对直径而言较小,故可将此圆管壁近似为平壁即:
12111khh3183.56111015000390852()Wmk383.5624(50045)10912.5kAtKW若k2h100kkk8583.561.7283.56因为:
1211hh?
,21h?
即:
水侧对流换热热阻及管壁导热热阻远小于燃气侧对流换热热阻,此时前两个热阻均可以忽略不记。
18略5第一章导热理论基础思考题与习题(24P)答案:
1略2已知:
10.62()WmK、20.65()WmK、30.024()WmK、40.016()WmK求:
R、R解:
231241242242592101.1460.620.650.016mKRW232232560.265/0.650.024RmkW由计算可知,双Low-e膜双真空玻璃的导热热阻高于中空玻璃,也就是说双Low-e膜双真空玻璃的保温性能要优于中空玻璃。
34略56已知:
50mm、2tabx、200a、2000b/m2、45()WmK求:
(1)0xq、6xq
(2)vq?
解:
(1)00020xxxdtqbxdx3322452(2000)5010910xxxdtWqbxmdx
(2)由220vqdtdx?
62332245(2000)218010vdtWqbmdx?
7略8略9取如图所示球坐标,其为无内热源一维非稳态导热故有:
22tatrrrr00,tt0,0trr,()ftrRhttr10解:
建立如图坐标,在x=x位置取dx长度微元体,根据能量守恒有:
xdxxQQQ
(1)xdtQdx()xdxddtQtdxdxdx4()bbQEAEATUdx代入式
(1),合并整理得:
2420bfUdtTdx该问题数学描写为:
2420bfUdtTdx00,xtT,0()xldtxldx假设的4()bexldtfTfdx真实的7第二章稳态导热思考题与习题(P51-53)答案1.略2.略3.解:
(1)温度分布为121wwwttttx(设12wwtt)其与平壁的材料无关的根本原因在coust(即常物性假设),否则t与平壁的材料有关
(2)由dtqdx知,q与平壁的材料即物性有关4.略5.解:
2111222()0,(),wwwwddtrdrdrrrttttrrtt设有:
12124()11wwQttrr21214FrrRrr6.略7.已知:
4,3,0.25lmhm115wt,25wt,0.7/()Wmk求:
Q解:
lh?
,可认为该墙为无限大平壁15(5)0.7(43)6720.25tQFW8.已知:
2220,0.14,15wFmmt,31.28/(),5.510WmkQW求:
1wt解:
由tQF得一无限平壁的稳态导热3125.510150.1415201.28wwQttF9.已知:
12240,20mmmm,120.7/(),0.58/()WmkWmkr1r2rtw1tw2Qtw1tw283210.06/(),0.2Wmkqq求:
3解:
设两种情况下的内外面墙壁温度12wwtt和保持不变,且12wwtt由题意知:
1211212wwttq122312123wwttq再由:
210.2qq,有121231212121230.2wwwwtttt得:
123312240204()40.06()90.60.70.58mm10.已知:
1450wt,20.0940.000125,50wtt,2340/qWm求:
解:
412,0.0941.25102wwtttqmm412120.0941.25102wwwwtttttmqq445050450500.0941.25100.14742340m即有2340/147.4qWmmm时有11.已知:
11120,0.8/()mmWmk,2250,0.12/()mmWmk33250,0.6/()mmWmk求:
3?
22131312tw1qtw211122tw1tw2q11122339解:
21213123112313,wwwwttttqq由题意知:
qq即有:
21213123112313wwwwtttt333220.6250505000.12mm12.已知:
1600wt,2480wt,3200wt,460wt求:
123,RRRRRR解:
由题意知其为多层平壁的稳态导热故有:
14122334123wwwwwwwwttttttttqRRRR112146004800.2260060wwwwRttRtt223144802000.5260060wwwwRttRtt33414200600.2660060wwwwRttRtt13.略14.已知:
1)11012,40/(),3,250fmmWmkmmt,60ft220112,75/(),50/()hWmkhWmk2)223,320/()mmWmk3)223030,70/()hWmk221313tw1tw23311tw1tw2231231tw1tw423tw1tw4tw2tw3R1R2R3R=R1+R2R3+qtf11tf2210求:
123123,qqqkkk解:
未变前的122030102250605687.2/1113101754050ffttqWmhh1)21311121129.96/()1112101754050kWmkhh21129.96(25060)5692.4/qktWm21105692.45687.25.2/qqqWm2)22321221129.99/()11131017532050kWmkhh22229.99(25060)5698.4/qktWm22205698.45687.211.2/qqqWm3)22330101136.11/()1113101754070kWmkhh23336.11(25060)6860.7/qktWm23306860.75687.21173.5/qqqWm321qqq?
,第三种方案的强化换热效果最好15.已知:
35,130ACBmmmm,其余尺寸如下图所示,1.53/(),0.742/()ACBWmkWmk求:
R解:
该空斗墙由对称性可取虚线部分,成为三个并联的部分11111132222,ABCABCRRRRRRRRR3321111311135101301020.1307()/1.531.53CABABCRRmkW332322222335101301020.221()/1.530.742CABABCRmkW2212115.0410()/1111220.13070.221RmkWRR16.已知:
121160,170,58/()dmmdmmWmk,2230,0.093/()mmWmk33140,0.17/(),300wmmWmkt,450wt求:
1)123,RRR;2)lq:
3)23,wwtt.解:
tw1112323tw41010A1B1C1A2A3B2C2B3C3R1R1R1R2R3R2R2R3R3RRR121)4211111170lnln1.66410()/2258160dRmkWd2222221117060lnln0.517()/220.093170dRmkWd223332222111706080lnln0.279()/2220.1717060dRmkWd132RRR?
2)2330050314.1/0.5170.279littqWmRRR3)由121wwlttqR得4211300314.11.66410299.95wwlttqR同理:
34350314.10.279137.63wwlttqR17.已知:
1221211,22mmdd求:
llqq解:
忽略管壁热阻010121020122211lnln222ddRdd010122010122211lnln222ddRdd,llttqqRR(管内外壁温13,wwtt不变)0101220101010121020122211lnln22222211lnln222llddqRddddqRddtw1d0dd12mm12tw3221301010010101001241lnln22241lnln22dddddddd由题意知:
1001011
(2)2mdddd2112011
(2)32mmmdddd即:
21010101232()mmddddd(代入上式)15ln3ln231.27715ln3ln23llqRqR即:
0.783llqq21.7%lllqqq即热损失比原来减小21.7%。
18.已知:
1,dmm32.2210/,lRm0.15/()Wmk1max65wt,240wt,0.5,mm求:
maxI解:
21max2max12ln2wwllttqIRdd11221max2max36540123.7()22.2210120.5lnln220.151wwlttIARdd19.已知:
121185,100,40/(),180wdmmdmmWmkt230.053/(),40wWmkt,52.3/lqWm求:
2解:
13222121122211lnln22wwltttqddRRddtw1tw2qltw1tw2tw3RR1221lnd2d122lnd2+2d214整理得:
221111804011002(ln)20.053(ln)2252.3240852100
(1)
(1)7222ldtqddeemm或:
21RR?
故有132222221ln2wwltttqdRd2222
(1)722ltqdemm20.已知:
)4.7715.273(,/6.199,30,3,35.01211wtkgkJrmmmmmmd325wt,210.03/(),16.3/(),1WmkWmkh求:
m解:
1231wwFFttQRR311211111112111111()()111144
(2)
(2)(22)2222wwttdddd2(25273.1577.4)111111()()16.30.350.3560.030.3560.416102.7W或:
12FFRR?
故有:
312232(25273.1577.4)0.03102.711111()()40.3560.416wwttQWrr102.73.61.85/199.6Qmkghr21.略22.略23.解:
fffttlxttxttmdxd2211222,0,0ltf,ht,tt12ft1t2tw1tw2tw3RRF1F241(1r1)-1r242(1r2)-1r315解微分方程可得其通解:
12mxmxcece由此得温度分布(略)24.已知:
25,lmm3,mm20140/(),75/(),80WmkhWmkt30ft,0xlq求:
lq解:
3222750.0250.4725140310hUhLhmllllAL18.9m0()0.472518.9(8030)()(0.4725)chmlxchxchmlch44.91(0.472518.9)chx3044.91(0.472518.9)tchx002()()lQhUhqthmlthmlLmLm275(8030)(0.4725)174.7/18.9thWm25.已知:
15,20,48.5/(),84lmmlmmWmkt,040t220/()hWmk求:
t解:
3200.12248.51.510hUhdhmllllAd00()()fllfttchmlchmltt0()84
(2)4099.93()1
(2)1lftchmltchtchmlch
(2)3.7622ch99.9384100%100%15.9%99.93flftttt26.已知:
00.8,160,60mmlmmt,16.3/()Wmk,其他条件同25题16求:
t解:
3201606.2716.30.810hmll0()84(6.27)6084.09()1(6.27)1lftchmltchtchmlch(6.27)264.24ch84.0984100%100%0.11%84.09flftttt27.已知:
3,16mmlmm2
(1)140/(),80/()WmkhWmk2
(2)40/(),125/()WmkhWmk求:
f解:
(1)332228016100.312140310hUhLhmllllAL()(0.312)0.970.312fthmlthml
(2)3322212516100.7340310hUhLhmllllAL()(0.73)0.8530.73fthmlthml28.已知:
1277,140,4,25,50/()dmmdmmmmPmmWmk2060/(),320hWmkt,75ft求:
lq解:
211()31.52ldd33.52cll2172ccrrl334221()410(7238.5)101.3410cfrrm17P1133223224226033.5100.821501.3410chlf21722.1533.5crr查图得:
0.78f每片肋片的散热量为1Q100()fffQQhFtt222102()()cffrrhtt2262(7238.5)100.7860(32075)266.7W每米肋片管的散热量为:
12
(1)lqnQnQ100014125n片/米41266.7401.4811kW2Q为两肋片间的表面的散热量210()fQdPtt3377102510(32075)1.48W29.略30.已知:
212132.2,0.3,0.56/(),0wllmmWmkt,230wt求:
lq解:
1113100.3AlLLSL2222.27.330.3AlLLSLl1l21830.54SL121,5ll123(224)lSSStQqLL,12wwttt(21027.3340.54)0.56(300)618.6/Wm31.已知:
1165,90wdmmt,21.5,1.05/(),6wHmWmkt220/()hWmk求:
lq解:
3lrHr22ln()lsHr22ln()lQstqtHllr21.05(906)154.2/21.5ln0.165/2Wm32.已知:
21210.520.52,0.42,0.023/(),30wllmHmWmkt214wt,34QW求:
解:
1211212,lllHSSll3410.54,0.54SHSl1234(444)QSSSSt1213440.520.5240.520.42344440.540.4240.540.520.023(3014)lllHQSSt23.621036.2mmm33.已知:
5,2.54,2,80mmmPMPat,180/()WmkHdtw1tw2l1l2Q底=0H19求:
ct解:
由2.54,2mPMPa,查表得,420.8810()/cRmkWctQR31ttt再由cctQR,22cABttt得4340.88108049510220.8810180cccRttR第三章非稳态导热1.略2.略3.略4.略5.已知:
3210.15,420/(),8400/,58/()pdmmcJkgkkgmhWmk22126/()hWmk求:
0102,解:
33012111484004200.15322101.52()3235842pppddcccVshFhdh同理:
302284004200.152100.7()323126pdcsh6.略7.已知:
300.5,8930/,400/(),25pdmmkgmcJkgkt,120ft2095/(),1%,22/()hWmkWmk(康铜)求:
t解:
由001%ffttttRct1t2At2Bt3t1t1At2Bt3x2000.01()1200.01(25120)119.05fftttt34950.510133.6100.10.123223RVhhFBivM故满足集总参数法的求解条件,有:
0VVBiFoe320189304000.5103lnln(110)14.43952pcVshF8.已知:
23,11,mmFm239/(),48.5/()hWmkWmk,0300t,20ft,6212.710/,ams50t求:
解:
33330101220.98100.10.148.53VhBiM满足集总参数法的求解条件,故有:
0phFcVe00lnlnpcVVhFhaF36348.511050202ln3283912.710130020s9.略10.已知:
080t,20,20fdmmt,12/,5min,34umst38954/,383.1/(),386/()pkgmcJkgkWmk求:
h解:
假设可使用集总参数法,故有:
0phFcVe3201208954383.110342022lnln83.2/()5608020pcVhWmkF21由3383.22010122.16100.10.123862VRhhFBivM满足集总参数法的计算,上述假设成立。
11.已知:
2,12minABABpApBABfABBccttthh0050%mAmB求:
A解:
11000,0.52AmAmBAAAABiBih查表得:
0.24ABFoFo即:
222212248minAAABBAABBBaa12.已知:
300.50.50.5,30abcmt,800ft,52/()Wmk220.063/,80/(),30minamhhWmk求:
mt解:
1522.61800.52Bih220.06330600.536000.25aFo对于正六面体有:
3000fmmmftttt平板由12.60.5BiFo,查图有:
00.9m平板3300800(80030)0.9239mmftt平板13.已知:
72040,510/,4/(),25mmamsWmkt,1260ft240/(),1hhWmk缺少2240.14ft已知:
、20wt、smu8.0、ml45.0、5105Rec、mx1.01、mx2.02、mx3.03lx4求:
x解:
30)(21wfmttt按30mt查表得:
42.5Pr、)(618.0kmw、smv271005.8由vxuxRe得mx1.0141094.9Rexmx2.0251099.1RexecR均为层流mx3.0351098.2Rexmx45.0451047.4Rex2131RePr332.0xxmx1.013.1136mx2.029.803)(2kmwmx3.038.655mx45.045.535x2图略15.已知:
ml3.0、smu9.0、25ft求:
max、)(yul解:
由25ft查表得smv2710055.9clvluRe1098.210055.99.03.0Re57mll35max1055.23.01098.264.4Re64.423uyyyul3maxmax)(21)(23)(333)1055.21(9.0211055.219.023yy371071.24.529)(yyyulsm图略16.略17.略18.已知:
smu10、80ft、30wt、ml8.0、5105Rec、mb1求:
cx、Q解:
55)(21wfmttt按55mt查表得:
smv2510846.1、697.0Pr、)(10865.22kmw由muvxvxucccc923.01010864.1105ReRe55cxl全板长均为层流3121PrRe664.02lll)(9.13697.010846.18.0108.010865.266