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。10.tRRA1tRRA2218.331012m11.qtconst直线const而为(t)时曲线12、略13.解:1211tqhh=18(10)45.9210.361870.611242Wm111()fwqhtt11137.541817.5787wfqtth℃2222()wfqhtt22237.54109.7124wfqtth℃45.922.83385.73qAW14.解:40.27.407104532tKRWAHL30.24.4441045tR2mKW3232851501030.44.44410tKWqmR3428515010182.37.40710ttKWR15.()iwfqhthttiwfqtth51108515573℃0.052.551102006.7iAqdlqW16.解:12441.21.2()()100100wwttqc44227350273203.96()()139.2100100Wm12''441.21.2()()100100wwttqc442273200273203.96()()1690.3100100Wm'21.21.21.21690.3139.21551.1Wqqqm17.已知:224Am、215000()WhmK、2285()WhmK、145t℃2500t℃、'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3即:12111khh=3183.56111015000390852()Wmk383.5624(50045)10912.5kAtKW若k2h'100kkk%8583.561.7283.56%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:2'31241242242592101.1460.620.650.016mKRW'232232560.265/0.650.024RmkW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。3.4.略5.6.已知:50mm、2tabx、200a℃、2000b℃/m2、45()WmK求:(1)0xq、6xq(2)vq解:(1)00020xxxdtqbxdx43322452(2000)5010910xxxdtWqbxmdx(2)由220vqdtdx2332245(2000)218010vdtWqbmdx7.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,xtT5,0()xldtxldx假设的4()bexldtfTfdx真实的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121(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,(),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63125.510150.1415201.28wwQttF℃9.已知:12240,20mmmm,120.7/(),0.58/()WmkWmk3210.06/(),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121231212121230.2得:123312240204()40.06()90.60.70.58mm10.已知:1450wt℃,20.0940.000125,50wtt℃,2340/qWm求:解:412,0.0941.25102wwtttqmm41212[0.0941.2510]244505045050[0.0941.2510]0.14742340m即有2340/147.4qWmmm时有11.已知:11120,0.8/()mmWmk,2250,0.12/()mmWmk733250,0.6/()mmWmk求:'3?解:'2121'3123112313,由题意知:'qq即有:2121'3123112313'333220.6250505000.12mm12.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:14122334123∴112146004800.2260060223144802000.526006033414200600.266006013.略14.已知:1)11012,40/(),3,250fmmWmkmmt℃,60ft℃220112,75/(),50/()hWmkhWmk2)223,320/()mmWmk83)2'23030,,70/()hWmk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)22330'101136.11/()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111132222,ABCABCRRRRRRRRR93321111311135101301020.1307()/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:1)4211111170lnln1.66410()/2258160dRmkWd2222221117060lnln0.517()/220.093170dRmkWd223332222111706080lnln0.279()/2220.1717060dRmkWd132RRR2)2330050314.1/0.5170.279littqWmRRR3)由121wwlttqR得4211300314.11.66410299.95wwlttqR℃同理:34350314.10.279137.63wwlttqR℃1017.已知:1221211,,22mmdd求:'llqq