传热学课后答案(完整版)

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f Q 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。(T T 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。(T T 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

2、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.t R R A = 1t R R A = 2218.331012m -= 11.q t = const =直线 const 而为=(t 时曲线12、略13.解:1211tq h h =+=18(1045.9210.361870.61124-=+2W m 111(f w q h t

3、 t =- 11137.541817.5787w f q t t h =-=-=222(w f q h t t =- 22237.54109.7124w f q t t h =+=-+=-45.92 2.83385.73q A W =14. 解:40.27.407104532t K R W A HL -= 30.2 4.4441045t R -=2m K W 3232851501030.44.44410t KW q mR -= 3428515010182.37.40710t tKW R -=15.(i w f q h t h t t =- i w f qt t h =+ 51108515573

4、=+=0.05 2.551102006.7i Aq d lq W =16.解:12441.2 1.2(100100w w t t q c =-44227350273203.96(139.2100100W m +=-=12''441.2 1.2(100100w w t t q c =-442273200273203.96(1690.3100100W m +=-='21.2 1.2 1.21690.3139.21551.1W q q q m =-=-=17.已知:224A m =、215000(W h m K =、2285(W h m K =、145t = 2500t =、

5、'2285(W k h m K =、1mm =、398=(W m K 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:12111k h h =+=3183.5611101500039085-=+2(W m k 383.5624(5004510912.5kA t KW -=-=若k 2h '100k k k -=%8583.56 1.7283.56-=% 因为:1211h h ,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(W

6、m K =、20.65(W m K =、30.024(W m K =、40.016(W m K =求:'R 、''R 解:2'3124124224259210 1.1460.620.650.016m K R W -=+=+= '"232232560.265/0.650.024R m k W =+=+= 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 =、2t a bx =+、200a =、2000b =-/m 2、45(Wm K = 求:

7、(10x q =、6x q = (2v q解:(100020x x x dt q bx dx =-=-=3322452(20005010910x x x dt W q bx mdx -=-=-=-= (2由220vq d t dx +=2332245(2000218010v d tW q b m dx =-=-=-=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: 22t a t r r r r = 00,t t = 0,0tr r = ,(f tr R h t t r =-=-10.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: x dx x Q Q

8、Q += (1x dtQ dx =-+ (x dx d dtQ t dx dx dx +=-+4(b b Q EA E A T Udx =代入式(1,合并整理得:2420b fU d tT dx -=该问题数学描写为:2420b fU d tT dx -=00,x t T =,0(x ldtx l dx =假设的 4(b e x ldtf T f dx =-=真实的第二章稳态导热思考题与习题(P 51-53答案1.略2.略3.解:(1温度分布为 121w w w t t t t x -=- (设12w w t t >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

9、平壁的材料有关(2由 dt q dx =- 知,q 与平壁的材料即物性有关 4.略 5.解: 2111222(0,(,w w w w d dt r dr drr r t t t t r r t t =>=设有: 12124(11w w Q t t r r =- 21214F r r R r r -=6.略7.已知:4,3,0.25l m h m = 115w t =, 25w t =-, 0.7/(W m k = 求:Q解: ,l h ,可认为该墙为无限大平壁 15(50.7(436720.25tQ F W -= 8.已知:2220,0.14,15w F m m t =-,31.28/(

10、, 5.510W m k Q W =求:1w t解: 由 tQ F = 得一无限平壁的稳态导热312 5.510150.141520 1.28w w Q t t F =+=-+=9.已知:12240,20mm mm =,120.7/(,0.58/(W m k W m k = 3210.06/(,0.2W m k q q = 求:3解: 设两种情况下的内外面墙壁温度12w w t t 和保持不变,且12w w t t >由题意知:1211212w w t t q -=+122312123w w t t q -=+再由: 210.2q q =,有121231212121230.2w w w

11、w t t t t -=+得: 123312240204(40.06(90.60.70.58mm =+=+= 10.已知:1450w t =,20.0940.000125,50w t t =+=,2340/q W m 求: 解: 412,0.094 1.25102w w t t tq m m +=+412120.094 1.25102w w w w t t t t tmq q+-=+ 445050450500.094 1.25100.14742340m +-=+= 即有 2340/147.4q W m mm 时有11.已知:11120,0.8/(mm W m k =,2250,0.12/(mm

12、 W m k =33250,0.6/(mm W m k =求:'3?= 解: '2121'3123112313,w w w w t t t t q q -=+由题意知:'q q = 即有:2121'3123112313w w w w t t t t -=+'33322=+ 0.6250505000.12mm =+= 12.已知:1600w t =,2480w t =,3200w t =,460w t = 求:123,R R R R R R 解:由题意知其为多层平壁的稳态导热 故有: w w w w w w w t t t

13、 t t t t t q R R R R -= 112146004800.2260060w w w w R t t R t t -=-223144802000.5260060w w w w R t t R t t -=- 33414200600.2660060w w w w R t t R t t -=- 13.略14.已知:111012,40/(,3,250f mm W m k mm t =,60f t = 220112,75/(,50/(h W m k h W m k = 2223,320/(mm W m k =32'23030,70/(h W m k =求:123123,q q

14、q k k k 解: 未变前的 122030102250605687.2/1113101754050f f t t q W m h h -=+ 121311121129.96/(1112101754050k W m k h h -=+ 21129.96(250605692.4/q k t W m =-= 21105692.45687.2 5.2/q q q W m =-=-= 222321221129.99/(11131017532050k W m k h h -=+ 22229.99(250605698.4/q k t W m =-= 22205698.45687.211.2/q q q W

15、 m =-=-= 3 22330'101136.11/(11131*k W m k h h -=+ 23336.11(250606860.7/q k t W m =-= 23306860.75687.21173.5/q q q W m =-=-=321q q q > ,第三种方案的强化换热效果最好 15.已知:35,130A C B mm mm =,其余尺寸如下图所示, 1.53/(,0.742/(A C B W m k W m k = 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,A B C A B C R R R R R R R R R =+

16、=+3321111311135101301020.1307(/1.53 1.53C A B A B C R R m k W -=+=+= 332322222335101301020.221(/1.530.742C A B A B C R m k W -=+=+= 2212115.0410(/1111220.13070.221R m k W R R -=+16.已知:121160,170,58/(d mm d mm W m k =,2230,0.093/(mm W m k =33140,0.17/(,300w mm W m k t =,450w t = 求:1123,R R R ; 2 l q

17、: 3 23,w w t t . 解nln 1.66410(/2258160d R m k W d -= 2222221117060lnln 0.517(/220.093170d R m k W d +=223332222111706080lnln 0.279(/2220.1717060d R m k W d +=+132R R R < 2 2330050314.1/0.5170.279l i t t q W m R R R -=+ 3由 121w w l t t q R -=得 4211300314.1 1.66410299.95w w l t t q R -

18、=-=-= 同理:34350314.10.279137.63w w l t t q R =+=+=17.已知:1221211,22m m d d =求:'ll q q 解:忽略管壁热阻010121020122211lnln 222d d R d d +=+ '010122010122211lnln 222d d R d d +=+ '',l l t tq q R R = (管内外壁温13,w w t t 不变 01012'20101'010121020122211lnln 22222211ln ln 222l l d d q R d d d d

19、q R d d +=+01010010101001241lnln 22241ln ln 22d d d d d d d d +=+由题意知: 1001011(22m d d d d =+=+2112011(232m m m d d d d =+=+即:21010101232(m m d d d d d =+=+= (代入上式''15ln 3ln23 1.27715ln 3ln 23l l q R q R +=+ 即: '0.783l l q q ='21.7%l llq q q -=即热损失比原来减小21.7%。 18.已知:1,d mm =32.2210/,l

20、 R m -=0.15/(W m k =1max 65w t =,240w t =,0.5,mm =求:max I解: 21max 2max 12ln 2w w l l t t q I R d d-=+11221max 2max36540123.7(2 2.2210120.5ln ln 220.151w w l t t I A R d d - -= + 19.已知:121185,100,40/(,180w d mm d mm W m k t = 230.053/(,40w W m k t =,52.3/l q W m = 求:2 解: 13222121122211ln ln 22w w l t

21、 t tq d d R R d d -=+整理得:221111804011002(ln 20.053(ln 2252.3240852100(1(17222l d tq d d ee mm -=-=-= 或:21R R ,故有 132222221ln 2w w l t t tq d R d -=+ 2222(1722lt q demm =-=20.已知:4.7715.273(,/6.199,30,3,35.01211+-=w t kg kJ r mm mm mm d 325w t =,210.03/(,16.3/(,1W m k W m k h =求:m 解: 1231w w F F t t Q

22、 R R -=+311211111112111111(111144(2(2(222222w w t t d d d d -=-+-+2(25273.1577.4111111(16.30.350.3560.030.3560.416+-=-+-102.7W = 或:12F F R R ,故有: 312232(25273.1577.40.03102.711111(40.3560.416w w t t Q W r r -+-=-102.7 3.6 1.85/199.6Q m kg h r = 21.略 22.略23. 解: ff f t t l x t t x t t m dxd -=-=-=-221

23、1222,0,0解微分方程可得其通解: 12mx mx c e c e -=+ 由此得温度分布(略24.已知:25,l mm =3,mm =20140/(,75/(,80W m k h W m k t = 30f t =,0x l q = 求:,l q 解: 3222750.0250.4725140310hU h L h ml l l l A L -= 18.9m = 0(0.472518.9(8030(0.4725ch m l x ch x ch ml ch -=-44.91(0.472518.9ch x =- 3044.91(0.472518.9t ch x =+-002(l Q hU h

24、 q th ml th ml L mL m = 275(8030(0.4725174.7/18.9th W m =-=25.已知:15,20,48.5/(,84l mm l mm W m k t =,040t = 220/(h W m k =求:t 解: 3200.12248.5 1.510hU h d h ml l l l A d -= 00(f l l ft t ch ml ch ml t t -=-0(84(24099.93(1(21l f t ch ml t ch t ch ml ch -=-(2 3.7622ch =99.9384100%100%15.9%99.93f l ft t

25、t t -=26.已知:00.8,160,60mm l mm t =,16.3/(W m k =,其他条件同25题求:t 解: 320160 6.2716.30.810hml l -= 0(84(6.276084.09(1(6.271l f t ch ml t ch t ch ml ch -=-(6.27264.24ch = 84.0984100%100%0.11%84.09f l ft t t t -=27.已知:3,16mm l mm = 2(1140/(,80/(W m k h W m k = 2(240/(,125/(W m k h W m k = 求:f 解:/p>

26、00.312140310hU h L h ml l l l A L -= (0.3120.970.312f th ml th ml = (23322212516100.7340310hU h L h ml l l l A L -= (0.730.8530.73f th ml th ml = 28.已知:1277,140,4,25,50/(d mm d mm mm P mm W m k = 2060/(,320h W m k t =,75f t = 求:l q解: 211(31.52l d d =-=33.52c l l =+=2172c c r r l =+=334221(410(7238.5

27、10 1.3410c f r r m -=-=-=P (1133223224226033.5100.82150 1.3410c h l f -= 2172 2.1533.5c r r = 查图得: 0.78f = 每片肋片的散热量为1Q 100(f f f Q Q hF t t =-222102(c f f r r h t t =-2262(7238.5100.7860(32075266.7W -=-= 每米肋片管的散热量为:12(1l q nQ n Q =+- 100014125n =+=片/米 41266.740 1.4811kW =+= 2Q 为两肋片间的表面的散热量 210(f Q d

28、 P t t =-3377102510(32075 1.48W -=-= 29.略30.已知:21213 2.2,0.3,0.56/(,0w l l m m W m k t =,230w t =求:l q解: 1113100.3A l LLS L =222 2.27.330.3A l L LS L =30.54S L = 121,5l l >123(224l S S S tQ q L L+= ,12w w t t t =-(21027.3340.540.56(300=+- 618.6/W m =31.已知:1165,90w d mm t =,21.5, 1.05/(,6w H m W m

29、 k t = 220/(h W m k = 求:l q解: ,3l r H r >> 22ln(ls H r=l 1l 2Hdtw 1tw 222ln(l Q s t q t H l l r=2 1.05(906154.2/2 1.5ln 0.165/2W m =-= 32.已知:21210.520.52,0.42,0.023/(,30w l l m H m W m k t = 214w t =-, 34Q W = 求: 解: 1211212,l l l HS S l l =3410.54,0.54S H S l = 1234(444Q S S S S t =+1213440.52

30、0.5240.520.42344440.540.4240.540.520.023(3014l l l H Q S S t +=-+ 23.621036.2m mm -=33.已知:5, 2.54,2,80mm m P MPa t =,180/(W m k = 求:c t 解:由 2.54,2m P MPa =,查表得,420.8810(/c R m k W -= c tQ R =+ 31t t t =-再由 c ctQ R =,22c A B t t t =-得4340.88108049510220.8810180cc c R t t R -=+第三章 非稳态导热l 1l 2Q 底=0HRct

31、 1t 2At 2Bt 3t 1t 1A t 2Bt 3x1.略2.略3.略4.略5.已知:3210.15,420/(,8400/,58/(p d mm c J kg k kg m h W m k = 22126/(h W m k = 求:0102,解:33012111484004200.1532210 1.52(3235842p p p d d c c c V s h F h d h - =同理:302284004200.152100.7(323126p d c s h -= 6.略7.已知:300.5,8930/,400/(,25p d mm kg m c J kg k t =,120f

32、t =2095/(,1%,22/(h W m k W m k =(康铜 求:,t 解:由 001%f ft t t t -=- 00.01(1200.01(25120119.05f f t t t t =+-=+-=34950.51013 3.6100.10.123223R V hhF Biv M -=<=故满足集总参数法的求解条件,有:V VBi Fo e -=320189304000.5103ln ln(11014.43952p c V s hF -=-=-= 8.已知:23,11,mm F m =239/(,48.5/(h W m k W m k =,0300t =,20f t =

33、,6212.710/,a m s -=50t =求:解: 33330101220.98100.10.148.53V hBi M -=<= 满足集总参数法的求解条件,故有:phFc Ve -=00lnln p c VV hFhaF =-=- 36348.511050202ln 3283912.710130020s -=-=-9.略10.已知:080t =,20,20f d mm t =,12/,5min,34u m s t = 38954/,383.1/(,386/(p kg m c J kg k W m k = 求:h解:假设可使用集总参数法,故有: 0phFc Ve -=3201208

34、954383.110342022ln ln 83.2/(5608020p c V h W m k F -=-=-=-由 3383.2201012 2.16100.10.123862V Rhh F Biv M -=<= 满足集总参数法的计算,上述假设成立。11.已知:2,12min A B A B pA pB A B f A B B c c t t t h h =0050%mA mB = 求:A 解: 11000,0.52AmAmB A A A ABi Bi h -= 查表得:0.24A B Fo Fo =即: 222212248min A A A B BA AB B B a a = 12

35、.已知:300.50.50.5,30a b c m t =,800f t =,52/(W m k =220.063/,80/(,30min a m h h W m k =求:m t 解: 1522.61800.52Bi h -=220.06330600.536000.25a Fo =对于正六面体有:3000f mm m f t t t t -= -平板由 12.60.5Bi Fo -=, 查图有:00.9m = 平板3300800(800300.9239m m f t t =-=-= 平板13.已知:72040,510/,4/(,25mm a m s W m k t -=,1260f t =

36、240/(,1hh W m k =缺少40.14=f t 已知:、20=w t 、sm u 8.0=、m l 45.0=、5105Re =c 、m x 1.01=、m x 2.02=、m x 3.03=l x =4求:x 解:30(21=+=w f m t t t 按30=m t 查表得:42.5Pr =、(618.0k m w =、smv 271005.8-=由 vxu x =Re 得m x 1.01= 41094.9Re =x m x 2.02= 51099.1Re =x < ec R 均为层流m x 3.03= 51098.2Re =x m x 45.04= 51047.4Re =

37、x =2131Re Pr 332.0x x m x 1.01= 3.1136mx 2.02= 9.803(2k m wm x 3.03= 8.655m x 45.04= 5.535 x 2= 图略15. 已知:m l 3.0=、s m u 9.0=、25=f t 求:max 、(y u l解:由25=f t 查表得 smv 2710055.9-= c l v l u Re 1098.210055.99.03.0Re 57<=- m l l35max 1055.23.01098.264.4Re 64.4-=-=u y y y u l 3max max (21(23(=3331055.21(

38、9.0211055.219.023y y - 371071.24.529(y y y u l -= sm 图略16. 略 17. 略18. 已知:s m u 10=、80=f t 、30=w t 、m l 8.0=、5105Re =c 、m b 1=求:c x 、Q解:55(21=+=w f m t t t 按55=m t 查表得:smv 2510846.1-=、697.0Pr =、(10865.22k m w =-由 m u v x v x u c c c c 923.01010864.1105Re Re 55=- c x l < 全板长均为层流3121Pr Re 664.02ll l

39、=(9.13697.010846.18.0108.010865.2664.02312152k m w =-(w t F Q 5563080(8.019.13=-=19. 略 20. 略 21. 已知:y u u = 层流 求:x 解:由动量积分方程有:(1(1(202020=-=-=-=u dx d dy y y u dx d dy u u u u u dx d dy u u u dx d 式左式右=U udydU uy 0即: =U u U dx d (2即 const U = 即有:dx U v d U v dx d =6 两边积分有:=u v dx U v d x x62162021Re

40、 12x x=,vxU x =Re 22. 略23. 已知:w t cy by a t ,2+-= f t 求:x 解:由边界层特点知 0=y w t t = 0022=y xt d d=y f t t = 得:0=c 、 w t a = 、(f w t t b -=-=+-=-=b t t cy b t t d d t t fw y fw y yt fw x 002( 24. 略25. 略 dr r r r =001,t F Q =26. 略 dx l lx =01,t F Q =27. 略 0=-=y f w dydtt t 28. 已知:23m F =、m l 11=、1401=w t

41、、301=f t 、s m u 501=、w Q 150001=、m l 52=、202=w t 、702=f t 、s mu 82=求:2 解:8530140(21(21111=+=+=w f m t t t 457020(21(21222=+=+=w f m t t t 按 1m t 2m t 查表得:691.0Pr 1=、(1009.321k m w=-、sm v 2511016.2-=6985.0Pr 2=、(108.222k m w=-、smv 25210746.1-=有:6511111031.21016.2150Re =-v l u6522221031.210746.1508.8R

42、e =-v l u 知:1Re =2Re ,1Pr 2Pr ,且几何相似 得:21Nu Nu = 121121222=l l Nu l 而:1111t =F Q =22112l l 111t F Q (24.830140315000511009.3108.3222k m w =-=-( 29. 略30. 已知:mm d a 16=,mm d b 30=,b a G G 2=,b a t t =求:(1是否相似(2如何相似解:(1b a t t = ,且为同类流体 b a v v =,b a p p = 再:24d GFGu m =有:2222244ab b b baa ambB.C.D. A.

43、 从甲国贩卖布匹到乙国B. 从甲国贩卖木材到乙国C. 从乙国贩卖木材到甲国D. 从乙国贩卖布匹到甲国26.党的十八大报告提出,在发展平衡性、协调性、可持续性明显增强的基础上,2020年实现国内生产总值和城乡居民人均收入比2010年翻一番的目标。实现这一目标,应推进经济结构战略性调整,转变经济发展方式ma d u d u v d u v d u 22a b d d b ad d 12=a b d d 即:b a Re Re A.B.知:两者流态不相似(2若要相似,需b a Re Re = 即:11Re Re =bmb a ma b a d u du 而:22a b b a mb ma d G d

44、 G u u =,带入上式有:122=baa b b a d d d d G G 1583016=b a b a d d G G 即:要使两者相似,两者的质量流量之比应为15831. 略 32. 略 33. 略 34.略第六章 单相流体对流换热及准则关联式30.公安部提出一系列措施,进一步简化办事程序,方便群众办证、办户口。下列做法与材料体现的政府职能一致的是A.全国法院建立“失信者黑名单制度”,向社会公布失信被执行人名单B.中国人民银行授权中国外汇交易中心公布汇率中间价C.国家工商总局决定组织开展专项整治“霸王条款”活动D.某气象局利用自身的网络技术建立“农网”,为农民免费提供信息咨询31.

45、具有157年历史的东莞道生园是广东四大名园之一,日前被拆,引发广泛关注。不久,东莞市政府决定在原址复建。保护好文物古迹主要是因为它是民族历史文化的物质载体作为文化遗产是一个国家和民族历史文化成就的重要标志具有很高的经济价值是民族文化的集中展示和民族情感的集中表达A.B.C.D.32.2013年第二届岭南民俗文化节在广州举行,来自广东省内12个地市的30多支民俗队伍参加表演和巡游。举办民俗文化节是尊重文化多样性的表现促进了中华民族精神的丰富和发展有利于在文化的包容性中形成和保持文化特色是人们以自省、自律的方式,提高思想道德修养A.B.C.D.33.右边漫画给我们的哲学启示是A.关键部分对整体起决

46、定作用B.量变必然引起质变C.事物发展是前进性和曲折性的统一D.实践是认识的基础34.世界各国对海洋经济可持续发展理念,虽有不同的理解,但都认为是经济发展、环境以及社会之间的平衡。材料体现的唯物辩证法的哲理有矛盾的普遍性与特殊性相互联结辩证否定的实质是扬弃事物是普遍联系的认识事物的角度不同,其价值评价A.B.按f t 查表有:62219.6210/,Pr 0.695, 2.93710/(f f f m s W m K -=3621.0075/(, 1.041/,20.4110/f f f Cp kJ kg K kg m N s m -=42360.045,Re 5.510511020.4110

47、444m m f ffu dGGd d 材料三:十八届三中全会立足党和国家全局,提出坚持和完善基本经济制度、加快完善现代市场体系、构建开放型经济新体制等一系列新部署。全会要求全面深化改革必需坚持以马列主义、毛泽东思想、中国特色社会主义理论体系为指导,必须牢牢把握正确方向、掌握科学有效的改革方法,才能不断推动改革开放的伟大事业向前发展。(1从材料一反映了什么经济态势?(4分(2运用经济生活的知识,结合材料一、二,为我国全面提高开放型经济水平出谋划策。(12分(3分析材料三所体现的辩证唯物论道理。(15分37.(21分阅读下列材料,结合所学知识回答问题。材料一:广东省惠州市在“崇文厚德,包容四海,敬业乐群”的惠州精神指导下,深度激活文化生产力,大力发展文化产业,促进惠州经济蓬勃发展;扎实开展各项文化惠民工程,相继建成一大批公共文化场馆,充分尊重人民群众的主体地位,推动不同人群共享文化发展成果;以热情开放的客家文化为起点,大力