传热学第二章导热基础理论
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第一章导热理论基础2已知:10.62()W m K λ=∙、20.65()W m K λ=∙、30.024()W m K λ=∙、40.016()W m K λ=∙求:'R λ、''R λ 解:2'3124124224259210 1.1460.620.650.016m K R W λσσσλλλ-⨯⨯⨯⨯⎛⎫∙=++=++⨯= ⎪⎝⎭'"232232560.265/0.650.024R m k W λσσλλ⨯⎛⎫=+=+=⋅ ⎪⎝⎭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。
5.6.已知:50mm σ=、2t a bx =+、200a =℃、2000b =-℃/m 2、45()Wm K λ=∙求:(1)0x q =、6x q = (2)v q解:(1)00020x x x dtq bx dx λλ====-=-= 3322452(2000)5010910x x x dtW q bx m dx σσσλλ-====-=-=-⨯⨯-⨯⨯=⨯(2)由220vq d t dx λ+=2332245(2000)218010v d t W q b m dxλλ=-=-=-⨯-⨯=⨯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 τ∂∂∂⎛⎫= ⎪∂∂∂⎝⎭00,t t τ==0,0tr r∂==∂ ,()f tr R h t t rλ∂=-=-∂ 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 Q ε++= (1)x dt Q dx λ=-+()x dx d dtQ t dx dx dxλ+=-++∙ 4()b b Q EA E A T Udx εεεσ===代入式(1),合并整理得:2420b fU d t T dx εσλ-= 该问题数学描写为:2420b f U d t T dx εσλ-= 00,x t T == ,0()x ldtx l dx ===假设的 4()b e x ldtfT f dx λεσ=-=真实的 第二章稳态导热3.解:(1)温度分布为 121w w w t t t t x δ-=-(设12w w t t >)其与平壁的材料无关的根本原因在 coust λ=(即常物性假设),否则t 与平壁的材料有关 (2)由 dtq dxλ=- 知,q 与平壁的材料即物性有关5.解: 2111222()0,(),w w ww d dt r dr drr r t t t t r r t t===>==设有:12124()11w w Q t t r r πλ=-- 21214F r r R r r λπλ-=7.已知:4,3,0.25l m h m δ=== 115w t =℃, 25w t =-℃, 0.7/()W m k λ=⋅ 求:Q解: ,l h δ ,可认为该墙为无限大平壁15(5)0.7(43)6720.25tQ FW λδ∆--∴==⨯⨯⨯= 8.已知:2220,0.14,15w F m m t δ===-℃,31.28/(), 5.510W m k Q W λ=⋅=⨯ 求:1w t解: 由 tQ Fλδ∆= 得一无限平壁的稳态导热312 5.510150.141520 1.28w w Q t t F δλ⨯=+=-+⨯=⨯℃ 9.已知:12240,20mm mmδδ==,120.7/(),0.58/()W m k W m k λλ=⋅=⋅3210.06/(),0.2W m k q q λ=⋅=求:3δ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 >221313由题意知:1211212w w t t q δδλλ-=+122312123w w t t q δδδλλλ-=++再由: 210.2q q =,有121231212121230.2w w w w t t t t δδδδδλλλλλ--=+++得:123312240204()40.06()90.60.70.58mm δδδλλλ=+=⨯⨯+= 10.已知:1450w t =℃,20.0940.000125,50w t t λ=+=℃,2340/q W m ≤ 求:δ 解: 412,0.094 1.25102w w t t tq m m λλδ+∆==+⨯⨯41212[0.094 1.2510]2w w w w t t t t tmq qδλ+-∆==+⨯⋅ 44505045050[0.094 1.2510]0.14742340m +-=+⨯⨯⨯= 即有 2340/147.4q W m m mδ≤≥时有 11.已知:11120,0.8/()mm W m k δλ==⋅,2250,0.12/()mm W m k δλ==⋅33250,0.6/()mm W m k δλ==⋅求:'3?δ=解: '2121'3123112313,w w w w t t t t q q δδδδδλλλλλ--==+++由题意知:'q q =212tw 1tw 2q 11λ12λ23λ322即有:2121'3123112313w w w wt t t t δδδδδλλλλλ--=+++'33322λδδδλ=+ 0.6250505000.12mm =+⨯= 12.已知:1600w t =℃,2480w t =℃,3200w t =℃,460w t =℃ 求:123,,R R R R R R λλλλλλ解:由题意知其为多层平壁的稳态导热 故有: 14122334123w w w w w w w w t t t t t t t t q R R R R λλλλ----====∴112146004800.2260060w w w w R t t R t t λλ--===-- 223144802000.5260060w w w w R t t R t t λλ--===--33414200600.2660060w w w w R t t R t t λλ--===-- 14.已知:1)11012,40/(),3,250f mm W m k mm t δλδ==⋅==℃,60f t =℃ 220112,75/(),50/()h W m k h W m k λλ==⋅=⋅ 2)223,320/()mm W m k δλ==⋅ 3)2'23030,,70/()h W m k δδλλ===⋅求:123123,,,,,q q q k k k ∆∆∆ 解:未变前的122030102250605687.2/1113101754050f f t t q W m h h δλ---===⨯++++tw 1tw 4tw 2tw 3R 1R2R3R =R 1+R 2R3+t αt f221)21311121129.96/()1112101754050k W m k h h δλ-===⋅⨯++++ 21129.96(25060)5692.4/q k t W m =∆=⨯-= 21105692.45687.2 5.2/q q q W m ∆=-=-= 2)22321221129.99/()11131017532050k W m k h h δλ-===⋅⨯++++ 22229.99(25060)5698.4/q k t W m =∆=⨯-= 22205698.45687.211.2/q q q W m ∆=-=-= 3) 22330'101136.11/()131********k W m k h h δλ-===⋅⨯++++ 23336.11(25060)6860.7/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 δδδ===,其余尺寸如下图所示,1.53/(),0.742/()A C B W m k W m k λλλ==⋅=⋅求:R λ解:该空斗墙由对称性可取虚线部分,成为三个并联的部分R 1R 1R 1R2R3R 2R 2R3R311113222,A B C A B C R R R R RR R R R =++==++ 3321111311135101301020.1307()/1.53 1.53C A B A B C R R m k W δδδλλλ--⨯⨯∴=++=⨯+==⋅332322222335101301020.221()/1.530.742C A B A B C R m k W δδδλλλ--⨯⨯=++=⨯+=⋅2212115.0410()/1111220.13070.221R m k W R R λ-∴===⨯⋅⨯+⨯+16.已知:121160,170,58/()d mm d mm W m k λ===⋅,2230,0.093/()mm W m k δλ==⋅33140,0.17/(),300w mm W m k t δλ==⋅=℃,450w t =℃求:1)123,,R R R λλλ; 2) l q : 3) 23,w w t t . 解:1)4211111170lnln 1.66410()/2258160d R m k W d λπλπ-===⨯⋅⨯2222221117060lnln 0.517()/220.093170d R m k W d λδπλπ++===⋅⨯ 223332222111706080lnln 0.279()/2220.1717060d R m k W d λδδπλδπ++++===⋅+⨯+tw 1112323tw 4132R R R λλλ∴< 2) 2330050314.1/0.5170.279l i t t q W m R R R λλλ∆∆-====++∑ 3)由 121w w l t t q R λ-=得 4211300314.1 1.66410299.95w w l t t q R λ-=-=-⨯⨯=℃ 同理:34350314.10.279137.63w w l t t q R λ=+=+⨯=℃ 17.已知:1221211,,22m m d d δδλλ=== 求:'ll q q 解:忽略管壁热阻010121020122211ln ln 222d d R d d λδδδπλπλδ+++=++ '010122010122211ln ln 222d d R d d λδδδπλπλδ+++=++ '',l l t tq q R R λλ∆∆== (管内外壁温13,w w t t 不变)01012'20101'010*******22211lnln 22222211ln ln 222l l d d q R d d d d q R d d λλδδδπλπλδδδδπλπλδ+++++∴==+++++01010010101001241lnln 22241ln ln 22d d d d d d d d δδδδδδ++++=++++由题意知: 1001011[(2)]2m d d d d δδ=++=+ 2112011[(2)]32mm m d d d d δδ=++=+ 即:21010101232()m m d d d d d δδδ=⇒+=+⇒= (代入上式)3''15ln 3ln23 1.277ln 3ln 23l l q R q R λλ+∴===+ 即: '0.783l l q q ='21.7%l llq q q -∆==即热损失比原来减小21.7%。