3、=1/2,y=δ/2=const,等截面肋。n=0y=δ/2(x/H),三角形肋。n=1/3y=δ/2(x/H)1/2,凸拋物線n=∞,y=δ/2(x/H)2,凹拋物線邊界條件:x=0,肋端:(1)1stB.C:t=tf。(2)2ndB.C中絕熱邊界條件:dt/dx=0。(3)3rdB.C:-λdt/dx=h(t-tf)x=H,肋基:t=t0。3.2等截面直肋的導(dǎo)熱分析上式中:n=1/2,y=δ/2=const,等截面肋。換一下坐標(biāo)得:d2t/dx2–hU/(λA)(t-tf)=0令:θ=t-tf過余溫度。d2θ
4、/dx2–m2θ=0m2=hU/(λA)邊界條件:x=H,肋端:(1)1stB.C:θ=0。(2)2ndB.C中絕熱邊界條件解:dθ/dx=0。(3)3rdB.C:-λdθ/dx=h2θx=0,肋基:θ=θ0。通解:θ=c1e-mx+c2emx3.2.11stB.C解:c1e-mH+c2emH=0c1+c2=θ0c1=θ0emH/(emH-e-mH)c2=-θ0e-mH/(emH-e-mH)θ=θ0sh(m(H-x))/sh(mH)整個(gè)肋片散熱量:Φ=-λAdθ/dx」x=0=λAmθ0ch(mH)/sh(mH)
5、=(hUλA)1/2(t0-tf)ch(mH)/sh(mH)特例:H→∞θ=θ0e-mxθH=0→tH=tf整個(gè)肋片散熱量:Φ=-λAdθ/dx」x=0=λAmθ0=(hUλA)1/2(t0-tf)3.2.22ndB.C中絕熱邊界條件解:-c1e-mH+c2emH=0c1+c2=θ0c1=θ0emH/(emH+e-mH)c2=θ0e-mH/(emH+e-mH)θ=θ0ch(m(H-x))/ch(mH)整個(gè)肋片散熱量:Φ=-λAdθ/dx」x=0=λAmθ0sh(mH)/ch(mH)=(hUλA)1/2(t0-tf
6、)th(mH)特例:H→∞θ=θ0e-mxθH=0→tH=tf整個(gè)肋片散熱量:Φ=-λAdθ/dx」x=0=λAmθ0=(hUλA)1/2(t0-tf)結(jié)果與1stB.C解相同。3.2.33rdB.C解:-c1e-mH+c2emH=h2θ/(λm)c1+c2=θ0θ=θ0{[ch(m(H-x))+h2/(λm)sh(m(H-x))]/[ch(mH)+h2/(λm)sh(mH)]}整個(gè)肋片散熱量:Φ=-λAdθ/dx」x=0=λAmθ0{[sh(mH)+h2/(λm)ch(mH)]/[ch(mH)+h2/(λm)s
7、h(mH)]}=(hUλA)1/2(t0-tf){[th(mH)+h2/(λm)]/[1+h2/(λm)th(mH)]}特例:h2=h,可得h2=0,可得絕熱邊界條件解。h2=∞,可得1st邊界條件解。H→∞?θ=θ0e-mx整個(gè)肋片散熱量:?Φ=-λAdθ/dx」x=0=λAmθ0=(hUλA)1/2(t0-tf)3.2.4三種肋效率由上分析:溫度場(chǎng)變化特點(diǎn):a.過余溫度為指數(shù)(雙曲)曲線,肋基與換熱流體溫差大,肋端溫差小。肋各處換熱量不同,肋基處換熱量最大,肋端處換熱量最小。b.當(dāng)肋高趨向無窮大時(shí),溫度分布和
8、換熱量有下列趨勢(shì):θ=θ0e-mxΦ=-λAdθ/dx」x=0=λAmθ0=(hUλA)1/2(t0-tf)由特點(diǎn)a定義第一類肋效率(肋片有效度):η1=實(shí)際傳熱量/以肋基導(dǎo)熱面積為基準(zhǔn)的最大傳熱量(未裝肋時(shí)肋基傳熱量)。對(duì)絕熱邊界條件:η1=(hUλA)1/2(t0-tf)th(mH)/(hA(t0-tf))=th(mH)/(m(A/U))由特點(diǎn)a定義第二類肋效率(工程