4、比转速与几何形状的关系 Relation between specific speed and geometry 对水轮机、泵、风机,有以下关系: D 2008-5-1 11
2008-5-1 11 4、比转速与几何形状的关系 比转速与几何形状的关系 Relation between specific speed and geometry Relation between specific speed and geometry 对水轮机、泵、风机,有以下关系: 对水轮机、泵、风机,有以下关系: s p s p D D u u =
4、比转速与几何形状的关系 Relation between specific speed and geometry ■考察转轮的最优工况点 设低压侧为法向出口(进口), 基本方程为: Wi2 B2 Cm2 g M 设: △=0△B,=0 2008-5-1 12
2008-5-1 12 4、比转速与几何形状的关系 比转速与几何形状的关系 Relation between specific speed and geometry Relation between specific speed and geometry 考察转轮的最优工况点 考察转轮的最优工况点 设低压侧为法向出口(进口), 设低压侧为法向出口(进口), 基本方程为: 设: 1 H u th p up c g = ∆ = 0 β p ∆ = 0 β s W U C C W u2 2 m2 2 2 β2
4、比转速与几何形状的关系 Relation between specific speed and geometry Cu 由高压侧速度三角形可得 Cw=up-CmpctgBpe Cm πDnBp 低压侧(法向出(进) 回) Q=F,cm=πD,B,cms=πD,B,4,tgBo B2 πD,B.u.igB Cm2 mp xD,B。 C 2008-5-1 13
2008-5-1 13 4、比转速与几何形状的关系 比转速与几何形状的关系 Relation between specific speed and geometry Relation between specific speed and geometry 由高压侧速度三角形可得 由高压侧速度三角形可得 低压侧(法向出(进)口): 低压侧(法向出(进)口): C u up p mp pe = − c ctgβ mp p p Q c π D B = ms s se c u = tgβ Q Fs ms s s ms s s s se = = c π D B c = π β D B u tg s s s se mp p p D B u tg c D B π β π = C U C W C u m α β W U C C W u2 2 m2 2 2 β2
4、比转速与几何形状的关系 Relation between specific speed and geometry D,B,4,g阝.ctg Bpe Cup =up DB。 上式代入基本方程: D.B.u.tgB,.cig pe D.B. up D B. 2008-5-1 14
2008-5-1 14 4、比转速与几何形状的关系 比转速与几何形状的关系 Relation between specific speed and geometry Relation between specific speed and geometry 上式代入基本方程: 上式代入基本方程: s s s se pe up p p p D B u tg ctg c u D B β β = − ( ) 1 p p s s s se pe th p p D B D Bu tg ctg u u g H β β = − s p s p D D u u = ⎥⎥⎦⎤ ⎢⎢⎣⎡ = − pese ps p p s th tgtg BB DD gu H β 2 β 2 1 ( )
4、比转速与几何形状的关系 Relation between specific speed and geometry 将流量和水头分别代入比转速定义”,= H n,=3.13 60 - B。g阝e 设进出口过流面积相等,即 πD,B1=πD2B2 D D。D n =k k=31360(g D,gp。 2008-5-1 15
2008-5-1 15 4 、比转速与几何形状的关系 比转速与几何形状的关系 Relation between specific speed and geometry Relation between specific speed and geometry 将流量和水头分别代入比转速定义 将流量和水头分别代入比转速定义 设进出口过流面积相等,即 设进出口过流面积相等,即 3 4 q n Q n H η = 4 3 2 2 1 2 3 4 3 1 ( ) ( ) 60 3.13 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = pe se p s p s se s s p s s tg tg B B D D tg D B D D n g β β β π 4 3 4 3 2 1 ( ) 60 3.13 1 k g tg tg D D tg D B D D n k s pe se p s se p p p s s s π β β β = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = πD1B1 = πD 2 B 2