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WATERPURIFICATIONTECHNOLOGYVol.24No.22005V(,350002)V,,VAnalysisofDistributingUniformityinVPatternDistributingChannelChenShuidiZhangXiangzhongXieShangbinCaiWeiyiCaiWenliang(CollegeofcivialEng.,FuzhouUniversity,Fuzhou350002,China)AbstractThisarticleanalysdthewatersurfacelineofVpatterndistributingchannel,formulatedthecomputingexpression,andispresentedadesingmethodtomeettheneedofdistributinguniformity.KeywordsfilterVpatternchanneldistributingwateruniformityratioofdrillinghole[1]()V,,,,V,V,,,V,7cm,14cm,(qmin/qmax)[2]80%,V,1VV,i=0,,(15cm),,,,,,,,,:,V,,1V,[3]V,dx,Q,Q+dQ,dxQ0Ldx,Q=Q+dQ+Q0LdxdQ=-Q0Ldx(1)Q=-Q0Lx+C1,x=0,Q=00,C1=Q0Q=1-xLQ0(2)H+Q22gA2=H+dH+(Q+dQ)22gA+Q2A2C2Rdx72VVol.24No.22005dH=-QdQgA2-Q2A2C2Rdx(3)(1)(2)(3)dH=Q20gA2L1-xLdx-Q20A2C2R1-xL2dx,A,C,R,H=-Q202gA21-xL2+Q20L3A2C2R1-xL3+C2,x=0,H=H0,C2=H0+Q202gA2-Q2L3A2C2R,H=H0+Q202gA21-1-xL2-Q20L3A2C2R1-1-xL3=H0+V202g1212xL2223gLC2R1212xL3(4)f(x)=1-1-xL2-23gLC2R1-1-xL3(5)(4)H=H0+V202gf(x)(6)f(x),V,V;M=gLC2R,V,,V,M=gL1nR1/62R=n2gL/R4/3(7)VM1,,L=10m,n=0.013,M=0.016562/R4/3,M1,R0.0462m,,V0.10m,,,,f(x)f(x)=2L1-xL-2LM1-xL2(8),M1,f(x),f(x):x=L,f(x)=0,f(L)=1-23M,V,He=H0+1-23MV202g(9)H=He-H0=1-23MV202g=1-23MQ202gA2(10)2h0,1,h0+H,=q0qe=h0h0+H=11+H/h0(11)a,,k,q=Q0/k,h=q22g2a2=(Q0/k)22g2a2=Q202gk22a2(12)h0=h-12H=Q202g2k2a2-121-23MQ202gA2(13)(10)(13)(11)=6-22(3-2M)6+22(3-2M)(14)=kaA=V0u0(15),,,(16)=16(1-2)(3-2M)(1+2)(16)(16)VM,M(7),0.4V,V=0.82,V=0.62=0.82,M=0.4,=0.95,=0.4559VMV,M1M=0.340.46,0.820.62,95%,82Vol.24No.220051V(=0.95)M0.340.360.380.400.420.440.46(=0.82)0.44400.44840.45180.45590.46010.46440.4689(=0.62)0.58720.59230.59760.60300.60850.61420.6201AQ0V0,V0,0.600.80m/s,Aa,k=Aa,3VQ0=0.100m3/s,L=15.00m,V22VVV0=0.65m/sVA=Q0V0=0.1000.65=0.1538m2A=bH0+12H20tan400=0.1538m2,VH0=0.475m=b+H0+H0/cos400=1.215mVR=A=0.1266mVM=n2glR4/3=0.01329.815.000.12664/3=0.3911=0.95,V,=0.82,(16)=10.826(1-0.952)(3-20.391)(1+0.952)=0.4541V=A=0.45410.1538=0.06984m2d=0.025m,k=4d2=40.069840.0252=142.3,k=142L=lm=15.00142=0.106m=106mmu0=V0=0.650.4541=1.425m/sh=u202g2=1.425229.80.822=0.154mH=1-23MV202g=0.016mVH=H0+H2=0.475+0.0162=0.483mZ0=H-h=0.483-0.154=0.329mV,,=0.62,d=30mm,:=0.6005,=0.09236m2,k=130,L=115mm,u0=1.082m/s,h=0.155m,Z0=0.328m4(5)(6)V,,(11)(14),,(16)1.[M]..:,1995:2882892,.[M].(4).:,1999:3373383.[J].:,1996,22(6):510:200425217:,,1969,:059123709230(H)92
本文标题:V型滤池配水槽均匀性分析
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