3234×15.8×0.3pumd=5.55x10福保家盗等快雅Remer0.02×10-9DU=0.82ASuR图,得一查UmxBTTUmpFOEOLD2201F.u=0.82x15.8=13.0(m/s)0138.已知:孔板流量计测流量d=5cm,d=25mmR-220mmHg,p=1050kg/m,-0.6mpa.saoto012001010求:qvdo2=0.25解:面积比m=101x001010050d20设Re>Rea极限-8×10+1x88-90-10-9由图1-54,查得C=0.6250(开鑫2gR(e-p)..u.=p2×9.81×0.22x(13600-1050)2=4.5m/s是费童新=0.625,1050u=mu。=0.25×4.5=1.12m/s1050×1.1×0.05=9.8×10*>8×104mm0-b0pud验证:Re,-P0.6x10惠高隆滑A1因而假设成立md-u=0.785x0.052×l12=22x10-m/s=79m2/h)1qy=4,39.已知:转子流量计q=400-40001/h,P=2670kg/m,9-0求:qvo上限解:由转子流量计9g0000Par(pr-Peo)quorVPe,(p-pa)qui26
"papprPo,"ppPrParpt29M.qvP.=0.81224MquirVPo9(上限)-0.8124(上限)-0.812×40003248/h40.已知:B=1m,1-0.01m,u0.5m/s,8=0.001m,T=2500(du/dy)040.01m求:剪切率du/dy拉力F,功率P解:8-0.001m,近似认为模口中速度分布为线性剪切率duAu0.50.001m名=5001/sdy80.001剪应力=2500(±04=2500(500)=3.0×10*Pady拉力F=t:A=3.0×10*x1×0.01=3×10°N功率P=Fu=300×0.5=150W41.已知:q-6.28×10m/s,L-20m.d-0.1mt=0.05(du/dy)*p=1250kg/m求:h(0/kg)6.28×10qv解:U=0.8m/sInd0.785×0.14d.un.p0.1*5x0.8x1250-14310Re1+3n1+3×05#-88-X805K-f0.05x4n4x05h,=4rlu查得f=0.0044d2200.83h,=4x0.0044x1.13J/kg012列两截面间柏努利方程27
ui+huitheP2P+gz2++g+22pP4PP,u=,Z,Z,-6m,u-uu0.8+113=603J/kg+h,=981x6+得h。=g(z,-z)+220808dum20l00-m年片42.已知:V-411000dy本过膜中南#大店40℃下与du/dy关系为:1005010.05.01.00.5(du/dy)/s)65937712383.242.134T/Pa求:,与口。(最小二乘法)00盘吧2010100002小du解:Vi=yh,+vuW1>=100xX1-0E-A-7Lmdy02-00-00duB=VuA=yh,令Y=V,X-Vdym01X=5e6A)由最小二乘法原理得BN2x2xduS2月-7762其中之x-24.18111司Sdu,2xy.-瑞=460.08F=166.52xdy国道O80000ktsETT0到大库民用同费商校28
(小肉77622418良面M166.517990460.084341624.18414.336(d)166.5X24.18安文海生欧西品6出77.626之研世博460.0824.18883.632.13B=A24.186414.3324.18166.5T-A-1884Pa4=B-4.55Pa-S8元-0=tb1R-143.p=1200kg/m,d=10mm,L=1m,H=0.35m时齐末酱不再流动京工求:T解:对管径为d,管长为L的管内流体作力平衡:11APnd+pg--mdL=tydL44由静力学方程得AP-PgHddt,=pgH4+P840.010.01+1200x981x=1200×9.81×0.35x44×1=39.7PaAOduyu(2r)d层流时t=K圆管内44.已知:国21dyAnInR0求证:(U=2K1Rn+129
%RK(2)U=(2K13n+1KAP证明:(a)Tm-21duyT=K(-(b)drA两式联立得d2K7%dr49CRdu=r"dr2K1积分得A-(RUPn2K1nAPn1R"得证iD2K1n+1BRR22u(2元r)dr=u.RAS22nR明u-rdrRnNAO0Svnerrdru=TRJ.R32K7n+1R国R2AO2nnR3)Rn2R2K1n+13n+1的6出款140n得证Ro2K13n+1国中街单中销三O0RE大书.4口头30