T9AGXDELu专宾菜the.ZgT品2十4一9（真月adn3×10-3qvODOEOC=1.53m/sVur0.785x0.052CmE8S-id2005287.04150dly-153x(2=2.39m/su,=u(40d,选e=0.2mm20℃水，u=1mPa·S000120050=7.65x1044.1d,u,p0.05x1.53x1000rDxRe,1x10-H000ED0.04×2.39×1000dup=9.56×10*Re,1x10-00-人国密套u020.22=0.004查莫迪图得入-0.030入=0.031Seldr=0.0058/d,50403-0.18-0.75,闸阀1/2开，5-4.5突然缩小2=0.5查得：90°弯头，u3u4+55)+25+a+(2:Z-dz2gd,2g2g8RE1.5322.39380+5×0.75+05)+(0.03x2×9.810.052x9.812.39220（内+4.5+0.18)x+(0.031x2x9.810.04=12.4(m)n27，电知：9=0.10m/min，无缝钢管中38×3mm，H-10m，1-20m一个闸阀（全开），8个标准90°弯头，p=1830kg/m，μ=12mPa-S求：压缩空气P（表）(MPa）解：列1-2截面柏务利方程真祥庆味P.uP+hna+g++gz2210mP.D压缩空气Z-0,ZHu-0P,（表）=P-P=P,P216
U工天演，动至达内留空美=ha-2ad2P（表）-pgH+12p(a=+t+1)EEA店2dab0.10/60u,-9=2.07m/s0.785x0.0322Aa-orl0ROIXDECEEI0.032×207×1830Re-pud=1.01x10*12×10AaeTdu无缝钢管取=0.15e/d-0.15/32-0.005000x2式0x00A-DNOILL查莫迪图入-0.037套90弯头-0.75斤闸阀全开=0.17,缩小2-0.5TF20P（表）=1830×（0.037×+0.75×8+0.17+0.5+D)0.0322.07+1830×9.81×102=3.00x10°Pa=0.3MPa已知：μ=30mPa.S,p=900kg/m，d-40mm，1.=50m1=20m，阔全关28/P（表）=0.09MPa，P，（表）=0.045MPa，阀打开至1/4开度，1e=30m求：(1)q中大观2）阀打开时P，P如何变化？解：（D）取阀的高度Z-0A阀关闭时流体静止，由静力学方程可知光9-P+P（表）=1.013X105+0.09×10=1.91X10（N/m2）3-P+P2（表）=1.013×105+0.045×10=1.46×105（N/m）5e阀1/4开度时，列A-B截面柏努利方程N2+2h22野式标部pP日Zh, =2 u?U,=ug=0d217
现设管内为层流，则，入=64μ-pudA得932m2H89（）9d'pppo0toroPmTOsCE0.0X2870O-d1.91×105-1.46×105u=0.04232u2132×30×10×（50+20±30）YOFBrOSObiin=075m/s-O1XC14pud0.04×0.75×900管降教示2-9002000假设成立0#验Re=430×10-HTEOLO=X≥du=0785x004x0.75=9.42×10m/sq=A.u-1.03套00克403=339m/hr0-1108X20100X01-8SEO0（2）阀打开utCOS118020881由兰P+l+5)0E0-0-0082dpP不变，P变小100e2mo面由心？B星D02d2pp#L,且包括突然扩大损失心>1d价安用造生博花面构4-D元1al）即ud2#正学武白北国商品国光.P变大。0-01-0XmX501）99-81X210001XE101E%4929.已知：T-20℃（苯），H-5mP-P-Pa，Φ32X×3mme=0.05mml-100m求：qv41龙十鱼解：列两槽液面间柏努利方程只+82+号P+gz,+5+元22PPd2OUP,=P,P,u,=u,=0,Z-HZ-018
gH-元lEd20.05eld-5-0.00226LA0假设流动已进入阻力平方区，查莫迪图入-0.0232gHd2x9.81×5×0.026=1.05m/sFahsaGbku210.023x100DaDT-20℃苯p=880kg/m，u=0.67mPa·S浴880×105×0.026Repud大男同酒商-3.59×10*0.67×10-3A查莫迪图人-0.0282gHd2x9.81×5×0.026与假设入有差距，重新计算U=0.95m/sA0.028×100HORe=324×10查莫迪图入=0.028基保的香国版计算正确0400计=d*.u=0.785×0.026*×0.95=5.04×10~m*/s9=X=1.81ml/hr3g已知：d-2me./d-0.0004,paz=067kg/m±=0.026mPa.S.q-80000m/he=1.15kg/m，P（真）=0.2kPa通求：H解：列烟肉底部（1截面）与顶部（2截面）柏务利方程uuz+Zh-PP-+gz2++g2+22PaPa烟肉d-du,-Z-0,2-H数烧护UA8POP-PP（真）HX空气P-P-pagH19
Hu人-街入Zhre120LON80000/3600qv=7.08(m/s)U=0.785×22人国贵莫青习式平代国人石好雅时道blles2-3.65×100201:0200/180m)0.67×7.08×2Re=pudV0OLESOOV0.026x10-3u228.04.g086#100m/d-0.0004、查表得入=0.017260.0-201x088buas81-2截面间柏努利方程为OCZED1x100H+gH+Hu-P（真）-PagHT00人国邮真卖d2PEPa17.0820.2x1031.15×9.81+9.81+0.017xH生证重国鸡人好婚220.670.676.82H=-298.500人01X8H=43.8(m)烟肉得以排气的必要条件是P赠P若p+p时，P+0，即无法起到抽吸作用。3200x200=vH增加，P降低（即真空度增加），抽吸量增加。demtet31.已知：A=3m，h-2mH-4m1-10m，入=0.022,Φ32×3mmA求：a-?0-39m解：列1-1和2-2截面间柏努利方程uuPP+gzi++Zhn-2满#e（西#现电酒+g1t21+2pPHZ-H,u-0P,-P2Z-0,z.hr=alu?d2Hsu10H=a+=0.482u￥79-9+1)x(0.022)2×9.810.026d2gFg9:9-9对水槽作质量衡算：20