H1二x0.026xdt1m-dr=1山-dHA-u..Vo.48244H-{dt3924.90bebVHT=2104s=058hr韩国32/已知：q-2.5×10m/s，P(表）-02MPa，H-6mp=1100kg/mV由Φ40X3mmL-50m,入=0.024求：He(/N)8sbO解：列两槽面间柏努利方程ufHe=PPL+Z15+0025+2g2gd2gpgpgPCOP-Pa.P,Pa=0.2MPaZ-Z,=6m,uu02N2O2.5×10-32.75m/sT辽x0.034340.2×10%L2753五150.He+6+0.024x38/N1100×9.810.0342×9.81pvpn中关费33,已知：Z=6m，d.c=600mmlgc=3000mdg-dz-250mm.1-lc-2500mm人=0.04,忽略局部阻力求解：由连续性方程q,=qvnc=qyco+QycEd2-ud2+udm00OA10.0Y:ha-hrsPOw美bTbutb21
Hth.bt三人业tbxa0oEOVd2d210AI,=ly,d,=d,R20-2101Ssur=uyudly因而2dMS0-am01X-p51由A、A两截面列柏努利方程120.0-4m02-tmmtX04Zu-a+Gnold,2gd2g黑式的可面商购du+e-4dd,2g'd.25003000060.04136-0250.250.6A2×9.81u,=0.183(m/s)9一md-u=0.785x0.6x0.183=0.052(m2/s)=186(m2/hr)qv434.已知：lag=6md=41mm，lne-15mlan-24mdd-25m,-0.03求：(1)qvv2qva(2)D阀关闭，9v解（I）从B点至两管口列柏努利方程td,u豆uz-hshea22大t山+dyu3ah+iuu即（元之22d.d.1524-+Du+1)uz=(0.03x）A(0.03x0.0250.025ug=0.798u由连续性方程：9vi-qv2+qvud=ud+ud=1.798du22
0.025u0.669uu,=1.798x0041mb上u:=1.50u#由槽内液面至C阀出口处截面列柏努利方程：日68我通雅世d22d2白荐6u15心1+0.03x:10×981=0.03x(1.50u）+0.50u)0.04120.0252=23.6uu=2.04(m/s)u=1.5×2.04=3.06(m/s)u：=0.798x3.06=2.44(m/s)因而有：1md×u=0.785×0.041×2.04=2.69×10-(m/s)=9.70（m/h）4V=41md,xu，=0.785x0.025x2.44=1.20x10(m/s)=4.31(m/h)qv24gv=q-qv=9.70-4.31=5.39(m/b)（2）D闻关闭时：连续性方程：qy=qvs0.04lyu=2.69uu0.025由槽内液面至C阀出口处截面列柏努利方程：d,22d26+0.03x15x(269u,）+(2.69u,）：10x9.81=0.03x0.04120.02570.94uu=118(m/s)天保世市动得乐育化9m=v1mmdz-u=0.785x0.041xl.18Ak001-mgl0002=1.55x10(m2/s)=559(m/h)km8o4b1023
etl0000m35.已知：d=d.=20mm,l-2m,1=4m,入=0.0280元I-4求：（1）22-017时，qv/qvc钟顺商装收口出车面板内外电(2）H增加，22-24时，qv/qvc（3）均布条件解：（1）由汇点0至两管口截面列柏努利方程：吃+i-hehe+22m+50+00唱le.al+5g+12-23Vm07-106xk1-02d.dg2aantkae501e+5e+1d.青商园UB+58+1uc0d.dg=de1220038F0--0a1+50+1dimrtitl-otoap-appd.qv_UeUc元4+58+1qve#a（）d.代主量量A+0.17+10.028x0.02=13162+24+10.028x0.02面仓石副原子国动鲜包国（2）24时AS14+24+10.028x0.02qm=1.052qve+24+10.028x600=180:010.02qv与qye的比值与H的变化无关。ADOOTQlm81（3）以上计算说明流体均布是以能量损失为条件的。36.已知：9m-5000kg/h，L-100kmd-300mmP,=0.15MPa（绝压），T-20℃，入=0.016.p=0.85kg/m（标准状态）24
求：PEONERINET01x801800解：对等温流动GmP.P-P1G=0+2山面#2dP2PV11.013×10P29311TV273PP0.85TPPum一4m128×10P量刻图卡级充乐现店5000/3000G9mC19.66kg/ms0.785x03A(0.15×10))-P2P艺车19.66-lm0.15×10°2×P×128×10P果油海蛋粉50元100×10x19.66=0+0.016x2×0.320.0-90日InP-101x10-P+2.88×10°-0AR试差求得P（绝）=5.35×10°PaCOF-OOEDCCOVaGEOUeOl流量测量mn37.已知：d-300mmM=60T=40C.P=101.3kN/m，u=0.02mPa?S用毕托H管测流速，R=30mmHO，求：u5O8T0-1MPI60273Y=234kg/m解：P=224PT22.4313p量证5由毕托管流速计算式p2gR(p-p)U:P格配丽干馨街粉2x9.81×0.03(1000-234)2gR(pp)u.09-042.34P-9=15.8m/s25