1、工程流体力学习题021.如图所示,两圆筒用管子连接,内充水银,第一个圆筒直径d1=45cm,活塞上受力F1=3197N,压强表读数pe=9810Pa;第二个圆筒直径d2=30cm,活塞上受力F2=4945.5N,开口通大气。若不计活塞质量,求平衡状态时两活塞的高度差h。(已知水银密度=13600kg/m3)解:p1=p2,p1=F1/A1+pe+gh=4F1/(d12)+pe+gh,p2=F2/A2=4F2/(d22)4F1/(d12)+pe+gh=4F2/(d22),h=4F2/(d22)4F1/(d12)pe/(g)h=44945.5/(3.140.32)43197/(3.140.452)
2、9810/(136009.8)=0.3m题1图题2图2.如图所示,活塞直径d=35mm,重15N。油的密度1=920kg/m3,水银密度=13600kg/m3。若不计活塞的摩擦和泄漏,当活塞底面和U形管中水银液面的高度差h=0.7m时,求U形管中水银两液面的高度差R。解:U形管左侧水银液面处压强p1=4G/(d2)+1gh,U形管右侧相同高度处的压强p2=gRp1= p2,R=4G/(d2)+1gh/(g)R=415/(3.140.0352)+9209.80.7/(136009.8)=0.164m题3图题4图3.图示一水平方向运动的木板,其速度为u=1m/s。平板浮在油面上,=10mm,油的动
3、力粘度=0.09807Pas,求作用于平板单位面积上的阻力。解:=du/dy=u/=0.098071/0.01=9.807N/m2。4.一底面积为4045cm,高为1cm的木块,质量为5kg,沿着涂有润滑油的斜面等速向下运动,已知v=1m/s,d=1mm,求润滑油的动力粘度。解:F=G5/13=59.85/13=245/13=Av/d,=245d/(13Av)=2450.001/(130.40.451)=0.105Pas5.某地大气压强为98.1kPa,求(1)绝对压强为117.7kPa时的相对压强及其水柱高度。(2)相对压强为7mH2O时的绝对压强。(3)绝对压强为68.5kPa时的真空度。
4、解:(1)p表=117.798.1=19.6kPa,h=19.61000/(10009.8)=2mH2O。(2)p绝=98.1+9.87=166.7kPa。(3)p真=98.168.5=29.6kPa。6.在封闭管端完全真空的情况下,水银柱差Z2=50mm,求盛水容器液面绝对压强p1和水面高度Z1。解:p1=HggZ2=136009.80.05=6664Pa,p1=水gZ1,Z1=p1/(水g)=6664/(10009.8)=0.68m。题6图题7图7.在封闭水箱中,水深h=1.5m的A点上安装一压力表,其中心距A点Z=0.5m,压力表读数为4.9kPa,求水面相对压强及其真空度。解:根据等压
5、面列方程:p0+水gh=p读数+水gZ,p0=p读数+水gZ水gh=4900+10009.8(0.51.5)=4900Pa水面相对压强为4900Pa,其真空度为4900Pa。8.封闭容器的水面的绝对压强p0=107.7kPa,当地大气压pa=98kPa,试求(1)水深h1=0.8m时,A点的绝对压强和相对压强。(2)压力表和酒精(=7.944kN/m3)测压计的读数h为多少?解:(1)A点的绝对压强p绝=p0+水gh1=107700+10009.80.8=115540PaA点的相对压强p表=p绝pa=11554098000=17540Pa(2)压力表的读数p读=p0pa=10770098000
6、=9700Pa 酒精测压计的读数:p读=h,h=p读/=9700/7944=1.22m题8图题9图9.测压管水银柱差h=100mm,在水深h1=2.5m处安装一测压表,试求测压表的读数。解:p0=Hggh=136009.80.1=13328Pa,p读=p0+水gh1=13328+10009.82.5=37828Pa。10.两高度差Z=20cm的盛水容器,当1为空气及油(1=9kN/m3)时,R均为10cm,试分别求两容器的压差。解:A-A为等压面,pA=pA,则p2水g(Z+R)=p11gR,p2p1=水g(Z+R)1gR(1)当1为空气时,由于其密度远远小于水,故忽略不计。p2p1=水g(Z
7、+R)=10009.8(0.2+0.1)=2940Pa。(2)当1为油时,p2p1=水g(Z+R)1gR=10009.8(0.2+0.1)90000.1=2040Pa。题10图题11图11.管路上安装一U形测压管,测得R=30cm,h=60cm,已知(1)为油(油=8.354kN/m3),1为水银;(2)为油(油=8.354kN/m3),1为水;(3)为气体,1为水,求A点的压强水柱高度。解:依题意:pAgh=1gR,pA=1gR+gh,(1)为油(油=8.354kN/m3),1为水银时,pA=1gR+gh=136009.80.3+83540.6=44996.4Pa,h=pA/水=44996.
8、4/9800=4.6mH2O(2)为油(油=8.354kN/m3),1为水时,pA=1gR+gh=10009.80.3+83540.6=7952.4Pa,h=pA/水=7952.4/9800=0.81mH2O(3)为气体,1为水时,气体的密度远小于水,故忽略不计,pA=1gR+gh=1gR=10009.80.3=2940Pa,h=pA/水=2940/9800=0.3mH2O题12图题13图12.已知图中Z=1m,h=2m,求A点的相对压强及测压管中气体压强的真空度。解:测压管中气体的真空度为:p真=水gh=10009.82=19600Pa,A点的相对压强:pA=p真+水gZ=19600+100
9、09.81=9800Pa。13.已知水箱真空表M的读数为0.98kPa,水箱与油箱的液面差H=1.5m,水银柱差R=0.2m,油=7.85kN/m3,求h为多少m?解:依题意得:pM+水(R+h+H)=油h+HgR,h=HgRpM水(R+H)/(水油)h=136009.80.2+9809800(0.2+1.5)/(98007850)=5.63m题14图题15图14.图为倾斜水管上测定压差的装置,测得Z=200mm,R=120mm,当(1)1=9.02kN/m3为油时,(2)1为空气时,分别求A、B两点的压差。解:pA水(m+R)=pB水(Z+m)1R,pApB=水(RZ)1R(1)1=9.02
10、kN/m3时,pApB=水(RZ)1R=9800(0.120.2)90200.12=1866.4Pa(2)1为空气时,由于空气密度与水相比很小,忽略不计,pApB=水(RZ)1R=水(RZ)=9800(0.120.2)=784Pa15.如图所示,求A、B两点的的压强差。解:1-1,2-2,3-3为等压面,故p1=p1,p2=p2,p3=p3,pA+水gh1Hgg(h1h2)=pB+水gh4+Hgg(h3h4)水g(h3h2)pApB=Hgg(h1h2+h3h4)+水g(h4h3+h2h1)=(Hg水)g(h1h2+h3h4)题16图题17图16.A、B两管的轴心在同一水平线上,用水银压差计测定
11、压差,测得R=13cm,当A、B两管通过(1)为水时,(2)为煤气时,试分别求压差。解:(1)当为水管时,pApB=(Hg水)gRpApB=(136001000)9.80.13=16052.4Pa(2)当为风管时,气体密度与水银相比忽略不计,pApB=HggR=136009.80.13=17326.4Pa17.在水泵的吸入管1和压出管2中安装水银压差计,测得R=120mm,问水经过水泵后压强增加多少?若为风管,则水泵换为风机,压强增加多少mH2O?解:(1)当为水管时,p2p1=(Hg水)gR=(136001000)9.80.12=14817.6Pa(2)当为风管时,气体的密度与水银相比忽略不
12、计,p2p1=HggR=136009.80.12=15993.6Pa,h=(p2p1)/水=15993.6/9800=1.632mH2O18.直径为150mm的给水管道,输水量为980kN/h,试求断面平均流速。解:G=gvA,v=G/(gA)=4G/(gd2)=4980000/(360010009.83.140.152)=1.57m/s19.断面为300mm400mm的矩形风道,风量为2700m3/h,求平均流速。如风道出口处断面收缩为150mm400mm,求该断面的平均流速。解:Q=vA,v1=Q/A1=Q/(a1b1)=2700/(36000.30.4)=6.25m/s,v2=Q/A2=
13、Q/(a2b2)=2700/(36000.150.4)=12.5m/s20.水从水箱流经直径为d1=10cm,d2=5cm,d3=2.5cm的管道流入大气中,当出口流速为10m/s时,求(1)容积流量及质量流量;(2)d1及d2管段的流速。解:(1)Q=v3A3=v3d32/4=103.140.0252/4=0.0049m3/s,M=Q=10000.0049=4.9kg/s。(2)v1=v3(d3/d1)2=10(2.5/10)2=0.625m/s,v2=v3(d3/d2)2=10(2.5/5)2=2.5m/s。21.设计输水量为2942.1kN/h的给水管道,流速限制在0.91.4m/s之间
14、。试确定管道直径,根据所选直径求流速,直径规定为50mm的倍数。解:G=gvA=gvd2/4,d=4G/(gv)0.5,dmin=4G/(gvmax)0.5=42942100/(360010009.81.43.14)0.5=0.28m,dmax=4G/(gvmin)0.5=42942100/(360010009.80.93.14)0.5=0.34m,取d=0.3m,则流速v=4G/(gd2)=42942100/(360010009.83.140.32)=1.18m/s。22.圆形风道,流量为10000m3/h,流速不超过20m/s,试设计直径,根据所选直径求流速,直径规定为50mm的倍数。解:
15、Q=vA=vd2/4,d=4Q/(v)0.5,d=4Q/(v)0.5=410000/(3600203.14)0.5=0.42m,取d=0.45m,则流速v=4Q/(d2)=410000/(36003.140.452)=17.47m/s。23.某蒸汽干管的始端蒸汽流速为25m/s,密度为2.62kg/m3。干管前段直径为50mm,接出直径40mm支管后,干管后段直径改为45mm。如果支管末端密度降低至2.30kg/m3,干管后段末端密度降低至2.24kg/m3,但两管质量流量相等,求两管末端流速。解:干管始端的质量流量为:M=vd2/4=2.62253.140.052/4=0.129kg/s干管
16、后端和支管的质量流量为:M1=M2=0.129/2=0.0645kg/s,干管后端流速为v=4M1/(d12)=40.0645/(2.243.140.0452)=18.11m/s支管流速为v=4M2/(d22)=40.0645/(2.303.140.0402)=22.33m/s24.空气流速由超音流过渡到亚音流时,要经过冲击波。如果在冲击波前,风道中速度为660m/s,密度为1kg/m3,冲击波后,速度降低至250m/s。求冲击波后的密度。解:1v1d2/4=2v2d2/4,2=1v1/v2=6601/250=2.64kg/m3。25.管路由不同直径的两管前后相连所组成,小管直径dA=0.2m
17、,大管直径dB=0.4m,水在管中流动时,A点压强pA=70kPa,B点压强pB=40kPa,B点流速vB=1m/s。试判断水在管中的流动方向。计算水流经两断面间的水头损失。解:在A,B间列能量方程:ZA+pA/(g)+vA2/(2g)=ZB+pB/(g)+vB2/(2g)+hW,pA=70000Pa,pB=40000Pa,vB=1m/s,vA=vB(dB/dA)2=1(0.4/0.2)2=4m/s,ZB=1m,ZA=0,hW=ZAZB+(pApB)/(g)+(vA2vB2)/(2g)=01+(7000040000)/9800+(4212)/19.6=2.83m由于所求hW为正值,所以水流方向
18、为AB。26.油沿管线流动,A断面直径为150mm,流速为2m/s,B断面直径为100mm,不计损失,求开口C管中的液面高度。题25图题26图解:在A,B间列能量方程:ZA+pA/(g)+vA2/(2g)=ZB+pB/(g)+vB2/(2g)+hW,ZB=0,ZA=1.2m,vA=2m/s,vB=vA(dA/dB)2=2(0.15/0.1)2=4.5m/s,pA/(g)=1.5m,hW=0,pB/(g)= ZAZB+pA/(g)+(vA2vB2)/(2g)=1.2+1.5(224.52)/19.6=1.87m27.水沿管线下流,若压力计的读数相同,求需要的小管直径d0,不计损失。解:在大小管压
19、力计所在截面间列能量方程:Z+p/(g)+v2/(2g)=Z0+p0/(g)+v02/(2g)+hW,Z0=0,Z=3m,v=3m/s,p=p0,hW=0,v02=2g(ZZ0)+v2=19.63+32=67.8m2/s2,v0=8.234m/sd0=(vd2/v0)0.5=(30.22)/8.2340.5=0.12m=120mm题27图题28图28.用水银比压计测量管中水流过流断面中点的流速u,测得比压计读数R=60mm,(1)求该点的流速u,(2)若管中流体是密度为0.8g/cm3的油,R不变,该点流速为多少?不计损失。解:u2=2p/,u2=2(Hg)gR/(1)当流体为水时,u2=2(
20、136001000)9.80.06/1000=14.8176m2/s2,u=3.85m/s(2)当流体为油时,u2=2(13600800)9.80.06/800=18.816m2/s2,u=4.34m/s29.水由图中喷嘴流出,管嘴出口d=75mm,1截面管径d1=125mm,2和4截面管径均为100mm,不计损失。计算H及p的大小。解:在1和2截面间列能量方程:Z1+p1/(g)+v12/(2g)=Z2+p2/(g)+v22/(2g),Z2=0,Z1=Z,(p1p2)/(g)=0.175Hg/Z0.175=0.17513.6Z0.175=2.205Z,v2=v1(d1/d2)2=v1(125
21、/100)2=1.5625v1,(v22v12)/(2g)=Z1Z2+(p1p2)/(g)=Z0+2.205Z=2.205,v1=5.48m/s,v2=8.56m/s,v3=v1(d1/d3)2=5.48(125/75)2=15.22m/s,在0和3截面间列能量方程:Z0+p0/(g)+v02/(2g)=Z3+p3/(g)+v32/(2g),Z0=H,Z3=0,p0=0,p3=0,v0=0,v3=15.22m/s,H=v32/(2g)=15.222/19.6=11.8m在0和4截面间列能量方程:Z0+p0/(g)+v02/(2g)=Z4+p4/(g)+v42/(2g),Z0=H,Z4=0,p0
22、=0,v0=0,v4=v2=8.56m/s,p4=gZ0v32/(2g)=10009.8(11.88.562/19.6)=79003.2Pa题29图题30图30.计算管线中水的流量,管出口d=50mm,求出A、B、C、D各点的压强。不计水头损失。解:在0和D间列能量方程:Z0+p0/(g)+v02/(2g)=ZD+pD/(g)+vD2/(2g),Z0=4m,ZD=0,p0=0,pD=0,v0=0,Z0=vD2/(2g),4=vD2/19.6,vD=8.85m/sQ=vDdD2/4=8.853.140.052/4=0.0174m3/s在0和A间列能量方程:Z0+p0/(g)+v02/(2g)=Z
23、A+pA/(g)+vA2/(2g),Z0=7m,ZA=0,p0=0,v0=0,vA=vD(dD/dA)2=8.85(50/150)2=0.98m/spA=gZ0vA2/(2g)=9800(70.982/19.6)=68119.8Pa在0和B间列能量方程:Z0+p0/(g)+v02/(2g)=ZB+pB/(g)+vB2/(2g),Z0=0m,ZB=0,p0=0,v0=0,vB=vA=0.98m/spB=gvA2/(2g)=98000.982/19.6=480.2Pa在0和C间列能量方程:Z0+p0/(g)+v02/(2g)=ZC+pC/(g)+vC2/(2g),Z0=0m,ZC=2m,p0=0,
24、v0=0,vC=vA=0.98m/spC=gZCvC2/(2g)=9800(20.982/19.6)=20080.2Pa题31图题32图31.如图,闸门关闭时的压力表读数为49kPa,闸门打开后,压力表读数为0.98kPa,由管进口到闸门的水头损失为1m,求管中水的平均流速。解:闸门打开后静压强下降,下降的部分一部分用来克服水头损失,一部分转化成了动能。故:p/(g)=v2/(2g)+hW,(49000980)/9800=v2/19.6+1,v=8.74m/s32.由断面为0.2m2和0.1m2的两根管子所组成的水平输水管系从水箱流入大气中:(1)若不计损失,(a)求断面流速v1及v2;(b)
25、求进口A点的压强。(2)若不计局部损失,只计沿程损失,第一段为4v12/(2g),第二段为3v22/(2g),(a)求断面流速v1及v2;(b)各段中点C和D的压强。解:(1)在水箱液面0和管子出口B间列能量方程:Z0+p0/(g)+v02/(2g)=ZB+pB/(g)+vB2/(2g),Z0=4m,ZB=0,p0=0,v0=0,pB=0,Z0=vB2/(2g),4=vB2/19.6,v2=vB=8.85m/s,v1=v2A2/A1=8.850.1/0.2=4.425m/s4.43m/s,在水箱液面0和A间列能量方程:Z0+p0/(g)+v02/(2g)=ZA+pA/(g)+vA2/(2g),
26、Z0=4m,ZA=0,p0=0,v0=0,vA=v1=4.43m/s,Z0=pA/(g)+vA2/(2g),pA=gZ0vA2/(2g)=9800(44.432/19.6)=29387.55Pa(2)在水箱液面0和管子出口B间列能量方程:Z0+p0/(g)+v02/(2g)=ZB+pB/(g)+vB2/(2g)+hW,Z0=4m,ZB=0,p0=0,v0=0,pB=0,hW=4v12/(2g)+3v22/(2g),Z0=vB2/(2g)+4v12/(2g)+3v22/(2g),v2=vB,v1=v2A2/A1=0.1/0.2v2=0.5v2,Z0=5v22/(2g),4=5v22/19.6,v
27、2=3.96m/sv1=v2A2/A1=3.960.1/0.2=1.98m/s,在水箱液面0和C间列能量方程:Z0+p0/(g)+v02/(2g)=ZC+pC/(g)+vC2/(2g)+2v12/(2g),Z0=4m,ZC=0,p0=0,v0=0,vC=v1=1.98m/s,Z0=pC/(g)+3vC2/(2g),pC=gZ03vC2/(2g)=9800(431.982/19.6)=33319.4Pa在水箱液面0和D间列能量方程:Z0+p0/(g)+v02/(2g)=ZD+pD/(g)+vD2/(2g)+4v12/(2g)+1.5v22/(2g),Z0=4m,ZD=0,p0=0,v0=0,vD
28、=v2=3.96m/s,4v12/(2g)=v22/(2g),Z0=pD/(g)+3.5vD2/(2g),pD=gZ03.5vD2/(2g)=9800(43.53.962/19.6)=11757.2Pa题33图题34图33.一压缩空气罐与文丘里式的引射管连接,气体从d2处喷入大气,d1,d2,h均为已知,问气罐压强p0多大时方才能将B池水抽出。损失不计。解:在水箱液面B和1截面间列能量方程:ZB+pB/(水g)+vB2/(2g)=Z1+p1/(水g)+v12/(2g),设B和1截面连接管中的速度为0,则ZB=0,pB=0,vB=0,v1=0,Z1=h,故:p1=水gh,在1和2截面间列能量方程
29、:p1/(g)+v12/(2g)=p2/(g)+v22/(2g)p1=水gh,p2=0,v2=v1(d1/d2)2,v12/(2g)=v22/(2g)+水gh/(g)v1=v2(d2/d1)2,v22(d2/d1)4=v22+2水gh/,v22=2水gh/(d2/d1)41在气罐0和1截面间列能量方程:p0/(g)+v02/(2g)=p1/(g)+v12/(2g)Z1=Z0,p1=水gh,v12/(2g)=v22/(2g)+水gh/(g),v0=0,p0=水gh+v22/2+水gh=v22/2=水gh/(d2/d1)4134.高层楼房煤气立管B和C两个供煤气点各供应Q=0.02m3/s的煤气量,煤气的密度为0.6kg/m3,管径为50mm,压强损失AB段用3v12/2,BC段用4v22/2计算,假定C点要求保持余压300Pa,求A点酒精(酒=7.9kN/m3)液面应有的高差(空气密度为1.2kg/m3)。解:在A和C间列能量方程:pA+vA2/2+(a)g(ZCZA)=pC+vC2/2+pfvA=v1=4Q总/(d2)=40.04/(3.140.052)=20.38m/s,ZA=0,ZC=60m,a=1.2kg/m3,pC=300Pa,vC=v2=4Q/(d2)=40.0
copyright@ 2008-2022 冰豆网网站版权所有
经营许可证编号:鄂ICP备2022015515号-1