02流体静力学-习题与讨论.

1、复习思考题试述静止流体中应力的特性。-怎样认识流体静力学基本方程Z +上-const的几何意义 和物理意义？兀绝对压强、相对压强真空度是怎样定义的?相互之间如 何换算？怎样绘制液体静压强分布图？何谓压力体？怎样界定压力体？判断dh实压力体有何实 际意义？-怎样计算作用在静止流体中潜体和浮体上的总压力？ 液体表面压强以相对压强计时， 怎样计算作用在平面或曲 面上的静水总压力？选择题(单选题)静止液体中存在:(a)压应力；0)压应力和拉应力；(c)压应力和切应力；個)压应力、拉应力和切应力。选样.題（单选题）相对压强的起点是：個）绝对真空；0）1个标准大气压；（0当地大气压；0）液面压强。金属压力

2、表的读值是：S）绝对压强；9）相对压强；（c）绝对压强加当地大气压；0）相对压强加当地大气压。选择.題 （单选题）某点的真空度为65（M）0Pa,当地大气压为O.IMpa,该点的绝对压强为:(n)65000Pa;(b)55000Pa;(c)35000Pa;(rf) 165000Pa。绝对压强几血与相对压强八真空度几、当 地大气压几之间的关系是：（a）Pab尸P+Pv；（b）p=pabs+pa；（C）P.-PuPabs；（d）p=p、+P2选挣题（单选题）在密闭容器上装有形水银测压计，其中1. 2、3点位于同一水平面上，其压强关(a) p1=p2=p3;(b)ppp2/h；(C) )PP2中3；

3、(d)p2p=p选挣题（单选题）用形水银压差计测量水管两点的压强差，水银面高产10cm,为(a)1333kPa；(ft)12.35kPa；(c)9 8kPa；(rf)6.4kPao系为选择愿(单选题)露天水池，水深5m处的相对压强为：(fl)5kPa; (6)49kPa; (c) 147kPa;(d)205kPao垂直放置的矩形平板挡水，水深3m,静水 总压力的作用点到水面的距离为:(a) 1.25m;(b) 1.5m;E -(c) 2m;e(rf) 2.5m o选择JS(单选题)满液体，顶箍与底箍所受张力之比为(a)1/2;(b)1.0; (c)2 0;(d)3.0在液体中潜体所受浮力的大小

4、:(a)与潜体的密度成正比； )与液体的密度成正比；(c)与潜体淹没的深度成正比; 個)与液体表面的压强成反比。形木桶，顶部及底部用环箍紧，桶内盛章后习题简解2=2；2T3;2r4；?r5；2-6;2z7；2r8；2r9；2 10；2-11; 2-12;2J3;2-14;2-15;2二16; 2-17：2-18(x)；2 19;2 20;2-21;2-22;2 23;2-24;2-25;2-26;2-27;2-28; 2-29;2-30;2-31;2-32;2=33；U4；2=35；236；2-37；2z38；2-39；2=40；2-41：2-42;2-43(x)；2 44; 2-45; 2

5、46; 2-47; 2-48。【解】(a)pA=x/r?t=l 000 x9.8x7=68.65xl(PPa=68.65kPa(b) pA=ppghApa=105+1000 x9.8x3-101325=28.1 Ox 1 (PPa=28.1 OkPa(c) pA=pg/iA=1000 x9.8x(-3)=-29.4x1 OTa=29 4kPapB=p( (liR=1000 x9.8x0=01000 x9.8x2= 19.6x 103Pa= 19.6kPa萊底习息简朋22 Z2=5(JhmxpI=?Z1=?【解Pi=PitgSi= 135959.81x50 xl( (r,= 6.67x1PaI=

6、 667kPaBrPnfiS6.67x105一1000 x9.81 = 0.6(bnH,O = 68(hi ni HO柬底习M健3ppp、【答】测压管1液面较高。章后习题简解24=9.07JlAi, ()p,= U7.7kPa.p = ?h = ?(2)p = ImH jOyp = ? (3)pr= (SkPapv= ?【解】(l)p = p,-p( (;=1117-9&07=19 63kPafp 19.63 “ch =-= - -= 2ni H、0Gwg 9.807 (2)(2)pf= pa+QH*029K07 + 9.807x7 = 166i7kPa(3)(3)p = Pa - p

7、 = 9&07-6K5 = 29.57k Pa且测压管1液面与容器 液面同高。2 5 =1.5/n, Z = 03/71, p = 4.9kPa,几=?p, =?【解】P = P + pg( (%-h)= 4900+1000 x 9.807x (0.5 -1.5)= -4,9xlPa章后习题简解2 6 p0= 1017kPa.pd=9&07kPa(l)/i = 0.8m,p = ?pr= ? (2)Z = 5m.力=?(3)Q渍绍=81 flk g/m, pu= ? h駅辅=?章后习题简解【解】( (1) p=几 + /% =10X7 + 9.807x ().8 = 11555

8、k Pa pup-Pa=11555-9&07=17.4SkPii测压管高度为心伊眾7林 测压管水头为Z + Z = Z+=5+l 78=6.7SmPR(3)pv=几一几=1077-9&07 = 9 63kPa= 1.21酒精9.630.81x9.807章后习题简解2-7 A/i = 10(hnu/i = 2.5m,pv【解】PM= M十pu6Ji= 9.807x(2.5 + l工595x0.1)= 37.2 7k Pa测压管水头线距水 面距离为h = SNiPS= 1X595x0.1=136m2 8 h = Hp =24QnmHg化=73(knmHg章后习题简解=1.2x10-

9、240+7301A6= 4011 mHg解】P =席(Z 力)=9.07x (1-2) = -9.80怎Pa测压管液面真实度为= 32*n mHg1.2x10116章后习Jfi简解2-10 h 1.22iii,q油=92(kg/m,zVi = 2O2hini, pv= ? p0= ?【解】Pv =PHM= 1A595X9.807x0.203=27kPaPo = -pHgZ - PM=-(13.595x 0.203+ 0.92x1.22) x 9.807= -3XkPa章后习愿简解h=3()cmJ/2=6(k?nK( (l)pf)il= 85kg/nipi= Hg;章后习题简解【解1）厂0 .=

10、% + S/i = 0.85x 0.6 +1 A595x 0.3 = 4.6mP仏:hy + hy= * x 0.6 + 0.3 - P、=P2Pl=Pzn2-132-15 Z = 2(kmptnl= 92(kg/m/t = 1 Ocm,?草币刁題徇解【解】Q 为空气时，3 = PA_P”=Pfi( (Z+h) =9.807x(0.2 + 0.1) =2.943Pa0为油时，切=/(Z+加-怎妙=9.807x |(0.2 + 0)一0.92x0.1 = 2.042kPa萊底习M健2- 6PM=0.98P/ = 1.5m,“2 =0.2ni,pw/=80(kg/m/1=?【解】由题意，列方程章后

11、习题简解一 PM+% +仏）-0农劝2一怎禺仏=（）PM一兀( (H十hj + pgh,阳一/W-(1 5 + 0 2)+136x0.2!_80010000.98=至章后习J8简解【解】水经过泵时，3 = P2-P =9m-pgh= (1A6-1)X9.807X0.12= 14.8kPa空气通过风机时， = P2-Pi =( (PH其中久=Pa+久讷|，PH= Pa一P誌H+久皿2= ().54kg/m章后习J8简解2-24 a = 2(f,/ = 10Qninpa/f)te/=81(kg/irf, = ?【解】M = PaziglHna= 810 x9.807x0.1sin20= 272N/

12、nrProal-ga-%)+ 如 gH=100(b)矩形形心在二分之一处，即压力中心在闸门中心处。2-28b= lm,/ =ABa= 45,/i, = 3m,% = 2m.章后习题筒解【解】用图解法（过程用板书讲解）。章后习愿简解2-2a = mjix= 201/刖=80(l = lm.【解】用图解法（用板书画图）。章后习愿简解Px=-pgh2b/sina = -x 0.8 x 9.807x l2xl/si n60 = 4.53k N 2 2P2=-/1/sina = ix9.807x2 xl/sin60 = 2Z6MN2 2=P/siiia = 0.8x9.807xlx2xl/sin60 =

13、 18.12kNP = Pi+B + E=453+2265+l&12=453(JkN _；方片+ ( +扌力i)骂+ (方+扌hx)P5“二P lxlx433+(l + x2)x2Z65+(l + |)xl&12=- =-=2.03m4530章后习题简解章后习题简解件=(几一 “) )A = (111.7 - 9& 07)x05x0.6 = 5 8咏NA =1x9.807x0.6x05 x0.6 = 0.8kN2 2P、= 9.807x0.8 x 03 x 0.6 = 2,35k NP =:片 + /： +1 = 5.89+0.88+235=9.12kN| x (5.8

14、9+235)4-x0.882-x0.6 = 0.3 Im 9.12【解】用图解法（用板书画图）。章后习题简解章后习题简解【解】P = /icA =9.807x3x-xl2=2AlkN4yc= = = 3.4 6ni八sina sin60Jz*兀14 /兀儿讥+而狄壮/ /儿4=3.464-12/(16X3.46) )= 3 48mF(儿匚乜/sina兰d/2) )Plxsin60拿后习题简解章后习慰简解2-32b= 2ni,m= 200(k曙曾=In】,* =2ni,7 = ?章后习愿简解【解】P =臥人=朋叭+1妙爲/sin45= 9.807x(l + )x2x2/siii45o= ll(1

15、95kN(% +A J阳力门“才彳厶。_ (1 +；x 2) )x9 807x2*2/Zsin245。_ 】11(1952x1恥訥加+善恁瓠“ _卽“)+竽v- 拿后习J8简解h；h：_hJsh -2/2 xl2-1x0.41-2x0.4=1.33m=PS拿后习题简解章后习题简解P = ( (Po-P+恁化=137.37-9&U7+9.807x(1+Z)sin60 x22= 223cNP137.37-9&07+9.807x si n60- - x22= 1.05mz、x2仏-几 +恁sin60)xl +兀 *rin603x222-34 Po = 137 37kPiU = 2x2

16、m2,p“ =9&07kP爲章后习题简解2-35 =3m=?力=?儿=?章后习愿简解【解】（板书画图）。,2 =(仏-九(h2-hx)(lh2+ 州)+ht(h2+ =|x(2.45-1.73)3x -x(2.45-1.73)x(2x2.45+1.73)+1.73x(2.45+1.73)=22my、=(一方2)；(一h2)(Ui+ /i,) +lb(/ +A,) 3=-x(3-2.45)x -x(3-245)x(2x3 + 2.45) +2.45x(3 + 2.45) 33=2.73m章后习题简解2-36 / = l(hi,D = 4m.HI=4m.H2=2m.Px= 2P：=?【解】

17、（用板书画受力图）。= 58HkN(-p3叫3= 9.807x-x4= 924kN(T)章后习J8简解2452 3d= 2m,/ = 5m,Px=?( =?【解】（用板书画受力图）。章后习）8简解P. =+=字d(彳M+sin6()x2x5x x2 + I42丿=；如=9 807x；x5 = 245kN9.8072= 119.49kN11949竺夕=782G46”【解】T = / = ?6%30 x44.74或= 44748,-a) rdr在()的顶盖处，令P) ),得2x9.807x0.5 0.4J2-*章后习J简解【解】由p = papg -a) )2r2-z可解得离心力水头为l?g2g(

18、 (2丿飲又据自由表面至容器口平面所围成的体积应与静止时柱体 积相等，即1 rd)2I)2a) )27d)2(H-h)24 8g4由上式可解得容器的旋转角速度为童后习J8简解AyfgiH-h) 4x79.807x(0.5-03)(D 11111= D03由此，得60 4yjg(H-h) _ 120 J9.807x(0 50 3)2/r D x *03=1 &67s1=17 &pm章后习题简解2-48D =6()Qn m, H = 50(inmji =0.4m, pol7= 0.8g/m nt章后习题简解p=0,解得c = pMgll,故压强分布公式为P = fg+ pltilgHL2g旋转前后油的体积相等，即 討H牛仍H-从 由此得；又油水分界面为等压面，则应有下列关系【解】在p = Xfy-W2r2Z