02流体静力学习题与讨论.docx

1、02流体静力学习题与讨论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。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

2、属压力表的读值是：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系为选挣题（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；

3、(C)PP2 中 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)98kPa；(rf)6.4kPao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)20; (d) 3.0

4、在液体中潜体所受浮力的大小:(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。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 46; 2

5、-47; 2-48。【解】(a)pA=x/r?t=l 000x9.8x7=68.65xl(PPa=68.65kPa(b)pA=ppghApa=105+1000x9.8x3-101325=28.1 Ox 1 (PPa=28.1 OkPa(c)pA =pg/iA=1000x9.8x(-3)=-29.4x1 OTa=294kPapB=p(liR= 1000x9.8x0=01000x9.8x2= 19.6x 103Pa= 19.6kPa萊底习息简朋 2 2 Z2=5(JhmxpI=?Z1=?【解Pi = PitgSi= 135959.81x50xl(r,= 6.67x1PaI= 667kPa BrPn

6、fiS6.67x105 一 1000x9.81 = 0.6(bnH,O = 68(hi ni HO3 ppp、【答】测压管1液面较高。且测压管1液面与容器 液面同高。章后习题简解2 4 =9.07JlAi, ()p, = U7.7kPa.p = ?h = ?(2)p = ImH jOyp = ? (3)pr = (SkPapv = ?【解】(l)p = p,-p(; =1117-9&07=1963kPaf p 19.63 “ch = = = 2ni H、0Gwg 9.807 (2)pf = pa +Qh*02 9K07 + 9.807x7 = 166i7kPa(3)p = Pa - p = 9

7、&07-6K5 = 29.57k Pa2 5 = 1.5/n, Z = 03/71, p = 4.9kPa,几=? p, =?【解】P = P + pg(%-h)= 4900+1000x 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, p u = ? h 駅辅=?章后习题简解【解】(1) p=几 + /% = 10X7 + 9.807x ().8 = 11555k Pa pup-Pa =11555-9&07

8、=17.4SkPii测压管高度为心伊眾7林 测压管水头为Z + Z = Z+ = 5+l78=6.7SmPR (3)pv =几一几=1077-9&07 = 963kPa= 1.21酒精章后习题简解2 8 h = Hp =24QnmHg化=73(knmHg章后习题简解= 1.2x10-240+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章后习Jfi简解2-10 h 1.22iii,q油=92(kg/m,zVi = 2O2hini, pv = ? p0 = ?【解】Pv = PhM= 1A595x 9.807x0.20

9、3=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 .= % + 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

10、.2 + 0.1) = 2.943Pa0为油时，切= /(Z+加-怎妙=9.807x |(0.2 + 0)一 0.92x 0.1 = 2.042kPa萊底习M健2- 6 Pm =0.98P/ = 1.5m,“2 =0.2ni,pw/ =80(kg/m/1 =?一 Pm +% +仏）-0农劝2一怎禺仏=（）0.98=至Pm 一兀(H 十hj + pgh,阳一/W-(15 + 02)+136x0.2!_ 8001000h = 12(him,zV? = ?【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 =(

11、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= 810x9.807x0.1sin20= 272N/nr拿后习简解2-25 H、b、p p2g =? 【解】以点为基准，建立右 环的平衡关系式-P 妙-卿=0 得以水箱自由表面为基准，建立 平衡关系式Pa-M +Pgb = PR 故 = Pa_Pb=M _PgH _阳d_)H =D D童后习Jfif筆126 ax = l(km2,/l2 = 100a-m2,A/i =

12、 lm,(1)P, =9K1N,A = 2(2) AP, = 19.6, =?加円呛“呛及？S+?WZPM=(叫沖+說丽=10791 N = 10.791 kN儿 .= (1O()Ox9.81x1+)xO.i_io791章后习遂简解= 196(N = le96kN2 2 / 方= 3m4 = 2m比=6md =5m,P = ?” =?【解法一】由dP = pbdh= pghbdh积分，得 師习题简解 P = dP = J： pHhdH- paHhdH= M尿-(% - )-； -(“2 f 卜 Mg-h2)=9.807x2x3x(6-5) = 5K4kNMpdM J： A 怨（-九 + 必 -

13、J：人您 （ -% +阳= pgbhh、-h2)/P= xO07x2x32x(6-1)/5 &84=lSm 即压力中心在闸门中心处。2-28 b = 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 =-pgh 2b/sina = -x 0.8 x 9.807x l2xl/si n60 = 4.53k N 2 2P2 =-/1/sina = ix9.807x2 xl/sin60 = 2Z6MN2 2=P/siii

14、a = 0.8x9.807xlx2xl/sin60 = 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【解】用图解法（用板书画图）。章后习题简解件=(几一 “)A = (111.7 - 9& 07)x05x0.6 = 58 咏 NA =1x9.807x0.6x05 x0.6 = 0.8kN2 2P、= 9.807x0.8 x 03 x 0.6 = 2,35k NP =:片 + /： +1 = 5.8

15、9+0.88+235=9.12kN| x (5.89+235)4-x0.88 2 x0.6 = 0.3 Im 9.12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4F (儿匚乜/sina兰d/2)Plxsin60=3.464- 12/(16x 3.46)= 348m章后习慰简解2-32 b = 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/siii4

16、5o = ll(195kN(% +A J阳力门“才彳厶。11(1952x1_ (1 + ； x 2)x9807x2*2/Zsin2 45。_ 】恥訥加+善恁瓠“ _卽“)+竽v- 章后习题简解P = (Po-P+恁化 =137.37-9&U7+9.807x(1+Z)sin60 x22 z 、 x2仏-几 +恁sin60)xl +兀 * rin603 x22= 223cNP137.37-9&07+9.807x si n60 x22 = 1.05m2-35 = 3m=?力=?儿=?章后习愿简解【解】（板书画图）。,2 =(仏-九(h2 -hx)(lh2 + 州)+ ht(h2 + = |x(2.4

17、5-1.73)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) 3 3=2.73m章后习题简解2-36 / = l(hi,D = 4m.HI =4m.H2=2m.Px = 2P： =?【解】（用板书画受力图）。= 58HkN(-p 3叫3= 9.807x-x4= 924kN(T)23 d = 2m,/ = 5m, Px =?( =?【解】（用板书画受力图）。

18、章后习）8简解9.8072P. = + =字 d(彳 M+sin6()x2x5x x2 + I4 2丿= 119.49kN【解】T = / = ?【解】由p = papg -a)2r2-z可解得离心力水头为l?g 2g (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)2 I)2a)2 7d)2(H-h)24 8g 4由上式可解得容器的旋转角速度为童后习J8简解=1 &67s1AyfgiH-h) 4x79.807x(0.5-03)(D 111 11 = D 03由此，得=17 &pm60 4yjg(H-h) _ 120 J9.807x(0503)2/r D x *

19、 03章后习题简解2-48 D = 6()Qn m, H = 50(in mji = 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p = p= 6.93+ 27.74=34.67k N二 (% -hjP)+(% 学)此Psln45；x(3-2)x693+(3-；)x2774 34.67sin45=245m或 hD = 1.73ni或 yD = 2.35m这说明压力中心比闸门形心低00 im。/cosa=(348-346+0小231=2杯_ yDP + ；h2tg45mg _ 1.65x1 Ift95+|x2xlx2x9.807= 仏45。 = 10L34kN亍 =碍二 U-h、当ha-h?时，闸门处于平衡状态；但当hDhc 时，闸门开启，则应有1 3“ h，故Iff 1 2