流体力学习题库

1．已知一流动的速度场为：vx = 2xy+x，vy = x2-y2-y，试证明该流动为有势流动，且存在流函数，并求速度势及流函数。

vyvx解：（1）∵ 2x2x，

yx

则 ωx= ωy= ωz = 0， 流动为无旋流动，

∴ 该流动为有势流动。

vxvy又 ∵ 2y12y10，即流动为不可压缩流体的平面流动，

xy

∴该流动存在流函数。

（2） ∵

ddxdyvxdxvydy(2xyx)dx(x2y2y)dy xy ∴ 速度势为： x2y3y2222(2xyx)dx(xyy)dyxyc 232

dxdyvydxvxdy(x2y2y)dx(2xyx)dy ∵ dxy

∴ 流函数为： x3222(xyy)dx(2xyx)dyxyxyc 3

2．如图所示，两圆筒内装的是水，用管子连接。第一个圆筒的直径d1= 45 cm，其活塞上受力F1=320 N，密封气体的计示压强为981.0 Pa；第二个圆筒的直径d2= 30 cm，其活塞上受力F2=490 N，开孔通大气。若不计活塞重量，求平衡状态时两活塞的高度差h。

F解： P11PePa水ghA1

F P22PaA2

d12 A14

d22 A24

∵ P1P2

∴ FF21p e22d2d1 h44水g0.402(m)

3．已知：一闸门如图，h1 = 2m，h2 =3m，h3 =5m，闸门宽B = 2m，γ1 =9806 N/m3，γ2 =12000 N/m3，γ3 =46000 N/m3。求作用在AB板上的合力，以及作用在B点的合力矩。 hF11ghc1A111h1B980612239224(N)解： 2 F 2 '1gh1A29806232117672(N)

F2''2ghc2A2120001.532108000(N) F33ghc3A3460002.5(23)2 1150000(N)

F合F3F1F2115000039224108000

885104(N) hhhhMBF33F1(h21)F2'2F2''2 3323 5233115000039224(3)1176721080001488324.2(N.m) 3323

4．图示为水自压力容器定常出流，压力表读数为10atm，H=3.5m，管嘴直径D1=0.06m，D2=0.12m，试求管嘴上螺钉群共受多少拉力？计算时管嘴内液体本身重量不计，忽略一切损失。

解：对容器液面和管嘴出口截面列伯努利方程： Pev12H g2g 2Pev2gH1 

2101.013105 29.83.51000

45.77(m/s)

22 D10.06v2 Dv10.1245.7711.44(m/s)2

选管嘴表面和管嘴进出口断面所围成的体积为控制体，列动量方程： Fpnxqvv1xv2xP2eA2Fx

Fxqvv1v2P2eA2

对管嘴的进出口断面列伯努利方程，得

2 P2ev12v2 2gg2g

P2e981986.77(Pa)

Fx100045.776663.6(N)40.06245.7711.44981986.7740.122 ∴ FFx6663.6(N)

5．如图示，水流经弯管流入大气，已知d1=100mm，d2=75mm，v2=23m/s，不计水头损失，求弯管上所受的力。

2解：由连续方程： v1d12v2d2 得： 22d275 v1v222312.94(m/s)2d1100

对弯管的进、出口截面列伯努利方程：

2 P1bP2bv12v2z1z2 g2gg2g其中，P2 b= 0，z1 = z 2，代入得：

2 v2v121000P1b23212.9421.808105(Pa) 22g2选弯管所围成的体积为控制体，对控制体列动量方程：

qvv2xv1xFpn1xFpn2xFpnbxP1bA1Fpnbx

qvv2yv1yFpn1yFpn2yFpnbyFpnby

 1000230.0752v2cos30v11.8081050.12Fpnbx44

 1000230.0752v2sin300Fpnby4

求得：Fpnbx= - 710.6 (N) ∴ Fx= - Fpnbx= 710.6 (N) Fpnby= 1168.5 (N) Fy= - Fpnby= -1168.5 (N)

FFx2Fy21367.6(N)

6．已知油的密度ρ=850 kg/m3，粘度μ=0.06 Pa.s，在图示连接两容器的光滑管中流动，已知H=3 m。当计及沿程和局部损失时，求：（1）管内的流量为多少？（2）在管路中安一阀门，当调整阀门使得管内流量减小到原来的一半时，问阀门的局部损失系数等于多少？（水力光滑流动时，λ= 0.3164/Re0.25）。 解：（1）对两容器的液面列伯努利方程，得： H hwhfhj

lv2v21.5

d2g2g 22340vv1.50.32g2g即： (1)

设λ= 0.03，代入上式，得 v = 3.27 m/s，则

vd8503.270.3Re13897.5

0.06

0.31640.3164

'0.0291 Re0.25138970.25 '0.030.02913%2% 0.03故，令λ=λ’=0.0291，代入（1）得：v=3.306（m/s）

vd8503.3060.3则

Re14050.5 0.060.31640.3164

'0.0291 Re0.2514050.50.25

d2∴ qvv0.323.3060.234(m3/s)44

10.234（2）

qv'qv0.117(m3/s) 22 q'0.1174vv21.655(m/s)2 d3.140.3 4vd8501.6550.3

Re7033.75 0.06

0.31640.31640.0345

Re0.257033.750.25 则 Hhwhfhj

lv2(1.5v)

d2g

401.6552 3(0.03451.5v)0.329.8

求得： v15.37

7．为确定鱼雷阻力，可在风洞中进行模拟试验。模型与实物的比例尺为1/3，已知实际情况下鱼雷速度vp=6 km/h，海水密度ρp=1200 kg/m3，粘度νp=1.145×10-6 m2/s，空气的密度ρ

3-52

（1）风洞中的模拟速度应为多大？（2）若m=1.29 kg/m，粘度νm=1.45×10 m/s，试求：

在风洞中测得模型阻力为1000N，则实际阻力为多少？ 解：已知 l1klm lp3（1）由Rep = Rem 得， kν = kv kl，

 km1.45105kv3386 ∴ klpkl1.14510

vm= kvvp= 38×6 =228 (km/h)

（2）由kF= kρkl2 kv2 得

22 Fmm11.29122k38380.1725F

Fpp312003

∴ FP = Fm/kF = 1000/0.1725 = 5798 (N)

7．流体通过孔板流量计的流量qv与孔板前、后的压差ΔP、管道的内径d1、管内流速v、孔板的孔径d、流体密度ρ和动力粘度μ有关。试用π定理导出流量qv的表达式。 （dimΔP =ML-1T-2， dimμ=ML-1T-1）。

解：设qv= f (ΔP, d1, v, d,ρ,μ)

选d, v, ρ为基本变量 qv1a1vb1dc1

P2a2vb2dc2 d13a3vb3dc3

4a4vb4dc4

上述方程的量纲方程为：

a1b1 L3T1ML3LT1Lc1

123a21b2c2MLTMLLTL

a3b3 LML3LT1Lc3

113a31b3c3MLTMLLTL

由量纲一致性原则，可求得：

a1=0 a2=1 a3=0 a4=1 b1=1 b2=2 b3=0 b4=1 c1=2 c2=0 c3=1 c4=1

Pqvd1∴ 12324vdv2dvd

Pd1 qvvd2fv2,d,vd

8．如图所示，由上下两个半球合成的圆球，直径d=2m，球中充满水。当测压管读数H=3m

时，不计球的自重，求下列两种情况下螺栓群A-A所受的拉力。（1）上半球固定在支座上；（2）下半球固定在支座上。 解：（1）上半球固定在支座上时 d22r3FgVpg4H3



10009.83.141323.141/3

112.89(kN)

（2）下半球固定在支座上时 d22r3FgVpg4H3



10009.83.141323.141/3

71.84(kN)

9. 新设计的汽车高1.5m，最大行驶速度为108km/h，拟在风洞中进行模型试验。已知风洞试验段的最大风速为45m/s，试求模型的高度。在该风速下测得模型的风阻力为1500N，试求原型在最大行驶速度时的风阻。

解：

v45 kvm1.5108vp 3.6 k1

根据粘性力相似准则， RemRep kvkl1

h11 klmkv1.5hp

hp

hm1(m) 1.5 又 FkFkkv2kl2m Fp

Fm1500F1500(N) p221kkvkl

10. 连续管系中的90º渐缩弯管放在水平面上，管径d1=15 cm，d2=7.5 cm，入口处水平均流速v1=2.5 m/s，静压p1e=6.86×104 Pa（计示压强）。如不计能量损失，试求支撑弯管在其位置所需的水平力。

解：由连续方程：

v1A1v2A2v1A1d1215v22v1()22.510(m/s)A27.5d2

由能量方程： 2p1ev12p2ev2 g2gg2g

1000 2p2ep1e(v12v2)6.86104(2.52102)22

21725(Pa)

X方向动量方程：

qv(v2xv1x)Fxp2eA2

Fxqv(v2xv1x)p2eA2



10002.50.152(100)217250.075244

537.76(N)

Y方向动量方程：

qv(v2yv1y)Fyp1eA1

Fyqv(v2yv1y)p1eA1 10002.50.152(02.5)6.861040.152 44 1322.71(N) 合力为:

FFx2Fy2537.7621322.7121427.85(N)

11. 小球在不可压缩粘性流体中运动的阻力FD与小球的直径D、等速运动的速度v、流体的密度ρ、动力粘度μ有关，试导出阻力的表达式。 （dimF =MLT-2， dimμ=ML-1T-1）。（15分）

解：设FD = f (D, v, ρ,μ)

选D、v、ρ为基本变量 FD1a1vb1Dc1

4a2vb2Dc2

上述方程的量纲方程为： abMLT2ML3LT111Lc1

113a21b2c2MLTMLLTL

由量纲一致性原则，可求得：

a1=1 a2=1 b1=2 b2=1 c1=2 c2=1

FD∴ 12v2d2vd

 FDv2d2fvd

12. 如图所示，一封闭容器内盛有油和水，油层厚h1=40 cm，油的密度ρo=850 kg/m3，盛有水银的U形测压管的液面距水面的深度h2=60 cm，水银柱的高度低于油面h=50 cm，水银的密度ρhg= 13600 kg/m3，试求油面上的计示压强（15分）。

 P w gh解： P 2 e e   o gh 1  2

  g

hg(h1h2h)

Pehgg(h1h2h)ogh1wgh2

136009.8(0.40.60.5)8509.80.4

10009.80.6

57428(Pa)

13. 额定流量qm=35.69 kg/s的过热蒸汽，压强pe=981 N/cm2，蒸汽的比体积为v=0.03067 m3/kg，经内径为227mm的主蒸汽管道铅垂向下，再经90º弯管转向水平方向流动。如不计能量损失，试求蒸汽作用给弯管的水平力。

解：由连续方程： 得： qmVA qmqmv35.690.030674V Ad20.2272427.047(m/s)

选弯管所围成的体积为控制体，对控制体列x方向动量方程：

qvV2xV1xFpn1xFpn2xFpnbxP2eA2Fpnbx

FpnbxqmV2xV1xP2eA2

0.2272435.69(27.0470)98110 4 3.98105(N)

14. 为测定90º弯头的局部阻力系数，在A、B两断面接测压管，流体由A流至B。已知管径d=50 mm，AB段长度LAB = 0.8 m，流量qv = 15 m3/h，沿程阻力系数λ=0.0285，两测压管中的水柱高度差Δh = 20 mm，求弯头的局部阻力系数ξ。（15分）

q解： 15vAvBvv22.12(m/s) d0.0523600 44 对A、B列伯努利方程：

22 PAvAPBvBzAzBhf 水g2g水g2g PAPzA1z1 水g水g

PBP2zz2 B水g水g 22P1vAP2vB z1z2hfg2gg2g水水

又vAvBv

ghPPPP2 hf1z12z21(z1z2)汞h水g水g水g水g

lv2v2

又hfhh d2g2g 2glv22g汞hl29.80.02136000.8(h)(h)(1)0.0285f 222vd2gv水d2.1210000.050.64

15. Pipe AB is of uniform diameter and h=10.5 m The pressure at A is 170 kPa and at B is 275 kPa. Find the direction of the flow, and what is the pipe friction head loss in meters of the fluid if the liquid has a specific gravity of (a) 0.85; (b) 1.45? (10分)

Solution: (a)

2 v2v2pAvA170000zA  10.530.89(m) H A   2g0.8598102g2g

2v2v2pv275000EE H   z   032.98(m)BE2g2g  2 g 0 .85  9810 ∵ HAHB ∴ the flow is from B to A 22vv  H h f  H(30.89)2.09(m)BA32.982g2g

2 (b) v2v2pAvA170000 z10.522.45(m) H A  A 2g1.4598102g2g 222vvpv275000 H  E E z019.33(m)BE2g1.4598102g2g

HAHB ∵

∴ the flow is from A to B 22vv h  H  H22.45(19.33)3.12(m)fAB2g2g

16. A curved pipe is attached to a tank, and liquid flows out into atmosphere through the curved pipe. Determine the resultant force on the curved pipe. All the significant data are given in the figure. Assume an ideal liquid with γ=55 lb/ft3. Neglect the weight of the liquid. (10分)

Solution: 22p1z1pvv13z332g2g2v330144352055232.2v377.6(fps)

vA377.69 v2343.6(fps)A162

22 p3v3p2v2z2z3 2g2g 22v3v277.6243.62p2(z3z2)55(2010)

2g232.2

24075.2(lb/ft)28.3(psi)

FxF2xF3xFbxQ(v3xv2x)

FyF2yF3yFbyQ(v3yv2y)

D32D22Fbxv3(v3cos20v2)p2165(lb)

44

FxFbx165(lb) 2D3 Fbyv3(v3sin200)173(lb)4

FyFby173(lb)

22

FFxFy239

17. An 180-mm-diameter pipeline (f=0.032) 150 m long discharges a 60-mm-diameter water jet into the atmosphere at a point that is 80 m below the water surface at intake. The entrance to the pipe is reentrant, with ke=0.9, and the nozzle loss coefficient k2=kn=0.055. Find the flow rate and the pressure at B. (15分)

Solution: (a)

22 p0v0p2v2z0z2hL 2g2g 22v12v2v2lv12z0z2fkek2

d2g2g2g2g

2v1D26021

v2D12180292v29v1v1150113.022280(0.0320.9920.05592)v10.1829.8129.81 (b) v29v133.54(m/s)22 pBvBp2v2zBz2hL

2g2g 222vvv2B2 pB(k2)2g2g

22v2v12v2

(k2)981059.785.86105(Pa)586(kPa) 2g2g

18. What minimum value of b is necessary to keep the rectangular stone from sliding if it weighs 160 lb/ft3, a=14ft, c=16ft, and the coefficient of friction is 0.45? With this minimum b value, will it also be safe against overturning? Assume that water dose not get underneath the stone block. (10分)

Solution1:

a142 FwwhcA62.4aB62.4B6115.2B(lb)22

FfsbcB0.45160b16B1152bB(lb) f

to keep the rectangular stone from sliding, then

FfFw

1152bB6115.2B

b5.31(ft)

Since a14MwFw6115.2B28537.6B(lb.ft) 33 b5.31MssbcB1605.3116B36091(lb.ft)Mw 22 ∴ it is also safe against overturning.

19. A flow is defined by u=2(1+t), v=3(1+t), w=4(1+t). What is the velocity of flow at the point (3,2,4) at t=2? What is the acceleration at that point at t=2? (10分)

Solution: At t=2

u2(1t)2(12)6

v3(1t)3(12)9

w4(1t)4(12)12

uuuuaxuvw2

txyz

vvvv ayuvw3txyz

wwww

azuvw4 txyz

222222Vuvw691216.16

222 aaxayaz2232425.39

20. Water flows over the spillway of constant section. Given that y1 = 4.2m and y2 = 0.7m, determine the resultant horizontal force on the spillway per meter of spillway width (perpendicular to the spillway section). Assume ideal flow. (10分)

Solution:

v1y1v2y2

22 p1v1p2v2y1y2

2g2g

将y1 = 4.2m, y2 = 0.7m, p1 = p2 , γ=9810 N/m2 , 代入得

4.2v10.7v2

22v1v2

4.20.729.8129.81

解得：v1 =1.4 , v2 =8.4

FxQ(v2xv1x)F1F2Fbv1y1B(v2v1)FbFby1yy1B2y2Bv1y1B(v2v1)224.220.72

To the right

21. Water and oil in an open storage tank. (a) Find the total forces exerted by the fluids on a tank wall, and (b) the location of the center of pressure. (10分) Solution: (a) FF1F2'F2''

F1oilhc1A1

1.5 0.89800(1.52)2

17640(N)

F2'oilh1A2 0.898001.5(12)

23520(N)

F2''waterhc2A2980012(12)9800(N)

FF1F2'F2''1764023520980050960(N)1.5(b)

yp111.5(m) 31yp2'0.5(m)

2

1yp2''0.333(m)

3 '''F.yF.yF.yF.yp2''176401.5235200.598000.333p1p12p2'2

41483.4

41483.441483.4

yp0.814(m) F50960

22. A diverging nozzle that discharges an 8-in-diameter water jet into the air is on the right end of a horizontal 6-in-diameter pipe. If the velocity in the pipe is 12fps, find the magnitude and direction of the resultant axial force the water exerts on the nozzle. Neglect fluid friction. (10分) Solution:

A162v2v12126.75(ft/s)A28

2v12p2v2z1z22g2gp1

262.4 p1(v2v12)(6.752122)95.38(lb/ft2)2g232.2

FxQ(V2V1)P1A1Fbx

FFPAQ(VV)95.38(6)262.412(6)2(6.7512)xbx112141232.2412

 5 . 24 ( lb ) to the right

23. A horizontal 100-mm-diameter pipe (f=0.027) projects into a body of water (ke=0.8) 1m below the surface. Considering all losses, find the pressure at a point 5 m from the end of the pipe if the velocity is 4 m/s and the flow is (a) into the body of water; (b) out of the body of water. (10分)

Solution: (a) 2p1v12p2v2 z1z2hL2g2g

p1v12v12lv12

z2z1f(k') 2gd2g2g p154210.027

98000.129.8

p120.6(kN/m2)

2(b) p1v12p2v2z1z2hL 2g2g

p1v12v12lv12z2z1fke

2gd2g2g

p154242 10.0271.898000.129.829.8

p115.4(kN/m2)

24. A rectangular plate 5 ft by 4 ft is at an angle of 30°with the horizontal, and the 5-ft side is horizontal. Find the magnitude of the force on one side of the plate and the depth of its center of pressure when the top edge is (a) at the water surface; (b) 1 ft below the water surface. (10分)

Solution:

4(a)

FhcA62.4sin30(54)1248(lb) 222

hph4sin301.333(ft) 334(b)

FhcA62.4(1sin30)(54)2496(lb) 2131bl543

Ic1412()4124.333(ft) ypyc14ycAsin3024(45)()A

sin302

hpypsin304.3330.52.17(ft)

1. 25. Water in a reservoir is discharged from a vertical pipe. If a=25 ft, b= 60 ft, c=40 ft, d= 2ft. All the losses of energy are to be ignored when the stream discharging into the air at E has a diameter of 4 in. What are pressure heads at B, C and D, if the diameter of the vertical pipe is 5 in? (10分)

22 pAvApEvEzAzE 2g2g2 vE0(256040)000 2g

vE232.212589.72(ft/s)

AE42vBvCvDvE289.7257.42(ft/s) AB5 22pAvApBvB zAzB2g2g

2 PBvB57.422a2526.2(ft) 2g232.2

22 pCvCpBvBzBzC 2g2g

PP CB(zBzC)26.26033.8(ft) 22pBvBpDvD zBzD2g2g

PP DB(zBzD)26.26040271.8(ft)

26. A nozzle that discharges a 60-mm-diamater water jet into the air is on the right end of a horizontal 120-mm-diameter pipe. In the pipe the water has a velocity of 4 m/s and a gage pressure of 400 kPa. Find the magnitude and direction of the resultant axial force the water exerts on the nozzle, and the head loss in the nozzle. (10分)

Solution:

A11202v2v1416(m/s)2A260 FQ(VV)PAFx2111bx

FFPAQ(VV)4000000.122100040.122(164)xbx112144

4523.9542.93981(N)

to the right

2 p1v12p2v2z1z2hL

2g2g

2222pvv40000416112 hL28.57(m)2g2g980029.829.8



27. A 450-ft-long pipeline runs between two reservoirs, both ends being under water, and the intake end is square-edged and nonprojecting. The difference between the water surface levels of the two reservoirs is 150ft. (a) What is the discharge(流量) if the pipe diameter is 12 in and f = 0.028? (b) When this same pipe is old, assume that the growth of tubercles has reduced the diameter to 11.25 in and that f=0.06. What will the rate of discharge be then? (10分)

Solution: (a) 22pvpv1122 z1z2hL2g2g

2v2 lvz1z2f(kek') d2g2g

450v2150(0.0280.51)

12232.2

12

v26.2(ft/s)

12 ()21220.6(cfs) Qv.A26.24 (b)

p12v12p2v2z1z2hL2g2g2v2lvz1z2f(kek')d2g2g450v2150(0.060.51)11.25232.212v17.8(ft/s)11.252()12Qv.A17.812.32(cfs)4