流体力学英文版课后习题答案

1、1.1 what will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below the surface? water = 1000 kg/m3, and patmosphere = 101kn/m2.solution:rearranging the equation 1.1-4set the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below the surface i

2、s absolute pressure of water at depth 12m 1.3 a differential manometer as shown in fig. is sometimes used to measure small pressure difference. when the reading is zero, the levels in two reservoirs are equal. assume that fluid b is methane（甲烷）, that liquid c in the reservoirs is kerosene (specific

3、gravity = 0.815), and that liquid a in the u tube is water. the inside diameters of the reservoirs and u tube are 51mm and 6.5mm , respectively. if the reading of the manometer is145mm., what is the pressure difference over the instrument in meters of water, (a) when the change in the level in the r

4、eservoirs is neglected, (b) when the change in the levels in the reservoirs is taken into account? what is the percent error in the answer to the part (a)? solution：pa=1000kg/m3 pc=815kg/m3 pb=0.77kg/m3 d/d=8 r=0.145mwhen the pressure difference between two reservoirs is increased, the volumetric ch

5、anges in the reservoirs and u tubes (1)so (2)and hydrostatic equilibrium gives following relationship (3)so (4)substituting the equation (2) for x into equation (4) gives (5)（a）when the change in the level in the reservoirs is neglected, （b）when the change in the levels in the reservoirs is taken in

6、to accounterror=1.4 there are two u-tube manometers fixed on the fluid bed reactor, as shown in the figure. the readings of two u-tube manometers are r1=400mm，r2=50mm, respectively. the indicating liquid is mercury. the top of the manometer is filled with the water to prevent from the mercury vapor

7、diffusing into the air, and the height r3=50mm. try to calculate the pressure at point a and b. figure for problem 1.4solution: there is a gaseous mixture in the u-tube manometer meter. the densities of fluids are denoted by , respectively. the pressure at point a is given by hydrostatic equilibrium

8、 is small and negligible in comparison withand h2o , equation above can be simplified= =10009.810.05+136009.810.05=7161n/m=7161+136009.810.4=60527n/mddpapahhfigure for problem 1.51.5 water discharges from the reservoir through the drainpipe, which the throat diameter is d. the ratio of d to d equals

9、 1.25. the vertical distance h between the tank a and axis of the drainpipe is 2m. what height h from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank a to the throat of the pipe? assume that fluid flow is a potential flow. the reservoir

10、, tank a and the exit of drainpipe are all open to air.solution:bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane:where p1=0, p2=0, and u1=0, simplification of the equation 1the relationship between the velocity at outlet and velocity uo at throat can

11、 be derived by the continuity equation: 2bernoulli equation is written between the throat and the station 2-2 3combining equation 1,2,and 3 givessolving for hh=1.39m1.6 a liquid with a constant density kg/m3 is flowing at an unknown velocity v1 m/s through a horizontal pipe of cross-sectional area a

12、1 m2 at a pressure p1 n/m2, and then it passes to a section of the pipe in which the area is reduced gradually to a2 m2 and the pressure is p2. assuming no friction losses, calculate the velocities v1 and v2 if the pressure difference (p1 - p2) is measured.solution: in fig1.6, the flow diagram is sh

13、own with pressure taps to measure p1 and p2. from the mass-balance continuity equation , for constant where 1 = 2 = ,for the items in the bernoulli equation , for a horizontal pipe,z1=z2=0then bernoulli equation becomes, after substituting for v2,rearranging,performing the same derivation but in ter

14、ms of v2,1.7 a liquid whose coefficient of viscosity is flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity v. show that the pressure loss in a length of pipe is .oil of viscosity 0.05 pas flows through a pipe of diameter 0.1m with a average vel

15、ocity of 0.6m/s. calculate the loss of pressure in a length of 120m.solution:the average velocity v for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area 1from velocity profile equation for laminar flow 2substituting equation 2

16、for u into equation 1 and integrating 3rearranging equation 3 givesfigure for problem 1.81.8. in a vertical pipe carrying water, pressure gauges are inserted at points a and b where the pipe diameters are 0.15m and 0.075m respectively. the point b is 2.5m below a and when the flow rate down the pipe

17、 is 0.02 m3/s, the pressure at b is 14715 n/m2 greater than that at a. assuming the losses in the pipe between a and b can be expressed as where v is the velocity at a, find the value of k.if the gauges at a and b are replaced by tubes filled with water and connected to a u-tube containing mercury o

18、f relative density 13.6, give a sketch showing how the levels in the two limbs of the u-tube differ and calculate the value of this difference in metres. solution: da=0.15m; db=0.075mza-zb=l=2.5mq=0.02 m3/s,pb-pa=14715 n/m2when the fluid flows down, writing mechanical balance equation0.295making the

19、 static equilibriumfigure for problem 1.91.9the liquid vertically flows down through the tube from the station a to the station b, then horizontally through the tube from the station c to the station d, as shown in figure. two segments of the tube, both ab and cd，have the same length, the diameter a

20、nd roughness.find:（1）the expressions of , hfab, and hfcd, respectively.（2）the relationship between readings r1and r2 in the u tube.solution:(1) from fanning equationandsofluid flows from station a to station b, mechanical energy conservation gives hence 2from station c to station dhence 3from static

21、 equationpa-pb=r1（-）g -lg 4pc-pd=r2（-）g 5substituting equation 4 in equation 2 ，thentherefore 6substituting equation 5 in equation 3 ，then 7thusr1=r21.10 water passes through a pipe of diameter di=0.004 m with the average velocity 0.4 m/s, as shown in figure. 1) what is the pressure drop dp when wat

22、er flows through the pipe length l=2 m, in m h2o column?lrfigure for problem 1.102) find the maximum velocity and point r at which it occurs. 3) find the point r at which the average velocity equals the local velocity.4）if kerosene flows through this pipe，how do the variables above change？（the visco

23、sity and density of water are 0.001 pas and 1000 kg/m3，respectively；and the viscosity and density of kerosene are 0.003 pas and 800 kg/m3，respectively）solution:1） from hagen-poiseuille equation2）maximum velocity occurs at the center of pipe, from equation 1.4-19so umax=0.42=0.8m3）when u=v=0.4m/s eq.

24、 1.4-174) kerosene:1.12 as shown in the figure, the water level in the reservoir keeps constant. a steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. one arm of the u-tube manometer is connected to the drainpipe at the position 15m away from the bottom o

25、f the reservoir, and the other is opened to the air, the u tube is filled with mercury and the left-side arm of the u tube above the mercury is filled with water. the distance between the upstream tap and the outlet of the pipeline is 20m. a) when the gate valve is closed, r=600mm, h=1500mm; when th

26、e gate valve is opened partly, r=400mm, h=1400mm. the friction coefficient is 0.025, and the loss coefficient of the entrance is 0.5. calculate the flow rate of water when the gate valve is opened partly. (in m/h)b) when the gate valve is widely open, calculate the static pressure at the tap (in gau

27、ge pressure, n/m). le/d15 when the gate valve is widely open, and the friction coefficient is still 0.025.figure for problem 1.12solution：(1) when the gate valve is opened partially, the water discharge isset up bernoulli equation between the surface of reservoir 11 and the section of pressure point

28、 22，and take the center of section 22 as the referring plane, then （a）in the equation (the gauge pressure) when the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of u tube).（b） where h=

29、1.5mr=0.6msubstitute the known variables into equation bsubstitute the known variables equation a9.816.66=the velocity is v =3.13m/s the flow rate of water is 2） the pressure of the point where pressure is measured when the gate valve is wide-open. write mechanical energy balance equation between th

30、e stations 11 and 3-3，then （c）since input the above data into equation c，9.81the velocity is: v=3.51 m/swrite mechanical energy balance equation between thestations 11 and 22, for the same situation of water level （d）since input the above data into equation d，9.816.66=the pressure is: 1.14 water at

31、20 passes through a steel pipe with an inside diameter of 300mm and 2m long. there is a attached-pipe (603.5mm) which is parallel with the main pipe. the total length including the equivalent length of all form losses of the attached-pipe is 10m. a rotameter is installed in the branch pipe. when the

32、 reading of the rotameter is 2.72m3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. the frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.solution： the variables of main pipe are denoted by a subscript1, and branch

33、 pipe by subscript 2. the friction loss for parallel pipelines isthe energy loss in the branch pipe isin the equation input the data into equation cthe energy loss in the main pipe isso the water discharge of main pipe istotal water discharge is1.16 a venturimeter is used for measuring flow of water

34、 along a pipe. the diameter of the venturi throat is two fifths the diameter of the pipe. the inlet and throat are connected by water filled tubes to a mercury u-tube manometer. the velocity of flow along the pipe is found to be m/s, where r is the manometer reading in metres of mercury. determine t

35、he loss of head between inlet and throat of the venturi when r is 0.49m. (relative density of mercury is 13.6).figure for problem 1.16solution:writing mechanical energy balance equation between the inlet 1 and throat o for venturi meter 1rearranging the equation above, and set (z2-z1)=x 2from contin

36、uity equation 3substituting equation 3 for vo into equation 2 gives 4from the hydrostatic equilibrium for manometer 5substituting equation 5 for pressure difference into equation 4 obtains 6rearranging equation 6 1.17.sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal

37、 diameter. a thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. assuming that the leads to the manometer are filled with the acid, calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure c

38、aused by the orifice.the coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m3solution:a)b) approximate pressure drop12066.3papressure difference due to increase of velocity in passing through the orifice pressure drop caused

39、 by friction loss 2.1 water is used to test for the performances of pump. the gauge pressure at the discharge connection is 152 kpa and the reading of vacuum gauge at the suction connection of the pump is 24.7 kpa as the flow rate is 26m3/h. the shaft power is 2.45kw while the centrifugal pump opera

40、tes at the speed of 2900r/min. if the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. calculate the mechanical efficiency of pump and list the performance of the pump under this operating condition.

41、solution: write the mechanical energy balance equation between the suction connection and discharge connection wheretotal heads of pump is efficiency of pump is since n=2.45kwthen mechanical efficiency the performance of pump is flow rate ，m/h26total heads，m18.41shaft power ，kw2.45efficiency ，%53.12

42、.2 water is transported by a pump from reactor, which has 200 mm hg vacuum, to the tank, in which the gauge pressure is 0.5 kgf/cm2, as shown in fig. the total equivalent length of pipe is 200 m including all local frictional loss. the pipeline is f573.5 mm , the orifice coefficient of co and orific

43、e diameter do are 0.62 and 25 mm, respectively. frictional coefficient l is 0.025. calculate: developed head h of pump, in m (the reading r of u pressure gauge in orifice meter is 168 mm hg)solution:equation(1.6-9)mass flow rate2) fluid flow through the pipe from the reactor to tank, the bernoulli e

44、quation is as followsfor v1=v2dz=10m dp/rg=7.7mthe relation between the hole velocity and velocity of pipefriction lossso h=7.7+10+5.1=22.8mhg2.3 . a centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. at the rated discharge,

45、 the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum. if losses in the suction pipe accounted for a head of 1.5m. what must be the least height of the liquid level in the condenser above the pump inlet?solution:from an energy balance,where po

46、=760-640=120mmhgpv=760-710=50mmhguse of the equation will give the minimum height hg as2.4 sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. calculate the drop in pressure. if the pressure drop falls by one half, what will the new flowrate be ? density of acid 1840kg/m3 v

47、iscosity of acid 2510-3 passolution:velocity of acid in the pipe:reynolds number:from fig.1.22 for a smooth pipe when re=6109, f=0.0085pressure drop is calculated from equation 1.4-9or friction factor is calculated from equation1.4-25if the pressure drop falls to 783.84/2=391.92kpasonew mass flowrat

48、e=0.785d2u=0.7850.02522.271840=2.05kg/s2.4 sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. calculate the drop in pressure. if the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ? density of a

49、cid 1840kg/m3 viscosity of acid 2510-3 pafriction factor for hydraulically smooth pipesolution:write energy balance equation:p=46.9218409.81=847.0kpa2.6 the fluid is pumped through the horizontal pipe from section a to b with the 382.5mm diameter and length of 30 meters, shown as figure. the orifice

50、 meter of 16.4mm diameter is used to measure the flow rate. orifice coefficient co0.63. the permanent loss in pressure is 3.5104n/m2, the friction coefficient =0.024. find: (1) what is the pressure drop along the pipe ab? (2)what is the ratio of power obliterated in pipe ab to total power supplied t

51、o the fluid when the shaft work is 500w, 60%efficiency? (the density of fluid is 870kg/m3 ) solution：u= (16.4/33)28.5=2.1m/s (2)sothe ratio of power obliterated in friction losses in ab to total power supplied to the fluid acknowledgements my deepest gratitude goes first and foremost to professor aaa , my supervisor, for her constant encouragement and guidance. she has walked me throug