1、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 gravity = 0.815), and that liquid A in the
2、 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 reservoirs is neglected, (b) when the chang
3、e 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 changes in the reservoirs and U tubes(1)RdxD
4、224so(2)2and hydrostatic equilibrium gives following relationship(3)gRxpgRAcc21so(4)cAc)(21substituting the equation (2) for x into equation (4) gives (5)gRgDdpcAc)(221 （a）when the change in the level in the reservoirs is neglected, PagcAcAc 26381.95104.)()(221 （b）when the change in the levels in th
5、e reservoirs is taken into accountPagRgDdpcAccc 8.21.981504.819514.05.6)(22221 error= 7.8.311.4 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，R 2=50mm, respectively. The indicating liquid is mercury. The top
6、 of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R3=50mm. Try to calculate the pressure at point A and B. Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by , respectively. The
7、pressure at point A is given by hydrostatic HgOg,2equilibrium gRRpggHA )(32232 is small and negligible in comparison with and H2O , equation above can be simplifiedg H=cA232HgOH=10009.810.05+136009.810.05=7161N/m=7161+136009.810.4=60527N/m1RpHgADBFigure for problem 1.41.5 Water discharges from the r
8、eservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 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 t
9、ank A to the throat of the pipe? Assume that fluid flow is a potential flow. The reservoir, 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:2211 ugzpugzpWhere p1=0, p2=0, and u1=0, simpli
10、fication of the equation1The relationship between the velocity at outlet and velocity uo at throat can be derived by the continuity equation:22Dduo22doBernoulli equation is written between the throat and the station 2-23Combining equation 1,2,and 3 givesD dpapaHhAFigure for problem 1.5uHg20upSolving
11、 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 A1 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
12、 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 shown 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 equat
13、ion , for a horizontal pipe,Then Bernoulli equation becomes, after substituting for 2,212ARearranging, 14.28910.25.11244 ghdDug1221ApPerforming the same derivation but in terms of 2,1.7 A liquid whose coefficient of viscosity is flows below the critical velocity for laminar flow in a circular pipe o
14、f diameter d and with mean velocity V. Show that the pressure loss in a length of pipe is .Lp23VOil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.Solution:The average velocity V for a cross section is
15、 found by summing up all the velocities over the cross section and dividing by the cross-sectional area1From velocity profile equation for laminar flow2substituting equation 2 for u into equation 1 and integrating 3rearranging equation 3 givesRRrdudAV02012014RrLpu0DVL2dLpPaV1520.06522 1.8. In a vert
16、ical 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 is 0.02 m3/s, the pressure at B is 14715 N/m2 greater than that at A. Assuming the losses in the pipe
17、between A and B can be expressed as where V is the velocity at A, find the gkvalue of k.If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ a
18、nd 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/m2 smdQVA/132.5.078.422sdBB/529.407.85.42When the fluid flows down, writing mechanical balance equation22ABBAA VkgzpVgzp13.5.41073.895.k68.26.240.295kmaking the static equilibrium g
19、RxglpRgxp HAB mlgH 7981.92605147Figure for problem 1.81.9The 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, t
20、he diameter and roughness.Find:（1）the expressions of , hfab, and hfcd, respectively.gpabcd（2）the relationship between readings R 1and R2 in the U tube.Solution:(1) From Fanning equationandsoFluid flows from station a to station b, mechanical energy conservation giveshence2from station c to station d
21、hence3From static equationpa-pb=R1（ -）g -l g 4pc-pd=R2（ -）g 5Substituting equation 4 in equation 2 ，thenFigure for problem 1.92VdlhfablfcdfcfabhfbplgfabbahlfcdcpfcdchfabhlR1）（therefore6Substituting equation 5 in equation 3 ，then7ThusR1=R21.10 Water passes through a pipe of diameter di=0.004 m with t
22、he average velocity 0.4 m/s, as shown in Figure. 1) What is the pressure drop P when water flows through the pipe length L=2 m, in m H2O column?2) 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
23、 through this pipe，how do the variables above change？（the viscosity 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） 1600.4Reudfrom Hagen-Poiseuille equation4232dLPgRhfab1ghfcd2LrFigure for p
24、roblem 1.10mgph163.08.912）maximum velocity occurs at the center of pipe, from equation 1.4-19max0.5Vuso umax=0.42=0.8m3）when u=V=0.4m/s Eq. 1.4-172max1wr*5.004.max2uV mr 284.714) kerosene: 03.4ReudPap81.6 mgh.0.9841.12 As shown in the figure, the water level in the reservoir keeps constant. A steel
25、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 of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side
26、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 the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient is 0.025, and the loss coeffic
27、ient 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 gauge pressure, N/m). le/d15 when the gate valve is widely open, and the friction coefficient is still 0.025
28、.Solution：(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 22，and take the center of section 22 as the referring plane, then（a）1,211 fhpugZpugZIn the equation (the gauge pressure)01 222
29、/39604.18104.891360 mNghRpOHg 021ZuWhen 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）gRhgHOH)(12where h=1.5mR=0.6mSubstitute the known variables into equation b 2222_
30、1, 13.)5.01.0()(6.50.36 VVKdlhmZcf Substitute the known variables equation a9.816.66= 223.1096the velocity is V =3.13m/sFigure for problem 1.12the flow rate of water is hmVdVh /5.813.204364360 32 2） the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanic
31、al energy balance equation between the stations 11 and 3-3，then（c）31,323121 fhpgZpgZsince m6.1310pu22_,81.4 5.0).(05VVKdlhcefinput the above data into equation c，9.81 22.6.the velocity is: V=3.51 m/sWrite mechanical energy balance equation between thestations 11 and 22, for the same situation of wat
32、er level （d）21,22121 fhpVgZpgZsince m6.12103.5/(page rsuu） kgJVKdlhcf /2.651.3)0.25.()22_1, input the above data into equation d，9.816.66= .61025.32pthe pressure is: 32970p1.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, i
33、s 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 pressure drop caused by the orifice.The coefficient of the orifice m
34、ay be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m3Solution:a) 2.0510D81.9)3016(.)(21gRpHskgVDm/187.0318.0.4220 b) approximate pressure drop588.6Pa *.9)6(.)(21gRpH2.1 Water is used to test for the performances of pump. The gauge pressure at the discha
35、rge 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 operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge conne
36、ction 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.Figure for problem 1.17mpCVo /69.61 3081.9)6(2.24240 Solution: Write the mechanical energy balance equatio
37、n between the suction connection and discharge connection 2_1,22121 fHgpuZHgpuZwhere m4.0120(Pa5.72_1,fHupresugp） ）total heads of pump is mH41.81.9002475.145efficiency of pump is Ne/since kWgQe 3.36082360N=2.45kWThen mechanical efficiency %1.504.231The performance of pump is Flow rate ，m/h 26Total h
38、eads，m 18.41Shaft power ，kW 2.45Efficiency ，% 53.12.2 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 5
39、73.5 mm , the orifice coefficient of Co and orifice diameter do are 0.62 and 25 mm, respectively. Frictional coefficient 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 rate skgSVmo /02.1025.413. 2
40、) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as followsfor V1=V2 fHzgpHz=10m Pa750103.76218.9504p/g=7.7mThe relation between the hole velocity and velocity of pipeFriction lossso H=7.7+10+5.1=22.8m10m2smRgDdCVf/6975.02 10)036(8.9.25016.1 4400 ）（smDdV/12.420gudlfH
41、f .58.90.5.22 2.3 . 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, the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum. If losses in the
42、 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=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 th
43、rough 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 Viscosity of acid 2510-3 PasSolution:Velocity of acid in the pipe: smdmpieofartinlcroswtvlumec /32.05.18475.038.422 Reynolds numb
44、er: 610910253.84.Reudfrom Fig.1.22 for a smooth pipe when Re=6109, f=0.0085pressure drop is calculated from equation 1.4-9 kgJudlfphf /45023.0685.42HgNPSHgpfvomNPSHgpfvo 5.3.18.9106)5.2（ kPap5.8271405or friction factor is calculated from equation1.4-25 kgJudludlfhf /4263.05619046.2Re46.2 20 kPap8.73
45、1406if the pressure drop falls to 783.84/2=391.92kPa 8.18.12.2.03 8.120. 97561584046. 46Re.92 uudldl so smu /7.6.489.10723818. new mass flowrate=0.785d2u=0.7850.025 22.271840=2.05kg/s2.6 The fluid is pumped through the horizontal pipe from section A to B with the 38 2.5mm diameter and length of 30 m
46、eters, shown as figure. The orifice 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 i
47、n pipe AB to total power supplied to the fluid when the shaft work is 500W, 60%efficiency? (The density of fluid is 870kg/m3 )solution： fAAAA hupgzwupgz 2202dlhfB47.03.162AosmgRCu /5.8870136.927.0632.100 u= (16.4/33) 28.5=2.1m/s 242 /768510.3.08724.0 mNhpfBA (2) WudWm.2.5.64Ne2sothe ratio of power obliterated in friction losses in AB to total power supplied to the fluid 610.538
