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1、天津大学化工流体课后题答案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, that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and that liquid
2、 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 instrumentIn meters of water, (a) when the change in the level in the reservoirs is neglected, (b) when t
3、he 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/m3pb=0.77kg/m3 D/d=8 R=0.145m When the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tub
4、es pp22Dx=dR (1) 44so dx=R (2) D2and hydrostatic equilibrium gives following relationship p1+Rrcg=p2+xrcg+RrAg (3) so p1-p2=xrcg+R(rA-rc)g (4) substituting the equation (2) for x into equation (4) gives dp1-p2=Rrcg+R(rA-rc)g (5) D2when the change in the level in the reservoirs is neglected, dp1-p2=R
5、rcg+R(rA-rc)gR(rA-rc)g=0.145(1000-815)9.81=263PaD2when the change in the levels in the reservoirs is taken into account dp1-p2=Rrcg+R(rA-rc)gDd=Rrcg+R(rA-rc)gD6.5=0.1458159.81+0.145(1000-815)9.81=281.8Pa51222error=281.8-263281.86.7 1.4 There are two U-tube manometers fixed on the fluid bed reactor,
6、as shown in the figure. The readings of two U-tube manometersareR1=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 diffusing into the air, and the height R3=50mm. Try to calculate the pressure at point
7、 A and B. Figure for problem 1.4 Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by rg,rH2O,rHg, respectively. The pressure at point A is given by hydrostatic equilibrium pA=rH2OR3g+rHgR2g-rg(R2+R3)g rgis small and negligible in comparison with
8、rHgand H2O, equation above can be simplified pApc=rH2OgR3+rHggR2 =10009.810.05+136009.810.05 =7161N/m pBpD=pA+rHggR1=7161+136009.810.4=60527N/m 1.5 Water discharges from the reservoir through the drainpipe, which thethroat diameter is d. The ratio of D to d equals 1.25. The verticaldistance h betwee
9、n 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, tank A and the exit of drainpipe are
10、 all open to air. pa H D d h pa Figure for problem 1.5 A Solution: Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane: p1+gz1+u122r=p2r+gz2+u222Where p1=0, p2=0, and u1=0, simplification of the equation 2u2 Hg = 1 2 The relationship between the velocit
11、y at outlet and velocity uo at throat can be derived by the continuity equation: u2uod= D22uoD=u2 2 dBernoulli equation is written between the throat and the station 2-2 + = 3 r22 Combining equation 1,2,and 3 gives p0u02u21hrg1210009.8129.81 Hg=442r10002.44-1(1.25)-1D -1 dSolving for H u2H=1.39m 1.6
12、 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 no friction los
13、ses, 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 equation , for a h
14、orizontal pipe, Then Bernoulli equation becomes, after substitutingu2=u1A1A2 for 2, Rearranging, u1p1-p22A1A-122rPerforming 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 of diameter d and wi
15、th mean velocity V. Show that the pressure loss in a length Dp32mVof pipe is . 2LdOil 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 found by su
16、mming up all the velocities over the cross section and dividing by the cross-sectional area RR11 V = udA = u 2 p 1 rdr2A0pR0 From velocity profile equation for laminar flow 2 u = p 0 - p L R 2 1 r 2 - 4mLR substituting equation 2 for u into equation 1 and integrating p0-pL2V=D 3 32mL rearranging equ
17、ation 3 gives Dp32mV= 2Ld 32mVL320.050.6120Dp=11520Pa22 d0.1 1.8. 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 is 0.02 m3/s, the pressure
18、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 kV22gwhere V is the velocity at A, find the Figure for problem 1.8 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
19、of 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=0.075m zA-zB=l=2.5m Q=0.02 m3/s, 2pB-pA=14715 N/m Q=VA=p4dAVAQ=220.020.7850.152p4=1.132m/sdAQ=VB=p4dBVBQ=220.020.7850.0752
20、p4=4.529m/sdBWhen the fluid flows down, writing mechanical balance equation pA+zAg+VA22r=pBr+zBg+VB22+kVA222.59.81+1.1322=147151000+4.5322+k1.132224.525+0.638=14.715+10.260+0.638k k=0.295 making the static equilibrium pB+Dxrg+Rrg=pA+lrg+Dxrg+RrHggR=(pB(r-pA)-lrgHg-rg)=14715-2.510009.81126009.81=-79m
21、m 1.9The 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 abandcd，have the same length, the diameter and roughness. Find: the expressions of Dp
22、abrg, hfab,Dpcdrg and hfcd, respectively. Figure for problem 1.9 the relationship between readings R1and R2 in the U tube. Solution: (1) From Fanning equation 2lV h=lfabd2 and 2lV hfcd=ld2so hfab=hfcdFluid flows from station a to station b, mechanicalenergy conservation gives pap+lg=b+hfab rr hence
23、pa-pb + lg = h fab 2 r from station c to station d pp c=d+hfcdrr hence p-pcd= h fcd 3 r From static equation pa-pb=R1g -lg 4 pc-pd=R2g 5 Substituting equation 4 in equation 2 ，then Rg-lrg1+lg=hfab rtherefore r-r h fab = R g 6 1rSubstituting equation 5 in equation 3 ，then r-r h fcd = R 2 g 7 rThus R1
24、=R2 1.10 Waterpasses 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 water flows through the pipe length L=2 m, in m H2O column? 2) Find the maximum velocity and point r at which L it occurs. 3) Find the point r at whi
25、chthe average velocity equals the local velocity. 4）if kerosene flows through this pipe，how do the variables above change？ solution: 1）Re=udr=0.40.00410000.001=1600 3r Figure for problem 1.10 mfrom Hagen-Poiseuille equation 32uLm320.420.001DP=d2=0.0042=1600 h=Dprg=160010009.81=0.163m 2）maximum veloc
26、ity occurs at the center of pipe, from equation 1.4-19 Vumax=0.5 so umax=0.42=0.8m 3）when u=V=0.4m/s Eq. 1.4-17 uumaxr=1-r w22Vr1-=0.5 * 0.004umaxr=0.0040.5=0.0040.71=0.00284m 4) kerosene: Re=udr=0.40.0048000.0030.0030.001=427 mDp=Dpmm=1600=4800Pa h=Dprg=48008009.81=0.611m 1.12 As shown in the figur
27、e, 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 theU-tubemanometeris connectedto the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air
28、, 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 the gate valve is opened partly, R=400mm, h=1400mm.Th
29、e friction coefficient is 0.025, and the loss coefficient of the entranceis 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 wid
30、ely open, and the friction coefficient is still 0.025. Solution： Figure for problem 1.12 (1) When the gate valve is opened partially, the water discharge is Set up Bernoulli equation between the surface of reservoir 11and the section of pressure point 22，and take the center of section 22as the refer
31、ring plane, then gZ1+u122+p1r=gZ2+u222+p2r+hf,12In the equation p1=0(the gauge pressure) p2=rHggR-rH2Ogh=136009.810.4-10009.811.4=39630N/m 2u10Z2=0 When the gate valve is fully closed, the height of water level in the reservoir can berelated rHg(Z1+h)=rHggR to h (the distance between the center of p
32、ipe and the meniscus of left arm of U tube). 2O where h=1.5m R=0.6m Substitute the known variables into equation b Z1=136000.61000f,1_2-1.5=6.66m+Kc)V2h=(lld2=(0.025150.1+0.5)V2=2.13V22Substitute the known variables equation a 9.816.66=V22+3963010002+2.13V the velocity is V =3.13m/s the flow rate of
33、 water is p2p3 Vh=3600dV=36000.123.13=88.5m/h 44 2） the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 11and 3-3，then gZ1+V122+p1r=gZ3+V322+p3r+hf,13since Z1=6.66m Z3=0u10p1=p3hf,1_3=(ll+led35+Kc)V22V2
34、=0.025( =4.81V20.1+15)+0.52input the above data into equation c， 9.816.66=V22+4.81V 2the velocity is: V=3.51 m/s Write mechanical energy balance equation between thestations 11and 22, for the same situation of water level gZ1+V122+p1r=gZ2+V222+p2r+hf,12since Z1=6.66m Z2=0u10u23.51m/sp1=0(page pressu
35、re）hf,1_2=(lld+Kc)V22=(0.025150.1+0.5)3.5122=26.2J/kg input the above data into equation d， 9.816.66=3.5122+p21000+26.2 the pressure is: p2=32970 1.17.Sulphuric acid of specific gravity 1.3 is flowing througha pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and th
36、e 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 may be taken as 0.61, the s
37、pecific gravity of mercury as 13.6, and the densityof water as 1000 kg/m Solution: a)D0D1=1050=0.2 3Figure for problem 1.17 p1-p2=R(rHg-r)g=0.1(13600-13000)9.81= V2= m=CoD01-D142(p1-p2)r=0.611-0.2420.1(13600-13000)9.81130000.610.0906=0.610.30=0.183m/sp4D0V2r=2p40.010.18313000=0.187kg/s 2b) approxima
38、te pressure drop p1-p2=R(rHg-r)g=0.1(13600-13000)9.81=588.6Pa * 2.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152kPaand 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
39、 is 2.45kw while the centrifugal pump operates 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
40、of the pump under this operating condition. Solution: Write the mechanical energy balance equation between the suction connection and discharge connection Z1+u1222g+p1rg+H=Z2+u22g+p2rg+Hf,1_2where Z2-Z1=0.4m p1=-2.4710Pa(gaugep2=1.5210Pa(gaugeu1=u2Hf,1_254pressure）pressure）0total heads of pump is H=
41、0.4+1.52105+0.24710510009.81=18.41m efficiency of pump is h=Ne/N since Ne=N=2.45kW Then mechanical efficiency h=1.32.45100%=53.1% QHrg3600=2618.4110009.813600=1.3kW The performance of pump is Flow rate ，m/h 26 Total heads，m Shaft power ，kW Efficiency ，% 2.2 Water is transported by a pump from reacto
42、r, which has 200 mm Hg vacuum, to the tank, in which the gauge pressure is 0.5 kgf/cm, 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 orifice diameter do are 0.62 and 25 mm, respectively
43、. 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) 218.41 2.45 53.1 10m21Solution: Equation(1.6-9) V=0 C0d1-0D42Rgr=0.62251-50420.1689.81(13600-1000)10000.626.44=4.12m/s =0.9375Mass flow rate m=VoSor=4.123.1440.02521000=2.02kg/s 2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V1=V2 H=p2-p1+Dz+HfrgDz=10m Dp=0.59.81104+2007601.013105
