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1、Unit Operations of Chemical Engineering1.1 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 1.1-4 pb=pa+rgh Set the pressure of atmosphere to be zero, then
2、 the gauge pressure at depth 12m below the surface is pb=pa+rgh=0+10009.8112=117.72kPa Absolute pressure of water at depth 12m pb=pa+rgh=101000+10009.8112=218720Pa=218.72kPa 1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressure difference. When the reading is zero
3、, 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 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 manom
4、eter 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 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=100
5、0kg/m3 pc=815kg/m3 pb=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 tubes p4so D2x=p42d2R (1) dx=R (2) Dand hydrostatic equilibrium gives following relationship p1+Rrcg=p2+xrcg+RrAg (3) so p1-p2=xrcg+R(rA-rc)
6、g (4) substituting the equation (2) for x into equation (4) gives dp1-p2=Rrcg+R(rA-rc)g (5) Dwhen the change in the level in the reservoirs is neglected, 2dp1-p2=Rrcg+R(rA-rc)gR(rA-rc)g=0.145(1000-815)9.81=263PaDwhen the change in the levels in the reservoirs is taken into account 2dp1-p2=Rrcg+R(rA-
7、rc)gDd=Rrcg+R(rA-rc)gD6.5=0.1458159.81+0.145(1000-815)9.81=281.8Pa51error=222281.8-2636.7 281.81.4 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 t
8、op 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. Figure for problem 1.4 Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denote
9、d 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 withrHgand H2O , equation above can be simplified pApc=rH2OgR3+rHggR2 =10009.810.05+136009.810.05 =7161N/m pBpD=pA+rHggR1=7161+136009.810.4
10、=60527N/m 1.5 Water discharges from the reservoir 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
11、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, pa tank A and the exit of drainpipe are all open to air. H d D h pa Figure for problem 1.5 A Solution: Bernoulli equation is written between stations 1-1 and 2-2, with
12、 station 2-2 being reference plane: p12u12p2u2+gz1+=+gz2+ r2r2Where p1=0, p2=0, and u1=0, simplification of the equation 2Hg=u22 1 The relationship between the velocity at outlet and velocity uo at throat can be derived by the continuity equation: u2uod=D 22Duo=u2 2 dBernoulli equation is written be
13、tween the throat and the station 2-2 22p0u0u + = 3 22 rCombining equation 1,2,and 3 gives u21hrg1210009.8129.81 Hg=442Dr10002.44-1(1.25)-1 -1 dSolving for H H=1.39m 1.6 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
14、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 s
15、hown with pressure taps to measure p1 and p2. From the mass-balance continuity equation , for constant where 1 = 2 = , V2=V1A1 A2For the items in the Bernoulli equation , for a horizontal pipe, z1=z2=0 Then Bernoulli equation becomes, after substitutingV2=V1A1 for V2, A2A12V2V1p1A12p20+=0+ 2r2r21Rea
16、rranging, A12rV(2-1)A1p1-p2= 221V1p1-p22A1A22r-1Performing the same derivation but in terms of V2, V2p1-p22A21-A12r1.7 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
17、in a length of pipeDp32mV is . Ld2Oil 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 summing up all the velocities over the cross secti
18、on and dividing by the cross-sectional area RR11 V = udA = u 2 p rdr 1 2A0pR0 From velocity profile equation for laminar flow 2p-pr u = 0 L R 2 1 - 2 R4mL substituting equation 2 for u into equation 1 and integrating p0-pL2V= D 3 32mL rearranging equation 3 gives Dp32mV= Ld2 32mVL320.050.6120Dp=1152
19、0Pa22 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 at B is 14715 N/m2 greater than that at A. Assumin
20、g the losses in the pipe between A and B can be V2expressed as kwhere V is the velocity at A, find the 2gFigure 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 of relative density 13.6, give a sketch showing ho
21、w 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, pB-pA=14715 N/m2 Q=pd24AVAVQ=0.02 A=p0.7850.152=1.132m/s4d2AQ=p4d2BVBVQ=0.02 B=p0.7850.0752=4.529m/s4d2BWhen the fluid flows down, writing
22、mechanical balance equation p2AVArzp2BVBV2+AAg+2=r+zBg+2+k2 1.132147154.5321.1322.59.81+2=1000+2+k2 24.525+0.638=14.715+10.260+0.638k k=0.295 making the static equilibrium pB+Dxrg+Rrg=pA+lrg+Dxrg+RrHggR=(p(B-pA)-)lrg2.510009.81rHg-rg=14715-126009.81=-79mm 1.9The liquid vertically flows down through
23、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 and roughness. Find: the expressions of Dpabrg, hDpfab, cdrg and hfcd, respectively. t
24、he relationship between readings R1and R2 in the U tube. Solution: (1) From Fanning equation 2hfab=llVd2Figure for problem 1.9 and lV2 hfcd=ld2so hfab=hfcdFluid flows from station a to station b, mechanical energy conservation gives p r a + lg = p r b + h fab hence p a - p b + lg = h fab 2 rfrom sta
25、tion c to station d pcpr=dr+hfcdhence p c - p d = h r fcd 3 From static equation pa-pb=R1g -lg 4 pc-pd=R2g Substituting equation 4 in equation 2 ，then Rg-lrg+lg=hfab rtherefore h R r-rfab = 1 g r6 Substituting equation 5 in equation 3 ，then h r-rfcd = R 2 g 7 rThus R1=R2 5 1.10 Water passes through
26、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 which the average velocity equal
27、s the local velocity. r 4）if kerosene flows through this pipe，how do the variables above change？ solution: 1）Re=udrm=0.40.0041000=1600 0.001from Hagen-Poiseuille equation DP=h=32uLm320.420.001=1600 d20.0042Dp1600=0.163m rg10009.812）maximum velocity occurs at the center of pipe, from equation 1.4-19
28、so umax=0.42=0.8m 3）when u=V=0.4m/s Eq. 1.4-17 V=0.5umaxuumaxr=1-rw2 2Vr1-=0.5 0.004umaxr=0.0040.5=0.0040.71=0.00284m 4) kerosene: Re=udrm=0.40.004800=427 0.003Dp=Dpm0.003=1600=4800Pa m0.001h=Dp4800=0.611m rg8009.811.12 As shown in the figure, the water level in the reservoir keeps constant. A steel
29、 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
30、 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 coeffi
31、cient 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 0.0
32、25. Figure for problem 1.12 Solution： (1) When the gate valve is opened partially, the water discharge is Set 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 2u12p1u2pgZ1+=gZ2+2+hf,12 2r2rIn
33、 the equation p1=0(the gauge pressure) p2=rHggR-rH2Ogh=136009.810.4-10009.811.4=39630N/m2 u10Z2=0 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). rHOg(Z1+h)=rHggR 2 w
34、here h=1.5m R=0.6m Substitute the known variables into equation b 136000.6-1.5=6.66m1000 22lV15Vhf,1_2=(ld+Kc)2=(0.0250.1+0.5)2=2.13V2Z1=Substitute the known variables equation a V2396309.816.66=+2.13V2 21000the velocity is V =3.13m/s the flow rate of water is Vh=3600p4d2V=3600p40.123.13=88.5m3/h 2）
35、 the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 11 and 3-3，then V32p3V12p1gZ1+=gZ3+hf,13 2r2rsince Z1=6.66m Z3=0u10 p1=p3l+leV2hf,1_3=(ld+Kc)235V2 =0.025(+15)+0.5 0.12 =4.81V2input the above data in
36、to equation c， V2+4.81V2 9.816.66=2the velocity is: V=3.51 m/s Write mechanical energy balance equation between thestations 11 and 22, for the same situation of water level V12p1V22p2gZ1+=gZ2+hf,12 2r2rsince Z1=6.66m Z2=0u10u23.51m/sp1=0(page pressure）hf,1_2lV2153.512=(l+Kc)=(0.025+0.5)=26.2J/kg d20
37、.12input the above data into equation d， p3.5129.816.66=+2+26.2 21000the pressure is: p2=32970 1.14 Water at 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
38、 length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the 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
39、 is 0.018 and 0.03, respectively. Solution： The variables of main pipe are denoted by a subscript1, and branch pipe by subscript 2. The friction loss for parallel pipelines is hf1=hf2Vs=VS1+VS2The energy loss in the branch pipe is 2l2+le2u2=l2 d22hf2In the equation l2=0.03 l2+le2=10m d2=0.053u2=3600
40、2.72p4=0.343m/s0.0532input the data into equation c hhf2100.3432=0.03=0.333J/kg 0.0532l1u12=l1=0.333 d12The energy loss in the main pipe is f1=hf2So u1=0.3330.32=2.36m/s 0.0182The water discharge of main pipe is Vh1=3600p40.322.36=601m3/h Total water discharge is Vh=601+2.72603.7m3/h 1.16 A Venturim
41、eter is used for measuring flow of water 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 2.5R m/s, where R is the manome
42、ter reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution: Writing mechanical energy balance equation between the inlet 1 and throat o for Venturi meter p1poVo2V12+z1g=+z2g+hf 1 r2r2rearrangin
43、g the equation above, and set (z2-z1)=x p1-porVo2-V12=+xg+hf 2 2Figure for problem 1.16 from continuity equation d1Vo=V1do5=V1=6.25V1 3 222substituting equation 3 for Vo into equation 2 gives p1-por39.06V12-V12=+xg+hf=19.03V12+hf=19.032.5R2=118.94R+xg+hf()2+xg+hf 4 from the hydrostatic equilibrium f
44、or manometer p1-po=R(rHg-r)g+xrg 5 substituting equation 5 for pressure difference into equation 4 obtains R(rHg-r)g+xrgrR(rHg-r)g=118.94R+xg+hf 6 rearranging equation 6 hf=r-118.94R=123.61R-118.94R=4.67R=2.288J/kg 1.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in th
