UnitOperationsofChemicalEngineering(化工单元操作).doc
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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 1.1-4ghpabSet the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below the sur
2、face is kPaab 72.18.910Absolute pressure of water at depth 12m kPaghpab 72.180. 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
3、 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 manometer is145mm., what is the pressure difference over the instrument In meters of water, (a)
4、 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=1000kg/m3 pc=815kg/m3 pb=0.77kg/m3 D/d=8 R=0.145mWhen the pressure difference between two rese
5、rvoirs is increased, the volumetric changes in the reservoirs and U tubes(1)RdxD224so(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 i
6、s neglected, | PagRggRDdp cAcAc 26381.95104.)()(221 (b)when the change in the levels in the 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
7、 manometers are R1=400mm,R 2=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 A and B. Solution: There is a gaseous mixture
8、in the U-tube manometer meter. The densities of fluids are denoted by , respectively. The pressure at point A is given by hydrostatic HgOg,2Figure for problem 1.4|equilibrium gRgRpHOA )(32232 is small and negligible in comparison with and H2O , equation above can be g Hgsimplified=cAp232gRHOH=10009.
9、810.05+136009.810.05=7161N/m=7161+136009.810.4=60527N/m1HgADB1.5 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 o
10、f 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 all open to air.Solution:Bernoulli equation is written between stations 1-1 and
11、 2-2, with station 2-2 being reference plane:2211 ugzpugzpWhere p1=0, p2=0, and u1=0, simplification of the equationD dpapaHhAFigure for problem 1.5|1The relationship between the velocity at outlet and velocity uo at throat can be derived by the continuity equation:22Dduo22doBernoulli equation is wr
12、itten between the throat and the station 2-23Combining 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 A1 m2 at a pressure p1 N/m2, and then it passes to a section of the
13、 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 shown with pressure taps to measure p1 and p2. From the mass-balance
14、continuity equation , for constant where 1 = 2 = ,2uHg20up 14.28910.25.11442 ghdDg|21AVFor the items in the Bernoulli equation , for a horizontal pipe,z1=z2=0Then Bernoulli equation becomes, after substituting for V2,12A2121200pAVpRearranging, 2)1(2121AVp1221AVPerforming the same derivation but in t
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