《UnitOperationsofChemicalEngineering(化工单元操作)》由会员分享，可在线阅读，更多相关《UnitOperationsofChemicalEngineering(化工单元操作)（26页珍藏版）》请在金锄头文库上搜索。
1、.1.1 What will be the the gauge pressure and 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 is Absol
2、ute 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 , and that liquid A
3、 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, when the change in the level in the reservoirs is neglected, when the change i
4、n the levels in the reservoirs is taken into account? What is the percent error in the answer to the part ?Solution：pa=1000kg/m3 pc=815kg/m3pb=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 so and hydrost
5、atic equilibrium gives following relationship so substituting the equation for x into equation gives awhen the change in the level in the reservoirs is neglected, bwhen the change in the levels in the reservoirs is taken into accounterror=1.4 There are two U-tube manometers fixed on the fluid bed re
6、actor, 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 a
7、t 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 is small and negligible in comparison withand H2O, equation above can be simplif
8、ied= =10009.810.05+136009.810.05=7161N/m=7161+136009.810.4=60527N/mDdpapaHhAFigure for problem 1.51.5 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 between the tank A and axis of the drainpipe is 2m.
9、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 all open to air.Solution:Bernoulli equation i
10、s 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 be derived by the continuity equation: 2Bernoulli equation is written between the t
11、hroat 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 A1 m2 at a pressure p1 N/m2, and then it passes to a section of the pipe in which the
12、 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 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 c
13、onstant 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 terms of V2,1.7 A liquid whose coefficient of viscosity is flows below the critical velocity for
14、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 velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.Solution:The average veloc
15、ity 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 for u into equation 1 and integrating 3rearranging equation 3 givesFigure for problem 1.81.8. In a vertical pipe carrying water, pressure gauges
