1、流体力学英文版课后习题答案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 1.1-4Set the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below
2、the surface is Absolute 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 kerose
3、ne (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) when the change in the l
4、evel 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 reservoirs is increased, the
5、volumetric changes in the reservoirs and U tubes (1)so (2)and hydrostatic equilibrium gives following relationship (3)so (4)substituting the equation (2) for x into equation (4) gives (5)（a）when the change in the level in the reservoirs is neglected, （b）when the change in the levels in the reservoir
6、s is taken into accounterror=1.4 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 top of the manometer is filled with the water to prevent from the m
7、ercury vapor diffusing into the air, and the height R3=50mm. Try to calculate the pressure at 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 hydrostati
8、c equilibrium is small and negligible in comparison withand H2O , equation above can be simplified= =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 the throat diameter is d. The ratio of
9、 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 tank A to the throat of the pipe? Assume that fluid flow is a potential flow.
10、 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:Where p1=0, p2=0, and u1=0, simplification of the equation 1The relationship between the velocity at outlet and velocity uo
11、 at throat can be derived by the continuity equation: 2Bernoulli equation is written between the throat 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-se
12、ctional 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 losses, calculate the velocities V1 and V2 if the pressure difference (p1 - p2) is measured.Solution: In Fig1.6, the flow
13、 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 horizontal pipe,z1=z2=0Then Bernoulli equation becomes, after substituting for V2,Rearranging,Performing the same derivat
14、ion but in terms of V2,1.7 A liquid whose coefficient of viscosity is flows below the critical velocity for laminar flow in a circular pipe of diameter d and 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
15、 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 section and dividing by the cross-sectional area 1From velocity profile equation for laminar flow 2substituti
16、ng 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 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
17、 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 between A and B can be expressed as where V is the velocity at A, find the value of k.If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube contai
18、ning mercury 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=0.075mzA-zB=l=2.5mQ=0.02 m3/s,pB-pA=14715 N/m2When the fluid flows down, writing mechanical balance equation0
19、.295making the static equilibriumFigure for problem 1.91.9The 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,
20、the diameter and roughness.Find:（1）the expressions of , hfab, and hfcd, respectively.（2）the relationship between readings R1and R2 in the U tube.Solution:(1) From Fanning equationandsoFluid flows from station a to station b, mechanical energy conservation gives hence 2from station c to station dhenc
21、e 3From static equationpa-pb=R1（-）g -lg 4pc-pd=R2（-）g 5Substituting equation 4 in equation 2 ，thentherefore 6Substituting equation 5 in equation 3 ，then 7ThusR1=R21.10 Water passes 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 dr
22、op P when water flows through the pipe length L=2 m, in m H2O column?LrFigure for problem 1.102) 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 through this pipe，how do the variables above chan
23、ge？（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） from Hagen-Poiseuille equation2）maximum velocity occurs at the center of pipe, from equation 1.4-19so umax=0.42=0.8m3）when u
24、=V=0.4m/s Eq. 1.4-174) kerosene:1.12 As shown in the figure, 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 the U-tube manometer is connected to the drainpipe at the position 15m away from
25、 the bottom of the reservoir, and the other is opened to the air, 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=150
26、0mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient is 0.025, and the loss coefficient 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
27、tap (in gauge pressure, N/m). le/d15 when the gate valve is widely open, and the friction coefficient is still 0.025.Figure for problem 1.12Solution：(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 pr
28、essure point 22，and take the center of section 22 as the referring plane, then （a）In the equation (the gauge pressure) 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).
29、 （b） where h=1.5mR=0.6mSubstitute the known variables into equation b Substitute the known variables equation a 9.816.66=the velocity is V =3.13m/s the flow rate of water is 2） the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equa
30、tion between the stations 11 and 3-3，then （c）since input the above data into equation c， 9.81the velocity is: V=3.51 m/sWrite mechanical energy balance equation between thestations 11 and 22, for the same situation of water level （d）since input the above data into equation d， 9.816.66=the pressure i
31、s: 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 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
