2010液压传动英文试题A答案

1、四 川 大 学 期 末 考 试 试 题（2010 2011 学年第 一 学期）课程号：302155020课序号： 课程名称：液压传动 任课教师：熊瑞平、傅波、刘苏 成绩：适用专业年级： 08级机制专业 学生人数： 印题份数： 学号： 姓名：考 试 须 知四川大学学生参加由学校组织或由学校承办的各级各类考试，必须严格执行四川大学考试工作管理办法和四川大学考场规则。有考试违纪作弊行为的，一律按照四川大学学生考试违纪作弊处罚条例进行处理。四川大学各级各类考试的监考人员，必须严格执行四川大学考试工作管理办法、四川大学考场规则和四川大学监考人员职责。有违反学校有关规定的，严格按照四川大学教学事故认定及处

2、理办法进行处理。Test of Fluid Power and Applications (A)No.123456TotalScore1. Fill out the blanks.( 26%)(1). Liquids are considered to be 不可压缩的 so that their volume does not change with pressure changes. On the other hand , gases are fluids which are 可压缩的 so that their volume will vary to fill the vessel co

3、ntaining them.(2%)(2) Pascals law states that 作用在密闭流体里的压力大小相同朝各个方向传递 . (1%)(3) The conservation of energy las states that 能量不会无故消失，也不会无故产生 .The total energy of fluid includes potential energy due to 压力 and 高度 and also kinetic energy due to 速度 . (4%)(4) The kinematic viscosity of a hydraulic oil is 1

4、00 cS. If the fluid is flowing in a 3 mm diameter pipes at a velocity of 10 mm/s , the Reynolds number is 0.3 . So the type of the fluid flow in pipe is 层流 . (2%)(5) The continuity equation states that for steady flow in a pipeline the weight flow rate is 相同的 for all cross sections of the pipe. (1%)

5、(6) The purpose of a 平衡阀 valve in hydraulic circuit is usually to maintain control of a vertical cylinder to prevent is from descending due to gravity. (1%) 注：1试题字迹务必清晰,书写工整。 本题 4 页，本页为第 1 页2题间不留空，一般应题卷分开。 教务处试题编号：3务必用A4纸打印。学号： 姓名：(7) Hydraulic motors are pushed upon by 压力液体 instead of pushing on th

6、e fluid as pumps. Hydraulic motors develop 转速 and produce 扭矩（或连续运动） . ( 3%)(8) Pressure-reducing valve (which is normally open) is used to maintain 减压 in specified locations of hydraulic systems. It is actuated by 出口（或下游） pressure and tends to close as this pressure reaches the valve setting.(2%)(9)

7、 Flow control valves are used to 调节速度 of hydraulic cylinders and motors controlling the flow rate to these actuators. There are two basic types of flow control valves: 节流阀（无压力补偿的流量阀） and调速阀（有压力补偿的流量阀） . ( 3%)(10) There are two basic types of flow in pipes, depending on the nature of the different fa

8、ctors which affect the flow. The first type is called 层流 , the second type is called 紊流 . (2%)(11) If the following data are obtained from the nameplate（铭牌）of a pump: pump speed n = 1450 rpm, rated flow rate Qn = 30 l/min, rated pressure Pn = 210 bar, the overall efficiency o = 0.8, then the power o

9、f the prime mover should be 13.125 KW. If the pump delivers 30 l/min at 40 bar and 1450 rpm in a hydraulic system, then the prime mover input power is 2.5 KW. If the outlet of a pump is connected directly to the tank (assuming no friction loss), then the pressure in pump outlet is 0 MPa. (5%)2. Answ

10、er the following questions. (26%) Name two undesirable results when using an oil with a viscosity that is too high (or too low).(5%)粘度太高：1. 流动阻力大；2. 压力损失大 （2.5分）粘度太低：1. 泄漏增大；2. 磨损加剧 （2.5分） What is the difference between gage and absolute pressure?(5%) 表压：大气压测出的压力 （2分） 绝对压力：相对于绝对真空测出的压力 （2分） 绝对压力 = 表

11、压 + 大气压 （1分）Fig. 1 Explain how the pressure-compensated flow control valve shown in Fig.1 works.(8%)1） 进口压力P1经过定差减压阀后，压力降为P2；2） 压力P2作用在定差减压阀阀芯的下端面；3）经过减压阀的液压油流经节流阀后接负载，压力降为P3；4）压力P3作用在定差减压阀阀芯的上端面（弹簧腔）；5）阀芯力平衡方程：（）A =， 式中A为阀芯面积，Fs为弹簧力；6）节流阀进出口压力差：P= P2- P3= Fs / A7）阀口开口远小于弹簧的预压缩量，因此Fs近似常数8）节流阀进出口压力差P

12、近似为常数，根据流量公式可知，调速阀的流量压力变化影响（每条一分） 本题 4 页，本页为第 2 页 教务处试题编号：学号： 姓名： Explain how the variable displacement pressure-compensated vane pump shown in Fig.2 works.(8%)Fig. 21) 组成：定子，转子，流量螺钉，压力螺钉（压力补偿器），泵壳等组成；（1分）2) 分析简图（2分）3) 弹簧预紧力：Fs = k x0（1分）4) 出口压力对定子产生的水平分力：Fx = p A x（1分）5) 当Fx Fs，弹簧压缩，偏心减小，流量减小。（1分）3

13、A hydraulic system as shown in fig. 3. The max flow rate produced by pump is 30 L/min. The motor has a dispalcement volume of 25cm3/r. The pressure relief valve setting is 7 Mpa. Find (assuming no power loss): ( 15%)1) The max output power, the max speed and the max torque of the motor?Fig. 32) If t

14、he output power of the motor is 2 KW, then how much is the max speed and the max torque of the motor.1) 2) 4. For the hydraulic system of Figure 4, the sequence of operations of the system is : rapid advance of cylinder I; slow feed of cylinder I; rapidly retract for cylinder I; slow feed of cylinde

15、r II; rapidly retract for cylinder II; Name the components marked with 1, 2, 3, 4, 8, 9 respectively. What function do the components act as in the hydraulic system, respectively? (12%) 本题 4 页，本页为第 3 页 教务处试题编号：学号： 姓名： 1 液压泵，将机械能转换为液压能2 溢流阀，溢流3 二位四通电磁换向阀，方向控制4 单向顺序阀（或平衡阀）， 顺序控制8二位三通电磁换向阀，快慢速转换9 节流阀，

16、速度调节（每个名称及作用各一分）Fig. 45. The following data are given for a pump: pr=200 bar, Q r = 20 l/min, v =0.95, find the theoretical flow rate Q t and the leakage flow rate Q l of the pump?(7%)Q t = Q r / v = 20 / 0.95 = 21.05 l/min （3.5分） Q l = Q t Q r = 21.05 20 = 1.05 l/min （3.5分）6. For the hydraulic syst

17、em of Fig.5, the following data are given:1)The pump is adding 3730W to the fluid. 2)Pump flow is 0.001896 m3/s. 3)The pipe has a 0.0254 m inside diameter. 4) The specific gravity of the oil is 0.9. 5) The elevation difference between stations 1 and 2 is 6.096m. Find the pressure available at the in

18、let to the hydraulic motor (station 2). The pressure at station 1 in hydraulic tank is atmospheric (0 Pa or 0 N/m2) . The head load HL due to friction between stations 1 and 2 is 9.144 m of oil. (14%)解：在位置1和2间写伯努力方程，得 Z1+P1/+v12/(2g)+HpHmHL= Z2 +P2/ + v2 2/(2g) （2分） 而H m =0, v1 =0. Z2 - Z1 = 6.096 m，HL =9.144 m， P1=0 得： Z1 + 0 + 0 + H p 0 9.144 = Z2 +P2/ + v2 2/(2g) （2分）解 P2/ , 得： P2/ = (Z1- Z2) + H p - v2 2/(2g) -9.144= H p - v2 2/(2g) - 15.24 （1分）而：H p = P (W)/Q(m3/s) Sg 9800 = 3730 /(0.001