2010-液壓傳動(英文)試題(答案)

精品文檔四 川 大 學 期 末 考 試 試 題（2010 2011 學年第 一 學期）課程號：302155020課序號： 課程名稱：液壓傳動 任課教師：熊瑞平、傅波、劉蘇 成績：適用專業(yè)年級： 08級機制專業(yè) 學生人數(shù)： 印題份數(shù)： 學號： 姓名：考 試 須 知四川大學學生參加由學校組織或由學校承辦的各級各類考試，必須嚴格執(zhí)行四川大學考試工作管理辦法和四川大學考場規(guī)則。有考試違紀作弊行為的，一律按照四川大學學生考試違紀作弊處罰條例進行處理。四川大學各級各類考試的監(jiān)考人員，必須嚴格執(zhí)行四川大學考試工作管理辦法、四川大學考場規(guī)則和四川大學監(jiān)考人員職責。有違反學校有關規(guī)定的，嚴格按照四川大學教學事故認定及處理辦法進行處理。Test of Fluid Power and Applications (A)No.123456TotalScore1. Fill out the blanks.( 26%)(1). Liquids are considered to be 不可壓縮的 so that their volume does not change with pressure changes. On the other hand , gases are fluids which are 可壓縮的 so that their volume will vary to fill the vessel containing them.(2%)(2) Pascals law states that 作用在密閉流體里的壓力大小相同朝各個方向傳遞 . (1%)(3) The conservation of energy las states that 能量不會無故消失，也不會無故產(chǎn)生 .The total energy of fluid includes potential energy due to 壓力 and 高度 and also kinetic energy due to 速度 . (4%)(4) The kinematic viscosity of a hydraulic oil is 100 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 層流 . (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%)(6) The purpose of a 平衡閥 valve in hydraulic circuit is usually to maintain control of a vertical cylinder to prevent is from descending due to gravity. (1%) 注：1試題字跡務必清晰,書寫工整。 本題 4 頁，本頁為第 1 頁2題間不留空，一般應題卷分開。 教務處試題編號：3務必用A4紙打印。學號： 姓名：(7) Hydraulic motors are pushed upon by 壓力液體 instead of pushing on the fluid as pumps. Hydraulic motors develop 轉速 and produce 扭矩（或連續(xù)運動） . ( 3%)(8) Pressure-reducing valve (which is normally open) is used to maintain 減壓 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) Flow control valves are used to 調節(jié)速度 of hydraulic cylinders and motors controlling the flow rate to these actuators. There are two basic types of flow control valves: 節(jié)流閥（無壓力補償?shù)牧髁块y） and調速閥（有壓力補償?shù)牧髁块y） . ( 3%)(10) There are two basic types of flow in pipes, depending on the nature of the different factors which affect the flow. The first type is called 層流 , the second type is called 紊流 . (2%)(11) If the following data are obtained from the nameplate（銘牌）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 of 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. Answer 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. 流動阻力大；2. 壓力損失大 （2.5分）粘度太低：1. 泄漏增大；2. 磨損加劇 （2.5分） What is the difference between gage and absolute pressure?(5%) 表壓：大氣壓測出的壓力 （2分） 絕對壓力：相對于絕對真空測出的壓力 （2分） 絕對壓力 = 表壓 + 大氣壓 （1分）Fig. 1 Explain how the pressure-compensated flow control valve shown in Fig.1 works.(8%)1） 進口壓力P1經(jīng)過定差減壓閥后，壓力降為P2；2） 壓力P2作用在定差減壓閥閥芯的下端面；3）經(jīng)過減壓閥的液壓油流經(jīng)節(jié)流閥后接負載，壓力降為P3；4）壓力P3作用在定差減壓閥閥芯的上端面（彈簧腔）；5）閥芯力平衡方程：（）A =， 式中A為閥芯面積，F(xiàn)s為彈簧力；6）節(jié)流閥進出口壓力差：P= P2- P3= Fs / A7）閥口開口遠小于彈簧的預壓縮量，因此Fs近似常數(shù)8）節(jié)流閥進出口壓力差P近似為常數(shù)，根據(jù)流量公式可知，調速閥的流量壓力變化影響（每條一分） 本題 4 頁，本頁為第 2 頁 教務處試題編號：學號： 姓名： Explain how the variable displacement pressure-compensated vane pump shown in Fig.2 works.(8%)Fig. 21) 組成：定子，轉子，流量螺釘，壓力螺釘（壓力補償器），泵殼等組成；（1分）2) 分析簡圖（2分）3) 彈簧預緊力：Fs = k x0（1分）4) 出口壓力對定子產(chǎn)生的水平分力：Fx = p A x（1分）5) 當Fx Fs，彈簧壓縮，偏心減小，流量減小。（1分）3A 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 the 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 cylinder 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%) 本題 4 頁，本頁為第 3 頁 教務處試題編號：學號： 姓名： 1 液壓泵，將機械能轉換為液壓能2 溢流閥，溢流3 二位四通電磁換向閥，方向控制4 單向順序閥（或平衡閥）， 順序控制8二位三通電磁換向閥，快慢速轉換9 節(jié)流閥， 速度調節(jié)（每個名稱及作用各一分）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 system 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 inlet 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間寫伯努力方程，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 /(0.0018