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1、附录A2对转向装置液压伺服系统的动态仿真摘要基于转向装置液压伺服系统的动态模型可通过力的曲线图来建立，然后在计算机上对其仿真。通过对已建立的模型在不同情况下的比较，这种比较是通过选择不同的参数并通过仿真实现的。在纸上的结果对于系统的设计者和分析者来说是有价值的。关键词: 组件的选择和匹配;力学图形；AEG转向装置的液压伺服系统。1.介绍一个液伺服系统被广泛地应用于板船上的转向装置。对于一个液压系统的设计者来说仅仅知道他提出的系统能够驱动负载从一种状态到达另一种状态或系统是可靠的是不的。他也应该知道怎样选择元件并正确的相互匹配以减少成本并节约能耗。这篇文章将通过仿真详细的介绍液压伺服系统。系统描
2、述系统可由图一描述。这个系统被应用于“YU CAI”号船上的转向装置，这只船1970年在德国建造。油的供应以及通过控制阀的流动和液压缸由图2所示。Fig. 1 系统 Fig. 2 阀及缸中的油液流动2.动态建模(1) 力学图形基于力流模型的概念，并根据键合图形结构的大体脉络，系统的键合图模型结构如图三所示。在这种结构中，阀的泄漏(从来源埠到尽头)和液压缸泄漏被忽略。 引动器磨擦片和活塞(包括活塞杆)的惯性Ia已经与负载(引动器负荷强行地被加倍)在一起计算。那线路和泵的容量已经被增加到过滤器器,被表示成 Cp.Fig. 3 键合图3.仿真为了要解决模型, 一个程序用ACSL进行。那第四个命令 R
3、unge-Kutta积分法运算法则已经在程序中被采用。基于初次的数值和系数以及输入Xv(t)是指定的, 仿真已经在计算机上被执行。图4表示输入Xv(t). 图5表演负载位移Xm(t)图 4 管阀的输入位移 图 5 负载位移Xm(t) 4.比较并讨论一经仿真进行,成象是一件十分简单的事情。正如像图5那样，调查在特性参数改变情况下系统的灵敏度。系统改变的动态变化是可以得到的。二个方法能被采用到变更一个或者较多的参数而且显示出所有变数的时间轨迹。一个将使用外部的运行时间 ACSL 的指令;另外的将在主要的样板程序中采用一个回路。 采用上述提到的方法,通过运行时间ACSL指令改变Vp和Kv，在图6到图
4、9中二个情况被显示。图6和图7表明如果泵和控制阀做一个完全的匹配,也就是说Vp 和Kv恰当地被描述(泵的流量略多于控制阀的额定流量),然后大部份的液体将会被抽到活塞工作;然而,如果泵和控制阀不匹配，也就是说泵的流量比控制阀允许通过的流量大，尽管控制阀保持开着,大部分液体将不得不通过卸荷阀被抽到油缸并且被消耗的能量没有做任何的有用工作全部转化成了热量。 这情况在图8和图9中显示出来。 Fig. 6 进入控制阀供给液压缸的流量 Fig. 7 经过卸荷阀到油缸的流量 Fig.8 进入控制阀供给液压缸的流量 Fig.9 经过卸荷阀到油缸的流量 5.结论系统的主要元件应该慎重地选择尤其选择泵和控制阀。使
5、用图表中的过程模型通过描绘仿真结果能解决问题。图表中列出的模型和方法也能够被应用于研究转向装置的动态伺服系统，可以是系统的设计也可以是动态的分析。附录B1New Concept for Hydrostatic Drive with Control of the Secondary Unit Connected to the Ring Main System.Introduction:This thesis proposes a new concept for Control of the Secondary Units Connected to the Hydraulic Ring Main
6、System. In this new concept, we study the ability to connect more than one variable displacement hydraulic motor to the hydraulic ring main system, and all motors can work at variable speeds, and drive different loads. The speed for any hydraulic motor can be changed individually at any time through
7、 the operation to any speed required, without any effect on the other. These hydraulic motors are controlled by hydraulic means for cost consideration, and energy saving.A major part of this work is directed towards investigation of the ability of this concept to meet the requirements of the drive c
8、oncept, the stability of the drives, and their ability to drive the loads maintaining the specific speed.Background information:Hydrostatic Drive is a fluid power technology, which has been used as means of transmitting power. The purpose of hydrostatic transmission is to convert mechanical power in
9、to hydraulic power, and convert it back into mechanical output power in a form, which matches the speed and torque demands of driven mechanisms or machines.In drive technology, two parameters are important to the power being transmitted: 1) Torque = M (Nm)2) Speed = n (RPM)These mechanical parameter
10、s correspond respectively to the following parameters in hydrostatic drives: 1) Pressure = P (bar)2) Flow rate = Q (m3/sec)In normal power sources e.g. a combustion engine or an electrical motor, the relation between the mechanical parameters is:P = M w (1.1)Where: P = Power (KW)M = Torque (Nm)w = A
11、ngular velocity (rad/sec)For the same power source and within efficiency considerations the power transmitted is constant and equal to the maximum power that can be produce by the power source, which is constant.Then equation (1.1) becomes: Pmax = M w = K (1.2)Where: K = constantM = K/wMa 1/w The re
12、lation between torque and speed is vice versa, that is, for any torque there is only one corresponding speed. In other words, in the case of variable speed, by increasing the speed from minimum to maximum speed, the torque will reduce from maximum to minimum torque for the same power being transmitt
13、ed,which is the maximum power produced by the power source, after efficiency considerations.In hydrostatic drives, the story is different.The equation (1.1) becomes: dP = Q p (1.3)Where: dp = pressure difference across the motorThe hydrostatic transmission is capable of maintaining a preset highpres
14、sure level in the hydraulic circuit,while changing the flow rate, by using a pressure regulated pump and a variable displacement motor.Then equation (1.3) becomes: P = Q dp = Q KWhere: K = constantP ; Q (1.4)From the equation (1.4), the speed in terms of flow rate (Q) has proportional relation with the power input to the hydrostatic transmission. In other words, in the case of variable speed, by increasing the speed in terms of flow rate (Q), from minimum to maximum speed, the power input to the hydrostatic transmission will increase
