液压泵站机械设计
喜欢这套资料就充值下载吧。资源目录里展示的都可在线预览哦。下载后都有，请放心下载，文件全都包含在内，图纸为CAD格式可编辑，有疑问咨询QQ:414951605 或 1304139763p
液压传动系统设计与计算
1 明确设计要求进行工况分析
在设计液压系统时，首先应明确以下问题，并将其作为设计依据。
主机的用途、工艺过程、总体布局以及对液压传动装置的位置和空间尺寸的要求；
主机对液压系统的性能要求，如自动化程度、调速范围、运动平稳性、换向定位精度以及对系统的效率、温升等的要求；液压系统的工作环境，如温度、湿度、振动冲击以及是否有腐蚀性和易燃物质存在等情况。
在上述工作的基础上，应对主机进行工况分析，工况分析包括运动分析和动力分析，对复杂的系统还需编制负载和动作循环图，由此了解液压缸或液压马达的负载和速度随时间变化的规律，以下对工况分析的内容作具体介绍。
1.1 运动分析
主机的执行元件按工艺要求的运动情况，可以用位移循环图(L—t)，速度循环图(v—t)，或速度与位移循环图表示，由此对运动规律进行分析。
1.1.1 位移循环图L—t
图1.1为液压机的液压缸位移循环图，纵坐标L表示活塞位移，横坐标t表示从活塞启动到返回原位的时间，曲线斜率表示活塞移动速度。
图1.1 位移循环图
1.1.2 速度循环图v—t(或v—L)
工程中液压缸的运动特点可归纳为三种类型。图1.2为三种类型液压缸的v—t图，第一种如图1.2中实线所示，液压缸开始作匀加速运动，然后匀速运动，
图1.2 速度循环图
最后匀减速运动到终点；第二种，液压缸在总行程的前一半作匀加速运动，在另一半作匀减速运动，且加速度的数值相等；第三种，液压缸在总行程的一大半以上以较小的加速度作匀加速运动，然后匀减速至行程终点。v—t图的三条速度曲线，不仅清楚地表明了三种类型液压缸的运动规律，也间接地表明了三种工况的动力特性。
1.2 动力分析
动力分析，是研究机器在工作过程中，其执行机构的受力情况，对液压系统而言，就是研究液压缸或液压马达的负载情况。
1.2.1 液压缸的负载及负载循环图
1.2.1.1 液压缸的负载力计算
工作机构作直线往复运动时，液压缸必须克服的负载由六部分组成：
(1.1)
式中：Fc为切削阻力；Ff为摩擦阻力；Fi为惯性阻力；Fg为重力；Fm为密封阻力；Fb为排油阻力。
1.2.1.2液压缸运动循环各阶段的总负载力
液压缸运动循环各阶段的总负载力计算，一般包括启动加速、快进、工进、快退、减速制动等几个阶段，每个阶段的总负载力是有区别的。
(1)启动加速阶段：这时液压缸或活塞处于由静止到启动并加速到一定速度，其总负载力包括导轨的摩擦力、密封装置的摩擦力(按缸的机械效率=0.9计算)、重力和惯性力等项，即：
(1.2)
(2)快速阶段：
(1.3)
(3)工进阶段：
(1.4)
(4)减速：
(1.5)
对简单液压系统，上述计算过程可简化。例如采用单定量泵供油，只需计算工进阶段的总负载力，若简单系统采用限压式变量泵或双联泵供油，则只需计算快速阶段和工进阶段的总负载力。
1.2.2 液压马达的负载
工作机构作旋转运动时，液压马达必须克服的外负载为：
(1.6)
1.2.2.1 工作负载力矩Me。工作负载力矩可能是定值，也可能随时间变化，应根据机器工作条件进行具体分析。
1.2.2.2 摩擦力矩Mf。为旋转部件轴颈处的摩擦力矩，其计算公式为：
(1.7)
式中：G为旋转部件的重量(N)；f为摩擦因数，启动时为静摩擦因数，启动后为动摩擦因数；R为轴颈半径(m)。
1.2.2.3 惯性力矩Mi。为旋转部件加速或减速时产生的惯性力矩，其计算公式为：
(1.8)
式中：ε为角加速度(r/s2)；Δω为角速度的变化(r/s)；Δt为加速或减速时间(s)；J为旋转部件的转动惯量()，。
式中：为回转部件的飞轮效应()。
各种回转体的可查《机械设计手册》。
根据式(1.6)，分别算出液压马达在一个工作循环内各阶段的负载大小，便可绘制液压马达的负载循环图
2 确定液压系统主要参数
2.1 液压缸的设计计算
2.1.1 初定液压缸工作压力
液压缸工作压力主要根据运动循环各阶段中的最大总负载力来确定，此外，还需要考虑以下因素：
2.1.1.1 各类设备的不同特点和使用场合。
2.1.1.2 考虑经济和重量因素，压力选得低，则元件尺寸大，重量重；压力选得高一些，则元件尺寸小，重量轻，但对元件的制造精度，密封性能要求高。
所以，液压缸的工作压力的选择有两种方式：一是根据机械类型选；二是根据切削负载选。
如表2.1、表2.2所示。
表2.1 按负载选执行文件的工作压力
负载/N
＜5000
500～10000
10000～20000
20000～30000
30000～50000
＞50000
工作压力/MPa
≤0.8～1
1.5～2
2.5～3
3～4
4～5
＞5
表2.2 按机械类型选执行文件的工作压力
机械类型
机 床
农业机械
工程机械
磨床
组合机床
龙门刨床
拉床
工作压力/MPa
a≤2
3～5
≤8
8～10
10～16
20～32
2.2 液压马达的设计计算
2.2.1 计算液压马达排量
液压马达排量根据下式决定：
(2.1)
式中：T为液压马达的负载力矩(N·m)；为液压马达进出口压力差()；为液压马达的机械效率，一般齿轮和柱塞马达取0.9～0.95，叶片马达取0.8～0.9。
2.2.2 计算液压马达所需流量液压马达的最大流量
(2.2)
式中：Vm为液压马达排量(m3/r)；nmax为液压马达的最高转速(r/s)。
3 液压元件的选择
3.1 液压泵的确定与所需功率的计算
3.1.1 液压泵的确定
3.1.1.1 确定液压泵的最大工作压力。液压泵所需工作压力的确定，主要根据液压缸在工作循环各阶段所需最大压力p1，再加上油泵的出油口到缸进油口处总的压力损失ΣΔp，即
(3.1)
包括油液流经流量阀和其他元件的局部压力损失、管路沿程损失等，在系统管路未设计之前，可根据同类系统经验估计，一般管路简单的节流阀调速系统为(2～5)×105Pa，用调速阀及管路复杂的系统为(5～15)×105Pa，也可只考虑流经各控制阀的压力损失，而将管路系统的沿程损失忽略不计，各阀的额定压力损失可从液压元件手册或产品样本中查找，也可参照表1.3选取。
表3.1 常用中、低压各类阀的压力损失(Δpn)
阀名
Δpn(×105Pa)
阀名
Δpn (×105Pa)
阀名
Δpn (×105Pa)
阀名
Δpn (×105Pa)
单向阀
0.3～0.5
背压阀
3～8
行程阀
1.5～2
转阀
1.5～2
换向阀
1.5～3
节流阀
2～3
顺序阀
1.5～3
调速阀
3～5
3.1.2 确定液压泵的流量qB
泵的流量qB根据执行元件动作循环所需最大流量qmax和系统的泄漏确定。
3.1.2.1多液压缸同时动作时，液压泵的流量要大于同时动作的几个液压缸(或马达)所需的最大流量，并应考虑系统的泄漏和液压泵磨损后容积效率的下降，即
(3.2)
式中：K为系统泄漏系数，一般取1.1～1.3，大流量取小值，小流量取大值；为同时动作的液压缸(或马达)的最大总流量(m3/s)。
3.1.2.2选择液压泵的规格：根据上面所计算的最大压力pB和流量qB，查液压元件产品样本，选择与pB和qB相当的液压泵的规格型号。
表3.2 液压泵的总效率
液压泵类型
齿轮泵
螺杆泵
叶片泵
柱塞泵
总效率
0.6～0.7
0.65～0.80
0.60～0.75
0.80～0.85
按上述功率和泵的转速，可以从产品样本中选取标准电动机，再进行验算，使电动机发出最大功率时，其超载量在允许范围内。
3.2 阀类元件的选择
3.2.1 选择依据
选择依据为：额定压力，最大流量，动作方式，安装固定方式，压力损失数值，工作性能参数和工作寿命等。
3.2.2 选择阀类元件应注意的问题
3.2.2.1 应尽量选用标准定型产品，除非不得已时才自行设计专用件。
3.2.2.2 阀类元件的规格主要根据流经该阀油液的最大压力和最大流量选取。选择溢流阀时，应按液压泵的最大流量选取；选择节流阀和调速阀时，应考虑其最小稳定流量满足机器低速性能的要求。
3.3 蓄能器的选择
3.3.1 蓄能器用于补充液压泵供油不足时，其有效容积为：
(3.3)
式中：A为液压缸有效面积(m2)；L为液压缸行程(m)；K为液压缸损失系数，估算时可取Ｋ＝1.2；qB为液压泵供油流量(m3/s)；t为动作时间(s)。
3.3.2 蓄能器作应急能源时，其有效容积为：
(3.4)
当蓄能器用于吸收脉动缓和液压冲击时，应将其作为系统中的一个环节与其关联部分一起综合考虑其有效容积。
根据求出的有效容积并考虑其他要求，即可选择蓄能器的形式。
3.4 管道的选择
3.4.1 油管类型的选择
液压系统中使用的油管分硬管和软管，选择的油管应有足够的通流截面和承压能力，同时，应尽量缩短管路，避免急转弯和截面突变。
3.4.1.1 钢管：中高压系统选用无缝钢管，低压系统选用焊接钢管，钢管价格低，性能好，使用广泛。
3.4.1.2 铜管：紫铜管工作压力在6.5～10MPa以下，易变曲，便于装配；黄铜管承受压力较高，达25MPa，不如紫铜管易弯曲。铜管价格高，抗震能力弱，易使油液氧化，应尽量少用，只用于液压装置配接不方便的部位。
3.4.2 油管尺寸的确定
3.4.2.1 油管内径d按下式计算：
d= (3.5)
式中：q为通过油管的最大流量(m3/s)；v为管道内允许的流速(m/s)。一般吸油管取0.5～5(m/s)；压力油管取2.5～5(m/s)；回油管取1.5～2(m/s)。
3.4.2.2 油管壁厚δ按下式计算：
(3.6)
式中：p为管内最大工作压力；n为安全系数，钢管p＜7MPa时，取n=8；p＜17.5MPa时，取n=6；p＞17.5MPa时，取n=4。
根据计算出的油管内径和壁厚，查手册选取标准规格油管。
3.5 油箱的设计
油箱的作用是储油，散发油的热量，沉淀油中杂质，逸出油中的气体。
3.5.1 油箱设计要点
3.5.1.1 油箱应有足够的容积以满足散热，同时其容积应保证系统中油液全部流回油箱时不渗出，油液液面不应超过油箱高度的80%。
3.5.1.2 吸箱管和回油管的间距应尽量大。
3.5.1.3 油箱底部应有适当斜度，泄油口置于最低处，以便排油。
3.6 滤油器的选择
选择滤油器的依据有以下几点：
3.6.1 承载能力：
按系统管路工作压力确定。
3.6.2 过滤精度：
按被保护元件的精度要求确定。
3.6.3 通流能力：
按通过最大流量确定。
3.6.4 阻力压降：
应满足过滤材料强度与系数要求。
4 液压系统性能的验算
为了判断液压系统的设计质量，需要对系统的压力损失、发热温升、效率和系统的动态特性等进行验算。
4.1 管路系统压力损失的验算
当液压元件规格型号和管道尺寸确定之后，就可以较准确的计算系统的压力损失，压力损失包括：油液流经管道的沿程压力损失、局部压力损失和流经阀类元件的压力损失，即：
(4.1)
系统的调整压力：
(4.2)
式中：P0为液压泵的工作压力或支路的调整压力；P1为执行件的工作压力。
如果计算出来的比在初选系统工作压力时粗略选定的压力损失大得多，应该重新调
整有关元件、辅件的规格，重新确定管道尺寸。
4.2 系统发热温升的验算
系统发热来源于系统内部的能量损失，如液压泵和执行元件的功率损失、溢流阀的溢流损失、液压阀及管道的压力损失等。
系统发热功率P的计算:
(4.3)
式中：PB为液压泵的输入功率(W)；η为液压泵的总效率。
若一个工作循环中有几个工序，则可根据各个工序的发热量，求出系统单位时间的平均发热量：
(4.4)
式中：T为工作循环周期(s)；ti为第i个工序的工作时间(s)；pi为循环中第i个工序的输入功率(W)。
4.3 系统效率验算
液压系统的效率是由液压泵、执行元件和液压回路效率来确定的。
液压回路效率nc一般可用下式计算：
(4.5)
式中：p1，q1；p2，q2；……为每个执行元件的工作压力和流量；pB1，qB1；pB2，qB2为每个液压泵的供油压力和流量。
液压系统总效率：
(4.6)
式中：为液压泵总效率；为执行元件总效率；为回路效率。
5 绘制正式工作图和编写技术文件
经过对液压系统性能的验算和必要的修改之后，便可绘制正式工作图，它包括绘制液压系统原理图、系统管路装配图和各种非标准元件设计图。
正式液压系统原理图上要标明各液压元件的型号规格。对于自动化程度较高的机床，还应包括运动部件的运动循环图和电磁铁、压力继电器的工作状态。
5.1 确定液压系统参数
由工况分析中可知，工进阶段的负载力最大，所以，液压缸的工作压力按此负载力计算，根据液压缸与负载的关系，选p1=40×105Pa。本机床为钻孔组合机床，为防止钻通时发生前冲现象，液压缸回油腔应有背压，设背压p2=6×105Pa，为使快进快退速度相等，选用差动油缸，假定快进、快退的回油压力损失为Δp=7×105Pa。
5.2 选择液压元件
5.2.1 选择液压泵和电动机
5.2.1.1 确定液压泵的工作压力。
前面已确定液压缸的最大工作压力为40×105Pa，选取进油管路压力损失Δp=8×105Pa，其调整压力一般比系统最大工作压力大5×105Pa，所以泵的工作压力PB＝(40＋8＋5)×105＝53×105Pa
这是高压小流量泵的工作压力。
液压缸快退时的工作压力比快进时大，取其压力损失Δp′＝4×105Pa，则快退时泵的工作压力为：
PB=(16.4＋4)×105＝20.4×105Pa
这是低压大流量泵的工作压力。
5.2.1.2 液压泵的流量。快进时的流量最大，其值为30L/min，最小流量在工进时，其值为0.51L/min，取K＝1.2，
则： qB＝1.2×0.5×10-3=36L/min
由于溢流阀稳定工作时的最小溢流量为3L/min，故小泵流量取3.6L/min。
根据以上计算，选用YYB-AA36/6B型双联叶片泵。
5.2.1.3 确定管道尺寸:根据工作压力和流量，按式(3.5)、式(3.6)确定管道内径和壁厚。(从略)
5.2.1.4 确定油箱容量油箱容量可按经验公式估算，取V＝(5～7)q。
本例中：V＝6q＝6(6＋36)＝252L有关系统的性能验算从略。
1Hydraulic actuation system design and computation1clear about the design request to carry on the operating mode analysis.When design hydraulic system below, first should be clear about thequestion, and takes it as the design basis.Main engine use, technological process, overall layout as well as tohydraulic gear position and spatial size request; The main engine to thehydraulic system performance requirement, like the automaticity, the velocitymodulation scope, the movement stability, the commutation pointingaccuracy as well as the request which to the system efficiency, warmpromotes; Hydraulic system working conditions, like temperature, humidity,vibration impact as well as whether has situation and so on corrosiveness andheat-sensitive material existence.In in the above work foundation, should carry on the operating modeanalysis to the main engine, the operating mode analysis including themovement analysis and the mechanical analysis, also must establish the loadand the operating cycle chart to the complex system, from this understood thehydraulic cylinder or the oil motor load and the speed change as necessary therule, below makes the concrete introduction to the operating mode analysiscontent1.1 movements analysesThe main engine functional element according to the technologicalrequirement movement situation, may use the displacement circulation chart(Lt), the speed circulation chart (vt), or the speed and the displacementcirculation chart indicated, from this carries on the analysis to the movementrule.1.1.1 displacements circulation attempts LtThe chart 1.1 is the hydraulic press hydraulic cylinder moves thecirculation chart, the y-coordinate L expression piston moves, the2x-coordinate t expression starts from the piston to the reposition time, the rateof curve expression movement of plunger speed.Chart 1.1 displacements circulation chart1.1.2 speeds circulation chart vt (or vL)In the project the hydraulic cylinder movement characteristic may induceis three kind of types. The chart 1.2 is three kind of types hydraulic cylindersv t chart, the first kind of like chart 1.2 center solid lines show, thehydraulic cylinder starts to make the uniform accelerated motion, thenuniform motion,Chart 1.2 speeds circulation chartFinally uniform retarded motion to end point; The second kind, thehydraulic cylinder preceding partly makes the uniform accelerated motion inthe overall travelling schedule, in another one partly makes the uniformretarded motion, also the acceleration value is equal; The third kind, thehydraulic cylinder one most above makes the uniform accelerated motion inthe overall travelling schedule by a smaller acceleration, then uniformdecelerates to the travelling schedule end point. Vt chart three velocitycurve, not only clearly has indicated three kind of types hydraulic cylindersmovement rule, also indirectly has indicated three kind of operating modesdynamic performance.31.2 mechanical analyses1.2.1 hydraulic cylinders loads and duty cycle chart1.2.1.1 hydraulic cylinders load strength computationsWhen the operating mechanism makes the straight reciprocating motion,the hydraulic cylinder must overcome the load is composed by six partsbmgifcFFFFFFF+=(1.1)In the formula: FcIn order to resistance to cutting; FfIn order to frictiondrag; FiFor inertia resistance; FgFor gravity; FmIn order to seal the resistance;FbIn order to drain the oil the resistance.1.2.1.2 hydraulic cylinders cycle of motion various stages overall loadstrengthThe hydraulic cylinder cycle of motion various stages overall loadstrength computation, generally includes the start acceleration, quickly enters,the labor enters, quickly draws back, decelerates applies the brake and so onseveral stages, each stage overall load strength has the difference.(1) starts the acceleration period: By now the hydraulic cylinder or the pistonwere in from static enough to starts and accelerates to the certain speed, itsoverall load strength including guide rail friction force, packing assemblyfriction force (according to cylinder mechanical efficiency m=0.9computation), gravity and so on item, namely:bmgifFFFFFF+=(1.2)(2) fast stage:bmgfFFFFF+=(1.3)(3) the labor enters the stage:bmgfcFFFFFF+=（1.4）(4) decelerates:bmgifFFFFFF+=(1.5)To the simple hydraulic system, the above computation process maysimplify. For example uses the single proportioning pump to supply the oil,4only must calculate the labor to enter the stage the overall load strength, if thesimple system uses the limiting pressure type variable displacement pump ora pair of association pumps for the oil, then only must calculate the fast stageand the labor enters the stage the overall load strength.1.2.2 oil motors loadWhen the operating mechanism makes the rotary motion, the oil motormust overcome the outside load is:ifeMMMM+=(1.6)1.2.2.1 operating duties moment of force Me. The operating duty moment offorce is possibly a definite value, also possibly as necessary changes, shouldcarry on the concrete analysis according to the machine working condition.1.2.2.2 friction moments. In order to revolve the part journal place frictionmoment, its formula is:)(MNGFRMf=(1.7)In the formula: G is revolves the part weight (N); F is the rubbing factor,when the start for the factor, after the start for moves the rubbing factor; R isthe journal radius (m).1.2.2.3 moment of inertiaMi. The moment of inertia which in order to revolvethe part acceleration or decelerates when produces, its formula is:)(MNtJMi=(1.8)In the formula: Is the angle acceleration (r/s2);tis the acceleration ordecelerates the time (s); J is revolves the part rotation inertia(2mKg ),GGDJ412=In the formula:2GDIn order to rotate the part the flywheel effect (2MN ).Each kind may look up According to the type (1.6), separately figures out the oil motor in a operatingcycle various stages load size, then may draw up the oil motor the duty cyclechart2 determinations hydraulic system main parameter52.1 hydraulic cylinders design calculations2.1.1 initially decides the hydraulic cylinder working pressureIn the hydraulic cylinder working pressure main basis cycle of motionvarious stages biggest overall load strength determined, in addition below, butalso needs to consider the factor:2.1.1.1 each kind of equipment different characteristic and use situation.2.1.1.2 considerations economies and the weight factor, the pressure electslowly, then part size big, the weight is heavy; The pressure chooses highsomewhat, then part size small, the weight is light, but to the part manufactureprecision, the sealing property requests high.Therefore, the hydraulic cylinder working pressure choice has two ways:One, elects according to the mechanical type; Two, according to cuts the loadto elect.If the table 2.1, the table 2.2 shows.The table 2.1presses the load to choose the execution file the workingpressureLoad/N50005001000010000200002000030000300005000050000Workingpressure/MPa0.811.522.53344556The table 2.2presses the mechanical type to choose the execution file theworking pressureMechanicaltypeEngine bedFarmmachineryProjectmachineryGrinderAggregatemachine-toolDragonGatedigsthe bedBroachingmachineWorkingpressure/MPaa2358810101620322.2 oil motors design calculation2.2.1 computations oil motor displacementUnder oil motor displacement according to the type decided that,)(28. 63minrmPTVmm=(2.1)In the formula: T is the oil motor load moment of force (N m);PmForoil motor import and export pressure difference (n/m3);is the oil motormechanical efficiency, the common gear and the plunger motor takes 0.9 0.95, the leaf blade motor takes 0.8 0.9.2.2.2 computations oil motor needs the current capacity oil motor themaximum current capacity)(3maxmaxsmnVqm=(2.2)In the formula:Vmis the oil motor displacement (m3/r);nmaxis the oilmotor highest rotational speed (r/s).3 hydraulic pressure parts choice3.1 hydraulic pumps determinations with need the power the computation73.1.1.1 determines the hydraulic pump the biggest working pressure. Thehydraulic pressure pumping station must the working pressure determination,mainly acts according to the hydraulic cylinder in the operating cycle variousstages to have most tremendous pressure p1, in addition the oil pump losesSigma Delta p the oil mouth to the cylinderplace always pressurep,namely+=PPPB1(3.1)Ploses, the pipeline including the oil after the flow valve and otherparts local pressures along the regulation loss and so on, before systempipeline design, may act according to the similar system experience toestimate, common pipeline simple throttle valve velocity modulation systemp is (2 5) 105Pa, with the velocity modulation valve and pipelinecomplex systemPis (5 15) 105Pa,Palso may only considerflows after various control valves pressure loss, but ignores the circuitry alongthe regulation loss, various valves rated pressure loses may searches from thehydraulic pressure part handbook or the product sample, Also may refer to thetable 1.3 selectionsThe table 3.1 is commonly used, the low pressure each kind of valve pressureloses (pn)Valvepn(105Pa)Valvepn(105Pa)Valvepn(105Pa)Valvepn(105Pa)Cone-wayvalve0.30.5Cone-wayvalve38Cone-wayvalve1.52Cone-wayvalve1.52Crossvalve1.53Crossvalve23Crossvalve1.53Crossvalve353.1.2 determines the hydraulic pump current capacityqB8Pumps the current capacityqBbasis functional element operating cyclemust the maximum current capacityqmaxand the system divulges thedetermination3.1.2.1 At the same time when more thanhydraulic cylinders movement,the hydraulic pump current capacity must be bigger than the maximumcurrent capacity which at the same time the movement several hydrauliccylinders (or motor) needs, and should consider the system divulging wearsthe volumetric efficiency drop after the hydraulic pump, namely)()(3maxsmqKqB=(3.2)In the formula: K is the system leakage coefficient, generally takes 1.1 1.3, the great current capacity takes the small value, the small current capacitytakes the great value )max(q; For at the same time movement hydrauliccylinder (or motor) is biggest (m3/s).3.1.2.2 chooses the hydraulic pump the specificationTable 3.2 hydraulic pumps overall effectiveness indicesHydraulicpump typeGear pumpThe screwrod pumpsVane pumpRam pumpOveralleffectivenessindex0.60.70.650.800.600.750.800.85Rotational speed and pumps which according to the above power, mayselect the standard electric motor from the product sample, again carries on,causes when the electric motor sends out the maximum work rate,inpermission scope.3.2 valves class parts choice3.2.1 choices bases9The choice basis is: Rated pressure, maximum current capacity,movement way, installment fixed way, pressure loss value, operatingperformance parameter and working life and so on.3.2.2 selector valves class parts should pay attention question3.2.2.1 should select the standard stereotypia product as far as possible,only if does not have already time only then independently designsspecial-purp3.2.2.2 valves class parts specification main basis class after this valve fatliquor most tremendous pressure and maximum current capacity selection.When chooses the overflow valve, should according to the hydraulicpump maximum current capacity selection; When chooses the throttlevalve and the velocity modulation valve, should consider its minimumstable current capacity satisfies the machine low-speed performance therequest3.3 accumulators choices3.3.1 accumulators use in to supplement when the hydraulic pump suppliesthe oil insufficiency, its dischargeable capacity is)(3mtqKLAVBii=(3.3)In the formula: A is the hydraulic cylinder active surface (m2); L is thehydraulic cylinder travelling schedule (m); K is the hydraulic cylinder losscoefficient, when the estimate may take K = 1.2; Supplies the oil currentcapacity for the hydraulic pump (m3/s); T is the operating time (s).3.3.2 accumulators make the emergency energy, its dischargeable capacity is:)(3mtqLAVBii=(3.4)When the accumulator uses in absorbs the pulsation to relax thehydraulic pressure impact, should take it as in the system a link if to beconnected partially together synthesizes considers its dischargeable capaciAccording to the dischargeable capacity which extracts and considered otherrequests, then chooses the accumulator the form103.4 pipelines choices3.4.1 drill tubings types choiceIn the hydraulic system uses the drill tubing divides the hard tube and thehose, the choice drill tubing should have enough passes flows the section andthe bearing pressure ability, simultaneously, should reduce the pipeline as faras possible, avoids the extreme turn and the section sudden change.3.4.1.1 steel pipes: Center the high tension system selects the seamlesssteel pipe, the low pressure system selects the welded steel pipe, the steelpipe price lowly, performance good, the use is widespread3.4.1.2 copper pipes: The copper tube working pressure below 6.5 10MPa,the instable tune, is advantageous for the assembly; Yellow copper pipewithstanding pressure higher, reaches 25MPa, was inferior to the copper tubeis easy to be curving. Copper pipe price high, earthquake resistance abilityweak, is easy to cause the fat liquor oxidation, should as far as possible littleuse, only uses in the hydraulic unit to match meets not the convenient spot.3.4.2 drill tubings sizes determination3.4.2.1 drill tubings inside diameters d presses down the type computationd=(3.5)In the formula: Q is passes the drill tubing the maximum current capacity(m3/s); V speed of flow which permits for the pipeline in (m/s). The commonoil suction pipe takes 0.5 5 (m/s); The pressure oil pipe takes 2.5 5 (m/s);The oil return pipe takes 1.5 2 (m/s).3.4.2 drill tubings sizes determination)(2dp(3.6)In the formula: P is in the tube the biggest working pressure; When n isthe safety coefficient, steel pipe p 7MPa, takes n=8; When p 17.5MPa, takes n=4.11According to drill tubing inside diameter and wall thickness which calculates,looks up the handbook selection standard specification drill tubing3.5 fuel tank designThe fuel tank function is the oil storage, disperses the oil discharge thequantity of heat, in the precipitation oil the impurity, is leisurely in the oil thegas3.5.1 fuel tanks designs main point3.5.1.1 fuel tanks should have the enough volume to satisfy the radiation,simultaneously its volume should guarantee in the system the fat liquorcompletely flows when the fuel tank does not seep out, the fat liquor liquidlevel should not surpass the fuel tank highly 80%.3.5.1.2 suction boxes tubes and the oil return pipe spacing should be as faras possible big3.5.1.3 fuel tanks bases should have the suitable ascent, releases the oil mouthto set to the most low spot, in order to drains the oil3.6 oil filters choicesChooses the oil filter the basis to have following several3.6.1 bearing capacitiesAccording to system pipeline working pressure determination.3.6.2 filters the precision:According to is protected the part the precision request determination3.6.3 flow the ability:According to through maximum current capacity determination.3.6.4 resistance pressure drops:Should the satisfied filter material intensity and the coefficient request.4 hydraulic systems performance12In order to judge the hydraulic system the design quality, needs to lose tothe system pressure, to give off heat , the efficiency and system dynamiccharacteristic and so on4.1 circuitries pressure losesAfter hydraulic pressure part specification model and pipeline sizedetermination, may the more accurate computing system pressure loss, thepressure loss include: The oilloses ,LPthe local pressure after the pipelineCPalong the regulation pressure damagesflows after the valve class partpressure lossVP, namely:VCLPPPP+=( (4.1) )System adjustment pressure:PPP+10(4.2)In the formula: P0For hydraulic pump working pressure or legadjustment pressure; P1In order to execution working pressure.If calculatesPin the primary election system working pressure timethe is sketchier than designation pressure to lose is much bigger than, shouldremove entire related part, auxiliary specification, again definite pipeline size.4.2 systems give off heatThe system gives off heat originates from the system interior energy loss,like the hydraulic pump and the functional element power loss, the overflowvalve overflow loses, the hydraulic valve and the pipeline pressure loss and soon.The system gives off heat the power P computation)(1(WPPB=(4.3)In the formula: PB is the hydraulic pump power input (W); Is thehydraulic pump overall effectiveness indexIf in a operating cycle has several working procedures, then may actaccording to each working procedure the calorific capacity, extracts thesystem unit time the average calorific capacity:13)()1 (11WtPTPinibi=（4.4）In the formula: T is the operating cycle cycle (s); tiFor i workingprocedure operating time (s); piis in the circulation the i working procedurepower input (W).4.3 systems efficiencyThe hydraulic system efficiency is by the hydraulic pump, the functionalelement and the hydraulic pressure return route efficiency determinedThe hydraulic pressurecreturn route efficiency generally may use thetype to calculate:2221.2211bbbbcqpqpqpqp+=(4.5)In the formula: p1，q1；p2，q2； For each functional element workingpressure and current capacity; pB1，qB1；pB2，qB2is each hydraulic pumpsupplies the oil pressure and the current capacity.Hydraulic system overall effectiveness index:cmB+=(4.6)In the formula:BFor hydraulic pump overall effectiveness index;mIn order to functional element overall effectiveness index;cFor returnroute efficiency5 draws up the regular worker mapping and the compilation technologydocumentPasses through after the hydraulic system performance and the essentialrevision, then may draw up the regular worker mapping, it including planhydraulic system schematic diagram, system pipeline assembly drawing andeach kind of non- standard part design drawing.In the official hydraulic system schematic diagram must mark varioushydraulic pressure part the model specification. Regarding automaticityhigher engine bed, but also should include the movement part the cycle ofmotion chart and the electro-magnet, the pressure switch active status.145.1 determinations hydraulic system parameterMay know by the operating mode analysis in, the labor enters the stagethe load strength to be biggest, therefore, the hydraulic cylinder workingpressure according to this load strength computation, according to thehydraulic cylinder and the load relations, p1=40105Pa. This engine bed forthe drill hole aggregate machine-tool, for prevented drills through beforewhen occurs flushes the phenomenon, the hydraulic cylinder oil dischargecavity should have the back pressure, 、p2=6105Pa, for causes quickly toenter quickly draws back the speed to be equal, selects212AA =thedifferential motion cylinder, the hypothesis quickly enters the oil dischargepressure which, quickly draws back to lose forp=7105Pa.5.2 choices hydraulic pressure part5.2.1 chooses the hydraulic pump and the electric motor5.2.1.1 determines the hydraulic pump the working pressure.Front had determined the hydraulic cylinder the biggest workingpressure for 40105Pa, selects the intake pipe road pressure to losep=8105Pa, its adjustment pressure is generally bigger than the systembiggest working pressure 5105Pa, therefore pumps working pressure PB=(40 + 8 + 5) 105 = 53105PaThis is the working pressure which the high-pressured small currentcapacity pumps.The hydraulic cylinder quickly draws back when the working pressurequickly enters when is bigger than, takes its pressure to lose Delta p =4105Pa, then quickly draws back time pumps the working pressure is:PB=(16.44)10520.4105PaThis is the working pressure which the low pressure great currentcapacity pumps.5.2.1.2 hydraulic pumps current capacities. Quickly enters when the currentcapacity is biggest, its value is 30L/min, the quantity enters when the labor,its value is 0.51L/min, takes K = 1.2,15Then:q
收藏