喜歡這套資料就充值下載吧。資源目錄里展示的都可在線預覽哦。下載后都有，請放心下載，文件全都包含在內，圖紙為CAD格式可編輯，有疑問咨詢QQ:414951605 或 1304139763p 液壓傳動系統(tǒng)設計與計算 1 明確設計要求進行工況分析 在設計液壓系統(tǒng)時，首先應明確以下問題，并將其作為設計依據。 主機的用途、工藝過程、總體布局以及對液壓傳動裝置的位置和空間尺寸的要求； 主機對液壓系統(tǒng)的性能要求，如自動化程度、調速范圍、運動平穩(wěn)性、換向定位精度以及對系統(tǒng)的效率、溫升等的要求；液壓系統(tǒng)的工作環(huán)境，如溫度、濕度、振動沖擊以及是否有腐蝕性和易燃物質存在等情況。 在上述工作的基礎上，應對主機進行工況分析，工況分析包括運動分析和動力分析，對復雜的系統(tǒng)還需編制負載和動作循環(huán)圖，由此了解液壓缸或液壓馬達的負載和速度隨時間變化的規(guī)律，以下對工況分析的內容作具體介紹。 1.1 運動分析 主機的執(zhí)行元件按工藝要求的運動情況，可以用位移循環(huán)圖(L—t)，速度循環(huán)圖(v—t)，或速度與位移循環(huán)圖表示，由此對運動規(guī)律進行分析。 1.1.1 位移循環(huán)圖L—t 圖1.1為液壓機的液壓缸位移循環(huán)圖，縱坐標L表示活塞位移，橫坐標t表示從活塞啟動到返回原位的時間，曲線斜率表示活塞移動速度。 圖1.1 位移循環(huán)圖 1.1.2 速度循環(huán)圖v—t(或v—L) 工程中液壓缸的運動特點可歸納為三種類型。圖1.2為三種類型液壓缸的v—t圖，第一種如圖1.2中實線所示，液壓缸開始作勻加速運動，然后勻速運動， ? 圖1.2 速度循環(huán)圖 最后勻減速運動到終點；第二種，液壓缸在總行程的前一半作勻加速運動，在另一半作勻減速運動，且加速度的數值相等；第三種，液壓缸在總行程的一大半以上以較小的加速度作勻加速運動，然后勻減速至行程終點。v—t圖的三條速度曲線，不僅清楚地表明了三種類型液壓缸的運動規(guī)律，也間接地表明了三種工況的動力特性。 1.2 動力分析 動力分析，是研究機器在工作過程中，其執(zhí)行機構的受力情況，對液壓系統(tǒng)而言，就是研究液壓缸或液壓馬達的負載情況。 1.2.1 液壓缸的負載及負載循環(huán)圖 1.2.1.1 液壓缸的負載力計算 工作機構作直線往復運動時，液壓缸必須克服的負載由六部分組成： 　　 (1.1) 式中：Fc為切削阻力；Ff為摩擦阻力；Fi為慣性阻力；Fg為重力；Fm為密封阻力；Fb為排油阻力。 1.2.1.2液壓缸運動循環(huán)各階段的總負載力 液壓缸運動循環(huán)各階段的總負載力計算，一般包括啟動加速、快進、工進、快退、減速制動等幾個階段，每個階段的總負載力是有區(qū)別的。 (1)啟動加速階段：這時液壓缸或活塞處于由靜止到啟動并加速到一定速度，其總負載力包括導軌的摩擦力、密封裝置的摩擦力(按缸的機械效率=0.9計算)、重力和慣性力等項，即： (1.2) (2)快速階段：  　(1.3) (3)工進階段：  　(1.4) (4)減速： 　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　 (1.5) 對簡單液壓系統(tǒng)，上述計算過程可簡化。例如采用單定量泵供油，只需計算工進階段的總負載力，若簡單系統(tǒng)采用限壓式變量泵或雙聯泵供油，則只需計算快速階段和工進階段的總負載力。 1.2.2 液壓馬達的負載 工作機構作旋轉運動時，液壓馬達必須克服的外負載為：  　　　 (1.6) 1.2.2.1 工作負載力矩Me。工作負載力矩可能是定值，也可能隨時間變化，應根據機器工作條件進行具體分析。 1.2.2.2 摩擦力矩Mf。為旋轉部件軸頸處的摩擦力矩，其計算公式為： (1.7) 式中：G為旋轉部件的重量(N)；f為摩擦因數，啟動時為靜摩擦因數，啟動后為動摩擦因數；R為軸頸半徑(m)。 1.2.2.3 慣性力矩Mi。為旋轉部件加速或減速時產生的慣性力矩，其計算公式為： (1.8) 式中：ε為角加速度(r/s2)；Δω為角速度的變化(r/s)；Δt為加速或減速時間(s)；J為旋轉部件的轉動慣量()，。 式中：為回轉部件的飛輪效應()。 各種回轉體的可查《機械設計手冊》。 根據式(1.6)，分別算出液壓馬達在一個工作循環(huán)內各階段的負載大小，便可繪制液壓馬達的負載循環(huán)圖 2 確定液壓系統(tǒng)主要參數 2.1 液壓缸的設計計算 2.1.1 初定液壓缸工作壓力 液壓缸工作壓力主要根據運動循環(huán)各階段中的最大總負載力來確定，此外，還需要考慮以下因素： 2.1.1.1 各類設備的不同特點和使用場合。 2.1.1.2 考慮經濟和重量因素，壓力選得低，則元件尺寸大，重量重；壓力選得高一些，則元件尺寸小，重量輕，但對元件的制造精度，密封性能要求高。 所以，液壓缸的工作壓力的選擇有兩種方式：一是根據機械類型選；二是根據切削負載選。 如表2.1、表2.2所示。 表2.1 按負載選執(zhí)行文件的工作壓力 負載/N ＜5000 500～10000 10000～20000 20000～30000 30000～50000 ＞50000 工作壓力/MPa ≤0.8～1 1.5～2 2.5～3 3～4 4～5 ＞5 表2.2 按機械類型選執(zhí)行文件的工作壓力 機械類型 機 床 農業(yè)機械 工程機械 磨床 組合機床 龍門刨床 拉床 工作壓力/MPa a≤2 3～5 ≤8 8～10 10～16 20～32 2.2 液壓馬達的設計計算 2.2.1 計算液壓馬達排量 液壓馬達排量根據下式決定： 　　　 (2.1) 式中：T為液壓馬達的負載力矩(N·m)；為液壓馬達進出口壓力差()；為液壓馬達的機械效率，一般齒輪和柱塞馬達取0.9～0.95，葉片馬達取0.8～0.9。 2.2.2 計算液壓馬達所需流量液壓馬達的最大流量 　　　　 (2.2) 式中：Vm為液壓馬達排量(m3/r)；nmax為液壓馬達的最高轉速(r/s)。 3 液壓元件的選擇 3.1 液壓泵的確定與所需功率的計算 3.1.1 液壓泵的確定 3.1.1.1 確定液壓泵的最大工作壓力。液壓泵所需工作壓力的確定，主要根據液壓缸在工作循環(huán)各階段所需最大壓力p1，再加上油泵的出油口到缸進油口處總的壓力損失ΣΔp，即  (3.1) 包括油液流經流量閥和其他元件的局部壓力損失、管路沿程損失等，在系統(tǒng)管路未設計之前，可根據同類系統(tǒng)經驗估計，一般管路簡單的節(jié)流閥調速系統(tǒng)為(2～5)×105Pa，用調速閥及管路復雜的系統(tǒng)為(5～15)×105Pa，也可只考慮流經各控制閥的壓力損失，而將管路系統(tǒng)的沿程損失忽略不計，各閥的額定壓力損失可從液壓元件手冊或產品樣本中查找，也可參照表1.3選取。 表3.1 常用中、低壓各類閥的壓力損失(Δpn) 閥名 Δpn(×105Pa) 閥名 Δpn (×105Pa) 閥名 Δpn (×105Pa) 閥名 Δpn (×105Pa) 單向閥 0.3～0.5 背壓閥 3～8 行程閥 1.5～2 轉閥 1.5～2 換向閥 1.5～3 節(jié)流閥 2～3 順序閥 1.5～3 調速閥 3～5 3.1.2 確定液壓泵的流量qB 泵的流量qB根據執(zhí)行元件動作循環(huán)所需最大流量qmax和系統(tǒng)的泄漏確定。 3.1.2.1多液壓缸同時動作時，液壓泵的流量要大于同時動作的幾個液壓缸(或馬達)所需的最大流量，并應考慮系統(tǒng)的泄漏和液壓泵磨損后容積效率的下降，即 　　(3.2) 式中：K為系統(tǒng)泄漏系數，一般取1.1～1.3，大流量取小值，小流量取大值；為同時動作的液壓缸(或馬達)的最大總流量(m3/s)。 3.1.2.2選擇液壓泵的規(guī)格：根據上面所計算的最大壓力pB和流量qB，查液壓元件產品樣本，選擇與pB和qB相當的液壓泵的規(guī)格型號。 表3.2 液壓泵的總效率 液壓泵類型 齒輪泵 螺桿泵 葉片泵 柱塞泵 總效率 0.6～0.7 0.65～0.80 0.60～0.75 0.80～0.85 ? 按上述功率和泵的轉速，可以從產品樣本中選取標準電動機，再進行驗算，使電動機發(fā)出最大功率時，其超載量在允許范圍內。 3.2 閥類元件的選擇 3.2.1 選擇依據 選擇依據為：額定壓力，最大流量，動作方式，安裝固定方式，壓力損失數值，工作性能參數和工作壽命等。 3.2.2 選擇閥類元件應注意的問題 3.2.2.1 應盡量選用標準定型產品，除非不得已時才自行設計專用件。 3.2.2.2 閥類元件的規(guī)格主要根據流經該閥油液的最大壓力和最大流量選取。選擇溢流閥時，應按液壓泵的最大流量選?。贿x擇節(jié)流閥和調速閥時，應考慮其最小穩(wěn)定流量滿足機器低速性能的要求。 3.3 蓄能器的選擇 3.3.1 蓄能器用于補充液壓泵供油不足時，其有效容積為： 　　　 (3.3) 式中：A為液壓缸有效面積(m2)；L為液壓缸行程(m)；K為液壓缸損失系數，估算時可?。耍?.2；qB為液壓泵供油流量(m3/s)；t為動作時間(s)。 3.3.2 蓄能器作應急能源時，其有效容積為： 　 (3.4) 當蓄能器用于吸收脈動緩和液壓沖擊時，應將其作為系統(tǒng)中的一個環(huán)節(jié)與其關聯部分一起綜合考慮其有效容積。 根據求出的有效容積并考慮其他要求，即可選擇蓄能器的形式。 3.4 管道的選擇 3.4.1 油管類型的選擇 液壓系統(tǒng)中使用的油管分硬管和軟管，選擇的油管應有足夠的通流截面和承壓能力，同時，應盡量縮短管路，避免急轉彎和截面突變。 3.4.1.1 鋼管：中高壓系統(tǒng)選用無縫鋼管，低壓系統(tǒng)選用焊接鋼管，鋼管價格低，性能好，使用廣泛。 3.4.1.2 銅管：紫銅管工作壓力在6.5～10MPa以下，易變曲，便于裝配；黃銅管承受壓力較高，達25MPa，不如紫銅管易彎曲。銅管價格高，抗震能力弱，易使油液氧化，應盡量少用，只用于液壓裝置配接不方便的部位。 3.4.2 油管尺寸的確定 3.4.2.1 油管內徑d按下式計算： d= (3.5) 式中：q為通過油管的最大流量(m3/s)；v為管道內允許的流速(m/s)。一般吸油管取0.5～5(m/s)；壓力油管取2.5～5(m/s)；回油管取1.5～2(m/s)。 3.4.2.2 油管壁厚δ按下式計算： (3.6) 式中：p為管內最大工作壓力；n為安全系數，鋼管p＜7MPa時，取n=8；p＜17.5MPa時，取n=6；p＞17.5MPa時，取n=4。 根據計算出的油管內徑和壁厚，查手冊選取標準規(guī)格油管。 3.5 油箱的設計 油箱的作用是儲油，散發(fā)油的熱量，沉淀油中雜質，逸出油中的氣體。 3.5.1 油箱設計要點 3.5.1.1 油箱應有足夠的容積以滿足散熱，同時其容積應保證系統(tǒng)中油液全部流回油箱時不滲出，油液液面不應超過油箱高度的80%。 3.5.1.2 吸箱管和回油管的間距應盡量大。 3.5.1.3 油箱底部應有適當斜度，泄油口置于最低處，以便排油。 3.6 濾油器的選擇 選擇濾油器的依據有以下幾點： 3.6.1 承載能力： 按系統(tǒng)管路工作壓力確定。 3.6.2 過濾精度： 按被保護元件的精度要求確定。 3.6.3 通流能力： 按通過最大流量確定。 3.6.4 阻力壓降： 應滿足過濾材料強度與系數要求。  4 液壓系統(tǒng)性能的驗算 為了判斷液壓系統(tǒng)的設計質量，需要對系統(tǒng)的壓力損失、發(fā)熱溫升、效率和系統(tǒng)的動態(tài)特性等進行驗算。 4.1 管路系統(tǒng)壓力損失的驗算 當液壓元件規(guī)格型號和管道尺寸確定之后，就可以較準確的計算系統(tǒng)的壓力損失，壓力損失包括：油液流經管道的沿程壓力損失、局部壓力損失和流經閥類元件的壓力損失，即： (4.1) 系統(tǒng)的調整壓力： (4.2) 式中：P0為液壓泵的工作壓力或支路的調整壓力；P1為執(zhí)行件的工作壓力。 如果計算出來的比在初選系統(tǒng)工作壓力時粗略選定的壓力損失大得多，應該重新調 整有關元件、輔件的規(guī)格，重新確定管道尺寸。 4.2 系統(tǒng)發(fā)熱溫升的驗算 系統(tǒng)發(fā)熱來源于系統(tǒng)內部的能量損失，如液壓泵和執(zhí)行元件的功率損失、溢流閥的溢流損失、液壓閥及管道的壓力損失等。 系統(tǒng)發(fā)熱功率P的計算: 　　(4.3) 式中：PB為液壓泵的輸入功率(W)；η為液壓泵的總效率。 若一個工作循環(huán)中有幾個工序，則可根據各個工序的發(fā)熱量，求出系統(tǒng)單位時間的平均發(fā)熱量： (4.4) 式中：T為工作循環(huán)周期(s)；ti為第i個工序的工作時間(s)；pi為循環(huán)中第i個工序的輸入功率(W)。 4.3 系統(tǒng)效率驗算 液壓系統(tǒng)的效率是由液壓泵、執(zhí)行元件和液壓回路效率來確定的。 液壓回路效率nc一般可用下式計算： (4.5) 式中：p1，q1；p2，q2；……為每個執(zhí)行元件的工作壓力和流量；pB1，qB1；pB2，qB2為每個液壓泵的供油壓力和流量。 液壓系統(tǒng)總效率： (4.6) 式中：為液壓泵總效率；為執(zhí)行元件總效率；為回路效率。 5 繪制正式工作圖和編寫技術文件 經過對液壓系統(tǒng)性能的驗算和必要的修改之后，便可繪制正式工作圖，它包括繪制液壓系統(tǒng)原理圖、系統(tǒng)管路裝配圖和各種非標準元件設計圖。 正式液壓系統(tǒng)原理圖上要標明各液壓元件的型號規(guī)格。對于自動化程度較高的機床，還應包括運動部件的運動循環(huán)圖和電磁鐵、壓力繼電器的工作狀態(tài)。 5.1 確定液壓系統(tǒng)參數 由工況分析中可知，工進階段的負載力最大，所以，液壓缸的工作壓力按此負載力計算，根據液壓缸與負載的關系，選p1=40×105Pa。本機床為鉆孔組合機床，為防止鉆通時發(fā)生前沖現象，液壓缸回油腔應有背壓，設背壓p2=6×105Pa，為使快進快退速度相等，選用差動油缸，假定快進、快退的回油壓力損失為Δp=7×105Pa。 5.2 選擇液壓元件 5.2.1 選擇液壓泵和電動機 5.2.1.1 確定液壓泵的工作壓力。 前面已確定液壓缸的最大工作壓力為40×105Pa，選取進油管路壓力損失Δp=8×105Pa，其調整壓力一般比系統(tǒng)最大工作壓力大5×105Pa，所以泵的工作壓力PB＝(40＋8＋5)×105＝53×105Pa 這是高壓小流量泵的工作壓力。 液壓缸快退時的工作壓力比快進時大，取其壓力損失Δp′＝4×105Pa，則快退時泵的工作壓力為： PB=(16.4＋4)×105＝20.4×105Pa 這是低壓大流量泵的工作壓力。 5.2.1.2 液壓泵的流量?？爝M時的流量最大，其值為30L/min，最小流量在工進時，其值為0.51L/min，取K＝1.2， 則： qB＝1.2×0.5×10-3=36L/min 由于溢流閥穩(wěn)定工作時的最小溢流量為3L/min，故小泵流量取3.6L/min。 根據以上計算，選用YYB-AA36/6B型雙聯葉片泵。 5.2.1.3 確定管道尺寸:根據工作壓力和流量，按式(3.5)、式(3.6)確定管道內徑和壁厚。(從略) 5.2.1.4 確定油箱容量油箱容量可按經驗公式估算，取V＝(5～7)q。 本例中：V＝6q＝6(6＋36)＝252L有關系統(tǒng)的性能驗算從略。 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