1.5兆牛擺動剪切機(jī)構(gòu)設(shè)計【6張CAD圖紙+PDF圖】

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(yù)覽哦。下載后都有，請放心下載，文件全都包含在內(nèi)，【有疑問咨詢QQ:1064457796 或 1304139763】 譯文軋制過程中的熱傳遞一 熱帶軋制的溫度變化板坯再加熱到所要求的溫度后進(jìn)行軋制。一個典型的熱帶軋制工藝包括以下幾個主要步驟：（1） 板坯軋制前用高壓水除鱗系統(tǒng)除鱗，有時采用立輥軋機(jī)同時除鱗。（2） 粗軋成1940mm后的中間料。粗軋過程通常伴隨立輥和道次間的除鱗操作。（3） 將中間料從粗軋機(jī)運(yùn)至安裝在精軋機(jī)前的飛剪處。飛剪用來剪切料頭和料尾。（4） 中間料在進(jìn)精軋機(jī)組前的除鱗。（5） 精軋至所要求的厚度。機(jī)架間可能進(jìn)行除鱗，有時也可能進(jìn)行帶鋼冷卻。（6） 軋材在輸出輥道上的空冷和水冷。（7） 軋材的卷取。在軋制工藝過程中，軋件向其周圍物質(zhì)進(jìn)行各種熱傳遞。一些損失的熱量由軋件變形所產(chǎn)生的熱予以彌補(bǔ)。熱帶軋制過程中，軋件溫度降低和升高的主要因素通?？梢詤^(qū)分如下：（1） 熱輻射引起的溫降。（2） 熱對流引起的溫降。（3） 水冷引起的溫降。（4） 向工作輥和輥道熱傳遞導(dǎo)引起的溫降。（5） 力學(xué)加工和摩擦引起的溫升。 關(guān)于這些因素的分析簡述如下。二 熱輻射引起的溫降 采用兩種方法進(jìn)行熱輻射引起的溫降公式的推導(dǎo)。第一種方法忽略了材料內(nèi)部的溫度提督，利用斯蒂芬-玻爾茲曼定律計算輻射到環(huán)境中的熱量為：q=S 式中 輻射體的表面積，m2； q從物體輻射的熱量，J； S斯蒂芬-玻爾茲曼常數(shù)； T軋件在t時刻的溫度，K; Ta環(huán)境溫度，K； t時間，s； 輻射系數(shù)。 物體損失的熱量由下式給定： q= 式中 c軋件質(zhì)量熱容，J/（kgK）； Vr輻射體的體積，m3 軋件的密度，kg/m3。 考慮到熱平衡條件q=q及式1-1和式1-2，可以計算出溫降速度ar： ar= 通常假設(shè)TaT，并簡化某些方程以達(dá)到協(xié)調(diào)形式，得出輻射溫度差速度公式，總結(jié)見表1-1所示。在推導(dǎo)這些公式時，未考慮溫度對參數(shù)S、及c的影響。不過實際上這些常數(shù)隨溫度的變化可能都是很大的，所以，式1-3的最終形式將取決于這些常熟選擇的平均值。 輻射時間tr內(nèi)的溫降可以通過對微分方程幾分進(jìn)行計算： = 第二種計算輻射引起溫降的方法考慮到沿材料厚度方向上的熱傳遞。若z是物體內(nèi)部至其表面的距離，則從傅里葉公式可得： 式中 a軋件的熱擴(kuò)散率，m2/s。微分方程1-5可以利用有限差分法進(jìn)行數(shù)值求解。這些計算的目的是要建立一個影響軋制過程軋件平均溫度T和可測量的軋件表面溫度T之間的關(guān)系。三 熱對流引起的溫降熱帶軋制時的對流傳熱與軋件周圍空氣的運(yùn)動有關(guān)。這種運(yùn)動不斷地帶入新的空氣粒子與軋件接觸。取決于該內(nèi)部運(yùn)動是強(qiáng)制的，還是自然的，將熱傳遞區(qū)分為強(qiáng)制對流和自然對流。在熱帶軋制中通常出現(xiàn)后一種情形。在計算對流引起溫降時的一個重要方面是確定傳熱系數(shù)。該系數(shù)取決于材料溫度、環(huán)境溫度、材料質(zhì)量熱融合密度以及空氣流的動態(tài)粘度及其特性，即自然、強(qiáng)制層流或紊流等情況。對于此關(guān)系所得出的數(shù)學(xué)描述有很大爭議，實際計算不宜采用。部分研究人員一致認(rèn)為，對流引起的溫降應(yīng)當(dāng)表示為輻射引起溫降的莫以分?jǐn)?shù)：=（） 這里，是對流和輻射引起溫降間的比率，根據(jù)不同的研究結(jié)果，其值在0.010.22之間變化。四 水冷引起溫降若假定在軋件向冷卻水傳熱石傳導(dǎo)起著重要作用，就可以計算出水冷引起的溫降。因此，當(dāng)冷卻沿軋件款度方向連續(xù)地接觸其一側(cè)表面時，通過軋件表面所傳遞的熱量就可以用公式表示為： 式中 k表層導(dǎo)熱系數(shù)，W/（mK）； 通過軋件外表面所傳遞的熱量，J； b冷卻水接觸長度，m； 軋件寬度，m； Tw冷卻水溫度，K； tw冷卻水接觸時間，s。 由軋件釋放的熱量由下式給定： 式中 V冷卻水所冷卻的軋件體積，m3；水冷引起的溫降，K。根據(jù)熱平衡條件 ，式1-7和式1-8，并考慮到： tw= 式中 v軋件速度，m/s。和另一條件： 我們得到水冷引起的溫降為： = 冷卻水所吸收的熱量可以表示為： = 式中 水的密度，kg/m3； 水的質(zhì)量熱容，J/（kgK）； Vw水的吸熱體積，m3； 水的溫升，K。根據(jù)熱平衡條件=，式1-8、式1-11和式1-12，并考慮到： 式中 d帶鋼單位寬度上的水流量，m3/（ms）。我們得到下列冷卻水溫升公式：= 式1-11并沒有明確地給出溫降與冷卻水流速和壓力的關(guān)系。然而，冷卻水的流速和壓力卻大大地影響著隔開軋件于冷水的表面成的導(dǎo)熱系數(shù)k。事實上，表面層中包含有充當(dāng)屏障作用的氧化鐵皮和沸騰水。隨著冷卻水流速和壓力的提高，該屏障作用將在很大程度上被削弱。五 因工作輥熱傳到引起的溫降如果假設(shè)兩個初始穩(wěn)定溫度分別為T和Tr的物體相互擠壓，并假設(shè)平面的界面處在又有氧化層的阻力，則可以計算出因工作輥熱傳導(dǎo)引起的溫降。在作出上述這些假設(shè)之后，則可以用以下的熱平衡方程進(jìn)行過程的描述。根據(jù)沙科的研究，通過鋼板的兩個最晚層的總熱量可以根據(jù)下式計算： 式中 Ac軋件和工作輥的接觸面積，m2； k軋件氧化成的導(dǎo)熱系數(shù)，W/（mK）； 由于熱傳遞工作輥所獲的熱量或軋件所失去的熱量，J； Tr軋輥溫度，K； a軋件的熱擴(kuò)散率，m2/s。輥縫處軋件損失的熱量由下式給定： 式中 軋件與工作輥接觸而產(chǎn)生的溫降，K。根據(jù)熱平衡條件=，式1-15和式1-16，并考慮到： 及 式中 R軋輥半徑，m； 軋件平均厚度，m。我們得出下列因工作輥熱傳導(dǎo)引起的溫降公式： = 通過簡化某些方程以達(dá)到協(xié)調(diào)形式，得出與輥接觸引起的溫降公式，總結(jié)見表1-2，繪制成曲線如圖1-3 所示。不同的溫降計算公式之間的顯著差異主要是由于在預(yù)測導(dǎo)熱系數(shù)k時的誤差造成的，該系數(shù)之取決于軋輥和軋件件氧化層接觸阻力的大小。原文Heat Transfer During the Rolling Process1.1WORKPIECE TEMPERATURE CHANGE IN HOT STRIP MILLAfter reheating a slab to a desired temperature, it is subjected to rolling. A rolling cycle in a typical hot strip mill includes the following main steps: 1.Descaling of the slab prior to flat rolling by using high-pressure water descaling system in combination, in some cases, with edging.2.Rough rolling to a transfer bar thickness which may vary from 19 to 40 mm. The rough rolling is usually accompanied by edging and inter pass descaling.3.Transfer of the transfer bar from roughing mill to a flying shear installed ahesd of finishing mill. The shear is usually designed to cut both head and tail ends of the bar.4.Descaling of the transfer bar prior to entering the finishing mill.5.Finish rolling to a desired thickness with a possible use of interstand descaling and strip cooling.6.Air and water cooling of the rolled product on run-out table.7.Cliling of the rolled product.Various types of heat transfer from the rolled workpiece to its surrounding matter occur during the rolling cycle. Some of the lost heat is recovered by generating heat inside the workpiece during its deformation.The main components of the workpiece temperature loss and gain in hot strip mill are usually identified as follows:1.loss due to heat radiation,2.loss due to heat convection,3.loss due to water cooling,4.loss due to heat conduction to the work rolls and table rolls,5.gain due to mechanical work and friction.The analytical aspects of these components are briefly described below.1.2TEMPERATURE LOSS DUE TO TADIATIONTwo methods have been employed to derive equations for temperature loss due to radiation.In the first method, the temperature gradient within the material is assumed to be negligible. The amount of heat radiated to the environment is then calculated using the Stefan-Boltzmann law:q=S Where surface area of body subjected to radiation, m2; qamount of heat radiated by a body,J; SStefan-Boltzmann constant; Ttemperature of rolled material at time,K; Taambient temperature,K; ttime,s; emissivity.The amount of heat lost by a body q is give by: q= Where cspecific heart of rolled material, J/（kgK）; Vrvolume of body subjected to radiation, m3 density of rolled material, kg/m3。The rate of temperature loss ar can be calculated by considering the heat balance condition q=q, and Eqs.1-1 and 1-2: ar= Equations for the rate of temperature loss due to radiation which have been obtained by reducing some of the known equations to a compatible form with an assumption that TaT are summarized in Table 1-1. In the derivation of these equations, the dependency of the parameters S、 and c on temperature is not taken into account. However, the variations of these constants with temperature may be significant and,therefore, the final from of 1-3 will depend on the average values selected for these constants.The temperature loss during radiation time tr can be calculate by intergrating the differential equation: = The second method of calculating temperature loss due to radiation takes into account the heat transfer along the thickness of the material. If z is the distance from the center of the body toward its surface, then from a Fourier equation we obtain: Where athermal diffusivity of rolled material ,m2/sThe differential equation 1-5 can be solved numerically by the method of finite differences.The goal of these calculations is to establish a relationship between the average temperature of the material Tave which would affect the rolling deformation process and the material surface temperature Tsurface which could be measured.1.3TEMPERTURE LOSS DUE TO CONVECTIONIn the hot strip mill, heat transfer by convection is related to the motion of air surrounding a workpiece. This motion continuously brings new particles of air into contact with the workpiece. Depending upon whether this internal motion is forced, or free, the heat transfer is referred to as either forced or free convection. The latter is a usual case in the hot strip mills.A key factor in the calculation of temperature losses due to convection is to determine the heat transfer coefficient, which depends on the material temperature, ambient temperature, material specific heat and density, and the dynamic viscosity of the air flow and its characteristic, i.e., free, enforced laminar, turbulent, etc. The known mathematical interpretations of this relationship are too controversial to be recommended for practical calculation. A consensus among some research workers is that the temperature loss due to convection should be expressed as a certain percentage of the temperature loss due to radiation:=（） Here is the ratio between the temperature loss due to convection and radiation and varies between 0.01 and 0.22 according to different studies.1.4TEMPERATURE LOSS DUE TO WTER COOLINGThe temperature loss due to water cooling can be calculated by assuming that conduction plays a major role in heat transfer from a workpiece to water. Therefore, when water contacts one side of the workpiece continuously across its width, the amount of heat passing through the outer surface of the workpiece may be expressed by the formula: Where kthermal conductivity of the surface layer, W/（mK）； amount of heat passing through outer surface of the workpiece,J; bwater contact length, m; wworkpiece width, m; Twwater temperature, K; twwater contact time,s.The amount of heat released by a workpiece is given by: Where vvolume of workpiece cooled by the water,m3; temperature loss due to water cooling, K.From the heat balance condition =,Eqs.1-7 and 1-8, and taking into account that tw= where Vworkpiece velocity, m/sand We obtain that the temperature loss due to water cooling is equal to = The amount of heat absorbed by cooling water may be expressed as:= Where density of water ，kg/m3； specific heat of water，J/（kgK）； Vwvolume of water absorbing heat，m3；From hert balance =, Eqs.1-8, 1-11, and 1-12, and also taking into account that Where dwater flow per unit of strip width, m3/（ms）.We obtain the following formula for the temperature rise of water:= Equation 1-11 does not show an explicit dependence of the temperature loss on the flow rate and pressure of cooling water. The flow rate and pressure, however, may substantially affect the thermal conductivity k of the surface layer that separates the body of workpiece from cooling water. Indeed, the surface layer consists of scale and boiled water, which work as a thermal barrier. This barrier will be weakened to a greater degree with increase of both the flow rate and pressure of cooling water.1.5TEMPERATURE LOSS DUE TO CONDUCTION TO WORK ROLLSTemperature loss due to heat conduction to the work roll can be calculated if it is assumed that two bodies of uniform unitial temperature T and Tr are pressed against each other and that, at the interface, considered to be plane, there is contact resistance formed by oxide layer.Under these assumptions, the process can be described with the following heat balance equations. According to Schack, the total amount of heat passing through two outer surfaces of the plate may be calculated from the formula Where Accontact area between rolled material and work rolls，m2； kthermal conductivity of the workpiece oxide layer，W/（mK）； heat gained by work roll or heat lost by body due to thermal conduction，J； Trroll temperature，K； athermal diffusivity of workpiece，m2/s。The amount of heat lost by the rolled metal in the roll bite is given by: Where temperature loss by rolled material due to contact with work rolls，K。From the heat balance condition =，Eqs 1-15 and 1-16, and also taking into account that and where Rwork roll radius, m. haaverage workpiece thickness, m.we obtain the following formula for the temperature loss due to conduction to work rolls:= Equation for temperature loss due to contact with rolls which have been obtained by reducing some of the known equations to a compatible form are summarized in the Table 1.2 and are plotted in Fig.1.3. The substantial discrepancies in temperature losses calculated from different equations are due mainly to the uncertainty in estimating thermal conductivity k which depends on the contact resistance resistance of the oxide layer between the roll and the rolled material.