外文翻译=轧制过程中的热传递=2000字符

译文 轧制过程中的热传递 一 热带轧制的温度变化 板坯再加热到所要求的温度后进行轧制。一个典型的热带轧制工艺包括以下几个主要步骤： （ 1） 板坯轧制前用高压水除鳞系统除鳞，有时采用立辊轧机同时除鳞。 （ 2） 粗轧成 19 40mm 后的中间料。粗轧过程通常伴随立辊和道次间的除鳞操作。 （ 3） 将中间料从粗轧机运至安装在精轧机前的飞剪处。飞剪用来剪切料头和料尾。 （ 4） 中间料在进精轧机组前的除鳞。 （ 5） 精轧至所要求的厚度。机架间可能进行除鳞，有时也可能进行带钢冷却。 （ 6） 轧材在输出辊道上的空冷和水冷。 （ 7） 轧材的卷取。 在轧制工艺过程中，轧件向其周围物质进行 各种热传递。一些损失的热量由轧件变形所产生的热予以弥补。 热带轧制过程中，轧件温度降低和升高的主要因素通常可以区分如下： （ 1） 热辐射引起的温降。 （ 2） 热对流引起的温降。 （ 3） 水冷引起的温降。 （ 4） 向工作辊和辊道热传递导引起的温降。 （ 5） 力学加工和摩擦引起的温升。 关于这些因素的分析简述如下。 二 热辐射引起的温降 采用两种方法进行热辐射引起的温降公式的推导。 第一种方法忽略了材料内部的温度提督，利用斯蒂芬 -玻尔兹曼定律计算辐射到环境中的热量为： d qr =S dtTTA ar )( 44 式中 rA 辐射体的表面积， m2； d qr 从物体辐射的热量， J； S 斯蒂芬 -玻尔兹曼常数； T 轧件在 t 时刻的温度， K； Ta 环境温度， K； t 时间， s； 辐射系数。 物体损失的热量由下式给定： d qr = dTcVr 式中 c 轧件质量热容， J/（ kg K）； Vr 辐射体的体积， m3 轧件的密度， kg/m3。 考虑到热平衡条件 d qr =d qr 及式 1-1 和式 1-2，可以计算出温降速度 ar： ar= )( 44arr TTcVASdtdT 通常假设 TaT，并简化某些方程以达到协调形式，得出辐射温度差速度公式，总结见表 1-1 所示。在推导这些公式时，未 考虑温度对参数 S、 、 及 c的影响。不过实际上这些常数随温度的变化可能都是很大的，所以，式 1-3 的最终形式将取决于这些常熟选择的平均值。 辐射时间 tr 内的温降 rT 可以通过对微分方程几分进行计算： rT =trrdta0第二种计算辐射引起温降的方法考虑到沿材料厚度方向上的热传递。若 z 是物体内部至其表面的距离，则从傅里叶公式可得： 2dzdTadtdT 式中 a 轧件的热扩散率， m2/s。 微分方程 1-5 可以利用有限差分法进行数值求解。 这些计算的目的是要建立一个影响轧制过程轧件平均温度 T平均和可测量的轧件表面温度 T表面之间的关系。 三 热对流引起的温降 热带轧制时的对流传热与轧件周围空气的运动有关。这种运动不断地带入新的空气粒子与轧件接触。取决于该内部运动是强制的，还是自然的，将热传递区分为强制对流和自然对流。在热带轧制中通常出现后一种情形。 在计算对流引起温降时的一个重要方面是确定传热系数。该系数取决于材料温度、环境温度、材料质量热融合密度以及空气流的动态粘度及其特性，即自然、强制层流或紊流等情况。对于此关系所得出的数学描述有很大争议，实际计算不宜采用。部分研究人员一致认为，对流引起的温降cvT应当表示为辐射引起温降的莫以分数： cvT=cvk（ rT ） 这里，cvk是对流和辐射引起温降间的比率，根据不同的研究结果，其值在0.01 0.22 之间变化。 四 水冷引起温降 若假定在轧件向冷却水传热石传导起着重要作用，就可以计算出水冷引起的温降。因此，当冷却沿轧件款度方向连续地接 触其一侧表面时，通过轧件表面所传递的热量就可以用公式表示为： atTTkbq www )(2 式中 k 表层导热系数， W/（ m K）； wq 通过轧件外表面所传递的热量， J； b 冷却水接触长度， m； 轧件宽度， m； Tw 冷却水温度， K； tw 冷却水接触时间， s。 由轧件释放的热量由 下式给定： )(dw TcVq 式中 V 冷却水所冷却的轧件体积， m3； dT 水冷引起的温降， K。 根据热平衡条件 wq wq，式 1-7 和式 1-8，并考虑到： tw=vb式中 v 轧件速度， m/s。 和另一条件： hVbw 1我们得到水冷引起的温降为： dT=avbTTchkw )(2 冷却水所吸收的热量可以表示为： wq= )(wwww TVc 式中 w 水的密度， kg/m3； wc 水的质量热容， J/（ kg K）； Vw 水的吸热体积， m3； wT 水的温升， K。 根据热平衡条件 wq= wq，式 1-8、式 1-11 和式 1-12，并考虑到： hvdVVw 式中 d 带钢单位宽度上的水流量， m3/（ m s）。 我们得到下列冷却水温升公式： wT=abvTThckwww )(2 式 1-11 并没有明确地给出温降与冷却水流速和压力的关系。然而，冷却水的流速和压力却大大地影响着隔开轧件于冷 水的表面成的导热系数 k。事实上，表面层中包含有充当屏障作用的氧化铁皮和沸腾水。随着冷却水流速和压力的提高，该屏障作用将在很大程度上被削弱。 五 因工作辊热传到引起的温降 如果假设两个初始稳定温度分别为 T 和 Tr 的物体相互挤压，并假设平面的界面处 在又有氧化层的阻力，则可以计算出因工作辊热传导引起的温降。 在作出上述这些假设之后，则可以用以下的热平衡方程进行过程的描述。根据沙科的研究，通过钢板的两个最晚层的总热量可以根据下式计算： atTTkAq crcc )(4 式中 Ac 轧件和工作辊的接触面积， m2； k 轧件氧化成的导热系数， W/（ m K）； cq 由于热传递工作辊所获的热量或轧件所失去的热量， J； Tr 轧辊温度， K； a 轧件的热扩散率， m2/s。 辊缝处轧件损失的热量由下式给定： )(ccc TcVq 式中 cT 轧件与工作辊接触而产生的温降， K。 根据热平衡条件 cq= cq，式 1-15 和式 1-16，并考虑到： vRtc 及 acc hVA 1式中 R 轧辊半径 ， m； ah 轧件平均厚度， m。 我们得出下列因工作辊热传导引起的温降公式： cT=avRTTchkra 5.0)()(4 通过简化某些方程以达到协调形式，得出与辊接触引起的温降公式，总结见表 1-2，绘制成曲线如图 1-3 所示。不同的温降计算公式之间的显著差异主要是由于在预测导热系数 k 时的误差造成的，该系数之取决于轧辊和轧件件氧化层接触阻力的 大小。 原 文 Heat Transfer During the Rolling Process 1.1WORKPIECE TEMPERATURE CHANGE IN HOT STRIP MILL After 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 TADIATION Two 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: d qr =S dtTTA ar )( 44 Where rA surface area of body subjected to radiation, m2； d qr amount of heat radiated by a body,J； S Stefan-Boltzmann constant； T temperature of rolled material at time,K； Ta ambient temperature,K； t time,s； emissivity. The amount of heat lost by a body d qr is give by: d qr = dTcVr Where c specific heart of rolled material, J/（ kg K） ； Vr volume 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 d qr =d qr , and Eqs.1-1 and 1-2: ar= )( 44arr TTcVASdtdT 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 cTduring radiation time tr can be calculate by intergrating the differential equation: rT=trrdta0The 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: 22dzTdadtdT Where a thermal diffusivity of rolled material ,m2/s The 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 CONVECTION In 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 cvTshould be expressed as a certain percentage of the temperature loss due to radiation: cvT=cvk（rT） Here cvTis 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 COOLING The 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: atTTkbq www )(2 Where k thermal conductivity of the surface layer, W/（ m K）； wq amount of heat passing through outer surface of the workpiece,J； b water contact length, m； w workpiece width, m； Tw water temperature, K； tw water contact time,s. The amount of heat released by a workpiece is given by: )( dw TcVq Where v volume of workpiece cooled by the water,m3； dT temperature loss due to water cooling, K. From the heat balance condition wq= wq,Eqs.1-7 and 1-8, and taking into account that tw=vbwhere V workpiece velocity, m/s and hVbw 1We obtain that the temperature loss due to water cooling is equal to dT=avbTTchkw )(2 The amount of heat absorbed by cooling water may be expressed as: wq = )( wwww TVc Where w density of water ， kg/m3； wc specific heat of water， J/（ kg K）； Vw volume of water absorbing heat， m3； From hert balance wq= wq, Eqs.1-8, 1-11, and 1-12, and also taking into account that hvdVVw Where d water flow per unit of strip width, m3/（ m s） . We obtain the following formula for the temperature rise of water: wT=abvTThckwww )(2 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 ROLLS Temperature 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 atTTkAq crcc )(4 Where Ac contact area between rolled material and work rolls， m2； k thermal conductivity of the w