北京科技大学 北科大 03 04年传热学 考研真题及答案解析

8-9 A horizontal hot water pipe passes through a large room. The rate of heat loss from the pipe by natural convection and radiation is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric pressure is I atm. 4 The temperature of the outer surface of the pipe is constant. Properties The properties of air at 1 atm and the film temperature of (T S +T ∞)/2 = (65+22)/2 = 43.5°C = 316.5 K are (Table A-15)k = 0.0272 C m W D ⋅/ 521.7210/m s ν−=× Pr=0.710 100316.0316511−===K KT f β Analysis (a) The characteristic length in this case is the outer diameter of the pipe, δ= D = 0.06 m. Then,32132522()(9.8/)(0.00316)(6522)(0.06)Pr (0.710)690,298(1.7210/)s g T T m s K K m Ra m s βδν−∞−−−===×1.13}])710.0/559.0([!)298,690(387.06.0{}]Pr)/559.0([!387.06.0{227816961227816961=++=++=Ra NuC m W m C m W Nu kh D D ⋅=⋅==2/94.5)1.13(06.0/0272.0δ 251.1)8)(06.0(m m m DL A ===ππW C m C m W T T hA Q s 7.385)2265)(51.1)(/94.5()(22=−⋅=−=∞⋅D D(b) The radiation heat loss from the pipe is44282444()(0.8)(1.51)(5.6710/)[(65273)(22273)]375s surr Q A T T m W m k K K Wεσ⋅−=−=×⋅+−+= 8-108-17 A circuit board is cooled by a fan that blows air upwards. The average temperature on the surface of the circuit board is to be determined for two cases.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gaswith constant properties. 3 The atmospheric pressure at that location is 1atm.Properties The properties of air at 1 atm and 1 atm and the anticipatedfilm temperature of K C T T s 5.3205.472/)3560(2/)(==+=+∞Dare (Table A-15)k = 0.0275C m W D ⋅/ 521.7710/m s ν−=× Pr = 0.710 100312.05.32011−===K KT f β AnalysisWe assume the surface temperature to be 60°C. We will check this assumption later on andrepeat calculations with a better assumption, if necessary. The characteristic length in this case is the length of the board in the flow (vertical) direction, δ = 0.12 m. Then the Reynolds number becomes 52(0.5/)(0.12)Re 33901.7710/V m s m m sδν∞−===× which is less than critical Reynolds number (5x]05 ). Therefore the flow is laminar and the forced convection Nusselt number and h are determined from5.34)710.0()3390(664.0Pr Re 664.0315.0315.0====L khl Nu C m W m C m W Nu kh D D ⋅=⋅==2/9.7)5.34(12.0/0275.0δ 2024.0)2.0)(12.0(m m m A == Then622103)710.0()024.0)(/9.7()05.0)(100(35)(×=⋅+=+=→−=⋅∞∞⋅m C m W W C hA Q T T T T hA Q s s D D which is sufficiently close to the assumed value in the evaluation of properties. Therefore, there is no need to repeat calculations.(b) The Rayleigh number is321362522()(9.8/)(0.00312)(6035)(0.12)Pr (0.710)310(1.7710/)s g T T m s K K m Ra m s βδν−−∞−−−===×× 5.24)103(59.059.041641=×==−Ra NuThis is an assisting flow and the combined Nusselt number is determined from 2.38)5.245.34()(3133=+=+=n natural n forced combined Nu NuNu Then C m W mC m W Nu kh combined D D ⋅=⋅==2/8.8)2.38(12.0/0275.0δ And C m C m W W C hA Q T T T T hA Q s s D D D 8.58)024.0)(/8.8()05.0)(100(35)(22=⋅+=+=→−=⋅∞∞⋅Therefore, natural convection lowers the surface temperature in this case by about 2°C.。

北京科技大学2003年招收攻读硕士学位研究生入学考试试题一、简要回答下列问题（每小题8分）1．写出下列准则的表达式，并注明各量的物理意义:Bi Fo Gr Nu Re Eu2．在工业用换热器中，有的将肋片装在圆管的内表面，圆管外表面为光管；有的换热器正好相反，为什么？举例说明。

3．为什么称大容器饱和沸腾中的临界热流为沸腾危机或烧毁点？ 4．试述强化管内流体对流换热常采用的方法，并简述理由。

5．什么是灰体？在实际工程计算中我们把物体表面当作灰体处理应满足什么条件？而又为什么要满足这样的条件？二、（20分）外径为mm 60 ,表面温度为℃200的蒸汽管道，外绝缘层为mm 100的正方形截面，绝热材料导热系数为()K m W/1.0⋅,绝热层外表温度为℃50。

试计算每米长管道的热损失。

若用同样多的材料做成圆形截面，热损失又是多少？（可假设两种情况下绝热层外表面与环境的对流换热系数相同）三、（20分）如图中所示，位于导热系数分别为A k 和B k 两种材料界面处的节点O 。

试推导出稳态状况下，无内热生成的有限差分方程，若划分网格时令y x ∆=∆，则此导热问题的有限差分方程可简化为何种形式。

四、（25分）一套管式换热器，饱和蒸汽在内管中凝结，使内管的外壁温度保持在℃100。

初温为℃25、质量流量为s kg /8.0的水从套管换热器的环形空间流过，换热器外壳绝热良好。

环形夹层内管外径为mm 40，外管内径为mm 60，试确定把水加热到℃55时所需的套管长度及管子出口截面处的局部热流密度。

（不考虑温差修正，水的物性参数为：()K m W ⋅=/635.0λ ，s Pa ⋅⨯=-6103.653μ，()K kg J C p ⋅=/4174，31.4Pr =）五、（25分）如图所示，由半径为m 1的1/4圆的两直径及圆周所组成的通道，在垂直纸面方向无限长。

表面1的温度和发射率分别为℃2001=t ，2.01=ε，表面2温度和发射率分别为℃272=t ，7.02=ε,表面3的发射率为5.03=ε，单位长度热流W/m 1000q 3=。

北京科技大学2011年硕士学位研究生入学考试试题=============================================================================================================试题编号： 814 试题名称：材料科学基础（共 3 页）适用专业：材料科学与工程说明：所有答案必须写在答题纸上，做在试题或草稿纸上无效。

=============================================================================================================一、名词解释(5分/题，共40分)1) 空间点阵2) 临界分切应力3) 滑移系4) 堆垛层错5) 调幅分解6) 脱溶7) 上坡扩散8) 再结晶温度二、分别给出下列离子晶体的布拉菲点阵类型和下面晶胞中正、负离子的个数。

(下图中的点阵参数均为a=b=c，α=β=γ=90º) (15分)NaCl CsCl ZnSCaF2CaTiO3三、写出面心立方结构和体心立方结构金属的密排面(或相对密排面)的晶面指数、画出密排面(或相四、组元A和组元B的熔点分别为1000℃和700℃，室温时B在A的固溶体α中的固溶度是x B=0.05，A在B的固溶体β中的固溶度是x A=0.10；在700℃时有一个三相平衡，在此温度α固溶体的成分是x B=0.1，一个成分为x B=0.30的合金在稍高于700℃时存在50%α相和50%液相，在稍低于700℃时则存在液相和化合物A3B两相；在500℃时存在另一个三相平衡，液相(x B=0.65)分解为化合物A3B和β固溶体(x B=0.85)两相。

试构造一个合理的A-B二元相图。

(15分)五、根据下面的Al-Cr-Si体系的局部液相面投影图，写出该图中的四相不变反应式。

北京科技大学2012年硕士学位研究生入学考试试题============================================================================================================= 试题编号： 811 试题名称：传热学（共 2 页）适用专业：动力工程及工程热物理、动力工程说明：所有答案必须写在答题纸上，做在试题或草稿纸上无效。

============================================================================================================= 一．（30分，每小题5分）简要回答下列问题：（1）对于第一类边界条件的稳态导热问题，其解析解与导热系数有没有关系？（2）采用集总参数法求解物体非稳态导热时，需满足什么条件？说明为什么要满足此条件。

（3）在用裸露热电偶测定气流的非稳定温度场时，怎样可以改善热电偶的温度响应特性？（4）温度均匀的空腔壁面上的小孔具有黑体辐射的特性，那么空腔内部壁面的辐射是否也是黑体辐射？（5）试述基尔霍夫对灰体的应用，简要说明该定律在辐射换热计算中的作用。

（6）有一台采暖用的散热器，用管内的热水来加热管外的空气。

为了提高散热器的散热效果，有人提出1）管外加装肋片；2）将原来的钢管换成铜管。

试从传热角度来评价这个方案。

二．（25分）考察一管长6m，内、外径分别为7.4 cm、8.0 cm，导热系数为14 W/m · °C 的压缩空气管道。

管的外表面由总功率为300W的电阻带均匀加热，外包绝热层，通过绝热层的散热损失为15％。

管内空气的平均温度为 10°C，管道内表面的对流换热系数为30 W/m2 · °C。

试：(1)建立管道一维稳态导热的微分方程和边界条件；(2)求解管道内的温度分布；(3)确定管道的内、外表面温度。

北京科技大学传热学第7章习题答案7-10 Hot engine oil flows over a flat plate. The total drag force and the rate of heat transfer per unitwidth of the plate are to be determined.Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is cr Re = 5x 105. 3Radiation effects are negligible.Properties The properties of engine oil at the film temperature of (s T + ∞T )/2 = (80+30)/2 =55°C = 328 K are (Table A-10)ρ=8673/m kg υ=123×10-6s m /2k = 0.141 W/m.°C Pr =1505Analysis Noting that L = 6 m, the Reynolds number at the end of the plate is5261046.1/10123)6)(/3(Re ×=×==?∞sm m s m LV L υ which is less than the critical Reynolds number. Thus we have laminar flow over the entire plate. Theaverage friction coefficient and the drag force per unit width are determined fromf C = 1.328 5.0Re ?L = 1.328(1.46 x 510)5.0 = 0.00347D F = N s m m kg m V A C f 3.812)/3)(/867()16)(00347.0(22322=×=∞ρ Similarly, the average Nusselt number and heat transfer coefficient are determined using the laminar flow relations for a flat plate,2908)1505()1046.1(664.0Pr Re 664.0315.05315.0=×===L k hl Nu C m W mC m W Nu l k hD D ?=?==2/3.68)2908(6/141.0 The rate of heat transfer is then determined from Newton's law of cooling to be 2020()(68.3W/m C)(61m )(8030)20.5kW s Q hA T T C ∞=?=?×?= 7-12 The top surface of the passenger car of a train in motion is absorbing solar radiation. Theequilibrium temperature of the top surface is to be determined.Assumptions l Steady operating conditions exist. 2 The critical Reynolds number , is Re cr = 5x 105. 3Radiation heat exchange with the surroundings is negligible.4 Air is an ideal gas with constantproperties.Properties The properties of air at 30°C ?300 K ar e (Table A-11)k =0.0261 W/m ?°C ν=1.57x10-5 m 2 /s Pr = 0.712Analysis The rate of convection heat transfer from the top surface of the car to the air must be equal tothe solar radiation absorbed by the same surface in order to reach steady operation conditions. TheReynolds number is[]652701000/3600/(8)Re 9.9101.5710/L m s m V L m sν∞?×===×× which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow.Using the proper relation for Nusselt number, the average heat transfer coefficient and the heat transferrate are determined to be270,12)712.0](871)109.9(037.0[Pr )871Re 037.0(318.063/18.0=?×=?==L khl Nu C m W m C m W Nu L k hD ?=?==2./40)270,12(8/0261.0The equilibrium temperature of the top surface is then determined by taking convection and radiationheat fluxes to be equal to each otherC C m w m w C h q T T T T h q q conv s s conv radD D D 35/40/20030)(22=?+=+=→?==?∞∞??7-14 A steam pipe is exposed to windy air. The rate of heat loss from the steam is to be determined.Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an idealgas with constant properties.Properties The properties of air at 1 atm and the film temperature of (∞+T T s )/2= (90+7)/2 =48.5°C = 321.5 K are (Table A-11)k =0.0275 W/m.°C ν=1.77x10-5 m 2 /s Pr = 0.710Analysis The Reynolds number is52[(50/)(1000/)/(3600/)](0.08)Re 62,7741.7710/V Dkm h m km s h m m sν∞?===× The Nusselt number corresponding to this Reynolds number is determined to be54854132315.0]28200Re (1[]Pr)/4.0(1[Pr Re 62.03.0+++==k hD Nu 54854132315.0])28200775,62(1[])710.0/4.0(1[)710.0()775,62(62.03.0+++==265.6 The heat transfer coefficient and the heat transfer rate becomec m W m c m W Nu Dk h D D ?=?==/3.91)6.265(08.0/0275.0 2251.0)1)(08.0(m m m DL A ===ππW C m c m W T T hA Q s conv1902)790)(251.0)(/3.91()(2=??=?=∞?D D (per m length) 7-16 Air is to be preheated before entering a furnace by hot water. Determine the rate of heattransfer per unit length of the tubes.Assumptions 1 Steady operating conditions exist. 2 Radiation effects is negligible. 3 The air is anideal gas.Properties The properties of fluid at the temperature of 15°C are (Table A-9)ρ= 1.225 kg/m 3 k = 0.02476C m W D/ 51.4710/kg m s ν?=×? C P =1007J/K g·℃ Pr = 0.7323The surface temperature T s =90℃,P rs =0.7132Analysis The velocity of fluid and the Reynolds number are s m s m D S V S V T T /55.61.25/8.35max =?×=?=?256.55 2.110Re 9,3571.4710m hV D ν??××===× which is greater than 1000 and Nu=8<16 (Table 7-2，7-3)7.74)7132.0/7323.0(7323.0935727.097.0)Pr (Pr/Pr Re 27.097.097.036.063.036.063.0=××××=×==s N Nu UDHeat transfer coefficient isC m W mC m W NuD k h h o o ?=?==2/5.88)7.74(021.0/02476.0 m DL N As22.4==π s kg L vNS m T /862.1.==ρThe exit temperature isCe eTi Ts Ts Te p c m h As °××=???=???=8.28)1590(90)()1007862.1/()5.8822.4()/()(.The rate of heat transfer is wTi Te m c Q p 25300)158.28(862.11007)(..=?××=??=7-19 Water is to be heated in a tube equipped with an electric resistance heater on its surface. Thepower rating of the heater and the inner surface temperature are to be determined.Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner surfacesof the tube are smooth.Properties The properties of water at the average temperature of (80+12) 12 = 46 °C ~ 45°C are (TableA-9)ρ= 992.1 kg/m 3 k = 0.631C m W D ?/ 62/0.65810/m s νμρ?==×C kg J C pD ?=/4179 Pr = 4.32Analysis The power rating of the resistance heater is33(992.1/)(0.008/min)7.937/min 0.132/m V kg m m kg kg s ρ?==== W C C Kg J s kg T T Cp m Q i e 37510)1280)(/4179)(/132.0()(=??=?=??D DThe velocity of water and the Reynolds number ares m m s m A V V c m /4244.04/)02.0(/)60/108(233=×==??π 62(0.4244/)(0.02)Re 12,8900.65810/m hV D m s m m sν?===× which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m m D L L h l h 2.0)02.0(1010==≈≈which is much shorter than the total length of the duct. Therefore, we can assume fully developedturbulent flow in the entire duct, and determine the Nusselt number from2.80)32.4()890,12(023.0Pr Re 023.04.08.03.08.0====khD Nu h Heat transfer coefficient isC m W mC m W NuD k h h D D ?=?==2/2530)2.80(02.0/631.0 Then the inner surface temperature of the pipe at the exit becomes)(,e e s T T hA Q ?=?C T m m C m W W sD D )80)](7)(02.0()[/2530(510,372??=πC T e sD 7.113,=7-20 Flow of hot air through uninsulated square ducts of a heating system in the attic is considered.The exit temperature and the rate of heat loss are to be determined.Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 Airis an ideal gas with constant properties. 4 The pressure of air is 1 atm.Properties We assume the bulk mean temperature for air to be 350 K since the mean temperature ofair at the inlet will drop somewhat as a result of heat loss through the duct whose surface is at a lowertemperature. The properties of air at 1 atm and this temperature are (Table A-1 I )ρ= 1.009 kg/m 3 k = 0.0297C m W D ?/ 522.0610/m s ν?=×C kg J C pD ?=/1008 Pr = 0.706Analysis The characteristic length that is the hydraulic diameter, the mean velocity of air, and theReynolds number are,m a aa P A D c h 15.04442==== s m m s m A V V c m /4244.04/)02.0(/)60/108(233=×==??π 62(0.4244/)(0.02)Re 12,8900.65810/m hV D m s m m sν?===× which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughl y m m D L L h l h 5.1)15.0(1010==≈≈which is much shorter than the total length of the duct. Therefore, we can assume fully developedturbulent flow in the entire duct, and determine the Nusselt number from84)706.0()362,32(023.0Pr Re 023.04.08.03.08.0====khD Nu h Heat transfer coefficient isC m W mC m W NuD k h h D D ?=?==2/6.16)84(15.0/0297.0 Next we determine the exit temperature of air,26)10)(15.0(44m m m al A ======s kg m m kg V m /1009.0min)/10.0)(/009.1(33ρ(16.6)(6)/()(0.1009)(1008)()70(7085)75.7p hA mC e s s i T T T T e e C =??=??=DThen the logarithmic mean temperature difference and the rate of heat loss from the air becomesC T T T T T T T i s e s i eD 5.985706.7570ln(856.75)ln(ln ===Δ 22ln(16.6/)(6)(9.5)941Q hA T W m C m C W ?=Δ=?=D DNote that the temperature of airdrops by almost 10°C as it flows in the duct as a result of heat loss.。

北京科技大学2004年硕士学位研究生入学考试量子力学试题答案--------------------------------------------------------------------------------------------------------------------- 物理常数：光速：812.99810c m s -=⨯⋅；普朗克常数：346.62610h J s -=⨯⋅；玻尔兹曼常数：231.38110/B k J K -=⨯；电子质量：319.10910e m kg -=⨯；碳原子质量：2612 2.00710C m u kg -==⨯；电子电荷：191.60210e C -=⨯---------------------------------------------------------------------------------------------------------------------1） 物质波（30分）：1924年，德布洛意提出物质波概念，认为任何实物粒子，如电子、质子等，也具有波动性，对于具有一定动量p 的自由粒子，满足德布洛意关系：_____________________________（10分）；假设电子由静止被150伏电压加速，求加速后电子的的物质波波长：_____________________________（10分，保留1位有效数字）；对宏观物体而言，其对应的德布洛意波波长极短，所以宏观物体的波动性很难被我们观察到，但最近发现介观系统（纳米尺度下的大分子）在低温下会显示出波动性。

计算1K 时，60C 团簇（由60个C 原子构成的足球状分子）热运动所对应的物质波波长：_____________________________（10分，保留1位有效数字）。

解：德布洛意关系（10分）：h pλ=电子物质波波长：34101.010h m p λ--====≈⨯ 60C 物质波波长：34101.010m λ--===≈⨯2） 薛定谔方程（40分）：质量为m 的一个粒子在边长为a 的立方盒子中运动，粒子所受势能(,,)V x y z 由下式给出：()()()0,0,;0,;0,(,,),x a y a z a V x y z others ∈∈∈⎧⎪=⎨∞⎪⎩；（i ）列出定态薛定谔方程，并求系统能量本征值和归一化波函数（10分）；（ii ）假设有两个电子在立方盒子中运动，不考虑电子间相互作用，系统基态能是多少？并写出。

目　录2014年北京科技大学819化工原理概论考研真题2013年北京科技大学819化工原理概论考研真题2003年北京科技大学化工原理考研真题2014年北京科技大学819化工原理概论考研真题一、填空题（每空2分，15空，共30分）1．并联管路各支管相等。

2．请写出两种变压头流量计：和。

3．离心泵流量的大小取决于泵的、和。

4．旋转半径，颗粒进入离心场的切线速度，离心分离效果更好。

5．热传导特点：。

6．流体的粘度，对流传热系数愈低。

7．在一单壳单管程无折流挡板的列管式换热器中，用冷却水将热流体由100℃冷却至40℃，冷却水进口温度15℃，出口温度30℃，在这种温度条件下，逆流的平均温度差为℃和并流的平均温度差为℃。

就提高传热推动力而言，逆流并流及其他形式流动。

8．采用折流和其他复杂流动的目的是为了提高传热系数，其代价是平均温度差相应。

9．在逆流吸收塔中，用清水吸收混合气中的溶质。

已知液气比L/V 为2.9，平衡关系可表示为Y=1.8X (X、Y为摩尔比)，吸收率为90%，则液气比与最小液气比之比值为。

二、判断题，正确的在题末括号内填“Ö”、错误的在题末括号内填“´”。

（每空2分，10空，共20分）1．从物料中脱除水分的过程称为去湿。

（ ）2．静止流体中颗粒的沉降过程可分为两个阶段，起初为加速段而后为等速段。

（ ）3．换热器的平均传热系数K必须和所选择的传热面积相对应，选择的传热面积不同，总传热系数的数值不同。

工程上大多以平均表面积为计算基准。

（ ）4．采用折流和其他复杂流动的目的是为了提高传热系数，其代价是平均温度差相应增大。

（ ）5．气体吸收可以亨利定律表示为，，。

难溶气体，溶解度小，E大，H小，m。

（　）6．一般情况下，气相中的分子扩散系数D的范围：10-5～10-4m2/min。

（ ）7．精馏塔塔釜加热，与间接蒸汽加热相比，直接蒸汽加热需要的理论板数略有增多（　）8．相对湿度代表湿空气的不饱和程度，其数值越低，表明该空气偏离饱和程度越远，干燥能力越大。

2004年北京科技大学471工程热力学考研真题
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