流体力学与传热 ：4-7 Heat-Exchange Equipment
换热器流体力学与传热特性分析
换热器流体力学与传热特性分析换热器是工业生产过程中常见的设备之一，用于将热量从一个介质传递到另一个介质中。
换热器的工作原理就是通过流体力学的传输，将热能从一个流体中传输到另一个流体中，以达到降温、升温或保持恒温的目的。
在使用换热器时，了解流体力学与传热特性对于设备的性能和效率都有着重要的影响。
首先，我们来了解换热器的流体力学。
流体力学是研究流体运动规律和压力变化规律的学问，是研究换热器中流体传输规律的基础。
对于换热器而言，流体力学有以下几个方面的内容：第一，流体运动状态。
流体运动状态有两种类型，一种是层流运动，另一种是湍流运动。
在工业生产中，湍流运动相对较为常见。
在换热器中，流体运动的状态将直接影响到传热的效率。
第二，管道截面形状。
管道截面形状的不同，会导致流体的流动状态不同，进而影响到传热效果。
比如，圆形管道截面形状对于流体的流速是稳定的，而矩形截面形状则会导致流速的变化。
第三，流体的黏性。
黏性是测量流体抵抗剪切运动的能力的一种物理量。
高黏度的流体会产生大的阻力，使得流体的运动速度减慢，进而影响到传热效率。
其次，我们来了解换热器的传热特性。
传热特性是指当换热器中有流体传热时，影响换热器传热效能的因素。
了解传热特性可以帮助我们评估设备的性能以及选取最适合的换热器。
第一，温升与传热系数。
温升指的是流体通过换热器时，温度的变化量。
传热系数则表示传热的速率，也就是流体单位时间内传热的量。
通常情况下，温升与传热系数呈现反比例关系。
第二，换热面积。
换热面积指的是换热器中流体相互传热的表面积。
在实际使用过程中，一般会通过增加换热面积来提高传热效率。
第三，热传导方程。
热传导方程是研究物体中传热规律的数学公式。
可以通过热传导方程来描述换热器中流体传热的规律，从而实现提高传热效率的目的。
综上所述，流体力学和传热特性是换热器中非常重要的基础知识。
在实际使用过程中，我们应该充分了解流体力学的基本原理，以确保设备的正常运行。
(完整版)流体力学与传热学试题及答案
流体力学与传热学试题及参考答案一、填空题：（每空1分）1、对流传热总是概括地着眼于壁面和流体主体之间的热传递，也就是将边界层的 和边界层外的 合并考虑，并命名为给热。
答案：热传导；对流传热2、在工程计算中，对两侧温度分别为t1,t2的固体，通常采用平均导热系数进行热传导计算。
平均导热系数的两种表示方法是 或 。
答案；221λλλ+=-；221t t +=-λ3、图3-2表示固定管板式换热器的两块管板。
由图可知，此换热器为 管程，管程流体的走向为 或 。
1 2 3图3-2 3-18 附图答案：4；2→4 →1→5→3；3→5→1→4→24、黑体的表面温度从300℃升至600℃，其辐射能力增大到原来的 倍. 答案: 5.39分析: 斯蒂芬-波尔兹曼定律表明黑体的辐射能力与绝对温度的4次方成正比, 而非摄氏温度,即4273300273600⎪⎭⎫⎝⎛++=5.39。
5、3-24 用0.1Mpa 的饱和水蒸气在套管换热器中加热空气。
空气走管内，由20℃升至60℃，则管内壁的温度约为 。
答案：100℃6、热油和水在一套管换热器中换热，水由20℃升至75℃。
若冷流体为最小值流体，传热效率0.65，则油的入口温度为 。
答案：104℃ 分析： ε=2020751--T =0.65 ∴1T =104℃1 2 37、因次分析法的基础是 ，又称因次的和谐性。
答案：因次的一致性8、粘度的物理意义是促使流体产生单位速度梯度的_____________。
答案：剪应力9、如果管内流体流量增大1倍以后，仍处于滞流状态，则流动阻力增大到原来的 倍。
答案：210、在滞流区，若总流量不变，规格相同的两根管子串联时的压降为并联时的 倍。
答案：411、流体沿壁面流动时，在边界层内垂直于流动方向上存在着显著的_______________，即使____________很小，____________仍然很大，不容忽视。
答案：速度梯度；粘度；内摩擦应力 12、雷诺数的物理意义实际上就是与阻力有关的两个作用力的比值，即流体流动时的______ 与__ ____ 之比。
流体力学与传热电子教案--chapter10
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Basic law of conduction
Fourier’s law (傅立叶定律)
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Chapter 15-11Heat-Exchange Equipment 天大化工原理上册英文版课件
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9. Design of heat exchanger
• Classification of exchanger problems • (1)Design : • Given: mh, Tha,Thb (or Tca, Tcb) , and design AT,
13
4. 2-4 Exchanger
• FIGURE 15.4 A 2-4 exchanger.
14
5. Temperature patterns in multipass
fluent heat exchange热交换模型介绍
fluent heat exchange热交换模型介绍
"Fluent" 是一种计算流体力学（CFD）软件，而"heat exchange" 则指的是热交换，即在流体中传递热量的过程。
在Fluent 中，可以使用不同的模型和方法来模拟和分析流体中的热交换过程。
热交换模型在Fluent 中涉及到流体流动、传热和传质等多个方面。
以下是一些常见的Fluent 中用于热交换模拟的模型和方法：
1.传热模型：Fluent 提供了多种传热模型，包括传导、对流和辐射传热。
用户可以选择
适当的传热模型，根据系统的特点来模拟热量的传递。
2.壁面热通量：可以在Fluent 中设置不同表面的壁面热通量，以模拟具体区域的热交
换情况。
这对于热交换器、散热器等设备的仿真很重要。
3.热源和热汇：用户可以设置热源和热汇，模拟系统中的加热或散热过程。
这对于热交
换系统的设计和优化非常有用。
4.多相流和相变：在一些热交换系统中，可能涉及到多相流动和相变过程，如蒸发、冷
凝等。
Fluent 支持多相流和相变模型，以更全面地模拟系统中的热交换。
5.换热器模块：Fluent 中有专门的换热器模块，用于更方便地建模和分析换热器的性能，
包括壁面传热系数、温度分布等。
使用Fluent 进行热交换模拟需要用户详细了解系统的几何形状、边界条件、材料属性等信息，并选择合适的模型和参数。
通过模拟，用户可以获得系统内部的流动、温度场等信息，帮助设计和优化热交换设备。
流体力学与传热(英文版)chapter4(a)4.1~4.2
Note that the linear dependence on the temperature driving force ts-tf is the same as that for pure conduction in a solid of constant thermal conductivity.
Chapter 4 Heat Transfer and Its Applications
4.1 Introduction
It’s hot summer, you are going to have a lunch, a bowl of noodles.
Nature of heat flow
When two objects at different temperatures are brought into thermal contact , heat flows from the object at the higher temperature to that at the lower temperature. The net flow is always in the direction of the temperature decrease.
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流体力学与传热：第四章 干燥第1,2课时
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湿比容随其温度和湿度的增加而增大
5、湿比热 cH (humid heat)
以1kg绝干空气为基准的湿空气的比热。
值随空气的温度t及湿度Ｈ而变。
7、干、湿球温度
干球温度：在空气流中放置一支普通温度计，所测得空气的温 度为t。（dry bulb temperature)
湿球温度（wet bulb temperature) ：
❖ 用水润湿的纱布包裹温度计的测温头，纱布用水保持湿润, 即成为一湿球温度计。
❖ 将它置于一定温度和湿度的流动的空气中，达到稳态时所
图中对应不同的t，可作出许多等t线。 各种不同温度的等温
线，其斜率为(1.88t+2492)，故温度愈高，其斜率愈大。因此，
这许多成直线的等t线并不是互相平行的。
4 等相对湿度线（等φ线）
H 0.622 ps P ps
当湿空气的湿度Ｈ为一定值时，温度愈高，其相对湿度φ
值愈低，即其作为干燥介质时，吸收水汽的能力愈强，故湿空 气进入干燥器之前必须经过预热器预热提高温度，目的除了提 高湿空气的焓值使其作为载热体外，也是为了降低其相对湿度 而作为载湿体。
第一节 湿空气的性质
●湿空气：含有湿分的空气，是常用的干燥介质。 通常干燥
是在常压或减压下进行的，因此，可把这种状态下的湿空 气作为理想气体来处理。 ●基准：干燥过程中，干空气的质量不变，故干燥计算以单 位质量干空气为基准。
流体力学与传热习题参考解答(英文)(完整资料).doc
【最新整理，下载后即可编辑】1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+U 1=012P =P10Z =W=60j/kg f h 30/kg = 2U =1m/s 2(60300.5)/g 3m Z =--= 21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U UZ g+W Z g+h 22ρρP P ++=++10P =5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*.1Z 0=2Z 15=422.67x101.97W 15x9.81120246.88J /kg 12002=-+++=N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation: h f =6.5U 2where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m 3 /h.22112212f U UZ g+Z g+h 22ρρP P +=++1U 0=12P =P1Z 6m =2Z 0=2f h 6.5U=22U 6x9.81 6.5U 2=+U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure thepressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A =2b U 2.5*(33/47)1.23m /s ==22aa b b a b f U U Z g+Z g+h 22ρρP P +=++a b Z =Z22abb a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tank and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocalU U h h h 4f k d 22l ∑=+=+∑31180kg /m ρ=300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ=k=0.025mm k/d=0.025/25=0.001c l r k =0.4 k =1 k =2x0.07=0.14 el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u dRe 2.78x10ρμ==f 0.063=2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be taken as 0.61, what is the flow rate of water?0o o2gR()u c ρρρ-=20o 02gx0.6x(136001000)V u s 0.61x /4x0.0001x1000π-==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have22112212f 1-2u u gZ gZ h 22ρρP P ++=+++2212Hg H o 0 g(R h)39630N/m ρρP =P =-=2212f1-2c u u u 0 Z =0 h 4f +k 2.13ud 22===lWe can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have 22331113f1-3u u gZ gZ h 22ρρP P ++=+++311Z 0 Z 6.66m u =0==22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81ud 20.01l l ++=+==229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++112120 Z 6.66 Z 0 u 0 u 3.51P =====22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+=22229.81x6.66 3.15/226.2N 32970m ρP=++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe linefA fB total A B 22A fAA 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg =22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /hπ∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied? a.2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K443u d10x0.02x1.06 Re=1.06x10100.02x10ρμ-==> 12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr=()()0.81/3081/34Nu 0027Re Pr0.027x 1.06x10x 0.69239.66==.=.()2i ii h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1)444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====>()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=.(2)12w 2w =4421Re Re /2=2x1010=>0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k ==(3) 44333u d 2000x0.01Re 2x10100.001ρμ===>0.81/3Nu 0.023Re Pr =()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease d The first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find: (1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturatedvapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why? (a)0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21 ()20h 642.9w/m k =12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca)300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m (c) 8020LMTD=86.514020ln14080-=--℃1122L t 70/86.5L t ∆==∆ 2L 0.81L1 4.4m == (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)oi h h Uo 111+= (1)o i o h h U 1'1'1+= (2) (1)-(2)=0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=-C=2859 io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet andoutlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected) ()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆ ()()h h 12c c 21W C T -T 'W C t 't =- 2150100401515080t 15--=-- 2t 50=℃212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T)80/()20ln(20802---=∆h h m T T TTh=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p 1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln 10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =-()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭ We can see K=0.015 qe=0.026For p=3.5Kg/cm 21-s K=2k ∆P 1-s K'=2k '∆P1s K 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E e V KA 2V+V d d θ⎛⎫= ⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+q=240 V=qA=240*0.1=24 (2) K2k=∆P=∆P K'2k'∆P=∆P K'2K500'2==()()2q'+10500x249.60.4=+q’=343.6 v=34.36。
流体力学与传热电子教案--chapter10
Section 3 Heat Transfer summarize1.Heat Transfer and its application2.Nature of heat flow (three type)3.Typical heat-exchange equipment•1.Heat Transfer and its application •The purpose of heat transfer involve:(1) heat up or cooling ,(2) heat exchange(3)heat preservation, reduce heat lossThe mechanisms by the heat flows are three:Conduction Convection Radiation----热传导----对流--热辐射Nature of heat flowIn gases,---conduction occurs by the randommotion of moleculesz In metals,.--conduction results from the motion of free electronsConvection对流convective flow of heat---a particle of fluid crosses a specific surface, such as the boundary of a control volume, it carries with quantity of enthalpy.对流---流体内部质点发生相对位移的热量传递过程There are two types of convection :z natural convection 自然对流由于流体内温度不同造成的浮升力引起的流动z force convection 强制对流流体受外力作用而引起的流动。
流体力学与传热(英文版)Chapter4(d)4.5
t tLstw Nhomakorabeas
D
t
w
(a)
(b)
(c )
• （a) and (b) film type, (c) dropwise
In film condensation
the liquid condensate forms a film, or continuous layer, of liquid that flows over the surface of the tube under the action of gravity.
• Friction losses in a condenser are normally small, so that condensation is essentially a constant-pressure process.
• The condensing temperature of a single pure substance depends only on the pressure, and therefore the process of condensation of a pure substance is isothermal.
4.5.1.1 Dropwise and film-type condensation A vapor may condense on a cold surface in one of two ways, which are well described by the terms dropwise and film-type.
Phase change covers condensation, boiling, melting, crystallization and sublimation (升华, 凝华)
流体输送与传热技术
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（3）不可压缩流体
液体--一般可当作不可压缩流体 气体--低速（标准状态，一般v<68m/s）气流可按不可压缩流体处理，v=102m/s 时，不考虑压缩性引起误差2.3%。
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理想气体(完全气体)状态方程
R——气体常数 空气R=8312/29=287J/kg·K 道尔顿分压定律 例题2 1kg氢气，温度-40 ℃，V=0.1m3，求p=?
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流体输送与传热技术 Fluid conveying and heat transfer
单击此添加副标题
202X
202X
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第一部分 流体输送技术
单击此添加副标题
202X
202X
引言 流体输送的应用
油品输送、化工过程中，流体物料需要从一个设备送到另一个设备（设备间移动），或从一个工序送往另一个工序（工序间转移），逐步完成各物理过程和化学过程。常见输送方式包括：高位槽送料、真空抽料、压缩空气送料和流体输送机械送料等。
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理解流体流动的基本概念、流动阻力产生的原因；
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第一章 流体及其物理性质
1-1 流体的特性、连续性假设
易变形性和流动 流体不能抵抗任何剪切力作用下的剪切变形趋势，这是流体与固体在宏观力学行为方面的主要差异；微观上，流体分子间吸引力小。 气体与液体的区别
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(1)流体质点无线尺度，无热运动，只在外力作用下作宏观平 移运动;
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1-2 流体的主要物理性质
密度
常温下取 ρ水 =1000 kg / m3
ρ空气= 1.2 kg/m3
混合液：
流体的密度（density）
均匀流体： 一般流体：
流体力学与传热(英文版)Chapter4(f)4.7
4.7 heat-exchange equipment
In industrial processes heat energy is transferred by a variety of methods. Heat exchangers are of importance for heat energy transport. There are lots of exchangers. Including conduction-convection in exchangers, boilers, and condensers; radiation in furnaces and radiant heat dryer.
Multipass construction increases the fluid velocity, with a corresponding increase in the heat-transfer coefficient.
An even number of tube-side passes are used in multipass exchangers. The shell side may be either single-pass or multipass.
Flash p343
The disadvantages of multipass are that
(1) the exchanger is slightly more complicated ; (2) Some sections in the exchanger have parallel flow, which limits the temperature approach; (3) the friction loss rough the equipment is increased because of the larger velocities and multiplication (增大） of exit and entrance losses.
流体力学与传热电子教案--chapter11
Tha: inlet temp. of hot fluid Thb: outlet temp. of hot fluid
Tca: inlet temp. of cold fluid Tcb: outlet temp. of cold fluid
parallel-current flows ---two fluids enter at the same direction to the exchanger
３．temperature of the fluid in the tubes increases continuously as the fluid flows through the tubes.
Figure: (1) temperature of the condensing vapor ---- tube length
(2) temperature of the liquid against tube length
Temp of condensing vapor T
Δt
Δt2
Δt1
Temp of cool fluid
Temperature ºC
Length of tube m
in a single-pass exchanger： the counterflow is commonly used. Parallel flow is rarely used ．
（３）Both sides with phase change
qh=mh λ= qc=mc λ
(4) Superheated vapor condensate
过热蒸汽冷凝
For colΒιβλιοθήκη side q= mcCpc (tc2 - tc1)
流体力学与传热课件Heat Transfer and Its Applications
On the other hand, k is a function of temperature, but not a strong one.
The negative sign reflects the physical fact that heat flow occurs from hot to cold and the sign of the gradient is opposite that of the heat flow.
In using equation it must be clearly understood that the area A is that of a surface perpendicular to the flow of heat and distance n is the length of path measured perpendicularly to area A.
• k vary over a wide range. They are highest for metals and lowest for finely powdered materials from which air has been evacuated.
Fourie’s law states that k is independent of the temperature gradient.
4.2 Heat Transfer by Conduction
Conduction is most easily understood by considering heat flow in homogeneous isotropic solids because in these there is no convection and the effect of radiation is negligible.
流体力学与传热习题参考解答(英文)
1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2U =1m/s 2(60300.5)/g 3m Z =--=21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U UZ g+W Z g+h 22ρρP P ++=++10P = 5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*. 1Z 0= 2Z 15=422.67x101.97W 15x9.81120246.88J /kg 12002=-+++=N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:h f =6.5U 2where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m 3 /h.22112212f U UZ g+Z g+h 22ρρP P +=++ 1U 0= 12P =P 1Z 6m = 2Z 0=2f h 6.5U = 22U 6x9.81 6.5U 2=+ U 2.9m /s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A = 2b U 2.5*(33/47)1.23m/s == 22aa b b a b f U U Z g+Z g+h 22ρρP P +=++ a b Z =Z 22a b b a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tankand delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocal U U h h h 4f k d 22l ∑=+=+∑31180kg /m ρ= 300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ= k=0.025mm k/d=0.025/25=0.001 c l r k =0.4 k =1 k =2x0.07=0.14 el re k 4x0.75 3 k 1.5-2.2===2u v/A 2/(3600x /4x0.025)1.13m/s π===4u d Re 2.78x10ρμ== f 0.063=2f 2h 4x0.0063x300/0.025x1.13/2+ (0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be takenas 0.61, what is the flow rate of water?o u c =20o 0V u s 0.61x /4x0.0001π==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have22112212f1-2u u gZ gZ h 22ρρP P ++=+++2212Hg H o 0 g(R h)39630N/m ρρP =P =-=2212f1-2c u u u 0 Z =0 h 4f +k 2.13ud 22===lWe can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have22331113f1-3u u gZ gZ h 22ρρP P ++=+++311Z 0 Z 6.66m u =0==22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81u d 20.01l l ++=+== 229.81x6.66u /2 4.81u =+ u 3.51m /s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++112120 Z 6.66 Z 0 u 0 u 3.51P =====22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+= 22229.81x6.66 3.15/226.2N32970mρP=++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe linefA fB total A B22A fA A 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg =22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /hπ∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied?a. 2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K443u d 10x0.02x1.06 Re=1.06x10100.02x10ρμ-==>12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr=()()0.81/3081/34Nu 0027Re Pr 0.027x 1.06x10x 0.69239.66==.=.()2i i i h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find: a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain? Hint: for laminar flow, Nu=1.86[Re Pr]1/3 for turbulent flow Nu=0.023Re 0.8 Pr 1/3(1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====>()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=.(2) 12w 2w = 4421Re Re /2=2x1010=>0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭ 0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k ==(3) 44333u d 2000x0.01Re 2x10100.001ρμ===>0.81/3Nu 0.023RePr = ()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease d The first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes throughthe pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ? (2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why?(a)0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21()20h 642.9w /m k= 12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln14080-=--℃1122L t70/86.5L t ∆==∆ 2L 0.81L 14.4m == (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored) o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)= 0.80.80.80.81211111121152660u C u C 1C 1.5C -=-=-C=2859 io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected)()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆()()h h 12c c 21W C T -T 'W C t 't =-2150100401515080t 15--=-- 2t 50=℃212m 1112m 2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m 1m 2L 1.85m L 1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T Tm m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T)80/()20ln(20802---=∆h h m T T TTh=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p 1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln10-∆=()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t pD g U 18D ρρμμρP -==()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+ 5 min 22e 12V 0.1x5K +=(1) 10min 22e 1.62x1.6V 0.1x10K +=(2) From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------ pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3 660 9.10⨯10-33.50 17.1 2.27⨯10-3 233 9.10⨯10-3 Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭We can see K=0.015 qe=0.026For p=3.5Kg/cm21-s K=2k ∆P 1-s K'=2k '∆P1sK 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E e V KA 2V+V d d θ⎛⎫=⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows: (q+10)2 = 250(t+ 0.4) where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+q=240 V=qA=240*0.1=24 (2) K 2k =∆P K'2k '=∆P'2∆P =∆P K '2K 500== ()()2q'+10500x 249.60.4=+ q ’=343.6 v=34.36。
流体力学与传热(英文版)Review
Velocity distribution of laminar flow
r ur 1 umax rw
Hale Waihona Puke 2Where umax=-pri2/4L
ur is a local velocity
Laminar Flow
If the flow is laminar, the following is satisfied. 1 parabolic profile of velcity
The
relationship between gauge pressure and absolute pressure as well as vacuum
Fluid-flow phenomena
Ideal
fluid
Boundary layer
Two
kinds of flow patterns
Laminar flow Turbulent flow Criterion of flow pattern:
Re d u
Newtonian fluid
The shear stress is proportional to the shear rate (Newtonian Law of viscosity)
Unit
Example:
what is the dimension and unit of viscosity, in SI system?
Unit
conversion: P=1atm=?mmHg=?mH2O=?N/m2
Fluid
Statics and Its Applications
V4-第七章-相变对流传热-2014
珠状凝结的表面换热系数 >> 膜状凝结，但是一般无法长久保持。
2.55×105 5000～25000
传热学 Heat Transfer
7-2 膜状凝结分析解及实验关联式 层流膜状凝结
努塞尔纯净饱和蒸汽层流膜状凝结理论分析解： 液体膜层的热阻为主要因素。 基本假设： 1. 二维、稳态、常物性、层流； 2. 蒸汽静止，汽液界面无对液膜的粘滞力； 3. 忽略惯性力，液膜的运动仅取决于重力和粘滞力； 4. 壁温tw＝const，汽液界面无温差 tδ＝ts 5. 液膜内部无对流而只有导热，温度分布为线性； 6. 忽略液膜的过冷度，即认为液膜仅存在潜热；
u v x y 0 u u dp 2u l (u x v y ) dx l g l 2 y t t 2t u x v y al 2 y
2u 0 l g l 2 y 2 t 0 2 y
传热学 Heat Transfer
7-1 凝结传热的模式 凝结传热：蒸汽与低于其饱和温度的壁面接触时，将汽化潜热释放给壁面的过程。 凝结传热产生的必要条件：
tw ts
tw ts
tw ts
g
膜状凝结
g
珠状凝结
凝结模式源于气液界面的接触角θ（图7-1）
传热学 Heat Transfer
7-1 凝结传热的模式 凝结传热：蒸汽与低于其饱和温度的壁面接触时，将汽化潜热释放给壁面的过程。 珠状凝结
竖壁雷诺数
伽利略数
4hl(t s t w ) Re r l
竖壁临界雷诺数＝1600
传热学 Heat Transfer
7-2 膜状凝结分析解及实验关联式 理论分析解在一定 的假设条件下获得 实验结果修正 实验关联式
流体力学中的流动与传热耦合
流体力学中的流动与传热耦合引言流体力学是研究流体运动规律的科学，而传热是指热量从一个物体传递到另一个物体的过程。
在许多实际应用中，流体力学和传热是密切关联的，两者之间存在着耦合关系。
本文将探讨流体力学中流动与传热的耦合问题，包括热传递的基本原理、耦合方程的建立以及数值模拟方法。
热传递的基本原理热传递是指热能从高温区域传递到低温区域的过程，其基本原理可归纳为三种传热方式：导热、对流和辐射。
导热导热是指热能通过物质内部的分子传递的过程。
根据傅里叶热传导定律，导热速率正比于温度梯度。
在流体力学中，热传导的数学模型可以表示为：$$ \\mathbf{q} = -k\\cdot \ abla T $$其中，$\\mathbf{q}$为热流密度，k为热导率，ablaT为温度梯度。
这个方程描述了流体中的热传导过程。
对流对流是指热能通过流体的流动传递的过程。
对流传热由于流体的运动而产生。
在流体力学中，对流传热的数学模型可以表示为：$$ \\mathbf{q} = h\\cdot (T-T_f) $$其中，$\\mathbf{q}$为热流密度，ℎ为对流换热系数，T为物体表面的温度，T f为流体的温度。
这个方程描述了流体中的对流传热过程。
辐射辐射是指热能通过电磁波的辐射传递的过程。
辐射传热不需要介质的存在，可以在真空中传递。
在流体力学中，辐射传热的数学模型可以表示为：$$ \\mathbf{q} = \\sigma\\cdot\\epsilon\\cdot (T^4-T_f^4) $$其中，$\\mathbf{q}$为热流密度，$\\sigma$为斯特藩-玻尔兹曼常数，$\\epsilon$为辐射率，T为物体表面的温度，T f为流体的温度。
这个方程描述了流体中的辐射传热过程。
耦合方程的建立在流体力学中，流动和传热是密切关联的，两者之间存在耦合关系。
当流体中存在温度梯度时，热量会通过流动而传递。
同样地，当流体中存在流动时，流体颗粒之间的热量也会通过对流传递。
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limitation. Because of the parallel-flow pass, the exchanger is unable to bring the exit temperature of one fluid very near to the entrance temperature of the other.
multipass exchanger
The 1-1 exchanger has limitations, because when the tube-side flow is divided evenly among all the tubes, the velocity may be quite low, giving a low heat transfer coefficient.
• Sometimes the design is governed by considerations that have little to do with heat transfer, such as the space available for the equipment or the pressure drop that can be tolerated in the fluid streams.
An even number of tube-side passes are used in multipass exchangers. The shell side may be either single-pass or multipass.
In multipass exchangers, floating heads are frequently used.
Single-pass 1-1 exchanger
This exchanger, because it has one shell-side pass and one tube-pass, is a 1-1 exchanger.
In an exchanger the shell-side and tube-side heat-transfer coefficients are of comparable important, and both must be large if a satisfactory overall coefficient is to be attained.
From material and energy balances, the required heat-transfer rate is calculated. Then, using the overall coefficient and the average T, the required heat-transfer area is determined.
Multipass construction increases the fluid velocity, with a corresponding increase in the heat-transfer coefficient.
The disadvantages for a multipass construction are that (1) the exchanger is slightly more complicated ;
The final design is nearly always a compromise, based on engineering judgment, to give the best overall performance in light of the service requirements.
The velocity and turbulence of the shell-side liquid are as important as those of the tubeside fluid.
To promote crossflow and raise the average velocity of the shell-side fluid, baffles are installed in the shell.
4.7 Heat-Exchange Equipment
In industrial processes heat energy is transferred by a variety of methods.
Including conduction-convection in exchangers, boilers, and condensers; radiation in furnaces and radiant heat dryer.
(2) Some sections in the exchanger have parallel flow, which limits the temperature approach;
(3) the friction loss through the equipment is increased because of the larger velocities and multiplication of exit and entrance losses.
In simple devices these quantities can be evaluated easily and with considerable accuracy,
but in complex processing units the evaluation may be difficult and subject to considerable uncertainty.