Heat transfer heat pipe design (传热热管设计)

1、Student course work 2.Student Name: Zhang Miao Student ID: 4114731E-mail: Date: 09.01.2011Introduction & Problem descriptionIntroduction of heat pipe technologyA heat pipe is a simple device that can quickly transfer heat from one point to another. It consists of a sealed aluminum or copper cont

2、ainer whose inner surfaces have a capillary wicking material (a wick material is to aid the flow of the liquid 1). It differs from a thermosyphon by virtue of its ability to transport heat against gravity by an evaporation-condensation cycle with the help of porous capillaries that form the wick.2Fi

3、g.1 Working principle of heat pipe 3Problem descriptionThis design report aims to solve a task concerning a heat transfer solution, which is based on designing cooling system by using a heat pipe system to dissipate the heat from a hot component on the PCB (see the red-colored component of 50 W in F

4、igure 1 shown below), and to make sure the temperature of this component will not exceed 85°C.Fig2. Component of the PCBThermal specifications of the component are shown below:Ambient temperature: 35°CPower load of the component: 50 WComponent dimension: 40mm x 40mm x 2mmSpace restriction

5、of the cooling system: 320mm (L) x 95mm (W) x 25mm (H)The purpose of the design focus on the following issues:Choose the suitable working fluid for the heat pipes and give explanations.Determine the diameter, condensation section and evaporation section of the heatpipe(s).Complete the heat pipe(s) c

6、alculation and determine the numbers of heat pipe(s).Cooling system designCalculate the heat transfer area of fin stackMethodContainerThe function of the container is to isolate the working fluid from the outside environment. It has to therefore be leak-proof, maintain the pressure differential acro

7、ss its walls, and enable transfer of heat to take place from and into the working fluid. 4The heat pipe in this design will be made by copper (the thermal conductivity is of 380 W/m K) with porous wick structure.Choice of suitable working fluidsOperating vapour temperature range is the first thing t

8、o be considered in theidentification of a suitable working fluid. Within the approximate temperature band, several possible working fluids may exist, and a variety of characteristics must be examined in order to determine the most acceptable of these fluids for the application considered, several va

9、rious working fluids are shown in appendix 5.Qualification of the working fluids: Compatibility with wick and wall materials; good thermal stability; wettability of wick and wall materials; vapour pressure not too high or low over the operating temperature range; high latent heat; high thermal condu

10、ctivity; low liquid and vapour viscosities; high surface tension; acceptable freezing or pour point.In this design of the heat pipe, water is chosen as the working fluid for the following advantages:1. Water is the most common heat pipe working fluid with no risk and low price, the former article ha

11、ve suggested that as working fluids, water is recommended to be used with copper material by past successful usage6;2. The actual temperature controlled in this task range from 35°C to 85°C, while the operating temperature of water is from 1°C to 325°C which can be examined from

12、the table in appendix7. In other words, the temperature range of water is totally enough for this working environment.3. Water is a working fluid with high specific heat capacity; low density with 1000kg/m3 under 0; good thermal stability; appropriate vapour pressure (not too high or low over the op

13、erating temperature range); low vapour density of 0.60kg/m3 high latent heat about 2258kJ/kg; high surface tension about 0.059N/m; acceptable boiling point of 373 K ,etc8Wick materialsThe wick material has been chosen and relevant information is as follows: Table 1. Wick informationFinBased on Newto

14、ns law of cooling, Q=AT, for a given heat transfer coefficient and given fluid and wall temperatures, the heat transfer can be increased by increasing the heat transfer area. One way of doing this is to add fins or studs project into the fluid on the heat transfer surface (or on one side of a heat e

15、xchanger); the effective heat transfer rate is thus increased.In this design, fins are installed on the condenser section to dissipate Q from the heat pipe.Fig.3 Fin stack installed on the condenser section 9Design calculation and explanationDesign description:1. The cooling system is designed in an

16、d out of the given box including evaporatorsection (inside the box) and condenser section (outside the box);2. Fin section is designed to cover condenser section as be shown in the figure 5.3. The adiabatic section of the heat pipe is inside the box of which the length is60mm (does not included in t

17、he calculation part for its isothermal character).4. The heat pipes are installed above the hot temperature area using seccotine andsimple bracket.5. Two heat pipes are assumed at the first stage of this task, for each one thedimension is 18mm, which is satisfied by the given condition that the comp

18、onent dimension is 40mm x 40mm. So 18mm ×2= 36mm < 40mm, which means that both two pipes will straight above the component area, and the height is also allowed by the 23mm (25mm-2mm) of the box.6. The main regions of the heat pipe are shown below:Fig4. The main regions of the heat pipeCalcul

19、ationAssumption on the design: steady state rounded.TTair,out1. QHP=air,in RHPKnown that heat from the hot component on the PCB: QHP= 50WThe highest temperature on the surface of the component: Tair,in=85°CThe Ambient temperature: Tair,out = 35°CThus, RHP= TQHPair,inTair,out=8535=1 50W2.Ca

20、lculate of thermal resistanceFig5. Thermal resistance along heat pipeThermal resistance applied to a heat pipe:RHP=Rconv,air,in+Rcond,hp,in+Rcond,wick+Revap+Rconden+Revap,wick+Rcond,hp,out+Rconv,air,out=1air,indo,HPLevap1evapdi,HPLevapln(ro,HP/ri,HP)HP2Lcon+ln(ro.HP/ri,HP)HP2Levap1+ln(ro,wick/ri,wic

21、k)wick2Lcond+Where:+condi,HPLcon1+ln(ro,wick/ri,wick)wick2diLevap+air,outdo,HPLcon(1) 𝑅𝑐𝑜𝑛𝑣,𝑎𝑖𝑟,𝑖𝑛- Convection in the evaporator region. 𝑅𝑐𝑜𝑛𝑣,𝑎𝑖𝑟,𝑖

22、𝑛=𝛼1𝑎𝑖𝑟,𝑖𝑛𝜋𝑑𝑜,𝐻𝑃𝐿𝑒𝑣𝑎𝑝Design assumption:1. According to the reference6-8 in the book principle of chemical engineering in page231 10, the thermal convection value of

23、 the device: air,in=260W/ (m2K) 2. The length of the evaporation section: Levap=80mm = 0.08m 3. The outer diameter of each heat pipe: do,HP=18mm=0.018m4. Assume there are expected 2 heat pipes are able to be designed in the cooling system;Rconv,air,in=1/(260×× 0.018×0.08) = 0.85(2) &#

24、119877;𝑐𝑜𝑛𝑑,𝑝,𝑖𝑛- Conduction across the heat pipe wall (evaporator region) 𝑅𝑐𝑜𝑛𝑑,𝑝,𝑖𝑛=𝑙𝑛(𝑟𝑜.𝐻𝑃/𝑟𝑖,𝐻𝑃

25、)𝜆𝐻𝑃2𝜋𝐿𝑒𝑣𝑎𝑝Design assumption:1. Known that the thermal conductivity is of 380 W/m K. HP=380W/mK;2. The thickness of the heat pipe is 2mm, so the inner diameter of the heat pipe is di=16mm= 0.016m Rcond,hp,in=380 ×2×

26、20587;×0.08= 6.2× 10-4ln0.018(3) 𝑅𝑐𝑜𝑛𝑑,𝑤𝑖𝑐𝑘-Conduction across the wick material. 𝑅𝑐𝑜𝑛𝑑,𝑤𝑖𝑐𝑘=𝑙𝑛(𝑟𝑜,𝑤⻕

27、4;𝑐𝑘/𝑟𝑖,𝑤𝑖𝑐𝑘)𝜆𝑤𝑖𝑐𝑘2𝜋𝐿𝑤𝑖𝑐𝑘Design assumption:1. FM1308 will be used as the wick material according to the former research11 2. The model of the wick s

28、ection will be parallel model;3. Here it is assumed that the wick and working fluid are effectively in parallel; The formula of the thermal conduction is given as:wick= f+ (1)s Where:f is the thermal conductivity of the working fluid. In our design, the working fluid is water, assume the water will

29、be evaporated at the temperature of 20, so f=0.5985W/m2K 12 ;s is the thermal conductivity of the wick material, s72 W/m2K is voidage fraction: w=0.52wick=0.52×0.5985+(10.52)×72 =34.9 W/m2K 4. The outer diameter of the wick: do,wick=16mm5. The thickness of the wick is 3 mm, so the inner di

30、ameter of the wick: di,wick=13mm6. The length of the wick of condense region is designed to be Lcond=100mm=0.1mRcond,wick=ln(0.016/0.013)/34.9×2××0.01=0.09(4) 𝑅𝑒𝑣𝑎𝑝- Wick thermal conductivity (evaporator region). 𝑅𝑒𝑣w

31、886;𝑝=𝛼Design assumption:1. The heat transfer coefficient at the evaporator outer surfaces: evap=4000W/(m2K)Revap=1/(4000××0.016×0.08)= 0.06(5) 𝑅𝑐𝑜𝑛𝑑𝑒𝑛-Wick thermal conductivity (condenser region). 𝑅Ү

32、88;𝑜𝑛𝑑𝑒𝑛=𝛼Design assumption:1. The heat transfer coefficient at the condenser outer surfaces: con=4000W/(m2K)According to information listed above,Rconden=1/(4000××0.016×0.1)=0.05(6) Revap,wick - Conduction across the wick material. 

33、9877;𝑒𝑣𝑎𝑝,𝑤𝑖𝑐𝑘=𝑙𝑛(𝑟𝑜,𝑤𝑖𝑐𝑘/𝑟𝑖,𝑤𝑖𝑐𝑘)𝜆𝑤𝑖𝑐𝑘2𝜋𝐿𝑒𝑣𝑎

34、;𝑝1𝑐𝑜𝑛1𝑒𝑣𝑎𝑝𝜋𝑑𝑖,𝐻𝑃𝐿𝑒𝑣𝑎𝑝𝜋𝑑𝑖,𝐻𝑃𝐿𝑐𝑜𝑛𝑅𝑒𝑣𝑎𝑝,ү

35、08;𝑖𝑐𝑘=(7) 𝑅𝑐𝑜𝑛𝑑,𝑝,𝑜𝑢𝑡ln(16/13)=0.01 - Conduction across the heat pipe wall (condenser region).𝑙𝑛 𝑟𝑜,𝐻𝑃𝑖,𝐻𝑃𝑅𝑐⻖

36、0;𝑛𝑑,𝑝,𝑜𝑢𝑡= 𝜆𝐻𝑃2𝜋𝐿𝑐𝑜𝑛According to the information listed above,Rcond,hp,out=ln(0.018/0.016)/380×2××0.1=4.9×10-4(8) 𝑅𝑐𝑜𝑛𝑣,⻔

37、6;𝑖𝑟,𝑜𝑢𝑡- Convection in the condenser region. 𝑅𝑐𝑜𝑛𝑣,𝑎𝑖𝑟,𝑜𝑢𝑡=𝛼1𝑎𝑖𝑟,𝑜𝑢𝑡𝜋𝑑𝑜,𝐻𝑃&

38、#119871;𝑐𝑜𝑛Design assumption:According to the reference6-8 in the book principle of chemical engineering in page231, the thermal convection outside the condense region: air,out= 300 W/m2KRconv,air,out=1=0.59 The total R for each heat pipe designed:Rper= 0.85+6.2× 10-4 +0

39、.09 + 0.08 + 0.06 +0.05 +0.01+ 4.9×10-4+0.59 = 1.73From the former conclusion we know that the total thermal resistance applied to a heat pipe RHP = 1. In the design, 2 heat pipes are assumed, in this case, each pipe should have the ability to eliminate: Q1=For each pipe: Qper=25<50 RperQt2=

40、25W. , so Rper <2.As the conclusion above, Rper =1.73 which is satisfied by the assumption (<2). This calculation result is coincident with the expected number of heat pipe of the assumption stage, and the size of the heat pipe is satisfied by the given condition. So, the number of the heat pi

41、pes designed to meet the requirement is determined: N=2 As a result, there should be 2 heat pipes installed above the hot component on the PCB to dissipate the heat. It is thus concluded that this design is feasible and satisfied.Fin calculation:Given conditions:1. There are 2 heat pipes, each heat

42、pipe dissipate Q=25W;2. 𝛼𝑎𝑖𝑟,𝑜𝑢𝑡=300 W/m2K3. Total thermal resistance except𝑅𝑐𝑜𝑛𝑣,𝑎𝑖𝑟,𝑜𝑢𝑡 is R=1.730.59=1.144. Temperature outside Tair,out=355. Temperature

43、 inside can be calculated as: Tinside=85R=851.14=63 So the area of the fin can be determined: Q=ATA=T=300×(6335)=0.003m2Since there are 2 heat pipes, so the total area of the fin stack should be: A= 0.003×2=0.006 m2Q25Q25Design summary / ConclusionThis report accomplished the task to dissi

44、pate the 50W heat from the component area by designing a heat pipe system and a cooling system including evaporator section and condenser section. The accomplished designs are listed as follows:1. Water is chosen as the suitable working fluid for the heat pipes for its low price,no risk, appropriate operating temperature range and good thermal stability, high latent