Phase Changes Heat must be added or removed to change a substance from one phase to another. Phases and Phase Changes. Evaporation

Transcription

1 Applied Heat Transer Part One (Heat Phase Changes Heat must be added or remoed to change a substance rom one phase to another. Ahmad RAMAZANI S.A. Associate Proessor Shari Uniersity o Technology انتقال حرارت کاربردی احمد رمضانی سعادت ا بادی Autumn, 185 ( /28/2006 The quantity o heat required or a phase change is called the latent heat, which is dierent or each substance. Phase changes occur at a constant temperature (at the melting point or the boiling point. Eaporation has a Transer cooling with Phase eect Change on a liquid. 2 Phases and Phase Changes INCREASING TEMPERATURE SOLID LIQUID GAS PLASMA Adding or remoing energy rom matter may cause it to change rom one phase to another: Eaporationrom liquid to gas phase. Condensation rom gas to liquid phase. Freezing rom liquid to solid phase. Meltingrom solid to liquid phase. Sublimation rom solid to gas phase. Ionization rom gas to plasma phase. Eaporation Eaporation occurs at the surace o a liquid. Molecules whose kinetic energy is greater than the aerage escape the surace o the liquid most readily. As a result, the aerage kinetic energy o the molecules that remain in the liquid is lowered, and its temperature is lowered (eaporatie cooling. 4

2 Condensation Molecules can change rom the gas to the liquid phase by condensing. Only the more energetic gas molecules may penetrate the surace and become part o a liquid. Condensation is a warming process. Fogand clouds are ormed when slow-moing water molecules collide and coalesce (stick together, orming tiny droplets. 5 Boiling Eaporation may occur beneath the surace o a liquid, creating gas bubbles that rise to the surace due to buoyant orces. The tight lid o a pressure cooker holds pressurized water apor aboe the liquid, inhibiting boiling and raising the boiling temperature. Boiling results in a cooling o the liquid. At eleations aboe sea leel, water boils at a lower temperature (e.g. 96ºC in Reno. 6 Thermal and Phase Constants or Water Property Speciic heat o ice Heat o usion o ice Speciic heat o water Heat o aporization Speciic heat o steam Value 2090 J/(kg ºC 5,000 J/kg 4186 J/(kg ºC 2,255,000 J/kg 2010 J/(kg ºC Condensation Heat Transer Filmwise Condensation. Liquid wets the surace Decreasing heat transer rate due to resistance in ormed liquid ilm layer. Dropwise condensation. Liquid condensate as drops on suraces Higher heat transer rate due to large portion o the area o the plate exposed to the apor. 7 8

3 Comparison between Film and dropwise condensation Subjects o These Chapter Because o the higher heat-transer (more uncoered accessible surace, dropwise condensation is preerred, 10 times to ilm condensation. There are so many diiculties to maintain dropwise condensation. Various surace coatings and apor addities are used. Condensation Film Condensation Dropwise Condensation Condensation Number Film Condensation inside Horizontal Tubes Boiling Heat Transer (or boiling with water Pool Boiling Boiling Heat Transer in Pipes The Heat Pipe 9 10 General Approach to Condensation Analyzing Film condensation on a Vertical Wall Multi-tubes 11 12

4 (Calculation o mass low Viscous Force acting on luid element at y Force balance on an element o condensate Force acting by graity on an element o condensate (liquid 1 14 First time integration Boundary Conditions 15 16

5 Second time integration (Calculation o mass low Heat Transer modeling inside ilm (Calculation o h Rate o luid at δ δ 2 Mass low rate (Calculation o h Ater two times integration (Calculation o h Temperature change is linear in Condensate ilm Heat Transer at wall per unit area 19 20

6 (ilm thickness Calculation Mass added to element with length O x is equal to dierence between mass low rate at X and X+ x Mass added to element with length o x is equal to dierence between mass low at X and X+ x (Film thickness Calculation Heat Balance on element o length x Heat added by condensation + Heat added by mass low= Heat remoed by mass low + Heat remoed by conduction at wall 21 h l and h are enthalpy o liquid and apor phases, respectiely 22 Mixing relations or m and last relation o preious page C pl is speciic heat capacity o liquid (Film thickness Calculation (Film thickness Calculation I liquid is consider at temperature as apor h -h l = h g Regarding linear heat change in liquid ilm 2 24

7 (Film thickness And h Calculation (aerage h calculation (Eects o Nonlinear temperature proiles and subcooling eects o ilm temperature These both eects can be considered in heat transer coeicient h by replacing enthalpy o aporization h h : g with g on luid temperature h g = h g c( Tg Tw c= speciic heat o liquid Otherwise, Properties o liquid must be calculated at ilm temperature T or calculation o h and h aerage (equations 9-7, 9-10: T = (T g +T w /

8 A problem about ilm condensation A problem about ilm condensation 29 0 A problem about ilm condensation A problem about ilm condensation 1 2

9 A problem about ilm condensation A problem about ilm condensation 4 Film condensation heat transer coeicient on cylinders Turbulence Film condensation heat transer coeicient on cylinders Vertical Plates and Cylinders: The relation or aerage h on ertical plate can be also used or ertical cylinders suraces with luids with pr> o.5 and ct /h g < 1.0 c= speciic heat o liquid T g =T sat Aerage Heat Transer Coeicient on Horizontal Tubes (Nusselt relation For using this relation or n ertically tubes placed ertically its enough to replace d with nd For Isothermal Sphere its enough to change coeicient rom to p( p p ghgk h = 0.94 Lµ ( Tg Tw p( p p gh k h = µ d( Tg Tw g 1/ 4 1/ 4 Reynolds Number Calculation D p 4Ap H Re = = µ pµ H A = low area p = Shear, or "wetted," perimeter = aerage elocity in low considering Where Γ = Mass low per unit depth o plate For erticall plate with unit width, p p = Π d Re D = hydrauli diameter that critical m 4Γ m = pa Re = = pµ µ =. 1, and ertical tubes 5 6

10 Reynolds Number relation with h Reynolds Number Calculation q = h A( T sat T = m h g 4hL( Tsat TW Re = h µ W 4Ah( Tsat TW Re = h pµ g g. h A( T m = h Preiousely we hae shown that : Re For plate with A = LW and p = W. sat T g. W 4m = pµ Modiied orm o h or considering ripple (Way motion eects For Reynolds more than 0 experiments show alues that are about 20 percent more than alues that predict by ormula that we hae obtained or aerage o h. So, McAdams changed coeicient in the ormula or h. p( p p ghgk h = 1.1 ( Lµ Tg Tw 1/ Calculation o h or inclined plate I plate or cylinder has a angle o φ with the horizontal, to use the preious ormula it is enough to replace g = g sin Φ g with g p( p p g sin Φh k g h = 0.94 Lµ ( T T g w 1/ 4 The Condensation Number Reynolds alues can aect alues o h, so it seems good to rewrite h with Reynolds or an other dimensionless number. For an inclined plates or cylinder rom perious results one can easily write; / 4 1/ 4 p( p p sin g Φhgk h = c ( µ L Tg Tw Balance o energy on ilm also gies" mh & g h A( T T = mh & g w g Tg Tw = h A ReplacingT T in h : h g w p( p p sin / g Φk A L = c µ m& 1/

11 The Condensation Number (con. Soling the preious relation or h gies: 1/ p( p p gk µ A 4 / p 4sin Φ / p h = c µ m 2 4 & We deine the ollowing, dimensionless group C as "The Condensation Number": 1/ 2 µ C = h 0 k p p p g ( So, Using two preious equations, C can be rewritten as : C = c 0 4sin Φ A/ p L 4 / D p h where Re = µ 1/ Re 1/ 0 0 The Condensation Number (con. For Vertical plate A/PL=1 and so C 0 reduce to: C = 1.47 Re or Re p 1800 For a horizontal Cylinder A/PL 0 or Re p 1800 For turbulence condition experical correlation o Kirk - Bride ealtion can be used : 0 0 C = 1.51Re 1/ 1/ 4 C = Re or Re 1800 = Π and Film Condensation inside horizontal tubes Although it is important in so many application, as rerigeration and air-conditioning due to complexity o system there is not any analytical analysis or this type o heat transer. Chato present the ollowing experimental relation or condensation o rerigerants inside horizontal tubes 1/ 4 p( p p gk h g dg h = or low apor elocites, where Re = p 5000 µ d(t - T g w µ Where G is mass low o Vapor and its Reynolds Number must be ealuated at inlet condition to the tube. For high low rate h k = pr 1/ Re where dg Aboe relation or Re = µ For calculation o mass low rate o 0.8 m relation o Akers - Deans, and Crosser can be used; Re d p = Mixture Reynolds Number = [ G + G ( µ p ] 1/ 2 m Re apor G and liquid G all area dg = 5000 µ has about 50 percent accuracy. o tube must be considered 4 Example 9-2. Condensation on Tube bank One hundred tube o 0.5 in diameter are arranged horizontally in a square array and exposed to atmospheric steam. Calculate the mass o steam condensed per unit length o tubes or a tube wall temperature o 98 o C. 44

12 Example 9-2. Condensation on Tube bank (con. The necessary ormula or calculation h o one horizontal tube is presented already, or n tubes we can replace d with nd (where n=10 is the raw o tubes in that ormula. p ( p p gh k g h = nd ( Tg T µ w Fluid proprties must be calculated at ilm temperature o T = = 98 C 2 4 µ = kg / m. s 2 p ( p p p T sat o = 100 C h g p = 960 kg / m o k = 0.68W / m. C 1/ = 2255KJ / Kg 4 Example 9-2. Condensation on Tube bank (con. We consider laminar low and use ormula obtained or one tube, but we replace d with nd where n=10 is number o tube in each ertical line 2 6 (960 (9.8( (0.68 h = ( (10(0.0127( o 2 o = W / m. C [2209 Btu/h.t. C] The total surace area is : A 2 = nπd = (100 Π( =.99m / m L q A( T T g W So the heat transer per length is = h L L = (12540(.99( = KW / m The total mass low o condensate is then 5 m& q / L = = 6 L h g = Boiling Heat Transer Boiling Heat Transer 47 48

13 Boiling Heat Transer Boiling Heat Transer Boiling Heat Transer Boiling Heat Transer P b -P = 51 52

14 Boiling Heat Transer Bubble Size Change Ater Separation o Boiling Interace Bubble Size dose not change i p b -p = p= 2 σ/r Bubble grows i p b increase Bubble also grows i temperature o luid is more than apor temperature (bubble temperature Bubble collapse i temperature o luid is less than apor temperature (bubble temperature 5 Nucleate Pool Boiling Heat Transer Experimental Correlation o Rohsenow h C T g l x s pr l = C s q / A µ lh g g cσ g( p l p 0. Where C l =Speciic heat o Saturated Liquid Pr l = Prandtl number o saturated liquid T x = temperature excess = T w -T sat C s = Constant, must be determined rom experimental data, or waterplatinum combination is 0.01, or other combinations see table 9-2 Holman. S= 1.0 or water and 1.7 or other liquids g c = Conersion actor 54 Boiling Heat Transer Boiling Heat Transer Fig Heat Flux data or water on a platinum wire, at dierent pressures Fig Veriication o Correlation or pool-boiling at dierent pressure 55 56

15 Boiling Heat Transer Example 9-: Boiling on Brass Plate. A heated brass (67 percent copper and percent zinc alloy is submerged in a container o water at atmosphere pressure. The plate temperature is 242 O F. Calculate the heat transer per Unit area o Plate. Gien inormation and data: Boiling heat transer (water on a brass plate T w = 242 O F T Sat = 212 o F, 100 o C (Atmospheric Pressure Required: q/a 57 Example 9-. Boiling on Brass Plate (Con. Solution: To sole this problem we can use directly Correlation o Rohsenow. h C T g l x s pr l = C s q / A µ lh g g cσ g( p l p This needs to put so many data in this relation and inally be soled it or q/a. 0. A more eectie way is to write this relation or another system whose q/a can be Easily obtained rom igures or tables such as Platinum-water at the same condition, and diide it with the system whose q/a must be calculated. ( q / A water brass Cs water platinium = ( ( q / A water platinium Cs water brass At this way we only need to know q/a or water-platinum system and C s o this system and water-brass system. T w -T sat = 0 o F Figure 9-8 (q/a water-platinum = *10^5 Btu/(h.t 2 = kw/m 2 58 Table 9-2 Example 9-. Boiling on Brass Plate (Con. ( q / A water brass Cs = ( ( q / A water platinium C (Cs water-brass = (C s water-platinium = 0.01 water water brass (q/a water-brass = (* 10^5 (0.01/0.006^ =.4 810^6 Btu/h.t^2 = 1.072*10^7 w/m^2 s platinium Forced-Conection Boiling Heat Transer When a liquid moes orcibly on a surace whose temperature is greater than saturation temperature o liquid, a orced conection boiling heat transer happens. In this case, total heat transer can be consider as heat transerred by two mechanisms (q/a total = (q/a boiling + (q/a orced conection The aboe relation can be used when bulk liquid is in subcooled temperature (only local orced-conection boiling 59 To calculate orced conection eects in smooth tubes one can use relation o Dittus-Boelter (Eq. 6.4 with replacing coeicient 0.02.by Nu d = (Re d 0.8 Pr

16 Forced-Conection Boiling Heat Transer Fully deeloped nucleate boiling in tubes: In this case eects o orce conection reduce and heat lux dose not depend on rate o conection For low pressure MacAdams proposed the ollowing relation q/a = 2.259( T x or 0.2<p<0.7 MPa For higher pressure Ley recommended the ollowing relation q/a = 28.2 p 4/ ( T x or 0.7<p<14 MPa T must be in degree Celsius and p in megapascal Forced-Conection Boiling Heat Transer Forced-Conection Vaporization in tubes: I boiling is maintained or a suiciently long length o tube, the majority o the low area will be occupied by apor. In this instance the apor may low rapidly in the central portion o the tube while a liquid ilm is aporized along the outer surace. This heat transer is called Forced-Conection Vaporization. This situation must be treated as two phase low with heat transer In this case also the peak o heat transer can be obsered. For calculation o peak o heat transer coeicient some researcher proposed the preiously proposed relation or orce conection in which q/a is replaced with its maximum alue; q/atotal= (q/aboiling (Max + (q/aorced conection The Peak Heat Flux in Pool Boiling Heat Transer There is a maximum point at cures o Heat lux ersus T x This point is indicated with a at igure 9. (Holman Zuber hae deeloped a analytical correlation to calculate Heat lux at this point. q ( A 1/ 4 Π σ g( pl p p 1/ 2 max = hg p ( p pl Where σ is the apor-liquid surace tension Types o material does not aect the peak heat lux, But dirty surace can increase about 15% in the peak alues 6 The Peak Heat Flux in liquid droplet impinge on hot surace In this case, experimental data show that temperature excesses ( T w at maximum o heat transer is about 165 o C or dierent liquids and it is independent on luid types. Experimental data also show that the peak lux in this case is unction o the luid properties and the normal component o the impact elocity (V: 2 2 Q max 10 P V d l ( = 1.8 Pd λ Where p σg l c Q max = maximum heat transer per drop p l = density o liquid droplet P = apor density ealuated at ilm temperature σ = surace tension d=droplet diameter λ=modiied heat o aporization = h g +c p (T w -T sat /2 Heat transer coeicient in this case is ery high and more than 50% o droplet is eaporated during ery short time o droplet contact. When droplets contact with temperature higher than saturated temperature but with zero elocity still droplets eaporate ery ast. This called Leidenrost Phenomenon

17 The Peak Boiling Heat Flux on Horizontal Cylinder Sun and Lienhard presented the ollowing relation or the peak boiling heat transer on horizontal cylinder which is in good agreement with experimental data q max = exp(.44 R or R > 0.15 q max F where R is a g(pl - p R = R[ ] σ and q = is the peak heat lux on an ininite horizontal plate. q max F max F = 0.11 dimensionless radius deinied by : p h [p -p ] g l 1/4 65 Heat transer coeicient in stable ilm-boiling on a horizontal tube Bromley suggested the ollowing relation or calculation o heat transer coeicient in stable ilm-boiling region on a horizontal tube: kg p( pl p g( h g + 0.4c p Tx 1/ 4 hb = 0.62[ ] ; d = tube diameter dµ Tx The aboe relation only consider conection through ilm and does not consider the eects o radiation. The total heat transer coeicient may be calculated rom the empirical relation hb 1/ h = hb ( + hr h where h is the radiation heat transer coeicient and is calculated by assuming σε ( w hr = T T σ = Stean_boltzman Constant; Properties o the apor in aboe and h g r T w 4 T sat an emissiity o 4 sat (enthalpy o aporization is to unity or the liquid. Thus; ε = Emissiity relation must be be ealuated at the saturation temperature surace ealuated at ilm temperature 66 Simpliied Relation or Boiling Heat Transer with Water Some simpliied relation or calculation o heat transer coeicient on the outside o submerged surace at atmospheric pressure Simpliied Relation or Boiling Heat Transer outside submerged surace at any pressure (Con. To consider inluence o pressure, the calculated h at atmospheric pressure can be modiied by ollowing relation; p h p = h1 ( p Where h p =Heat transer coeicient at some pressure p h 1 =heat coeicient at atmospheric pressure as determined rom table 9- (Preious table p= system pressure p 1 = Standard atmospheric pressure For orced conection local boiling inside ertical tubes the ollowing relation is recommended 2 o h = 2.54( T exp( p /1.551 W/m. C; or 5 atm p 170 atm wher p = pressure T x x in MPa = temperature dierence between surace and saturated liquid 68

18 Example 9-4. Flow Boiling Inside Tube The Heat Pipe Water at 5 atm lows inside a tube o 1 in [2.54 cm] diameter under local boiling conditions where the tube wall temperature 10 o C aboe saturation temperature. Estimate the heat transer in a a 1.0 m length o tube (q/a. Gien: Boiling low inside tube with T x = 10 o C P= 5 atm Requirement: q/a=? Solution. We can used the last preious relation to sole this problem. A ixed-conductance heat pipe: It acts with only one condensing luid and its thermal resistance is not a strong unction o heat load h = 2.54( T x exp( p /1.551 p= 5 (1.012 * 10^5 N/M 2 = MPa h=2.54 (10 ^exp(0.5066/1.551= 521 W/m 2. o C A=πdL= p(0.0254(1.0=0.0798m 2 q= h A(T w -T sat = 521 ( (10= 2810 W/m The Heat Pipe A Variable-conductance heat pipe: The gas in reseroir can change contact surace o apor with wick depend on apor pressure Applications o The Heat pipe 71 72

19 Structure o The Heat Pipe 7 Example 9-6. Heat Flux Comparison Using the data o table 9-4 (page 541 o Holman book, compare the axial heat lux in a heat pipe using water as the working luid (at about 200 o C with heat lux in a solid copper bar 8 cm long experiencing a temperature dierential o 100 o C. Gien inormation. Copper bar L= 8 cm T x = 100 o C Requirement: Comparison o q/a or heat pipe and a copper bar Solution. q/a or copper road; From table A-2 K copper = 74 W/m.oC q/a = -k( T/ x = -(74 (-100/0.08=467.5 kw/m 2 = kW/cm 2 And q/a or heat pipe: Pipe heat working with water Table 9-4 (q/a axial = 0.67 KW/cm 2 q/a heat pipe/( q/a copper bar = 0.67/ =14. Which show that why the heat pipe could hae wide applications 74