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DATE DUE DATE DUE DATE DUE 0,0,1'152005 1007 It 5 6/01 c:/CIRC/DateDue.p65-p. 15 SOLUTIONS TO THE FULLY DEVELOPED CONVECTION HEAT TRANSFER PROBLEM IN CORE ANNULAR FLOWS By SriharSha Chunduru A THESIS Submitted to Michigan State University in partial fulfillment Of the requirements for the degree Of MASTER OF SCIENCE Department Of Mechanical Engineering 2003 ABSTRACT SOLUTIONS TO THE FULLY DEVELOPED CONVECTION HEAT TRANSFER PROBLEM 1N CORE ANNULAR FLOWS By Sriharsha Chunduru Heat transfer in circular pipes with two concentric regions (core annular flows) is a problem that arises in materials processing and the chemical industries. Examples Of this problem include the flow Of Oil/water for a certain range Of Reynolds numbers and flow through packed bed chemical reactors. In this work, the convection heat transfer problem is solved for a core-annular flow that is both thermally and hydrodynamically fully developed. Solutions are Obtained for an outer surface Of the pipe that is subjected tO three different boundary conditions: uniform heat flux, constant wall temperature, and convective heat flux. The velocity and the temperature profiles are calculated analytically and Nusselt numbers are evaluated for the three cases. The analytical results are validated by comparison with the solutions for the single-phase fluid. Dedicated to all my friends and professors at Michigan State University iii ACKNOWLEDGEMENTS I am thankful tO my advisor, Dr.Andre Benard for having given me such an interesting and challenging topic to work for my Masters’ thesis. I am thankful to Dr.Craig Somerton for taking SO much Of his valuable time tO help me on my thesis. I am thankful to Dr.Charles Petty for kindly agreeing to be a member Of my thesis panel. I must also thank all my friends at Michigan State University who have provided me SO much Of moral support all along. Last but not the least, I thank my family who have always striven hard in providing me the best education throughout my life. iv TABLE OF CONTENTS LIST OF TABLES ................................................................................... vi LIST OF FIGURES ................................................................................. vii KEY TO SYMBOLS .............................................................................. viii INTRODUCTION .................................................................................... 1 CHAPTER 1 PHYSICAL MODEL ................................................................................. 2 CHAPTER 2 CONSTANT HEAT FLUX PROBLEM .......................................................... 9 Non-dimensionalization ................................................................... 12 Inner fluid calculations .................................................................... 16 Outer fluid calculations .................................................................... 17 The nusselt number at the outer wall .................................................... 19 Results and validation ..................................................................... 19 CHAPTER 3 CONSTANT TEMPERATURE ANALYSIS ................................................ 26 Inner fluid calculations .................................................................... 28 Outer fluid calculations .................................................................... 31 Results and Validation ..................................................................... 40 CHAPTER 4 CONVECTIVE BOUNDARY CONDITION ................................................... 47 Results and Validation ..................................................................... 53 CHAPTER 5 SUMMARY AND CONCLUSIONS ............................................................. 60 APPENDIX ........................................................................................... 63 REFERENCES ...................................................................................... 69 LIST OF TABLES Properties Of the two fluids ......................................................................... 23 Dependency Of Nuo on Nu... ........................................................................ 54 vi LIST OF FIGURES Physical model for two- region flow ................................................................ 4 Energy balance on an element of length dz ...................................................... 11 Nusselt number at the interface as a function of the radius ratio .............................. 21 Nusselt number at the wall as a function of radius ratio ........................................ 22 Temperature of the system as a function of radial position for different radius ratios. ....24 Temperature of the system as a function of radial position for extreme radius ratios. . ....25 Energy balance on an element of length dz ...................................................... 26 Flowchart to solve the nusselt number for the two-fluid case ................................. 37 Nuo vs. 77,- ........................................................................................... 42 Figure 3.4: Energy in the fluids vs. 77,- ........................................................... 43 Nul-J vs. 77,- .......................................................................................... 44 Temperature vs. radial position for different values of 77,- ..................................... 45 Temperature VS. radial position for extreme values of 77,- ..................................... 46 Nuo VS. Nu... for different radius ratios ............................................................ 55 Temperature profiles for extreme radius ratios .................................................. 58 Temperature profiles for different radius ratios ................................................. 59 vii KEY TO SYMBOLS C p Heat Transfer Coefficient D Diameter H Pressure gradient Pr Prandlt number T Temperature Nu Nusselt number g Non Dimensional pressure gradient h Heat transfer coefficient k Thermal Conductivity q Heat transfer r Radius u Velocity v Non-Dimensional Velocity GREEK ALPHABETS OLThermal diffusivity y Ratio of Thermal Conductivities n Non-Dimensional Radius A Ratio of Dynamic Viscosities it Dynamic Viscosity v Kinematic Viscosity 6 Non-Dimensional Temperature p Density (1) Ratio of Heat Capacities Subscripts i Inner Fluid 0 Outer Fluid m Mean 0 outer wall m,l Mean Property of the inner fluid m,2 Mean Property of the outer fluid viii INTRODUCTION Important engineering problems in the materials processing and chemical industries involve heat transfer in a two-region channel flow. Examples include the flow of two immiscible fluids and the flow through a packed bed chemical reactor. One of the most important engineering applications of two-region flow is the flow of water and oil in a pipe. Pipeline transport of highly viscous oil needs enormous pumping pressures to overcome the high viscosity and the corresponding wall shear stress. It can be affected by heating the oil and insulating the pipeline [1]. However these operations involve considerable capital investments and operating expenditures. As discussed by Joseph et al. [2], oil companies have had an intermittent interest in the technology of water- lubricated transport of heavy oil since 1904. In the past, experiments have been carried out to examine the possibility of water- lubricated transportation of oil as discussed by Chaves et al. [1]. A number of flow patterns were observed, such as water drops in oil, concentric oil-water flow and oil drops in water. It was found out that of all the observed flow patterns, the flow of the highly Viscous oil as a core, with the water flowing only in the annular space between the pipe and the core walls was the most desirable for simultaneous flow [1]. The pressure drop measured over the pipe indicated that the addition of water greatly reduced the pressure gradient. Ooms et al. [3] discussed theoretical models for core annular flows of a very viscous oil core and a water annulus through a horizontal pipe.-AS long as oil core was supplied at a velocity above a certain critical value, a water film remained between the oil-core and the pipe wall. However the amount of water used to transport the oil is determined by various stability analyses. If the water used is large, there is always a problem of dewatering and if the water used is less, there is a problem of fouling. Using cement-lined pipes as discussed by Amey et al. [4] can reduce fouling to a considerable extent. Joseph et al. [5] discussed the stability of annular flows. The annular flow is stable or can be stabilized when the fluid having the higher effective viscosity occupies the core region and lower viscosity fluid is in the annulus. Joseph et al. [6] also study the instability of the core annular flow as the thickness of the lubricating fluid in the system increases. Thus, volume ratio is a crucial factor in determining the stability. There is also a limitation on the velocities of the fluids. A minimum velocity exists below which the core-annular flow is unstable and results in oil slugs in water and a maximum velocity above which the flow is replaced by water emulsions in oil as discussed by Prezioski et al. [7]. Extent of the annular flow is determined by the location of the break-up point. Hason et al. [8] discovered that the wall-film broke up at low flow rates and thus destroys annular flow abruptly. A practical aspect of this technology is the question of how to guarantee that a certain oil-water system will operate in core-flow pattern. In practical situations, this flow mode is brought about by using special inlet nozzles in combination with physiochemical agents to facilitate wetting of the pipe wall by water and to stabilize the oil-water interface [9]. Shut down and restart procedures must also be known. For the technique to be practical, it should be possible to restart a pipeline from the stratified oil-water situation to the core flow pattern in a reasonable amount of time without the requirement of excessive pumping power [9]. In the late nineteenth century, development of Moody’s chart involving Reynolds number and Fanning’s Friction factor facilitated the design and development of Single-fluid pipeline systems. Ames et al. [10] extended this analysis to lubricated pipelining. Thus core-annular flow has been a topic of research for quite some time. However most of these works focus on the “fluid” part of the analysis. Most of the literature neglects the thermal analysis of the core-annular flow. The possibility of using these systems in cold regions such as on the ocean floors or in the arctic regions require a detailed heat transfer analysis for a variety of boundary conditions. Although the uniform wall heat flux problem is rather trivial, the constant wall- temperature problem is a challenging one in the case of a single-fluid system. The case of a convection boundary conditions applied on the outside of the pipe is also non-trivial. Solution to the heat transfer problem of a single fluid flowing down a pipe has been solved long ago by Graetz in 1885 [11]. Graetz neglected the variations of the properties of the fluid with the temperature and found a mathematical solution for the problem that was rediscovered by Nusselt in 1910[11]. In this work, a heat transfer analysis of the core-annular flow is performed using an approach different than the one used by Graetz to compute the temperature profiles and Nusselt numbers of the system for three boundary conditions on the outer wall- constant wall heat flux, constant wall temperature and convective heat flux CHAPTER 1 PHYSICAL MODELS The physical model for the two-region pipe flow and heat transfer problem is Shown in Figure 1. Each region 1 or 2 contains a fluid that is immiscible with the fluid in the other region. Region 2 Figure 1.1: Physical model for two region flow The following assumptions will be employed in the development of the solution to the heat transfer problem (a) The flow is steady and hydrodynamically fully developed. This will lead to the only non-zero velocity component being that in the streamwise direction and a constant pressure gradient (- H) in the streamwise direction. (b) The flow is thermally full developed. This will lead to a constant convective heat transfer coefficient on the outer surface and a constant temperature derivative in the streamwise direction. (c) Three types of boundary conditions have been analyzed- constant heat flux on the outer wall, constant wall temperature and convective boundary condition. The Continuity equation in the Cylindrical coordinates for steady and incompressible flow is [12]: li+li+3 The components of the momentum equation are as follows: r -momentum: Bu , a: 1 1 a +(U.V)u, 71.5 = —;d—’:+ g, +U(V2u, —————) (1.2) t9 -momentum: Bug u u 1 8p 2 2 Bu _u9 ————+ U.uV +_:__0:=__—+ 9+uv +—— —— 1.3 Z -momentum: ———z—+(U.V)u- =——1-:—p+gz+UV2u (1.4) where, the convective derivative is: a 1 a UV: u, -—+-— rug— 86 a 7 — 1.5 3r +u~ 32: ( ) The Laplacian operator is: re a 132 32 —r(—+— V2: ———+ rar 3" r 280—2— azz (1.6) Since a symmetric flow is assumed,u9 =0. The flow is assumed to be hydrodynamically . . . a . . fully developed, it, =0. Usrng this value of u, 1n Eq (1.1), a—(uz ) = 0. Since the flow IS 2 symmetric, there is no 0 dependency. SO “z is a function of r only. Consequently, when we use these above deductions in Eq (1.2), the r-momentum equation reduces to: 1 8p 0:—._—_ 1.7 ,0 dr ( ) which means that the pressure p = f (19, Z). The 6 momentum equation (Eq 1.3) reduces to the following form: O=—igg (1.8) ,0 d6 which implies that p = f (2) only, in other words the pressure in the pipe flow changes only in the axial direction. The z-momentum equation is au. 13p 2 r + U.Vu =———+ +vV u 1.9 at ( )Z pdz 82 Z ( ) Using the established facts that the flow is steady, p = f(z) only, gZ = Oand that uZ is a function of r only, Eq (1.9) can be reduced to the following form: 0=—l§’l+uv2u. (1.10) raz “ Equation (1.10) further Simplifies to, li[,d“z)=_fl (1.11) r dr dr ,1! where, H: — ip— (the pressure gradient) The energy equation [12] is analyzed next. 3T 2 [(2 2 2) 2 2 2] pcp[E+(U.V)T]=/ n i l ,m> 8Rti#1#2 “771 The next step involves the calculation of the temperature profiles: SO considering the energy Equation, Eq (2.27), 1 d d6,- 1 _d_ 77 d : 2 2 vi ’7 ’7 ’7 91,7771' V771,] + 917.1(1‘ ’71 Wm.” 15 (2.40) (2.41) (2.42) (2.43) (2.44) (2.45) (2.46) INNER FLUID CALCULATIONS: The non-dimensional temperature profile of the first fluid is calculated first using the above equation: 4 2 91 = E77 ‘ F77 (2.47) where, E = — gl , 2 2 64627177: VmJ +971,1(1—’71 )VmJI F: (-8177.'2 ’82 +8277?) 2 2 16(Ql,l77i va +9[1,1 (l—ni )val) This is converted to the dimensional form to get the following expression for the temperature of the first fluid: 2 Let m=QI,l7712V771,I '1' 911,10" 771' )val (2'48) : 4qu2(R§ —R,-2),u1 —(qu2(r2 -4R,?-) —32mk1R0Tcumfll),u2 32mk1Roum/1Lu2 T1 (2.49) Once the temperature profile of the inner fluid is known, its mean temperature is calculated. Rt qu = ——2——— Tlul 2717‘dr (2.50) 1' “772,1 0 The mean velocity of the inner fluid is: Ri umJ = —2 u127zrdr = 72' i O 711(211'02 #1 + R} (—2p1 + #2 )) 8#1#2 (2.51) The mean temperature of the first fluid is: 16 24HngRr'2/112 + 384mk1Rchumfl12/12 T 192mklRoRizTcumlul (27111 T #2 )flz + r 1 _ 417ng RSI/11 (—12#1 + 7772 ) + MIR-6047112 - 28711712 + 771%) m, _ 192mk1R0um #1272 (2R3 p, + R? (—2,u1 + M )) (2.52) The temperature at the interface of the two fluids is calculated as: Z. = Tl At r = R,- T. = 4114113113771 + 32mkiRoTcum#1#2 + 1‘1qu HM + 3m) (2 53) ' 32mkiRaum7t17tz The heat transfer at the interface is given by: dT ‘11' =—k1———1- At r=Ri dr 2 3 2 2 _ _ 8HqR0 Ri#1 — ZHCIRI #2 - ZHCIRI (4R7 (#1 — #2) + Rt #2) (2 54) l — . 32mRoumtul/l2 Finally, the Nusselt Number at the interface of the inner fluid of the inner fluid is given by: = (11 ,,. 2R1 (Ti T T771,1) kl Nu“ (2.55) with the values taken from Eq (2.52), (2.53), (2.54) OUTER FLUID CALCULATIONS: The next step involves calculating the temperature profile of the second fluid. The velocity and the mean velocity of the second fluid (outer fluid) are as follows: _H(R§—r2) (2.56) 4#2 “2 l7 R0 2 2 H R —R- umfl = 21 2 IuZZflrdr: ( 0 ') (2.57) ”(R0 TRi )Ri 4#2 Equation (2.46) is solved first to get the non-dimensionalzed temperature profile and then converted into the dimensional form to get the following result: r R- (ankzkf UREA + R? (4771 + 3772» + kl (4an(Log[-I-,—] — LogiR—'1)Ri4#2 0 0 + #1(4HqR02(mr2 - (m + 2(m — n)Log[RL] — 2(m — 71)L0g[%])R,-2 ) + 0 0 R- Hq(—mr4 + (m + 4(m — 2n)L0g[—REr—] — 4(m — 2n)Log[El—])R,-4) + 32mnk2 RoTcumflz) T2 = 0 0 32mnklk2R0u’nfll #2 (2.58) where 11: 97,27712 V771,] + 917.2(1‘ 7712 )VmJI m = 91,1777'2 VmJ + 977,1(1— 77:2)1’771,” (2'59) The mean temperature of the second fluid is given by: 1 R0 Tm,2 = 2 2 ITzuz 2717111”: ”(R0 —R1 )um,2 R1 (7h2mqk1ngl + 12717-77247:lesz p1 —36h2nqk1RgR,-2,u1 + 48h2quog(R./Ro )k1R06R1'2#1 -48h2nqLog(R.- /Ro )kIRSRim + 24h2nqk2R06R12 p1 —367127122111793R?”l + 84h2nqkleRi4pl — 24h2qu08(Ri/Ro)k1RgRi4#1 + 48h2nqLogk1R3R.-4m — 72722an<2R§R,-4pl + 20h2mqk1RgRi6fl1 —6Oh2nqk1R§R,-6,ul + 72h2nqk2RgR,-6,u1 ~3h2mqk1R,8,u1 + l2h2nqk1Ri8p1 — 24h2nqk2R,8,u1 — 18h2nqkle‘Ri4p2 + 24h2nqLog(R,- /R0)k1R;‘R,-“p2 +18h2nqk2R§pr2 + 24h2nqk1Rg'R16fl2 - 36h2nqk2RgRl-6fl2 — 6h2nqklRl-8,u2 + 18h2nqk2Ri8/12 + 192hmnk1k2Rchfl1y2um + 384hmnk1k2R03Ri2Tc 17117272,, + 192hmnk1k2R0 Rfrc #1 Mum 1921"""k1k2Ro (R3 — R12 )2 #1#2um 18 (2.60) THE NUSSELT NUMBER AT THE OUTER WALL: To find the Nusselt number at the outer wall, we need to compute the mean temperature of the system in total: Ri R0 Tm = —Tl—— Iu1T127zrdr + quT2 27zrdr (2.61) oum 0 RI The temperature of the outer wall is given by: Twa” =T2 AI r = R0 (2.62) The heat transfer at the interface is given by: dT qw : Tk2 drz at r = R0 (2.63) The Nusselt number at the outer wall is given by: 2R Nag: qw * 0 (2.64) (Tm T Twall ) k2 Where qw,Tm ’Twall are given by Eq (2.63), (2.61) and (2.62) RESULTS AND VALIDATION: Since there is no literature with which we can validate the Nusselt numbers for the 2 fluids case, we validate it using the standard one-fluid case. If we set 77,-=1. It means that the inner radius is equal to the outer radius, or in other words, there is only one fluid in the system and it’s the inner fluid. And the Nusselt numbers should converge to the Nusselt number of the one-fluid case. 19 The two Nusselt numbers, the Nusselt number at the interface of the first fluid, Nu“ and the Nusselt number at the outer wall, Nu” were evaluated for radius ratiozl. The Nusselt k number at the outer wall yielded a value of exactly igk—la We get an extra factor of the 2 ratio of the 2 thermal conductivities due to the way we define the Nusselt number. It has been defined with respect to the second fluid thermal conductivity. However there is no second fluid if the radius ratio is exactly equal to 1. Now, if it is scaled with respect to the first fluid (which is the only fluid present in the system), the Nusselt number Simplifies to a value of exactly 4.36 that is the Nusselt number of a single fluid flowing in a pipe with constant heat flux on the outer wall [12]. . . . . 48 . The Nusselt number at the interface for the first fluid also yields a value of exactly 1—1— If the radius ratio is equal to l, which is to be expected because the interface is at the wall in that condition. Another method in which the Nusselt number expression was validated was by assuming that both the fluids 1 and 2 are the same. In that case, the Nusselt number at the outer wall gave :11? irrespective of the value of the radius ratio. 20 Radius Ratio I 4 A A . NUM 8 0.2 0.4 0.6 0.8 1.0 7.9 7.8 1 7.7 7.6 Figure 2.2: Nusselt number at the interface as a function of the radius ratio AS can be seen from Figure 2.2, the Nusselt number at the interface approaches the value of 8 when the radius ratio approaches zero. This is because the inner fluid flow becomes a plug flow as the radius ratio becomes very very small. And the Nusselt number of a plug flow is 8. Although not exactly visible in the graph, when the radius ratio approaches 1, the N usselt number is equal to 4.36. 21 Nuo 1.5 0.2 0.4 0.6 0.8 1.0 Radius Ratio Figure 2.3: Nusselt number at the wall as a function of radius ratio From Figure 2.3, we can see that the Nusselt number at the outer wall becomes 4.36 when the radius ratio equals zero, in which case the only fluid present in the system is the inner fluid. When the radius ratio equals 1, the Nusselt number at the outer wall converges to a value of 1.029. From the program, it has been observed that the Nusselt number at the outer wall converges to 1.029 at a value of radius ratio of 0.9999999999. However when the radius ratio becomes 1, the only fluid present in the system is the inner one. Thus, the Nusselt number at the outer wall must be sealed with respect to the inner fluid by multiplying 1.029 by the ratio of thermal conductivities. In this case, the ratio of thermal conductivities is: k—1 — ——O'145 = 0.236 k2 T 0.613 The N usselt number at the outer wall becomes: N110 ‘—‘ -l-02—9 = 4.360 0.236 Thus, the results have been validated Following are the plots of Temperature as a function of the radial position for different values of radius ratios. An entry temperature of TC =1000c was assumed with the following properties of the fluids: Table 2.1:Properties of the two fluids Oil Water Density (,0) 884.1 997 kg / m3 Heat Capacity (cp ) 1909 4179 J / KgK Viscosity (#) 0.486 0.855*106-6 NS / m2 Thermal conductivity (k) 0.145 0.613 W / mK Thermal Diffusivity (a) 0.859*10e-7 l.471*10e-7 m2 / s Prandtl Number ( Pr) 6400 5.83 23 170 160 150 f 140 130 160 110 Theta / / . / ’ / _ _ 0.9 ’ — _— - 0.5 / 0.3 0.1 A i# A A l A A A 1 A A A 1 A A A l 0.2 0.4 0.6 0.8 1.0 Radial Position Figure 2.4: Temperature of the system as a function of radial position for different radius ratios. 24 Theta i 0.001 x" .. / 160 ----------- 0.999 X r- l/ / 150 I X/ ’ x " / 140 l x/ 130 f 120 E 110 : .4 - " A J A A A 1 A A . 1 A . A 1 A . A 1 0.2 0.4 0.6 0.8 1.0 Radial Position Figure 2.5: Temperature of the system as a function of radial position for extreme radius ratios. 25 CHAPTER 3 CONSTANT TEMPERATURE ANALYSIS Consider Eq (1.18) and Eq (1.19) 82 Tw T T771 37 _ dTw _ Tw -T dTw + TW —T dTm az dz Tw T Tm dz Tw T Tm dz (3.1) (3.2) The boundary condition of constant wall temperature reduces the above Eq (3.1) and Eq (3.2) to a_T_ TwTT dTm dz TW—Tm dz Figure 3.1 gives the energy balance on an element of length dz. Region 2 Figure 3.1: Energy balance on an element of length dz (3.3) ZHRqu(Z)dZ = (URI-214,,” )pf,lcp,ldTm + ITHROZ T Ri2 )um,2pf,2cp,2dTm dTm 261W R0 T 2 2 2 dz (pcp )1 R1 ”777,1 + (,OCP )2 (R0 T Ri )um,2 26 (3.4) (3.5) d7", 2qw"(x)R0 = (3.6) dz (pc )1 R2 R-2 ((4,. )2R31 P ”Bum + (1——':,—)u,,.,211 (pcp )2 R0 R0 Substituting the above expression from Eq (3.6) into Eq (3.3), we get: a_r : (Tw — T) 2q.."(x)R0 (3 7) 82 (T —T ) (pc MR2 182 ' W m {(pcp )2 R02[ P T£2TumJ + (IT TTZT)u771,2 ]} (pcp )2 R0 R0 2: = (1:17”) 2(Tw TT) (3 g) az <7 —T> (pc )1 R2 R2 ' W m {(pcp)2R02[—£_—12um,1 +(1T_%)um,2]} (pcp )2 R0 R0 But, 411111)— : h. (3.9) (Tw TTm) From the definition of Nusselt number (on the outer wall), 0 k2 Using Eq (3.9) in the above definition of the Nusselt number, Eq (3.8) reduces to: az 2R0 2 (pep )1 R,—2 R? ' {(pcp )2 R0 [TTT—TT Tunal + (1 T —2)um,2 ]} (“p )2 R0 R0 27; _ a2(Tw —T)Nu0 (3 12) az T (pc )1 R2 R-2 ' {R02 P —l§MI7r,l + (1 T Tl7)um,2 ]} (,0Cp )2 R0 R0 The energy equation Eq (1.17) l—a—(fl r] = u—Z—a—T reduces to the following form: r Br Or a Oz 27 —— r— = (3.13) rar Br (I 2(pcp)1§: R2 1 a[ 8T) uZ a2(Tw—T)Nuo {R u ,1+(1-—i—)u ’21} 0 (mp)2 R02 m R02 m Since there is only one velocity component, the z-component, uzis replaced by u for simplicity. The Boundary and the matching conditions are as follows: At r = 0 (at the centerline of the channel) dz- — O 3.14 dr ( ) At r = R1. (at the interface of the two regions) T1 2T2 (3.15) dTl dTo k — = k " 3.16 1 dr 2 dr ( ) At r = R0 (at the outer surface of the channel) T2 = Twall (3.17) INNER FLUID CALCULATIONS: Thus for the first fluid, the energy equation (3.14) becomes: ia 8T1 _u1 a2(Tw—T)Nu0 7570‘ dr)_; 2 (pcph R} R,-2 (3'18) {R0 2 um,1 + (1T7)um,2]} (pcp)2 R0 R0 2 u1 = 1hr,2(i——1—]+—h—ro— —i—r2(From Eq 2.38) 4 #1 #2 4#2 4#1 Defining new parameters: (9,-2T1—Tw Andn=7:—, 0 Equation (3.18) simplifies to: ii 77%. +[M—Nn2]Nu 9 =0 (3.19) ndn r177 01 a' 2 “'8' ‘j‘(81—82)77i + ’42 where M = . 2 2 . (3.20) (”i vm,lw+ (1-771. )Vm,2) 0’1 781 And N = 4 2 2 . (3.21) (771' Vme + (1 T 771' )vm,2) a,- = —- (3.22) And the other variables are same as defined in the constant heat flux problem. Equation (3.19) is solved by the Method of Frobenius [15]: Let (91 = 77“ Zann" (3.23) n=0 Substituting the value of 6, in Eq (3. 19) we get: 2(71 + s)(n + s --1)a,,775+"—2 + 2(n + s)a,,77"+s"2 + (M — N772)Nu0 Emlyn” =0 n=0 n=O n=0 (3.24) Equating the coefficient of the lowest power of 77 which is the coefficient of 775—2: s(s —1)ao + sao = O (3.25) which leads to s=0,0. (Double root) 29 If the indicial equation has 2 equal roots, as in Eq (3.25), at y = m1 and y =m2, the solution is given by [16] : d y = €10)",l + 62(1) (3.26) Mg Thus, the solution in our case is given by: 0161 6 =6 6 _ +c — 1 1(1)... Zia].-. (91 = Cl Zann" +c21n77 Zann" (3.27) n=0 n=0 From the boundary condition (3.14), 526—1 = 0 at 77 = 0 , we get c2 = 0. Eq (3.27) reduces 77 to the following form: 61 =c12ann" or (91 = ZAnn" (3.28) n=O n=0 Substituting in Eq (3.24), we get, Z(n)(n —1)A,,77"‘2 + Z(n)A,,77"T2 + (M — N772)Nu0 ZAnn" =0 n=2 n=1 ":0 (3.29) Equating the corresponding coefficients of 77 powers to zero: MN 770 :2A2 +2142 +MNu0A020 :3 [122— “0 A0 77‘:6A3+3A3 +MNu0A1 =0 :> A3 =0 30 [MA2 — NAG] Nuo 16 ”2:12A4 +4A4 +Nu0[MA2 —NA0]:O:> A4 :_ 773220145 +5A5 +Nu0[MA3 —NA1]=0 —_—> A5 =0 MA— A -[ 4 N2]Nu 27“: 30/16 +6A6 +Nu0[MA4 —NA2]=O:> A6 = 36 0 In General: A” = O for n 2 odd, Nuo = 2 [——MA,,_2 + NA,,_4] For n: even and n>2 n MN and, A2 = — 4“0 A0 (3.30) OUTER FLUID CALCULATIONS: For the second fluid (outer fluid), the energy equation (3.13) becomes: T - T N li(,§I§)=u_2 a2( W ) ”0 (3.31) rar 8r a2 2 (pcph R12 1.2 {(pcp)2Ro —2um,l + (1 T —2)um,2]} (pcp)2 R0 R0 2 h u2 = r0 — irz (From Eq 3.39) 4#2 4#2 Equation (3.31) simplifies to: 1 d d6 ——[n——2]+[P—Q272]Nu062 =0 77 dn d0 §2 A P = . 4 . And Q== 4 2 2 2 2 A (771' V171,] + (”(1 T 771' ”1112) (771' vm,l + (”(1 T 771' ”1212) Equation (3.32) is solved by the Method of Frobenius: 31 (3.32) La 62 = ”3 an77" 72:0 Substituting the value of 62 in Eq (3.32) we get: 71:0 77:0 202 + s)(n + s — 019,775+"—2 + Z (n + s)b,,n"“‘2 + (M — N772 )Nuo ann"+s = 0 (3.33) Equating the coefficient of the lowest power of 77 which is the coefficient of 773—2 in Eq (3.33) to zero, we arrive at the following indicial equation: s(s —1)b0 + sbo = O which leads to 5:0. (Double root) Thus, the solution in our case is given by: d62 92 = 63(92 )szo “4(1) 5:0 02 = C3 anfln + 641117] anfln 71:0 7120 From the boundary condition (3.15), (3.16), (3.17), we get, 62(7]=I)=0 91(fl=fii)=6’2(fl=7ii) d9 k1‘—1‘=k2—At77=777 d77 Equation (3.36) can be rewritten as: 92 = 237277” +2071 1117777" 71:0 72:0 Substituting in Eq (3.33), we get, 32 (3.34) (3.35) (3.36) (3.37) (3.38) (3.39) (3.40) Z(n)(n —1)13,,n"‘2 + Z(n)(n —1)D,, ln7777"T2 +Z(n)1),,77"‘2 +ZDn77"_2(n — 1) + ”:2 ”:2 n=1 "=0 Z(n)13,,77"‘2 +ZD,,27"‘2 +2001)" ln 7777"—2 +Z(P)B,,77"Nuo + 2(1),, )PNuO77" ln77 + n=l n=0 n=l n=0 n=0 -ZQNuon”+2 — 210,. ><2Nu..27"+2 In 77 = 0 (3.41) "=0 n=O Equating the corresponding coefficients of 77 powers to zero: 77’1 :2D1+Bl =0 770 : 482 + 402 + P80 Nuo = O (3.42) 77‘: 983 + 603 + PBlNuO =0 772 :16B4 + 804 + Nu0[P82 — Q30] = 0 OOOOOOOOOOOOOOOO Equating the coefficients of Logarithmic terms: 77'1 In 77: DI = O :70 1n 2; :41)2 + PDoNuO = 0 (3.43) 7711n7719D3 + PDlNuO = 0 7721;177:1604 + (PD2 —QDO)Nu0 = 0 From Eq (3.42) and Eq (3.43), we can deduce that all the odd D,- 's are equal to zero. In General, the values of D, 's are: D271+l:O ’ 33 PNuO Dz = — 4 DO Nu D271 : 2 )02 [QD2n-4 T PD2n—2] (3-44) ’1 where n varies from 0 to co and n is an integer. When the use the fact that all the odd Di's are equal to zero from Eq (4.44) in the set of Equation (3.42), we can deduce that all the odd B, 's are equal to zero. In General: B271+l = O NuO I B2n =__2[QB2n—4 T PB2n-2]T_D2n (3'45) 2n) n PN 32 = ”0 (Do ‘30) where n varies from 0 to co and n is an integer. Using the Boundary Condition (3.15) 49107 = 777') = 9207 = 771') 23/127177?" = 232717712” + ZD2n771'2” 1“ 771' (3-46) =0 n—0 n—0 Using the Boundary Condition (3.16) d klfl=k2_ (177 (177 yZ(2n)A2n77,-Z" l T 2(2n)827177i2n 1 + 20700272771" 1In 771‘ + 2132717” 71 71:1 72:1 77:1 n=0 (3.47) 34 From Eq (3.30), Eq (3.44), Eq (3.45), we realize that all Azi's are a function of A0 only, all D273 are a function of D0 only and all 827s are a function of D0 and B0 only. Equation (3.46) and Eq (3.47) are now solved to get D0 and B0 in terms of the third unknown which is A0. Now, the wall boundary condition (92 (77 = 1) = 0 to get the following equation (and arrive at the following equation: 0: i 82,1772" + i Dznnzn ln77 where 77 =1 n=0 n=0 (3.48) The values of DO and B0 are known in terms of A0. Thus Eq (3.48) takes the form: A0 [equation *1: 0 (3.49) The equation* in Eq (3.49) is solved to get the Nusselt number in terms of all the two- fluids parameters. The value of the Nusselt number varies depends on the number of terms in the summation. However, as we later observe, the value converges after 10 terms and the value doesn’t change appreciably upon increasing the number of terms than that. Following is the flowchart to solve the Nusselt number of a two-fluid system which is thermally and hydrodynamically fully developed and assuming that both the fluids are [m1 1 Input the value of radius ratio Newtonian. 35 Input the values of fluid properties Enter the pressure gradient Calculate various ratios such as a,7,/l,gl,gZ.m,n,p,q,w Calculate the value of A2,. 's in terms of AO Calculate the value of 9, Calculate the values of D2,. '3 in terms of DO 36 Calculate the values of 82,. 's in terms of B0 and D0 Calculate the value of 02 1 Apply interface conditions on 6, and 62 l Calculate the value of B0 and D0 in terms of A0 Apply boundary condition (outer wall) and calculate Nuo Stop Figure 3.2: Flowchart to solve the N usselt number for the two-fluid case 37 The point to be noted here is that the solution is not complete as yet. This is because we still have to find the temperature profiles of the two-fluids. And for that, we need to compute the values of A0 , Bo and D0. Since 30 and D0 are functions of A0, the two temperature profiles can be evaluated once A0 is known. Consider Eq (3.12) a: 0.720,, —-T)Nu0 82: T .2 .2 {R3 (““1 %um,1+(1-&2—)um.2]} (706,02 R0 R0 Equation (3.12) can be reduced to the following form: 17: = a2(Tw —T)Nu0 az {RozumlAl} pcp,2 um Ra2 R3 where A = u"! u ,2 2 2 ] m Jzam, Vm,l + (1-771. )vm,2 Let 6=T—Tw and {:FZT 0 Equation (3.50) reduces to the following form: 86 +Nu062az :0 R039" umRoA From Eq (3.28), (91 = ZAnn" =0 61=A0 +2127;2 +A4774 “16276 + .......... A2 2 A4 4 61=A0(1+—:-77 +—77 + ......... ) A0 A0 Equation (3.53) can also be written as: 38 (3.50) (3.51) (3.52) (3.53) 61=A0Y Where Y=(l+§l772+fl774+ ......... ) A0 A0 Substituting Eq (3.54) in Eq (3.52), we get the following equation: _Y_aAO + NuOAOYa2 —0 R0 8; ungA Or, 8A0 + Nquoaz :0 a; umROA Define a new non-dimensional quantity called the Peclet number, Pe : um.2R0 a2 Equation (3.56) reduces to the following form: 3A0 + 2Nu0A0 : O a; PeA —2Nu0§' = C * __ A0 mi J (3.54) (3.55) (3.56) (3.57) (3.58) (3.59) Assume an inlet mean temperature of 6,",0 at the point where the flow gets hydrodynamically fully developed (2:0). The mean temperature at the point f = 0 is 1 771 9771.0 =—‘2‘— I61v127md77 ”777' Vm,l O — 2N Using the value of A0 = C *exp(—P—:O£) from Eq (3.59) in Eq (3.60), e 6m,0 771' 22 fll+élnz+§i774+ ............ )(a—bnzpn A0 A0 771' vm,l 0 39 (3.60) (3.61) where, the velocity of the first fluid (inner fluid) is given: g 1 I g V1 :a—bnz, a:__2_+_g1”i2__g277i2,b:__l 4 4 4 4 Thus, the value of C is calculated from Eq (3.61) and the value of (91 is: 61 = A0 + A2772 + A4774 + A6776 + .......... (3.62) An = 0 For n = odd, N = “20 [—MAn_2 + NAn_4] For n: even and n>2 72 MN and, A2 = — 4140 A0 63- 2 a-g2 a- —'-(gi-82)fii +—'—— 481 4 4 , N = . 4 675711.10) + (1‘ 7772 )Vm,2) (77121071150 + (1" 7712 )Vm,2 ) — 2N And A0 = C * apt—530;] where, 6m,0 771' 2 (1+l—4-2—772+éi774+ ............ ](a—b772)d77 A0 A0 2 771' vm,l O RESULTS AND VALIDATION The Nusselt numbers were calculated at three locations, at the outer wall, at the interface for the first fluid and at the interface for the second fluid. The two fluids considered for the sake of analysis were engine oil at the inside and water at the outside. The properties of oil and water are taken from Table 2.1. 40 v = k—1=0.236 k2 ,1 = fl =568421 #2 a) = (pcp )1 =0.4051 (3.63) (pcp)2 ,- = $4712 “1 The Nusselt number at the outer wall (Nuo) was calculated for different values of radius ratios varying from 0 to 1 following the steps outlined in the flowchart in Figure 3.1. Because there was no existing solution for Nusselt number in a two-fluid case, the results were compared with the Nusselt number at the outer wall for a single fluid and constant temperature boundary condition. The value of Nu0 was evaluated at various values of radius ratio (771- = R,- / R0) close to zero. As the radius ratio approaches zero, it implies that the only major fluid in the system is the second fluid. Thus, the Nusselt number at the outer wall must approach the Nusselt number’s value of the single fluid case. The Nusselt number at the outer wall of a tube with a single fluid flowing in it and constant wall temperature boundary condition is 3.65679 [12]. The value of Nusselt number Nuoin the two-fluid case was 3.6575 for 77,. = 0.01 and 3.6568 for 77,. = 0.001. Thus, these results show that the Nusselt number expression for the two-fluid system is right as it approaches the value of that of a single fluid system with similar boundary conditions. The system also approaches the single-fluid system, if the radius ratio approaches 1. If 77,. is close to 1, it would imply that the only component present in the pipe is the 41 inner fluid (oil). Thus, the Nusselt number at the outer wall must converge to 3.65679. Nu0 Was evaluated for radius ratios close to 1. The value of Nu0 was 0.8648 for 77,. =1. However it must be noted that Nuo is evaluated with respect to the outer fluid. qw 2R0 (Tw T Tm) k2 N110 — Nu0 must now be calculated with the inner fluid because if 771:1, there is no outer fluid in the system. Thus 0.8648 is now scaled with respect to the inner fluid (the only fluid in the system). " 2R Nuo = qw . 0 (Tw TTm) kl k2 Thus, Nuo =O.8648* — k1 Nuo =0.8648* 0613 =3.656 0.145 4.5 — Nu" 4 4 3.5 ~ 3 —1 2.5 ~ 2 3 1.5 ~ 1 l 0.5 = O l T l I T fl 0 0.2 0.4 0.6 0.8 1 1.2 Radius Ratio Figure 3.3: Nuo vs. 77,- 42 From Figure 3.3, it can be observed that Nuo increases as radius ratio increases to a value of 0.4 and then drops. To understand this phenomenon, the total energy in the fluid was plotted as a function of the radius ratio. Figure 3.4 shows the energy of the inner fluid, energy in the outer fluid and the total energy in the fluid as a function of the radius ratio. Total e denotes the energy in the entire fluid and el and e2 denote the energy in the first and the second fluid respectively. It can be seen that the total energy in the fluid increases till the radius ratio increases to a value of 0.4. The nusselt number at the outer wall is dependent on the heat transfer that in turn is dependent on the total energy in the fluid. Thus, the Nusselt number at the outer wall increases till the radius ratio of 0.4 and then drops, showing the same pattern as the total energy in the fluid. 1.60E+06 — 1.40E+06 ~ 1.20E+06 ~ 1.00E+06 — +total e +91 +92 8.00E+05 ~ 6.00E+05 - 4.00E+05 . 2.00E+05 ~ 0.00E+00 0 0.2 0.4 0.6 0.8 1 Radius Ratio Figure 3.4: Energy in the fluids vs. 77,- 43 The next step was calculating the Nusselt number at the interface for the inner fluid. (Ii 2’7 = —— (3.64) (Ti TTmJ) kl NHL] dT1 6]: = ICE" |r=r,- T,- =Temperature of the interface TmJ = Mean Temperature of the inner fluid The Nusselt number at the interface is now verified again with the Nusselt number at the outer wall for the single fluid case. If the radius ratio goes to 1, 77,. = 1, the interface is now at the wall, and thus M111 should be same as that of the wall Nusselt number for the single fluid. The following chart shows how the radius ratio affects the Nusselt number at the interface. 9 Nu“ 8 a 7.1 O n 1 1 1 . ' 1 0 0.2 0.4 0.6 0.8 1 1.2 Radius Ratio Figure 3.5: Nu“ vs. 77,- As can be observed from Figure 3.5, the value of Nu“ is 8 as the radius ratio approaches 0. This is because as the radius ratio is close to 0, it would mean that the inner fluid flow is same as that of a plug flow. As can be seen in literature, the Nusselt number for a plug flow type problem is 8. And as the radius ratio approached 1, the value of Nu” is exactly 3.657 that again validate our expression for the Nusselt number at the interface. The next step that was done was to make the two fluids the same. If both the inner and the outer fluid are the same, the Nusselt number at the outer wall must be equal to 3.6567 for all values of radius ratio. The value of Nuo approached 3.6567 irrespective of the value of radius ratio thus validating the results. The following plots show the temperature variance with the Radial position. Theta 200 150 100 50 0 0.2 0.4 0.6 0.8 1.0 Radial Position Figure 3.6: Temperature vs. radial position for different values of 77,- 45 200 150 100 50 I- -. ~ ’ x l x )- \ \ - - - - - --0.999 \ p \ » ------------ 0001 >- \ \ \ \ __________ ~x~ K -. \“ ~ \‘ \ ~ \ \‘b x \\ \ \ \ Ts \ \ \ ~ \ T\ x‘ \ a x T'\ 5“ \ ~ 1 _L L 4 A L l 1 A 1 \g.‘ 0.2 0.4 0.6 0.8 1.0 Radial Position Figure 3.7: Temperature vs. radial position for extreme values of 77,- 46 CHAPTER4 CONVECTIVE BOUNDARY CONDITION One of our initial assumptions in solving this problem was that the flow is thermally fully developed. 8 T -T _ W _—_0 4.1 Equation (4.1) reduces to the following form: (4.2) 01 _ dTw _ Tw —T (17“,, _ dTm dz dz Tw — Tm dz dz From Eq. (3.4) and Eq (3.5), the energy balance on a element of length dz gives, 2HR0qw(x)dz = (1111317,,1 )prcdeTm + IT(R02 — R,-2 )umzpfgcp’szm (4.3) (1TH! 2g W R0 = (4.4) dz (,OCP)1 RizumJ + (pcp )2 (R62 TRi2 )“m,2 dTm ___ 2qW"(x)R0 (4 5) dz (,0c )1 R-2 R-2 . 2 P 1 1 {(90 )2R [———u ,1+(1——)u ,2]} p 0 (706,02 R3 m R2 m 0 The convective boundary condition implies that there is a fluid outside the pipe whose convective heat transfer coefficient is denoted by h... [17]. Therefore, the boundary condition is, Qw" : hoo (Too T Tw) : h0(Tw TTm) (4'6) Equation (4.6) reduces to N17000:» — Tw) = Nuo (Tw - Tm) where, 47 1100.2]? .2R Nam: 0 and Nu0=hO 0 1‘2 k2 (4.7) From Eq. (4.7) we can arrive at the value of the wall-temperature gradient in terms of the mean temperature and the two Nusselt numbers. dTw : Nuo dTm ( 4.8) dz N140 + Nuoo dz Substituting the above value from Eq (4.8) in Eq (4.2), we get, 0T _ NuO dTm + Tw —T Nuoo dTm az Nuo + Nuoo dz Tw —T,,, Nuo + Nuoo dz (4.9) From the definition of the heat transfer at the wall, " _ _ qw qw — hoo(T°o —TW) :> Tw —T°° -— h... (4.10) Substituting this value in Eq (4.10) _ £111: _ T 8T _ N110 dTm + 00 hoe Nuoo dTm az Nuo + Nu... dz Tw TTm Nuo + Nu... ‘13 (4.11) (IW : h_0 : NuO 8T _ Nuo dTm Nuo Nuoo dTm + T.>0 —T Nuoo dTm 0Z— T Nuo + Nuoo dz T Nu“, Nuo + Nuco dz Tw — Tm Nuo + Nu0° dz (4.13) in _ Too—T Nuoo dTm _ __ (4.14) az Tw —Tm Nuo + Nuoo dz From Eq (4.5), 48 dTm 2qw" (10110 dz (pc )1 R2 R-2 {(pcp)2R02[ P —%u77z,l+(lT—12')um,2]} (mp)2 R R 0 0 ( ) 2 u 2 u where, A = .m% "1’1 + (1_512_) m,2 (mp)l R0 um R0 u," 82 Tw -Tm NuO + Nuoo (pcp)” RoumA From the definition of heat transfer at the wall, qw : h0(Tw TTm) Using the expression from Eq (4.16) in Eq (4.15), we get, 2 91 = (T... —T) N”” ho aZ Nuo +Nu°o (pep)” RoumA ho = Nu0.k2 /2R0 So, from Eq (4.17) and Eq (4.18), we get, N o0N a—T=(T...—T> “ “0 “2 dz Nuo + Nuoo RozumA The energy equation for the flow is, 13( 8T]_u8T _ r_ ___ rdr Br aaz Using the value of $3): from Eq (4.19) in Eq (4.20), z _ r— —— r Br Br N N00 11( GT]: u @0004) uo u 1 um (I Nu0+Nu°° R311 49 261w (pcp )2RoumA (4.15) (4.16) (4.17) (4.18) (4.19) (4.20) (4.21) Non-dimensionalising Eq (4.21), (9 = T — T00 and 77 = R; , we get the following Eq (4.22) for the inner fluid. where And i—‘Z— nd—B‘ +[M—N772]Nu 6, =0 77d77 dn 6‘17 2 aigZ —- 81—82)” +— M 14 l 4 G712vm,lw + (1 T "1'2 )vm.2) fl N = . 4 81 ‘ 2 2 (771' vm,lw+(1T77i )vm,2) Nu = NuONuoo Nuo + Nuc,o (4.22) (4.23) (4.24) The above set of equations (4.22)-(4.24) are same as that in the constant temperature case except that Nu0 in Eq (3.19)-(3.22) is replaced with Nu . Similarly, for the outer fluid, we get the following Eq (4.25) ndn c177 g_2 _ 2 4 . “12 2 1 ”i vm,l +600 T77i )vm,2 Q Q: . 4 . (”£2va + (”(1— ”[2 )Vm,2) 50 li[nfl]+1P—anwu 62 =0 (4.25) (4.26) (4.27) Since the equations are of the same structure as the constant temperature case, the temperature profiles will also be similar. The boundary and interface conditions are given in Eq (4.28)-Eq (4.30) At 77 = 0, £131. = 0 (4.28) dn At 77 = m, 116; = L62 d7? 6177 91 = 62 (4.29) Atnzl, (162 + Nu°° 62 : 0 (4.30) d7] 2 Using the method of Frobenius, the expressions for thetal and theta2 are given by Eq (4.31) and Eq (4.32). 61 = ZAnn" (4.31) =0 AZ"+1 = O For n = odd, MNu A” = — A!) ' 4 7n : Nuoz [NA2n—4 — MA2n-7] ‘ (2n) " ' ' 62 = 23,177” +2 D" 1mm" (43?) n=0 n=0 51 D2n+l _O’BZn+l —O [UVu IUVu D2—_ Do 32: (Do Bo) 4 , 4 Nu Nu l : D7n— PD7Iz—7 ’ B7" = 0 B7n— _ PB”n—" D71: 2n (2”)2 [Q .. 4 _ _] _ (2)1)” [Q 4 ] _ Of, 61 = ZAZnn” And (92 = 232,37" +202n1mm" n=0 n=0 n=0 MNu PNu PNu A22" 4 A()’D2:- Do Bzz—(Do-Bo) Nu A2" : 2 02 [NA2n—4 —MA’2n-2] ( ’1) (4.33) Nu 1 2n=—— B7n— —PB"n—" __D"n .. (2n)2 [Q .. 4 .. -] n .. Nu 2n : Wmez—4 — PDZn-Z] Using the Boundary Condition (4.29) 6.07 =77.) =6207= 77.) 2A2n77i2n : 232.37.?" + 202,177.?" In 77.’ (4-34) n =0 n —0 n -0 Using the Boundary Condition (4.29) 6 6, k1 sz2 d 8 dry dry yZ(2n)A2n 771,211-1 z 29,1)an m2n—1 + Z (2”) 02,, m,2n-1 1n 77; + Z D2Iz77i2n_1 n=1 n=1 n=l n20 52 (4.35) From Eq (4.33) we realize that all A2143 are a function of A0 only, all Dzi's are a function of D0 only and all Bzi's are a function of D0 and 30 only. Eq (4.34) and Eq (4.35) are now solved to get D0 and 30 in terms of the third unknown which is A0. Now, the wall boundary condition is solved to get the following equation 0=§12n.B2,z + i D2,, + Elva—“’82,, (4.36) n=l n=0 n=0 2 The values of D0 and BO are known in terms of A0 . Thus Eq (4.36) takes the form: A0 [equation *] = 0 (4.37) The equation* in Eq (4.37) is solved to get the Nusselt number in terms of all the two- fluids parameters and Nusselt number of the fluid outside the tube. RESULTS AND VALIDATION The Nusselt numbers were calculated at the outer wall and at the interface. From Eq. (4.36), it can be seen that the Nusselt number at the outer wall, Nuo, a function of the Nusselt number of the fluid outside the pipe, Nu... The following table 4.1 shows the dependency of Nuo on Nu... for a radius ratio of 0.001 and 1. As can be seen from Table 4.1, the Nusselt number at the outer wall converges to the constant temperature case in the case of very high Nu... and to the constant heat flux case in the case of very low Nu... [17]. The same dependency was observed for other radius ratios also. Figure. 4.1 shows N usselt number at the outer wall for 3 different radius ratios as a function of Nu... 53 Table 4.1: Dependency of Nuo on Nu... "FD-0001 ”i=1 Nu.o Nuo Nu... Nuo 10° 3.65 106 O.86(3.65) 103 3.65 103 0.86(3.66) 102 3.66 102 0.87(3.66) 10 3.68 10 O.87(3.68) 1 4.26 1 l.00(3.98) 0.1 4.30 0.1 1.01(4.27) 0.01 4.35 0.01 102(433) 0.0001 4.36 0.0001 1.03(4.36) 10'“ 4.36 10'6 1.03(4.36) 54 0.3 ‘ ——0.5 4/\ —"0.9 7' .4 \ --.-.-.- .r/ 10'3 10‘2 10'l l 10 102 103 104 105 10" Figure 4.1: Nuo vs. Nu... for different radius ratios The values of A0 ,80 and Do need to be computed. Since Bo and D0 are functions of A0 , the two temperature profiles can be evaluated once A0 is known. Consider Eq (4.17) a_T : (T... _ T) Nuw 2h0 _a2(T,, —T)Nu az Nu0+Nu°° (pep)”ROumA— {RozumA} (4.38) 55 2 2 pCpJ um,1 R' R' “"12 2 2 where A = —'2 '1' 1" —12 = (0771' Vm,l + (1_ 771' )Vm,2 .0ch “m R0 R0 “m 7 &; Let 6=T—Too and f: 0 Equation (4.38) reduces to the following form: 86 +Nu6a2 _ R034” ungA From Eq (4.31), 61 = 2 Ann" n=0 61: A0 + A2772 + A4774 +A6776 + .......... 42 2 A4 4 61=AO(1+—77 +—77 + ......... ) 40 A0 Equation (4.40) can also be written as: A A 61=A0Y Where Y=(1+—ln2+—4—n4+ ......... ) 40 40 Substituting Eq (4.41) in Eq (4.39), we get the following equation: 1. 0A0 + NuAOYa2 _ 0 R0 0; ungA Or, 8A0 + Nil/1002 : 0 8: umROA Define a new non-dimensional quantity called the Peclet number, P8 = um '2R0 a2 56 (4.39) (4.40) (4.41) (4.42) (4.43) (4.45) Equation (4.43) reduces to the following form: 3A0 + 2mm" = 0 (4.46) a; PeA —2Nu{ = * 4.47 An C exp£ PeA ] ( ) Assume an inlet mean temperature of 6,",0 at the point where the flow gets hydrodynamically fully developed (2:0). The mean temperature at the point 4' = 0 is 771' ——1— Jalvl 27mm; (4.48) 6m,0 = 2 7277i vm,l 0 — 2N Using the value of A0 = C *expLP—uAOCJ from Eq (4.47) in Eq (4.48), e 6 C = ”“0 (4.49) 771' 22 m1+43772+i4774+ ............ J(a—bnz)d77 40 A6 771' vm,1 0 where, the velocity of the first fluid (inner fluid) is given: 80 1 2 1 v1 =a—b772, az—4-‘+Zgi771 _Zgznizvb'1-il- Thus, the value of C is calculated from Eq (4.49) and the value of 61 is: 61 = A0 + 42272 + A4774 + A6776 + .......... (4.50) A” = 0 For n = odd, _Nu n2 [—MA,,_2 + NAn_4] For n: even and n>2 MN and, A2 =-— 4qu 57 and _i aig2 fl 4 —4 81 g . ,N= L (81 T 82 )771'2 + 4 (nlzvaa) + (1 T 7712 )vm,2) M = . 2 2 (771' vaCI) + (1 T ”i )Vm,2) —2Nu§ = C ex where, 40 P[ P A ] e C = ”— 6m,0 l 22 I£l+§lnz+éin4+ ............ ](a—b772)dn 771' vm,1 0 A0 A0 The temperature profiles as shown below have been plotted with the Nusselt number Nu“, taken as 10. 100 80 60 40 20 Theta 0.2 0.4 0.6 0.8 1.0 Radial Position Figure 4.2: Temperature profiles for extreme radius ratios 58 Theta 100 0.1 80 _ _ _ - _ , _ 0.5 — — — 0.9 60 40 20 0 0.2 0.4 0.6 0.8 1.0 Radial position Figure 4.3:Temperature profiles for different radius ratios 59 CHAPTER 5 SUMMARY AND CONCLUSIONS Core-annular flow, which has a very important role in oil industry, has been analyzed so far in this work. The oil—water flow has been analyzed for three boundary conditions- constant wall heat flux, constant wall temperature and convective boundary condition. In all the cases analytical solutions have been derived for velocity profiles, temperature profiles and Nusselt numbers. The Nusselt numbers thus evaluated analytically were validated by comparison with the solutions for the single fluid case. The usefulness of water-lubricated oil flow can be found from the result that the mean velocity of the fluid system increased almost by a factor of 2000 for the same pressure gradient when the water in the system was increased from zero to 10%. Thus, the pumping pressure that is needed to pump the two—fluid flow, which has 10% water, is very less compared to the pumping pressure that would be needed to pump oil alone. The pumping pressure needed further decreases if the water in the pipe is increased. However the problem of dewatering limits the amount of water that can be used. In the constant heat flux case, the Nusselt number at the outer wall converged to 4.36 in three cases— (a) the radius ratio approached zero (b) the radius ratio approached one (c) the two fluid properties were made equal. The Nusselt number at the interface for the inner fluid converged to 4.36 when the radius ratio approached one and converged to 8 when the radius ratio converged to 0. This 60 implies that the inner fluid behaves like a slug flow when the inner radius becomes very small. Similar analysis was performed in constant temperature case. The method of Frobenius was used to get the analytical solutions for temperatures and hence the Nusselt numbers. Similar to the first boundary condition, the Nusselt number at the outer wall approached 3.65 in three cases (a) the radius ratio approached zero (b) the radius ratio approached one (c) the two fluid properties were made equal. Similar to the constant heat flux case, the Nusselt number at the interface approached 8 when the radius ratio approached 0 and 3.65 when the radius ratio approached 1. Figure 3.3 shows that Nusselt number at the outer wall increases with radius ratio till a value of 0.4 and then keeps decreasing. As can be seen from Figure 3.4, the energy in the two- fluid system rises till the radius ratio reaches a value of 0.4 and then falls. The nusselt number at the outer wall is dependent on the heat transfer that in turn is dependent on the total energy in the fluid. Thus, the oil-water system will have maximum heat transfer at the wall at a radius ratio of 0.4. The convective boundary condition was different from the above two conditions in that the Nusselt numbers are a function of the Nusselt number of the fluid outside the outer pipe (Nu...) As expected, the Nusselt number at the outer wall approached that of the constant heat flux case when the Nusselt number of the fluid outside the pipe (Nu...) was made zero (in other words, it was the zero wall heat flux case). The Nusselt number 61 at the outer wall approached the constant temperature case when Nu... was made very large. 62 APPENDIX The software “Mathematica” was used to solve for various parameters. Following are some of the important statements. The following equations solve for the velocity profiles. 1 DSOlve[%:: —91: V1177], ’7] 1 ; D[n D[V2['7] I n], 77] DSolve[%-- —gz. V207], n] 1 2 92 1 2 1 2 1 9201 Vl=—-— +—+—— -—— '+ ' '-—)t ‘+ 4U 91 4 491771 49201 Loglnl]771( 2 91m ) v2— 93 - ”292 +Log[n] n- (£1.3an 92'”) 4 4 1 2 1 hr2 1 1 1 hr2 u1—— +—hrE(—-—l+ o; 401 4 1141 112/ 4#2 hr2 hrg =- + 4H2 4H2 umeanl = Integrate[u127r r, {r, 0, ri}] / (7T r12) umean2= Integrate[u2*2* 71' *r, {r, r1, ro}]/ (7r (1302-1112)) 1 f .. l um: * u1*2*zr*rdlr+ u2*2*7r*rdr Him2 10 r' I The following set of equations solve for the temperature profiles in the case of constant heat flux boundary condition. The variable I denotes the radius ratio. 1 :7- D[0D[91[n] . n]. n] 63 V DSolve[96:= F1,6107], r7 1 ; D[0D[92[n], n], n] DSolve[%:= 173, 92177], r7] n _ n4 91 02 (-i2 g1+2i24 LOQli] 91-92+i292 -2i2109[i192) 64 m 16m T1[r_] ::el*2*ro*q/k1+Tc T2[r_1 ==92*2*ro*q/k2+Tc Temp[r_] :=u1[r] *T1[r] * 1:424:71 Integrate[Temp[r] , {1" 0. I'll] Tmeanl[r_] = %/ (mr12) Qinterface= -k (D[T1[r] , r]; r = i r = i ; Tinterface = T1[r_] Qinterface Tinterface - Tmeanl [r] Nuinterfacel = Temp2[r_] :=u2[r] *T2[r]* r*2*n Integrate[Taup2[r] , {r, r1, ro}] Tinean2[r_] = %/ (71* (roz—r12)) 'I‘rriean1*r12*mr\ean1+'1‘mean2*(r02—r12)*tmiean2 'Imeanz ; rozseum r: r0; 'IWall: T2[r] Nuo= qw 'IWall — 'Imean ' The following statements evaluate the nusselt number and the temperatures of the two fluids in the constant temperature boundary condition. A[2] = —m*A[0] *nuO/ 4; thetal = A[O] + A[2] * (27"2); DO[A[2*i] = (nuO/ (4*i*1)) * (n*A[2*i—4] — m*A[2*i—2]), {i, 2, 15, 1}]; Do[theta1= theta1+A[2*i] * (r)"(2*i)), {i, 2, 15, 1}]; Dee[2] = —p*Dee[0] *nu0/4; B[2] = -p* (B[O] —Dee[0]) *nuO/ 4; theta2 = B[O] +B[2] *r] *n+Dee[0] * Logtn] + DeeEZ] *Iogtn] * (17"2); Do[Dee[2*i] = (nuO/ (4*i* 1)) * (q*Dee[2*i—4] - p*Dee[2*i-2]), {i, 2, 15, 1}]; Do[B[2*i] = (nu0/ (44141)) * (q*B[2*i-4] -p*B[2*i-2]) — (1/i)*Dee[2*i], {i, 2, 15. 1}]; Do[theta2= theta2+B[2*i] *n"(2*i) + Dee[2*i] *Log[n] * (0"(2*i)). {i, 2.15.1}1; derthetal = D[theta1, n]; dertheta2=D[theta2, n]; n=0.5;ni=0.5; w: (884.1*1909) / (9974:4179); ni= 0.5; a: 1.471/0.859; y: 0.145/0.613; A: 0.486/ (0.8554: (109—6)); 65 g1: l*0.2*0.2/ (2*2*0.486); g2: 1*O.2*O.2/ (2*2*0.855*10A-6)7 va=g2/4-(9’2/8)*(1+(ni"2)); vm1= (—gl/8—g2/4+g1/4)*(ni"2) +g2/4; m: ((gl—g2) *ni"2+92) *a/ (4* (ni"2*vm1*w+ (1-ni"2) *vm2)); n=a*g1/ (4* (ni"2*vml*w+ (1—ni"2) *va)); p=g2/ (4* (ni"2*vml*w+ (l—ni"2) *Vm2)); CI=P: A[O] =1.0; eqn= {thetal— theta2 == 0, y * derthetal- dertheta2 == 0}; Solve[eqn, {B[0],Dee[0]}]: Dee[2] =-p*Dee[0] *nuO/4; B[2] = —p* (B[O] —Dee[0]) *nuO/ 4; theta2 = B[O] +B[2] *77 *n+Dee[0] *Log[n1+Dee[2] *LOgM] * (0‘2); Do[Dee[2*i] = (nuO/ (4*i*i)) * (quee[2*i—4] — p*Dee[2*i—2]), {it 2: 15: 1}]; D0[B[2*i] = (nuO/ (4*i* 1)) * (q*B[2*i—4] —p*B[2*i—2]) - (1/i) *Dee[2*i], {i, 2. 15. 1}]; Do[theta2= theta2+B[2*i] *n"(2*i) + Dee[2*il*1091’7]*(n“(2*i))l {i, 2, 15. 1}]; nuO = 3.85; theta2 nuO = 3 . 86; theta2 66 Plot[theta2, {nu0, 3.85, 3.86}] thetaminlet = 100; 0=-; A[2] = -m*A[0] *nu0/4; theta1= A[O] +A[2] * (ne2); Do[A[2*i] = (nuO/ (441141)) * (n*A[2*i—4] —m*A[2*i-2]) , {i, 2, 15, 1}]; Do[theta1= theta1+A[2*i] * (r}"(Z*i)), {i, 2, 15, 1}] 1 2 92 1 .2 1 -2 v1=—— 1+———+— 1n1-— 2n1 417g 4 4g 4g Integrate[(theta1*v1*n) , {r}, O, ni}] c=theta1nirflet*ni*ni*vml/ (2*%) a=w*vml*ni*ni+ (1—ni*ni) *vm2; A[O] = c«Exp[—2*nu0*§/ (pe*a)] A[2] = -m*A[0] *nu0/4; thetalz A[O] +A[2] * (0‘2); Do[A[2*i] = (nuO/ (44141)) * (n*A[2*i-4] -m*A[2*i-2]), {i, 2, 15, 1}]; Do[theta1= theta1+A[2*i] * (n"(2*i)), {i, 2, 15, 1}]; pe= 200; thetal B[O] = B[O] *A[0] ,- Dee[O] =Dee[0] *A[0]; Dee[2] = —p*Dee[0] *nu0/4; B[2] = -p* (B[O] —Dee[0]) *nuO/ 4; theta2 = B[O] +B[2] *r] *r;+Dee[O] * Log[r;] + Dee[2] *Logln] * (0A2); Do[Dee[2*i] = (nuO/ (4*i-k 1)) ~1- (q*Dee[2*i—4] — p*Dee[2*i-2]), {i, 2, 15, 1}]; D0[B[2*i] = (nuO/ (41:11: 1.)) * (q*B[2*i—4] —p*B[2*i—2]) —(1/i)*Dee[2*i], {i, 2, 15, 1}]; 67 Do[theta2 = theta2+B[2* i] *n"(2 * i) + Dee[2*i] *Loglnl * (77"(2*i)), {i, 2. 15. 1}]; The same procedure as above is repeated for convective boundary condition except that the term Nuo in the constant temperature condition is replaced by Nu . 68 10. 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