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Eb33766771

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Eb33766771

  1. 1. Ravi Kumar Singh, K. B. Sahu, Thakur Debasis Mishra / International Journal of EngineeringResearch and Applications (IJERA) ISSN: 2248-9622 www.ijera.comVol. 3, Issue 3, May-Jun 2013, pp.766-771766 | P a g eAnalysis of Heat Transfer and Flow due to Natural Convection inAir around Heated Square Cylinders of Different Sizes inside anEnclosureRavi Kumar Singh*, K. B. Sahu**, Thakur Debasis Mishra***1(Department of Mechanical Engineering, KIIT University, Bhubaneswar- 751024, India)ABSTRACTIn this paper, analysis of heat transferand flow due to natural convection in air aroundheated square cylinders of different sizes insidean enclosure has been computed. The squarecylinder is at higher temperature and the verticalwalls of the enclosure are at lower temperaturewith insulated horizontal walls. The variousgoverning equations such as continuity equation,momentum equation, and energy conservationequation are calculated by finite elements usingGalerkin method. Assuming constant propertiesof air, the Rayleigh number depends on the sizeof square enclosure for fixed high and lowtemperatures. The sizes of enclosures are takenas 20 mm, 40 mm, and 80 mm for differentRayleigh numbers. The size of the cylinder istaken accordingly for aspect ratio of 0.2, 0.3, and0.4. Results are displayed in the form ofIsotherms, streamline and velocity vectordiagrams and the heat transfer and flow aroundthe cylinder is analyzed.Keywords - Aspect ratio, Finite Element Method,Fluent, Galerkin methodNOMENCLATUREL Size of square enclosure filled with air, ma Size of heated square cylinder, mAR (In dimensionless form) Aspect Ratio (a/L)Nu, Pr, Ra (In dimensionless form) Nusseltnumber, Prandtl number, Rayleigh Number (air)Q (In dimensionless form) Heat flux of airT Temperature of the wall, KX, Y (In dimensionless form) Co-ordinate in x, y-directiong Gravitational force of air, m/s2k Thermal conductivity of air, W/m.Kx, y Co-ordinate in x, y-direction, mu, v Velocity of air constituents along x, y-direction, m/su*v*(In dimensionless form) Velocity of airconstituents along x, y-directionGreek Symbolν Kinematic viscosity of air, m2/sρ Density of air, kg/m3(1.007)α Thermal diffusivity of air, m2/sβ Volumetric co-efficient of thermal expansionof air, K-1 (In dimensionless form)Temperature of thewallω, ω*Vorticity of enclosure( s-1), (Indimensionless form) Vorticity of enclosureψ, ψ* Stream function of enclosure (m2/s), (Indimensionless form) Stream functionSubscriptsH Hot wall of heated square cylinderC Cold wall of enclosure filled with air1. INTRODUCTIONThe curiosity of natural convection inenclosures filled with air having cold vertical wallsand adiabatic horizontal walls has been theme ofresearch over the past years. The natural convectionissue is of concern in various engineering andtechnology uses such as solar energy collectors,cooling of electronics components etc. A. Dalal etal. [1] state that natural convection occur in thevicinity of tilted square cylinder in the range of (0 ≤ ≤ 45) inside an enclosure having horizontaladiabatic wall and cold vertical wall figure out bycell-centered finite volume method, which is used toreckoned two dimension Navier strokes equation forincompressible laminar flow. And taken the value ofRayleigh number is Ra= 105, Pr= 0.71. G. De VahlDavis [2] expresses that differentially heated sidewalls of square cavity are figure out for precisesolution of the equations. It has taken the Rayleighnumbers in the range of 103≤ Ra≤106. N. C.Markatos [3] deal with Buoyancy-driven laminarand turbulent flow is reckoned by computationalmethod. It has taken the Rayleigh numbers in therange of 103≤ Ra≤1016. Donor-cell differencing isapplied for this problem. The Rayleigh numbers ismore than 106for turbulence model. Demirdzic etal. [5] state that out of four issue of heat transfer andfluid flow only two orthogonal, boundary-fittedgrids are utilize. They have reckoned the issue ofthis work by using multigrid finite volume method.It has apply 320×320 grid for the control volume, afound the error approximately to be less than 0·1%.D. G. Roychowdhury et al. [6] deal with convectiveflow and heat transfer in the area of heated cylindersituated in square enclosure is reckoned by non-orthogonal grid based finite volume technique.
  2. 2. Ravi Kumar Singh, K. B. Sahu, Thakur Debasis Mishra / International Journal of EngineeringResearch and Applications (IJERA) ISSN: 2248-9622 www.ijera.comVol. 3, Issue 3, May-Jun 2013, pp.766-771767 | P a g eIn the current paper, natural convection in air aroundheated square cylinders of different sizes inside anenclosure having adiabatic horizontal and coldvertical walls. The continuity, momentum andenergy conservation equation are reckoned usingfinite element method. The impacts of aspect ratio atdifferent Rayleigh number on the air flow pattern,temperature distribution and heat transfer have beendebated.1. PROBLEM STATEMENTFigure1. Computational domain1.1. MODELAn air is filled in the square enclosure ofsize (L) is considered. The heated square cylinder(Th) of different sizes (a) placed inside the centre ofthe enclosure filled with air with the aspect ratio of0.2, 0.3, and 0.4. The left and right walls are coldwall (Tc) and top and bottom walls are adiabaticwall (Q=0). There is no slip condition (velocity) atall the walls.1.2. GOVERNING EQUATIONSThe governing Navier-stroke’s equation for steadylaminar flow of a boussinesq air in stream functionand vorticity form is given below2ψ = -ω (1)u∂ω∂x+ν∂ω∂y= ν2ω + g β∂T∂x(2)For constant air properties in the absence of internalheat generation and insignificant viscousdissipation, the conservation equation can berepresented as,u∂T∂x+ν∂T∂y= α∂2T∂x2 +∂2T∂y2 (3)The boundary conditions areΨ = 0, at all walls (4)Boundary conditions in dimension form areLeft wall T = TC, at x = 0; 0 ≤ y ≤ L (5)Right wall T = TC, at x = L; 0 ≤ y ≤ L (6)Top and bottom wall∂T∂y= 0; at y = 0, L; 0 ≤ x ≤ L(7)Left solid wall T = TH, at x =L−a2;L−a2≤ y ≤L+a2(8)Right solid wall T = TH, at x =L+a2;L−a2≤ y ≤L+a2(9)Bottom solid wall T = TH, at y =L−a2;L−a2≤ x ≤L+a2(10)Top solid wall T = TH, at y =L+a2;L−a2≤ x ≤L+a2(11)The following parameters are used for representingthe governing equations in dimension- less form.X =xLif x =L−a2then X =1−AR2where, (AR = a/L)X =xLif x =L+a2then X =1+AR2Y =yLif y =L−a2then Y =1−AR2Y =yLif y =L+a2then Y =1+AR2Ψ*=ψ𝛼, ω*=ωα/L2 , u*=uα/L,ν*=να/L,  =TTHBy using the above dimensionless quantities, thevarious governing equation are represented as,2ψ*= -ω*(12)u* ∂ω∗∂X+ν∂ω∗∂Y= Pr2ω + Pr Ra∂θ∂X(13)u* ∂θ∂X+ν* ∂θ∂Y= 2θ (14)The boundary conditions for all walls areΨ*= 0, at all walls (15)Boundary condition in dimension-less form is = 0.5; at X = 0; 0 ≤ Y ≤ 1 where,  =TCTH(16) = 0.5; at X = 1; 0 ≤ Y ≤ 1 (17)∂∂y= 0; at Y = 0, 1; 0 ≤ X ≤ 1 (18) = 1; X =1−AR2;1−AR2≤ Y ≤1+AR2(19) = 1; X =1+AR2;1−AR2≤ Y ≤1+AR2(20) = 1; Y =1−AR2;1−AR2≤ X ≤1+AR2(21) = 1; Y =1+AR2;1−AR2≤ X ≤1+AR2(22)2. COMPUTATIONAL ANALYSISIn the computational analysis first we haveused standard boussinesq air properties for differentlength of enclosure (L) of 20mm, 40, and 80mm toreckon the value of Rayleigh number. we havedepict our geometry according to given aspect ratio(AR) of 0.2, 0.3 and 0.4 at different Rayleighnumbers (Ra) of 1.93 ×104, 1.54 ×105and 1.23×106. Then we have done mesh and refinement at 3in vicinity of heated square cylinder, and taken meshsize 100, 40 for enclosure walls and heated squarecylinder walls respectively. After that we gaveboundary condition 373.16 K for all heated squarecylinder walls, 323.16K for both vertical diathermicwalls of enclosure and heat flux = 0 for horizontaladiabatic walls of enclosure. Then we have madesolution for this problem with respect to aspect ratio(AR) of 0.2, 0.3 and 0.4 at different Rayleighnumbers of 1.93 ×104, 1.54 ×105and 1.23 ×106toanalyze the heat transfer and flow.
  3. 3. Ravi Kumar Singh, K. B. Sahu, Thakur Debasis Mishra / International Journal of EngineeringResearch and Applications (IJERA) ISSN: 2248-9622 www.ijera.comVol. 3, Issue 3, May-Jun 2013, pp.766-771768 | P a g eThe Nusselt number of heated square cylinder wallsand cold vertical walls are find out after reachesconvergence by using following formula,Nu H =QHk ( TH −TC )=1AR 1−θCAR0∂θ∂XatX=0 dY (23)Nu C =QCk ( TH −TC )=1AR 1−θCAR0∂θ∂XatX=1 dY (24)(a)(b)(c)Figure2. Stream function, Isotherm and velocityvector diagram for Ra = 1.93* 104, AR = 0.2(a)(b)9.81E-069.81E-062.94E-052.94E-052.94E-052.94E-054.91E-054.91E-054.91E-054.91E-054.91E-054.91E-056.87E-056.87E-056.87E-056.87E-056.87E-056.87E-056.87E-059.81E-059.81E-059.81E-059.81E-059.81E-059.81E-059.81E-051.18E-041.18E-041.18E-041.18E-041.18E-041.37E-041.37E-041.37E-041.37E-043.27E+023.27E+023.27E+023.27E+023.27E+023.27E+023.27E+023.33E+023.33E+023.33E+023.33E+023.33E+023.33E+023.33E+023.40E+023.40E+023.40E+023.40E+023.40E+023.40E+023.47E+023.47E+023.47E+023.47E+023.47E+023.47E+023.53E+023.53E+023.53E+023.53E+023.53E+023.60E+023.60E+023.60E+023.60E+023.66E+023.66E+023.66E+023.06E-053.06E-059.18E-059.18E-059.18E-059.18E-051.53E-041.53E-041.53E-041.53E-041.53E-042.14E-042.14E-042.14E-042.14E-042.14E-042.14E-042.14E-042.14E-043.06E-043.06E-043.06E-043.06E-043.06E-043.06E-043.06E-043.67E-043.67E-043.67E-043.67E-043.67E-044.28E-044.28E-044.28E-044.28E-043.27E+023.27E+023.27E+023.27E+023.27E+023.27E+023.27E+023.33E+023.33E+023.33E+023.33E+023.33E+023.33E+023.33E+023.40E+023.40E+023.40E+023.40E+023.40E+023.40E+023.40E+023.47E+023.47E+023.47E+023.47E+023.47E+023.47E+023.53E+023.53E+023.53E+023.53E+023.53E+023.53E+023.60E+023.60E+023.60E+023.60E+023.60E+023.66E+023.66E+023.66E+023.66E+023.66E+02
  4. 4. Ravi Kumar Singh, K. B. Sahu, Thakur Debasis Mishra / International Journal of EngineeringResearch and Applications (IJERA) ISSN: 2248-9622 www.ijera.comVol. 3, Issue 3, May-Jun 2013, pp.766-771769 | P a g e(c)Figure3. Stream function, Isotherm and velocityvector diagram for Ra = 1.54* 105, AR = 0.3(a)(b)(c)Figure4. Stream function, Isotherm and velocityvector diagram for Ra = 1.23* 106, AR = 0.4(a)(b)6.39E-056.39E-056.39E-051.92E-041.92E-041.92E-041.92E-041.92E-043.19E-043.19E-043.19E-043.19E-043.19E-043.19E-044.47E-044.47E-044.47E-044.47E-044.47E-044.47E-044.47E-044.47E-046.39E-046.39E-046.39E-046.39E-046.39E-046.39E-046.39E-047.67E-047.67E-047.67E-047.67E-047.67E-048.94E-048.94E-048.94E-048.94E-043.27E+023.27E+023.27E+023.27E+023.33E+023.33E+023.33E+023.33E+02 3.33E+023.33E+023.33E+023.33E+023.40E+023.40E+023.40E+023.40E+02 3.40E+023.40E+023.40E+023.40E+023.47E+023.47E+023.47E+023.47E+023.47E+023.47E+023.47E+023.53E+023.53E+023.53E+023.53E+023.53E+023.53E+023.53E+023.60E+023.60E+023.60E+023.60E+023.60E+023.60E+023.60E+023.66E+023.66E+02 3.66E+023.66E+023.66E+023.66E+0200.20.40.60.810 2 4 6Enclosurewall(Y)Nusselt Number (Nu)AR = 0.2AR = 0.3AR = 0.40.30.40.50.60.72 12 22Cylinderwall(Right)Nusselt Number (Nu)AR = 0.2 AR = 0.3 AR = 0.4
  5. 5. Ravi Kumar Singh, K. B. Sahu, Thakur Debasis Mishra / International Journal of EngineeringResearch and Applications (IJERA) ISSN: 2248-9622 www.ijera.comVol. 3, Issue 3, May-Jun 2013, pp.766-771770 | P a g e(c)Figure5. Nusselt number variation along vertical walls ofenclosure & Right and top walls of cylinder for Rayleighnumber of 1.93×104(a)(b)(c)Figure6. Nusselt number variation along verticalwalls of enclosure & Right and top walls of cylinderfor Rayleigh number of 1.23×1063. COMPUTER CODE VALIDATIONA common benchmark solution for naturalconvection in differentially heated square cavitywith adiabatic horizontal walls and isothermalvertical walls with T= 1, for left wall and T= 0, forright wall [2]. In the present work, numericalpredictions have been acquired for Rayleighnumbers of Ra = 104, 105. The maximum, minimum,and average nusselt number on hot wall for differentRayleigh numbers are compared in table 1.Table1. Comparison of solution for naturalconvection in square cavityRa = 104a ba−ba×100NumaxNuminNu3.5280.5862.2433.4550.5672.1322.063.244.94Ra = 105a ba−ba×100NumaxNuminNu7.1170.7294.5197.2150.7564.417-1.37-3.702.25a, solution of de Vahl Davis [2]; b, present solution.4. RESULT AND ANALYSISThe analysis has been taken place of heattransfer and flow due to natural convection in airaround heated square cylinders of different sizesinside an enclosure having adiabatic horizontal anddiathermic vertical walls. We have taken size ofenclosure constant in range of 20mm, 40mm, and80mm with respect to aspect ratio of 0.2, 0.3, and0.4 to reckon the value of Rayleigh number for theconstant air properties. Thus we obtain the Rayleighnumbers for that size enclosure is 1.93 ×104, 1.54×105and 1.23 ×106. Then we have figure out theappropriate boundary condition for this problemwhich are represented above. After that we analyzethis problem to acquire the outcome in dimensionform of stream function, Isotherm and velocityvector diagram which are represented in fig. 2, 3and4.Table.2 Maximum velocity along X and Y direction& maximum stream function for Rayleigh numberof 1.93 ×104, 1.54 ×105and 1.23 ×106with respectto aspect ratio 0.2, 0.3 and 0.4 (In dimension-lessform)05100.3 0.5 0.7Nusseltnumber(Nu)Cylinder wall (Top)AR = 0.2AR = 0.3AR = 0.400.20.40.60.810 2 4Enclosurewall(Y)Nusselt Number (Nu)AR = 0.2AR = 0.3AR = 0.40.30.50.72 7 12 17Cylinderwall(Right)Nusselt Number (Nu)AR = 0.2 AR = 0.3 AR = 0.4024680.3 0.5 0.7Nusseltnumber(Nu)Cylinder wall (Top)AR = 0.2AR = 0.3AR = 0.4
  6. 6. Ravi Kumar Singh, K. B. Sahu, Thakur Debasis Mishra / International Journal of EngineeringResearch and Applications (IJERA) ISSN: 2248-9622 www.ijera.comVol. 3, Issue 3, May-Jun 2013, pp.766-771771 | P a g e4.1. STREAM FUNCTION, ISOTHERM ANDVELOCITY VECTOR DIAGRAMWe are estimating that how heat transferand flow occur inside an enclosure filled with air.This result has been exhibiting in dimension form ofstream function Isotherm and velocity vectordiagram. Figure 2, 3and 4. (a) Shows Stream linesincreases with increase of Rayleigh number butdecrease with rise of aspect ratio. Isotherm’scurvature increases with increases of aspect ratio atdifferent Rayleigh number. And velocity of fluidflow inside a computational domain rise withincrease of Rayleigh number for aspect ratio of 0.2,0.3, and 0.4 shown in fig. 2, 3and 4. (c). Table.2shows that velocity along X and Y direction andstream line increasing by rising the value ofRayleigh number.4.2. NUSSELT NUMBERWe have represented Nusselt number (Nu)in the dimension-less form of graph for the aspectratio (AR) of 0.2, 0.3 and 0.4 at different Rayleighnumbers of 1.93 ×104and 1.23 ×106. The figure 5, 6shows that Nusselt number versus enclosure walland cylinder walls (Right & Top) in dimension–lessform, and indicates that Nusselt number dependentof temperature. Because by increasing sizes ofheated square cylinder it gets closer to diathermicwalls of enclosure and transfer more heat to thatwalls, so that the Nusselt number (Nu) increaseswith increase of aspect but decrease with increasesof Rayleigh number for enclosure walls. While thenusselt number decreases with increase of aspectratio as well as Rayleigh number for heated squarecylinder walls. Hence it’s proved that heated squarecylinder can absorb more energy by increasingaspect ratio and transfer more heat to diathermicwalls.CONCLUSIONHeat transfer and fluid flow due to naturalconvection in air around heated square cylinders ofdifferent sizes inside an enclosure having adiabatichorizontal and diathermic vertical walls of size20mm, 40mm, and 80mm with respect to aspectratio of 0.2, 0.3, and 0.4 at different Rayleighnumbers of 1.93 ×104, 1.54 ×105and 1.23 ×106areanalyzed, and results are exhibited in the dimensionform of Isotherm, Stream function and Velocityvector diagram.It’s proved that Nusselt number dependent oftemperature and heated square cylinder can absorbmore energy by increasing the size and can transfermore heat to the diathermic walls for variouspurposes.REFERENCES[1] A.Dalal, V. Eswaran, G. Biswas, “Naturalconvection around a heated square cylinderplaced in different angles inside anenclosure,” Heat and mass transferconference, 2008, Paper No.45.[2] G. de Vahl Davis, “Natural convection ofair in a square cavity: A benchmarknumerical solution,” International journalfor numerical methods in fluids, 3, 1983,pp. 249-264.[3] N. C. Markatos, and K. A. Perikleous,“Laminar and turbulent natural convectionin an enclosed cavity,” International journalof heat and mass transfer, 27 (5), 1984, pp.755-772.[4] S. Ostrach, “Natural convection inenclosures,” Journal of heat transfer 50thanniversary issue, 110 (4-B), 1988, pp.1175-1190.[5] I. Demirdzic, Z. Lilek, M. Peric, “Fluidflow and heat transfer test problems fornon- orthogonal grids: Bench-marksolutions,” International journal fornumerical methods in fluids, 15, 1992, pp.329-354.[6] D. G. Roychowdhury, S. K. Das, T. S.Sundararajan, “ Numerical simulation ofnatural convective heat transfer and fluidflow around a heated cylinder inside anenclosure,” Heat and mass transfer, 38,2002, pp. 565-576.[7] V. Eswaran, S. Prakash, “A finite volumemethod for navier stroke equations,”Proceeding of third Asian CFD conference,Bangalore, India, 1998, PP. 127-136.[8] A. K. De, A.Dalal, “A numerical study ofnatural convection around a squarehorizontal heated cylinder placed in anenclosure,” International journal of heatand mass transfer, 49, 2006, pp. 4608-4623.[9] P. H. Oosthuizen, “Free convective flow inan enclosure with a cooled inclined uppersurface,” Computational mechanics 14,1994, pp. 420-430[10] K. B. Sahu, S. K. Mahapatra, S. Sen, A.Sarkar, “Natural convection indifferentially heated rectangular cavities ofvarious aspect ratios,” Heat and masstransfer conference, 2008, Paper No. NCH-1.Ra AR umax vmax Ψmax1.93 ×1040.2 13.57 20.75 5.490.3 11.93 18.04 4.220.4 9.68 13.93 2.821.54 ×1050.2 44.59 76.81 17.990.3 41.12 77.26 17.130.4 42.07 74.73 14.801.23 ×1060.2 131.16 205.65 32.310.3 124.33 216.26 32.460.4 120.08 217.52 35.79

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