Modelling of Distribution of Heat on a One-Dimensional Body - 1100 Words

Advanced Thermal Analysis: Modelling of Distribution of Heat on a One-Dimensional Body (Research Paper Sample) Content: Advanced Thermal AnalysisME 530Studentà ¢Ã¢â€š ¬s NameProfessorDate of SubmissionNomenclaturecp-specific heat capacity at constant pressureJKI-current AmpsL-length of slab [m]R-resistance ohmsS-source termt-time [sec]n-natural numbers [unitless]T-temperature [K]Ttr-transient temperatureTst-steady state temperatureTC-ambient temperature [K]ÃŽ-thermal diffusivity m2secà Ã‚ -density kgm3ÃŽÂ ºn-Eigenvalue nà Ã¢â€š ¬LAbstractThe aim of this project is to analytically solve the problem of thermal diffusion in one-dimensional conduction using Cartesian domain subjected to a steady source of heat due to a direct current (DC) Ohmà ¢Ã¢â€š ¬s heating. The current is connected to a slab, which lead to heat being transferred by conduction. The slab is of length, L, and has an ambient temperature Tc at x = 0 and x = L. There is no heat transfer due to convection. The slab has a constant conductivity along its length. The problem was solved and analyze to give a Fourier series solutionIntroductionOne-dimensional thermal diffusion in the Cartesian domain subjected to uniform heat generation due to direct current (DC) Ohmà ¢Ã¢â€š ¬s heating for a slab of length L and subjected to an ambient temperature TC at x=0 and x=L with constant properties, energy generation, no convection and constant conductivity. The problem was solved and analyzed to give a Fourier series solution.Problem FormulationThe problem can be modelled as shown below.1294765139700 z = Hz = 0///////////////////LTcTCx00 z = Hz = 0///////////////////LTcTCx2169795806450025184108763000244157550800S00S167894042545H00HFigure 1: One-dimensional Thermal Diffusion in the Cartesian Domain subjected to uniform heat.The governing equation is,à ¢Tà ¢t+Và ¢Ãƒ ¢T=ÃŽÃ ¢2T+Sà Ã‚ Cp (1) The governing equation is obtained by applying Fourier equation as followsIn figure 1, the temperature at the boundaries are set to an ambient temperature of TC.Also, ità ¢Ã¢â€š ¬s shown that the top and bott om of the slab are insulated resulting no heat transfer occurs in the z-direction. Equation 2 is the heat source over the density and specific heat.G=Sà Ã‚ Cp (2) We assume constant thermal conductivity. Then, we apply the assumption to the governing equation as the following,à ¢Tà ¢z=0 (3) à ¢2Tà ¢z2=0 (4) à ¢Tà ¢y=0 (5) à ¢2Tà ¢y2=0 (6) Therefore, the equation becomesà ¢Tà ¢t=ÃŽÃ ¢2Tà ¢x2+G (7) The below equation shows the relationship between resistance and voltage,W=VÃÆ'I=IÃÆ'RÃÆ'I=I2R (8) The source is uniformly distributed, as in the conversion from electrical to thermal energy (Ohmic heating):S=I2RV= I2RL3 Wm3 (9) Where V = Volume of the slabV=LÃÆ'LÃÆ'LMethod: Steady State solutionThe equation must be solved subject to specià ¯Ã‚ c boundary and initial conditions in order to determine T(x, t). Equation 3 can now be used to produce à ¯Ã‚ nite dià ¯Ã¢â€š ¬erence equations approximating the heat conduction problem. The equation may be integrated twic e to obtain the general solution.Tx,t=Tstx+Ttrt,xThe temperature at any given point has two components, i.e., the steady state and the transient temperature. (10) Subject to the boundary conditionx=0 : T=Tcx=L :T=Tc * The temperatures at the margins are the ambient temperaturesà ¢2Tà ¢t2=à ¢2Tà ¢x2+à ¢2Tà ¢y2+à ¢2Tà ¢z2 (11) Tst0+Ttrt,0=Tcà ¢Tst0=Tc (12) Tx,0=Tstx+Ttrt,0 (13) The initial conditiont=0 :T=TcTtr0,x=Tc-TstxWe have uniform heat generation (S per unit volume) within the domain. The general heat conduction equation reduces to,0=kà Ã‚ Cpà ¢2Tà ¢x2+Sà Ã‚ Cp (14) à ¢ à ¢2Tà ¢x2+Sk=0 (15) Integrating the above equation, we obtain the general solution asà ¢Tstxà ¢t=-Skx+C1 (16) Tst=-S2kx2+C1x+C2 (17) Both surfaces are maintained at a common temperature: (Refer Fig. 1) and the prescribed boundary conditions,x=0 : T=Tcx=L :T=TcUsing the boundary conditions as explained, the constants of integration take the values,C2=Tc ; C1=S2kLWhere C1 and C2 are constant of integration and can be determined according to boundary condition. The solution gives temperature distribution and heat transfer in a plane wall.Tst=-S2kx2+S2kLx+Tc (18) Tst=S2kL-xx+Tc (19) For steady state equation,Tst=S2kL-xx+Tc (20) Transient solutionSolving for transient,à ¢Tstxà ¢t+à ¢Ttrt,xà ¢t= à ¢Ã‚ Ãƒ ¢2Tstxà ¢x2+à ¢Ã‚ Ãƒ ¢2Ttrt,xà ¢x2 (21) à ¢ à ¢Ttrà ¢t=ÃŽÃ ¢2Ttrà ¢x2 (22) Transient Boundary Conditionsx =L : Ttr=0x=0 à ¢ Ttr=0Transient Initial Conditionst=0 : Tstx+Ttr0,x=Tct=0 à ¢ Ttr(0,x)=Tc-Tst(x)Transient Solution- Separation of VariablesTtrt,x=ÃŽÂ ¸tXx (23) ÃŽÂ ¸tXx=ÃŽÃŽÂ ¸(t)X''x ÃÆ'ÃŽ XxÃŽÂ ¸t (24) Since the left hand side is a function of X only and the right-hand side of the below equation is a function of ÃŽÂ ¸ only, both sides must be equal to a separation constant, ÃŽÂ º, i.e.,1ÃŽÃŽÂ ¸tÃŽÂ ¸t= X''xXx=-ÃŽÂ º2 (25) 1ÃŽÃŽÂ ¸ÃƒÅ½Ã‚ ¸=-ÃŽÂ º2 (26) 1ÃŽÂ ¸dÃŽÂ ¸dt=-ÃŽ ÃŽà ‚ º2 (27) dÃŽÂ ¸ÃƒÅ½Ã‚ ¸= -ÃŽÂ º2ÃŽ dt (28) Since ÃŽÂ º cannot be zero, C2 must be zero and the equation becomeslnÃŽÂ ¸=-ÃŽÂ º2ÃŽ t+C2 (29) ÃŽÂ ¸=eC3-ÃŽÂ º2à ¢Ã‚ t =eC3e-ÃŽÂ º2ÃŽt=Ae-ÃŽÂ º2à ¢Ã‚ t (30) ÃŽÂ ¸=Ae-ÃŽÂ º2ÃŽ t (31) 1x d2xdx2=-ÃŽÂ º2 (32) d2Xdx2=-ÃŽÂ º2X (33) d2xdx2+ÃŽÂ º2X=0 (34) ÃŽ2+ÃŽÂ º2=0 (35) Then, we obtain the equation,ÃŽ2=-ÃŽÂ º2 (36) =Â-k2 =Â-1 (37) ÃŽ ÃŽÂ º= Âi (38) X=B1eÃŽ1x+B2eÃŽ2x = B1eiÃŽÂ ºx+B2e-iÃŽÂ ºx (39) Therefore, Euler identities,eià =cosà +isinà  (40) e-ià =cosà -isinà  (41) For the real values,B2=B1* (42) B1=B1r+iB1i (43) Complex ConjugateB1*=B1r-iB1i (44) Then,B1+B1*=2B1r (45) B1B1*=B1r2-i2B1i2=B1r2+B1i2 (46) Solving for X as the following,X=B1eÃŽ1x+B2eÃŽ2x=B1eiÃŽÂ ºx+B1*e-iÃŽÂ ºx (47) X=B1cosÃŽÂ ºx+iB1sinÃŽÂ ºx+B1*cosÃŽÂ ºx+iB1*sinÃŽÂ ºx (48) X=B1+B1*cosÃŽÂ ºx+iB1+B1*sinÃŽÂ ºx (49) X=2B1rcosÃŽÂ ºx+i22B1isinÃŽÂ ºx (50) X=2B1rcosÃŽÂ ºx-2B1isinÃŽÂ ºx (51) X=b1sinÃŽÂ ºx+b2cosÃŽÂ ºx (52) Boundary Initial Conditionsx=0 : Ttr = 0= X0ÃŽÂ ¸t , X0=0L=0 : Ttr = 0= XLÃŽÂ ¸t , XL=0Ttr=0=XLÃŽÂ ¸tà ¢XL=0 (53) t=0 : Ttr=Tc-TstxTherefore, the equation becomesx=0 : 0=b1à ¢0+b2à ¢1=b2x=L : 0=b1sinÃŽÂ ºx+0From both equations we can see that,b2=0b1sinÃŽÂ ºx=0sinÃŽÂ ºL=0 (54) ÃŽÂ ºL=n à Ã¢â€š ¬ (55) ÃŽÂ º=n à Ã¢â€š ¬L for n=1,2,3, à ¢Ã¢â€š ¬ (56) Therefore,Xn=b1nsin(ÃŽÂ ºnX) (57) The solution of transient,Ttr=n=1à ¢Ã… ¾ÃƒÅ½Ã‚ ¸tXnx=n=1à ¢Ã… ¾Cne-ÃŽÂ ºn2ÃŽtsinnà Ã¢â€š ¬Lx (58) Integrating both sides,0Lsinmà Ã¢â€š ¬LxTc-Tstxdx=n=1à ¢Ã… ¾Cn0Lsinnà Ã¢â€š ¬Lx sinmà Ã¢â€š ¬Lxdx (59) Orthogonally Condition0Lsinnà Ã¢â€š ¬Lxsinmà Ã¢â€š ¬Lxdx=0 for mà ¢nL2 for m=n (60) Then, the solution becomes,0Lsinmà Ã¢â€š ¬Lx-S2kL-xxdx=n=1à ¢Ã… ¾Cn0Lsinnà Ã¢â€š ¬Lx sinmà Ã¢â€š ¬Lxdx (61) 0Lsinmà Ã¢â€š ¬Lx-S2kL-xxdx=n=1à ¢Ã… ¾CnL 2 for m=n (62) L2Cn=0Lsinnà Ã¢â€š ¬Lx-SxL2k+Sx22kdx (63) L2Cn= -SL2k0Lxsinnà Ã¢â€š ¬Lxdx+S2k0Lx2sinnà Ã¢â€š ¬Lxdx