Multidimensional Heat TransferTwo-D, Steady State CaseConduction Shape FactorExampleExample (cont.)Slide 6Multidimensional Heat TransferHeat Diffusion EquationcTtkTxTyTzq k T qp   ( ) 2222222• This equation governs the Cartesian, temperature distribution for a three-dimensional unsteady, heat transfer problem involving heat generation.• For steady state / t = 0• No generation• To solve for the full equation, it requires a total of six boundary conditions: two for each direction. Only one initial condition is needed to account for the transient behavior. q 0Two-D, Steady State CaseFor a 2 - D, steady state situation, the heat equation is simplified to it needs two boundary conditions in each direction.2 2TxTy2 20,There are three approaches to solve this equation:• Numerical Method: Finite difference or finite element schemes, usually will be solved using computers.• Graphical Method: Limited use. However, the conduction shape factor concept derived under this concept can be useful for specific configurations. • Analytical Method: The mathematical equation can be solved using techniques like the method of separation of variables. (review Engr. Math II)Conduction Shape FactorThis approach applied to 2-D conduction involving two isothermal surfaces, with all other surfaces being adiabatic. The heat transfer from one surface (at a temperature T1) to the other surface (at T2) can be expressed as: q=Sk(T1-T2) where k is the thermal conductivity of the solid and S is the conduction shape factor.• The shape factor can be related to the thermal resistance: q=Sk(T1-T2)=(T1-T2)/(1/kS)= (T1-T2)/Rt where Rt = 1/(kS)• 1-D heat transfer can use shape factor also. Ex: heat transfer inside a plane wall of thickness L is q=kA(T/L), S=A/L• Common shape factors for selected configurations can be found in Table 17-5ExampleAn Alaska oil pipe line is buried in the earth at a depth of 1 m. The horizontal pipe is a thin-walled of outside diameter of 50 cm. The pipe is very long and the averaged temperature of the oil is 100C and the ground soil temperature is at -20 C (ksoil=0.5W/m.K), estimate the heat loss per unit length of pipe.z=1 mT2T1From Table 17-5, case 1. L>>D, z>3D/2SLz Dq kS T TW     242 14 0 53 020 5 3 02 100 20181 21 2 ln( / )( )ln( / . ).( ) ( . )( . )( ). ( ) heat loss for every meter of pipeExample (cont.)TCmpTCmpIf the mass flow rate of the oil is 2 kg/s and the specific heat of the oil is 2 kJ/kg.K, determine the temperature change in 1 m of pipe length.q mC T TqmCCPP    ,.*. ( ) 181 22000 20 045Therefore, the total temperature variation can be significant if the pipe is very long. For example, 45C for every 1 km of pipe length.• Heating might be needed to prevent the oil from freezing up.• The heat transfer can not be considered constant for a long pipeGround at -20CHeat transfer to the ground (q)Length dx)( dTTCmpExample (cont.)Heat Transfer at section with a temperature T(x)q =2 k(dx)ln(4z / D)Energy balance: mCintegrate at inlet x = 0, T(0) = 100 C, C = 120T(x) = -20 +120P( ) . ( )( )( ). ( ) , . ,( ) ,..T T dxT q mC T dTmCdTdxTdTTdxT x CeePPxx        20 1 51 201 51 20 0200 000378200 0003780 0003780 1000 2000 3000 4000 500050050100T( )xx• Temperature drops exponentially from the initial temp. of 100C• It reaches 0C at x=4740 m, therefore, reheating is required every 4.7