Multidimensional Heat Transfer This equation governs the Cartesian, temperature distribution for a three-dimensional unsteady, heat transfer problem involving.

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Presentation transcript:

Multidimensional Heat Transfer 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.

Two-D, Steady State Case 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 Factor This 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 T 1 ) to the other surface (at T 2 ) can be expressed as: q=Sk(T 1 -T 2 ) 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(T 1 -T 2 )=(T 1 -T 2 )/(1/kS)= (T 1 -T 2 )/R t where R t = 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 17-5

Example An 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 (k soil =0.5W/m.K), estimate the heat loss per unit length of pipe. z=1 m T2T2 T1T1 From Table 17-5, case 1. L>>D, z>3D/2

Example (cont.) 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. Therefore, 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 pipe Ground at -20  C Heat transfer to the ground (q) Length dx

Example (cont.) 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 km.