Thermodynamics HW Solutions 103

Thermodynamics HW Solutions 103 - Chapter 2 Heat Conduction...

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Chapter 2 Heat Conduction Equation Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Properties The thermal conductivity is given to be k = 7.2 Btu/h ft °F. Analysis ( a ) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as 0 = dr dT r dr d and −= k dT r dr hT Tr () [( 1 1 ) ] T 22 160 == ° F ( b ) Integrating the differential equation once with respect to r gives Steam 250 ° F h =1.25 L = 15 ft T =160 ° F C = 1 r dT dr Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, dT dr C r = 1 C r C ln =+ 12 where C and C 1 2 are arbitrary constants. Applying the boundary conditions give r = r 1 : + k C r C r C 1 1 11 2 l n ) ] 2 C 2 r = r 2 : T l n 21 2 2 = Solving for C simultaneously gives 1
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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