Thermodynamics HW Solutions 156

Thermodynamics HW - k r g dr dT r dr d& and then integrating once with respect to r 1 2 2 C k r g dr dT r − =& Rearranging the differential

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 2 Heat Conduction Equation 2-136 A hollow pipe is subjected to specified temperatures at the inner and outer surfaces. There is also heat generation in the pipe. The variation of temperature in the pipe and the center surface temperature of the pipe are to be determined for steady one-dimensional heat transfer. 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 centerline. 2 Thermal conductivity is constant. Properties The thermal conductivity is given to be k = 20 W/m °C. Analysis The rate of heat generation is determined from [] 3 2 2 2 1 2 2 W/m 894 , 37 4 / ) m 12 ( m) 3 . 0 ( m) 4 . 0 ( W 000 , 25 4 / ) ( = π = π = = L D D W V W g Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as 0 1 = + k g dr dT r dr d r r 2 T 2 r r 1 T 1 g and C 60 ) ( 1 1 ° = = T r T C 80 ) ( 2 2 ° = = T r T Rearranging the differential equation 0 = =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: k r g dr dT r dr d & and then integrating once with respect to r , 1 2 2 C k r g dr dT r + − = & Rearranging the differential equation again r C k r g dr dT 1 2 + − = & and finally integrating again with respect to r , we obtain 2 1 2 ln 4 ) ( C r C k r g r T + + − = & where C and C 1 2 are arbitrary constants. Applying the boundary conditions give r = r 1 : 2 1 1 2 1 1 ln 4 ) ( C r C k r g r T + + − = & r = r 2 : 2 2 1 2 2 2 ln 4 ) ( C r C k r g r T + + − = & Substituting the given values, these equations can be written as 2 1 2 ) 15 . ln( ) 20 ( 4 ) 15 . )( 894 , 37 ( 60 C C + + − = 2 1 2 ) 20 . ln( ) 20 ( 4 ) 20 . )( 894 , 37 ( 80 C C + + − = Solving for C simultaneously gives C 2 C 2 1 and 2 . 257 34 . 98 2 1 = = C C Substituting C into the general solution, the variation of temperature is determined to be 1 and 2-79...
View Full Document

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.

Ask a homework question - tutors are online