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Unformatted text preview: PROBLEM 3.1 KNOWN: Onedimensional, plane wall separating hot and cold fluids at T and T ,1 ,2 , respectively. FIND: Temperature distribution, T(x), and heat flux, q x , in terms of T T h ,1 ,2 1 , , , h 2 , k and L. SCHEMATIC: ASSUMPTIONS: (1) Onedimensional conduction, (2) Steadystate conditions, (3) Constant properties, (4) Negligible radiation, (5) No generation. ANALYSIS: For the foregoing conditions, the general solution to the heat diffusion equation is of the form, Equation 3.2, ( ) 1 2 T x C x C . = + (1) The constants of integration, C 1 and C 2 , are determined by using surface energy balance conditions at x = 0 and x = L, Equation 2.23, and as illustrated above, ( ) ( ) 1 , 1 2 , 2 x=0 x=L dT dT k h T T 0 k h T L T . dt dx = = (2,3) For the BC at x = 0, Equation (2), use Equation (1) to find ( ) ( ) 1 1 , 1 1 2 k C h T C C + = + (4) and for the BC at x = L to find ( ) ( ) 1 2 1 2 , 2 k C h C L C T . + = + (5) Multiply Eq. (4) by h 2 and Eq. (5) by h 1 , and add the equations to obtain C 1 . Then substitute C 1 into Eq. (4) to obtain C 2 . The results are ( ) ( ) ,1 ,2 ,1 ,2 1 2 , 1 1 1 2 1 2 T T T T C C T 1 1 L 1 1 L k h h h k h h k = = + + + + + ( ) ( ) ,1 ,2 ,1 1 1 2 T T x 1 T x T . k h 1 1 L h h k = + + + + < From Fouriers law, the heat flux is a constant and of the form ( ) ,1 ,2 x 1 1 2 T T dT q k k C . dx 1 1 L h h k = = = + + + < PROBLEM 3.2 KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces of a rear window. FIND: (a) Inner and outer window surface temperatures, T s,i and T s,o , and (b) T s,i and T s,o as a function of the outside air temperature T ,o and for selected values of outer convection coefficient, h o . SCHEMATIC: ASSUMPTIONS: (1) Steadystate conditions, (2) Onedimensional conduction, (3) Negligible radiation effects, (4) Constant properties. PROPERTIES: Table A3 , Glass (300 K): k = 1.4 W/m K. ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.11 and 3.12, ( ) ,i ,o 2 2 o i 40 C 10 C T T q 1 L 1 1 0.004 m 1 h k h 1.4 W m K 65 W m K 30 W m K = = + + + + ( ) 2 2 50 C q 9 6 8 W m 0.0154 0.0029 0.0333 m K W = = + + . Hence, with ( ) i , i , o q h T T = , the inner surface temperature is 2 s,i ,i 2 i q 9 6 8 W m T T 40 C 7.7 C h 30 W m K = = = < Similarly for the outer surface temperature with ( ) o s,o ,o q h T T = find 2 s,o ,o 2 o q 9 6 8 W m T T 10 C 4.9 C h 65 W m K = = = < (b) Using the same analysis, T s,i and T s,o have been computed and plotted as a function of the outside air...
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This note was uploaded on 11/02/2009 for the course CHEM 378 taught by Professor Corti during the Spring '09 term at Purdue UniversityWest Lafayette.
 Spring '09
 Corti

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