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3200_solutions_hw13,14

# 3200_solutions_hw13,14 - r1=1.61/2"= 0.805 r2=1.9/2 = 0.95...

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r1=1.61/2”= 0.805 “ ksteel = 9.4 Btu/h F r2=1.9/2” = 0.95” kglass = 0.022 Btu/h F r3=r2+1” = 1.95” kmagnesia = 0.038 @ 40F and [email protected] 100F, use 0.039 r4=r3+1” = 2.95” --------------------------------------------------

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16.12 Here is the solution manual ‘solution’. It is ok until the integration step in part (a), after which there is a mistake in rewriting the integrand (the square is moved to the wrong position). Go to the next page for the corrections. Note that the A o = A 1 = π r 1 2 problem is here (1 + β x 2 ) should be (1 + β x) 2 Correction: - = + 2 1 1 0 2 ) 1 ( 1 T T L dT kA dx x q β ) ( | ) 1 ( 1 2 1 0 T T kA x q L - - = + - β β ) ( 1 ) 1 ( 1 1 2 1 T T kA L q - = - + β β ) ( ) 1 ( ) 1 ( ) 1 ( 1 1 2 1 T T kA L L L q - = + + - + β β β β ) ( ) 1 ( 1 2 1 T T kA L L q - = + - β ) ( ) 1 ( 2 1 1 T T L L kA q - + = β X
16.12 (b) dz dT x A aT k q 2 1 0 ) 1 ( ] [ β + - - = - - = + 2 1 0 1 0 2 ] [ ) 1 ( 1 T T L dT aT k A dx x q β The problem was worked with Fourier’s first law (FFL) of conduction as the starting point, since we didn’t need to know the detailed temperature profile. However, as I

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3200_solutions_hw13,14 - r1=1.61/2"= 0.805 r2=1.9/2 = 0.95...

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