ChemE324_prelim1_review_solutions

# ChemE324_prelim1_review_solutions - PROBLEM 3.33 KNOWN:...

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PROBLEM 3.33 KNOWN: Steady-state temperature distribution of convex shape for material with k = k o (1 + α T) where α is a constant and the mid-point temperature is T o higher than expected for a linear temperature distribution. FIND: Relationship to evaluate α in terms of T o and T 1 , T 2 (the temperatures at the boundaries). SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) α is positive and constant. ANALYSIS: At any location in the wall, Fourier’s law has the form () xo dT qk 1 T . dx α ′′ =− + (1) Since x q ′′ is a constant, we can separate Eq. (1), identify appropriate integration limits, and integrate to obtain 2 1 LT 0T qd x k 1 TdT =− + ∫∫ (2) 22 o 21 x2 1 TT k qT T . L2 2 αα     + +   (3) We could perform the same integration, but with the upper limits at x = L/2, to obtain 2 2 oL / 2 1 xL / 2 1 T T 2k T L22 + + (4) where 12 L/2 o L / 2 T . 2 + ==+ (5) Setting Eq. (3) equal to Eq. (4), substituting from Eq. (5) for T L/2 , and solving for α , it follows that ( ) o 2 o 2T . / 2 T T / = +− ++ <

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PROBLEM 3.68 KNOWN: Radius and heat dissipation of a hemispherical source embedded in a substrate of prescribed thermal conductivity. Source and substrate boundary conditions. FIND: Substrate temperature distribution and surface temperature of heat source. SCHEMATIC: ASSUMPTIONS: (1) Top surface is adiabatic. Hence, hemispherical source in semi-infinite medium is equivalent to spherical source in infinite medium (with q = 8 W) and heat transfer is one-dimensional in the radial direction, (2) Steady-state conditions, (3) Constant properties, (4) No generation. ANALYSIS: Heat equation reduces to () 22 1 2 12 1d d T r 0 r dT/dr=C dr dr r Tr C/ r+C.  =   =− Boundary conditions: () () os T T T r T ∞= = Hence, C 2 = T and s1 o 1 o s T C / r and C r T T . ∞∞ + = The temperature distribution has the form () ( ) so Tr T T T r/ r =+− < and the heat rate is o s q=-kAdT/dr k2 r T T r / r T T ππ  =   It follows that ( ) s -4 o q4 W T T 50.9 C 125 W/m K 2 10 m π −= = = s T 77.9 C. = < COMMENTS: For the semi-infinite (or infinite) medium approximation to be valid, the substrate dimensions must be much larger than those of the transistor.
PROBLEM 3.79 KNOWN: Wall of thermal conductivity k and thickness L with uniform generation q ± ; strip heater with uniform heat flux o q; ′′ prescribed inside and outside air conditions (h i , T ,i , h o , T ,o ). FIND: (a) Sketch temperature distribution in wall if none of the heat generated within the wall is lost to the outside air, (b) Temperatures at the wall boundaries T(0) and T(L) for the prescribed condition, (c) Value of q o required to maintain this condition, (d) Temperature of the outer surface, T(L), if o q=0 but q corresponds to the value calculated in (c).

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## This note was uploaded on 10/27/2008 for the course CHEM 3240 taught by Professor Stroock during the Fall '08 term at Cornell University (Engineering School).

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ChemE324_prelim1_review_solutions - PROBLEM 3.33 KNOWN:...

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