SolutionHW_C6 - Homework Set #6: Due date: 10/27/06, by...

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Homework Set #6: Due date: 10/27/06, by 11:00 am. Problem #1 : Solve problem 10.A.1 in BSL Answer: See Example 10.6-1 in BSL for the notation and the derivation of the energy equations in terms of temperature for individual layers. The final forms of these equations are as follow: ( ) () 10 0 1 00 0 1 01 21 1 2 1 2 12 32 23 0 0 2 3 0 0 23 Zone 01 (steel): Zone 12 (magnesia): Zone 23 (cor k): ln ln ln rr dT k r rq T T dr k dT T T dr k dT T T dr k =⇒ = = = Addition of these above equation gives 03 01 12 23 ln ln ln TT kkk = ++ Thus, the heat loss over the inner surface of the pipe with an area of 0 2 Lr π ( ) 0 01 12 23 2 2 ln ln ln LT T QL r q == Or the heat loss per hour per foot of the pipe
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( ) () 03 0 10 21 32 01 12 23 2 ln ln ln TT Q rr L kkk π = ++ The r i for this problem are 1 2 3 2.067 / 2 1.0335 in. 1.0335 0.154 1.19 in. 1.19 2 3.19 in. 3.19 2 5.19 in. o r r r r = = =+= =+ = = Insertion of numerical values into the above formula gives: 24.58 Btu/hr-foot o Q L = Problem #2 : Solve problem 10.B.1 in BSL Answer: a. Equation setup Due to the motionless body of fluid, heat conduction is the only mechanism of energy transfer accounting for the temperature distribution in the fluid. Obviously, this is a symmetric heat transfer problem where the heat is being steadily conducted in the r direction only. Therefore, consider a control volume of thickness r over which we make the energy balance.
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This note was uploaded on 12/14/2011 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas at Austin.

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SolutionHW_C6 - Homework Set #6: Due date: 10/27/06, by...

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