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Solution_HW6

# Solution_HW6 - Solutions for homework Set#6 Problem#1 Solve...

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Solutions for homework Set #6: 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 dr k dT 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   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   0 01 12 23 2 ln ln ln Q L

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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 : olve problem 10.B.1 in BSL . Equation setup ue to the motionless body of fluid, heat conduction is the only mechanism of energy transfer ccounting for the temperature distribution in the fluid. Obviously, this is a symmetric heat here 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. S Answer: a D a transfer problem w  2 2 44 0 rr r rq r r q   or   22 0 r know divide by r and then take the limit as r goes to zero We     lim r 0 r r r 0 We then use the definition of the first derivative to get r 2 0 r d dr
Insertion of Fourier’s law of heat conduction with constant thermal conductivity 22 0 0 dd T d d T rk r dr dr dr dr        or 2 0 T r dr dr b. Integration of this equation twice with respect to r gives 2 1 1 C 2 dT rC dr r TC 

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Solution_HW6 - Solutions for homework Set#6 Problem#1 Solve...

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