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problem4S.5

# problem4S.5 - (S 2 o)k(T A 1 T 2 Application M N S o A q...

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PROBLEM 4S.5 KNOWN: Long conduit of inner circular cross section and outer surfaces of square cross section. FIND: Shape factor and heat rate for the two applications when outer surfaces are insulated or aintained at a uniform temperature. m SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties and (3) onduit is very long. C ANALYSIS: The adiabatic symmetry lines for each of the applications is shown above. Using the flux plot methodology and selecting four temperature increments (N = 4), the flux plots are as shown below. For the symmetrical sections, S = 2S o , where S o = M A /N and the heat rate for each application is q =
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Unformatted text preview: (S 2 o / )k(T A 1- T 2 ). Application M N S o / A q ′ (W/m) A 10.3 4 2.58 11,588 B 6.2 4 1.55 6,975 < < COMMENTS: (1) For application A, most of the heat lanes leave the inner surface (T 1 ) on the upper portion. (2) For application B, most of the heat flow lanes leave the inner surface on the upper portion (that is, lanes 1-4). Because the lower, right-hand corner is insulated, the entire section experiences small heat flows (lane 6 + 0.2). Note the shapes of the isotherms near the right-hand, insulated boundary and that they intersect the boundary normally....
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