Solution HW#10.pdf - PROBLEM 6.14.FAM GIVEN A bottle containing a cold beverage is awaiting consumption During this period the bottle can be placed

# Solution HW#10.pdf - PROBLEM 6.14.FAM GIVEN A bottle...

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PROBLEM 6.14.FAM GIVEN : A bottle containing a cold beverage is awaiting consumption. During this period, the bottle can be placed vertically or horizontally, as shown in Figure Pr.6.14. Assume that the bottle can be treated as a cylinder of diameter D and length L . We wish to compare the surface-convection heat transfer to the bottle when it is (i) standing vertically or (ii) placed horizontally. For the vertical position, the surface-convection heat transfer is approximated using the results of the vertical plate, provided that the boundary-layer thickness δ α is much less than the bottle diameter D . D = 10 cm, L = 25 cm, T s = 4 C, T f, = 25 C. Neglect the end areas. Use the average temperature between the air and the surface to evaluate the thermo- physical properties of the air. SKETCH : Figure Pr.6.14 shows two positions of a beverage bottle. Cola D = 10 cm L = 25 cm Bottle T s = 4 o C < T f, Surrounding Air T f, = 25 o C Cola g (i) Vertically Arranged (ii) Horizontally Arranged Figure Pr.6.14 Thermobuoyant ﬂow and heat transfer from beverage bottles. (i) Standing vertically. (ii) Placed horizontally OBJECTIVE : (a) Determine the average Nusselt numbers Nu L and Nu D . (b) Determine the average surface-convection thermal resistances A ku R ku L [ C/(W/m 2 )] and A ku R ku D [ C/(W/m 2 )]. (c) Determine the rates of surface-convection heat transfer Q ku L (W) and Q ku D (W). SOLUTION : (i) Vertical Position: (a) The Rayleigh number is given by (6.88) as Ra L = ( T s T f, ) L 3 ν f α f . The properties for air at T δ = ( T s + T f, ) / 2 = 288 K are obtained from Table C.22: k f = 0 . 026 W/m-K, ν f = 14 . 60 × 10 6 m 2 /s, Pr = 0 . 69, and from (6.77) we have β f = 1 /T ave = 3 . 472 × 10 3 1/K. With ν f α f = ν 2 f / Pr, the Rayleigh number becomes Ra L = 9 . 81(m / s 2 ) × 3 . 472 × 10 3 (1 / K) × (25 4)(K) × (0 . 25) 3 (m) 3 (14 . 60 × 10 6 ) 2 (m 2 / s) 2 / (0 . 69) = 3 . 618 × 10 7 . 577 Since Ra L < 10 9 , from (6.91) the ﬂow is laminar. For thermobuoyant ﬂow over a vertical ﬂat plate, the average Nusselt number is obtained from (6.92) as a 1 = 4 3 0 . 503 [ 1 + ( 0 . 492 Pr ) 9 / 16 ] 4 / 9 = 0 . 5131 Nu L,l = 2 . 8 ln [ 1 + 2 . 8 a 1 Ra 1 / 4 L ] = 41 . 18 Nu L,t = 0 . 13Pr 0 . 22 (1 + 0 . 61Pr 0 . 81 ) 0 . 42 Ra 1 / 3 L = 33 . 88 Nu L = [ Nu L,l 6 + Nu L,t 6 ] 1 / 6 = 43 . 08 . (b) The average surface-convection thermal resistance is found from (6.49) as A ku R ku L = L k f Nu L = 0 . 25(m) 0 . 026(W/m-K) × 43 . 08 = 2 . 232 × 10 1 C / (W / m 2 ) . (c) The surface-averaged surface-convection heat transfer is found from (6.49) as Q ku L = A ku T s T f, A ku R ku L = π × 0 . 1(m) × 0 . 25(m) × 4( C) 25( C) 2 . 232 × 10 1 [ C / (W / m 2 )] = 7 . 390 W . (ii) Horizontal Position: (a) For the horizontal cylinder, the Rayleigh number is found from Table 6.5, i.e., Ra L = ( T s T f, ) D 3 ν f α f = 9 . 81(m / s 2 ) × 3 . 472 × 10 3 (1 / K) × [25( C) 4( C)](0 . 1) 3 (m) 3 (14 . 60 × 10 6 ) 2 (m 2 / s) 2 / (0 . 69) = 2 . 315 × 10 6 .  #### You've reached the end of your free preview.

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