c. The Grashof number is the ratio of buoyant forces to viscous forces acting on a fluid.

Transcription

1 QUESTION 1. (0 pts) With respect to free convection: a. What is an extensive, quiescent fluid? (4 points) b. What are the two major physical considerations or forces for free convection? (4 points) c. What is the Grashof number in words (ratio of forces) and mathematically? (6 points) d. What is the Rayleigh number in words and mathematically? (6 points) SOUTION a. As defined on page 561 of the text (Incropera, et.al., 4/e), an extensive medium is, in principle, an infinite medium. Since a quiescent fluid is one that is otherwise at rest. In other words, an extensive, quiescent fluid is one that has infinite bulk properties (extensive) with no forced convection, and by all practical purposes, is stagnant (quiescent). b. The two major physical considerations or forces for free convection are buoyant forces and viscous forces. Buoyant forces arise from a fluid density gradient and a body force proportional to the density at any given instant; this proportional force is often gravity. Viscous forces are always present and actually provide a counter-force situation to the buoyant force. This is similar to Newton s 3 rd aw of Motion. More discussion of these forces is provided in Section 9.1 of the text. c. The Grashof number is the ratio of buoyant forces to viscous forces acting on a fluid. d. The Rayleigh number is the product of the Grashof number and the Prandtl number, effectively being the ratio of buoyant forces to viscous forces multiplied by the ratio of momentum to thermal diffusivities.

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5 PROBEM 9. KNOWN: Interior air and wall temperatures; wall height. FIND: (a) Average heat transfer coefficient when T = 0 C and T s = C, (b) Average heat transfer coefficient when T = 7 C and T s = 37 C. SCHEMATIC: ASSUMPTIONS: (a) Wall is at a uniform temperature, (b) Room air is quiescent. PROPERTIES: Table A-4, Air (T f = 88K, 1 atm): β = 1/T f = K -1, ν = m /s, k = W/m K, α = m /s, Pr = 0.7; (T f = 305K, 1 atm): β = 1/T f = K -1, ν = m /s, k = W/m K, α = m /s, Pr = ANAYSIS: The appropriate correlation for the average heat transfer coefficient for free convection on a vertical wall is Eq h Ra Nu 0.85 = = + k ( 0.49/Pr) where Ra = g β T 3 /να, Eq. 9.5, with T = T s - T or T - T s. (a) Substituting numerical values typical of winter conditions gives m/s 3.47 K ( 0 ) K (.5m) Ra = = m / s 0.96 m / s ( ) Nu = = / 0.7 ( ) h = Nu k/ W/m K /.5m 3.03 W/m = = K. < Hence, ( ) (b) Substituting numerical values typical of summer conditions gives m/s 3.79 K ( 37 7 ) K (.5 m) Ra = = m / s m / s ( 1.30 ) Nu = = ( 0.49 / 0.706) Hence, h = Nu k/ = W/m K/.5m =.94 W/m K. < COMMENTS: There is a small influence due to T f on h for these conditions. We should expect radiation effects to be important with such low values of h.

6 PROBEM 9.78 KNOWN: Sphere with embedded electrical heater is maintained at a uniform surface temperature when suspended in various media. FIND: Required electrical power for these media: (a) atmospheric air, (b) water, (c) ethylene glycol. SCHEMATIC: ASSUMPTIONS: (1) Negligible surface radiation effects, () Extensive and quiescent media. PROPERTIES: Evaluated at T f = (T s + T )/ = 330K: Table A-4, Air (1 atm) Table A-6, Water Table A-5, Ethylene glycol ν 6, m /s k 3, W/m K α 6, m /s Pr β 3, K ANAYSIS: The electrical power (P e ) required to offset convection heat transfer is q conv = has ( Ts T ) = π hd ( Ts T ). (1) The free convection heat transfer coefficient for the sphere can be estimated from Eq using Eq. 9.5 to evaluate Ra D. Pr Ra 1/4 3 D gβ ΔT D Nu D = + Ra 4/9 D =. (,3) 9/16 να 1+ ( / Pr) Ra 11 D (a) For air 9.8m / s ( K 1 )( 94 0) K ( 0.05m) 3 Ra 4 D = = m / s m / s ( 4 ) 1/4 k W / m K h D = Nu D = +.6 W / m K D 0.05m 9/16 4/9 = 1 + ( / ) q conv = π.6 W / m K ( 0.05m) ( 94 0) K = 1.55 W. Continued..

7 PROBEM 9.78 (Cont.) (b,c) Summary of the calculations above and for water and ethylene glycol: Fluid Ra D hd ( W/m K) q(w) Air < Water < Ethylene glycol < COMMENTS: Note large differences in the coefficients and heat rates for the fluids.

8 PROBEM 9.90 KNOWN: Temperatures and dimensions of a window-storm window combination. FIND: Rate of heat loss by free convection. SCHEMATIC: ASSUMPTIONS: (1) Both glass plates are of uniform temperature with insulated interconnecting walls and () Negligible radiation exchange. PROPERTIES: Table A-4, Air (78K, 1 atm): ν = m /s, k = W/m K, α = m /s, Pr = 0.71, β = K -1. ANAYSIS: For the vertical cavity, 3 ( ) 9.8m / s ( K 1 )( 30 C)( 0.06m) 3 gβ T1 T Ra = = αν m / s m / s Ra 5 = With (H/) = 0, Eq. 9.5 may be used as a first approximation for Pr = 0.71, 0.3 1/ Nu 0.4 Ra 1/ Pr ( H / ) = = 0.4( 8.37 ) ( 0.71) ( 0) Nu = 5. k 0.045W / m K h = Nu = 5. =.1W/m K. 0.06m The heat loss by free convection is then ( ) q= ha T1 T q =.1W / m K 1.m 0.8m 30 C = 61W. ( )( ) COMMENTS: In such an application, radiation losses should also be considered, and infiltration effects could render heat loss by free convection significant. <

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