( )( ) PROBLEM 9.5 (1) (2) 3 (3) Ra g TL. h L (4) L L. q ( ) 0.10/1m ( C /L ) Ra 0.59/0.6m L2

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1 PROBEM 9.5 KNOWN: Heat transfer rate by convection from a vertical surface, 1m high by 0.m wide, to quiescent air that is 0K cooler. FIND: Ratio of the heat transfer rate for the above case to that for a vertical surface that is 0.m high by 1m wide with quiescent air that is 0K warmer. ASSUMPTIONS: Thermophysical properties independent of temperature; evaluate at 00K; () Negligible radiation exchange with surroundings, () Quiescent ambient air. PROPERTIES: Table A-4, Air (00K, 1 atm): ν = m /s, α = m /s. ANAYSIS: The rate equation for convection between the plates and quiescent air is q = h As T where T is either (T s - T ) or (T - T s ); for both cases, A s = w. The desired heat transfer ratio is then q1 h = 1. () q h To determine the dependence of h on geometry, first calculate the Rayleigh number, Ra g T = β / να () and substituting property values at 00K, find, Case 1: Ra 1 = 9. m/s (1/00K) 0K (1m) / m /s m /s = Case : Ra = Ra 1 ( / 1 ) = (0.m/1.0m) = Hence, Case 1 is turbulent and Case is laminar. Using the correlation of Eq. 9.4, h k Nu C n ( Ra ) h C Ra n = = = k (4) where for Case 1: C 1 = 0.10, n 1 = 1/ and for Case : C = 0.59, n =. Substituting Eq. (4) into the ratio of Eq. () with numerical values, find n 1/ 1 9 q 0.10/1m C 1/1 Ra = 1 = = 0.1 q n ( C / ) Ra 0.59/0.m ( ) ( ) COMMENTS: Is this result to be expected? How do you explain this effect of plate orientation on the heat rates?

2 PROBEM 9.7 KNOWN: Thin, vertical plates of length 0.15m at 54 C being cooled in a water bath at 0 C. FIND: Minimum spacing between plates such that no interference will occur between freeconvection boundary layers. ASSUMPTIONS: (a) Water in bath is quiescent, (b) Plates are at uniform temperature. PROPERTIES: Table A-, Water (T f = (T s + T )/ = (54 + 0) C/ = 10K): ρ = 1/v f = kg/m, µ = N s/m, ν = µ/ρ = m /s, Pr = 4., β = K -1. ANAYSIS: The minimum separation distance will be twice the thickness of the boundary layer at the trailing edge where x = 0.15m. Assuming laminar, free convection boundary layer conditions, the similarity parameter, η, given by Eq. 9.1, is y η = ( Gr x /4) x where y is measured normal to the plate (see Fig. 9.). According to Fig. 9.4, the boundary layer thickness occurs at a value η 5. It follows then that, where y bl = η x Gr x /4 g β Ts T x Gr x = ν Gr 9. m/s K K 0.15m / m /s x = =. Hence, y m / bl = = m =. mm and the minimum separation is d = ybl =. mm= 1. mm. COMMENTS: According to Eq. 9., the critical Grashof number for the onset of turbulent conditions in the boundary layer is Gr x,c Pr For the conditions above, Gr x Pr = = We conclude that the boundary layer is indeed turbulent at x = 0.15m and our calculation is only an estimate which is likely to be low. Therefore, the plate separation should be greater than 1. mm.

3 PROBEM 9.40 KNOWN: Horizontal, circular grill of 0.m diameter with emissivity 0.9 is maintained at a uniform surface temperature of 10 C when ambient air and surroundings are at 4 C. FIND: Electrical power required to maintain grill at prescribed surface temperature. ASSUMPTIONS: Room air is quiescent, () Surroundings are large compared to grill surface. PROPERTIES: Table A-4, Air (T f = (T + T s )/ = (4 + 10) C/ = 50K, 1 atm): ν = m /s, k = 0.00 W/m K, α = m /s, β = 1/Tf. ANAYSIS: The heat loss from the grill is due to free convection with the ambient air and to radiation exchange with the surroundings. q = A 4 4 s h( Ts T ) εσ ( Ts T + sur ). Calculate Ra from Eq. 9.5, Ra = gβ TsT c/ να where for a horizontal disc from Eq. 9.9, c = A s /P = (πd /4)/πD = D/4. Substituting numerical values, find 9.m/s 1/50K 10 4 K 0.5m/4 Ra = = m /s m /s Since the grill is an upper surface heated, Eq. 9.0 is the appropriate correlation, Nu h /k 0.54Ra = c = = = 17.7 h = Nu k/c = W/m K/ 0.5m/4 =.50W/m K. () Substituting from Eq. () for h into Eq., the heat loss or required electrical power, q elec, is W W 4 π q = 0.5m K K 4 m K m K q = 44.W+ 4.0W= 90.W. COMMENTS: Note that for this situation, free convection and radiation modes are of equal importance. If the grill were highly polished such that ε 0.1, the required power would be reduced by nearly 50%.

4 PROBEM 9.9 KNOWN: Dimensions of double pane window. Thickness of air gap. Temperatures of room and ambient air. FIND: (a) Temperatures of glass panes and heat rate through window, (b) Resistance of glass pane relative to smallest convection resistance. ASSUMPTIONS: Steady-state, () Negligible glass pane thermal resistance, () Constant properties. PROPERTIES: Table A-, Plate glass: k p = 1.4 W/m K. Table A-4, Air (p = 1 atm). T f,i = 7.K: ν i = m /s, k i = 0.05 W/m K, α i = m /s, Pr i = 0.710, β i = K -1. T = (T s,i + T s,o )/ = 7.K: ν = m /s, k = W/m K, α = m /s, Pr = 0.714, β = K -1. T f,o = 5.K: ν o = m /s, k o = 0.00W/m K, α = m /s, Pr = 0.71, β o = K -1. ANAYSIS: (a) The heat rate may be expressed as q = qo = hoh Ts,o T,o q = qg = hgh Ts,i Ts,o () ( ) q = qi = hi H T,i Ts,i () where h o and h i may be obtained from Eq. (9.), 0.7 Ra 1/ Nu H H = /7 9/1 1+ ( 0.49 / Pr) with Ra = gβ ( T T ) H / αν and β H o s,o,o o o 10 4 Ra 7 10, h g is obtained from 0. Nu = 0.4 Ra 1/ Pr ( H / ) where β yields Ra g Ts,i Ts,o / αν. RaH g i T,i Ts,i H / αν i i, = respectively. Assuming = A simultaneous solution to Eqs. () for the three unknowns Continued..

5 PROBEM 9.9 (Cont.) Ts,i = 9.1 C, Ts,o = 9. C, q = 5.7 W where h i =.9 W / m K, h o =.45 W / m K and hg = 1.90W/m K. Rcond p / kp m K / W, (b) The unit conduction resistance of a glass pane is = = and the R 1/ h 0.90 m = = K/W. Hence, smallest convection resistance is conv,o o R cond R conv,min and it is reasonable to neglect the thermal resistance of the glass. COMMENTS: Assuming a heat flux of 5.7 W/m through a glass pane, the corresponding temperature difference across the pane is T = q ( p / k p) = 0.15 C. Hence, the assumption of an isothermal pane is good. () Equations () were solved using the IHT workspace and the temperature-dependent air properties provided by the software. The property values provided in the PROPERTIES section of this solution were obtained from the software.