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39 Numerical coefficient 4600 49 COMMENTS: (1) For these condition, the convection coefficient for the water is nearly two orders of magnitude higher than for air. (2) Using the average-coefficient approach, the time-to-cool, to, values for both fluids is 15-20% faster than the more accurate numerical integration approach. Evaluating the average coefficient at sT results in systematically over estimating the coefficient. (3) The IHT code used for numerical integration of the energy balance equation and the correlations is shown below for the fluid water. // LCM energy balance - hDbar * As * (Ts - Tinf) = M * cps * der(Ts,t) As = pi * D^2 M = rhos * Vs Vs = pi * D^3 / 6 // Input variables D = 0.025 // Ts = 85 + 273 // Initial condition, Ts Tinf_C = 25 rhos = 8933 // Table A.1, copper, pure cps = 382 ks = 399 /* Correlation description: Free convection (FC), sphere (S), RaD<=10^11, Pr >=0.7, Churchill correlation, Eqs 9.25 and 9.35 . See Table 9.2 . */ NuDbar = NuD_bar_FC_S(RaD,Pr) // Eq 9.35 NuDbar = hDbar * D / k RaD = g * beta * deltaT * D^3 / (nu * alpha) //Eq 9.25 deltaT = abs(Ts - Tinf) g = 9.8 // gravitational constant, m/s^2 // Evaluate properties at the film temperature, Tf. Tf = Tfluid_avg(Tinf,Ts) // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x = 0 // Quality (0=sat liquid or 1=sat vapor) nu = nu_Tx("Water",Tf,x) // Kinematic viscosity, m^2/s k = k_Tx("Water",Tf,x) // Thermal conductivity, W/m·K Pr = Pr_Tx("Water",Tf,x) // Prandtl number beta = beta_T("Water",Tf) // Volumetric coefficient of expansion, K^(-1) (f, liquid, x = 0) alpha = k / (rho * cp) // Thermal diffusivity, m^2/s // Conversions Ts_C = Ts - 273 Tinf_C = Tinf - 273 PROBLEM 9.81 KNOWN: Temperatures and spacing of vertical, isothermal plates. FIND: (a) Shape of velocity distribution, (b) Forms of mass, momentum and energy equations for laminar flow, (c) Expression for the temperature distribution, (d) Vertical pressure gradient, (e) Expression for the velocity distribution. SCHEMATIC: ASSUMPTIONS: (1) Laminar, incompressible, fully-developed flow, (2) Constant properties, (3) Negligible viscous dissipation, (4) Boussinesq approximation. ANALYSIS: (a) For the prescribed conditions, there must be buoyancy driven ascending and descending flows along the surfaces corresponding to Ts,1 and Ts,2, respectively (see schematic). However, conservation of mass dictates equivalent rates of upflow and downflow and, assuming constant properties, inverse symmetry of the velocity distribution about the midplane. (b) For fully-developed flow, which is achieved for long plates, vx = 0 and the continuity equation yields zv z 0\u2202 \u2202 = < Hence, there is no net transfer of momentum or energy by advection, and the corresponding equations are, respectively, ( ) ( ) ( )2 2z c0 dp dz d v dx g gµ \u3c1= \u2212 + \u2212 < 0 = (dT2/dx2) < (c) Integrating the energy equation twice, we obtain T = C1x + C2 and applying the boundary conditions, T(-L) = Ts,1 and T(L) = Ts,2, it follows that C1 = -(Ts,1 - Ts,2)/2L and C2 = (Ts,1 + Ts,2)/2 \u2261 Tm, in which case, m s,1 s,2 T T x T T 2L \u2212 = \u2212 \u2212 < Continued... PROBLEM 9.81 (Cont.) (d) From hydrostatic considerations and the assumption of a constant density \u3c1m, the balance between the gravitational and net pressure forces may be expressed as dp/dz = -\u3c1m(g/gc). The momentum equation is then of the form ( ) ( )( )2 2z m c0 d v dx g gµ \u3c1 \u3c1= \u2212 \u2212 or, invoking the Boussinesq approximation, ( )m m mT T\u3c1 \u3c1 \u3b2\u3c1\u2212 \u2248 \u2212 \u2212 , ( )( )( )2 2z m c md v dx g g T T\u3b2\u3c1 µ= \u2212 \u2212 or, from the known temperature distribution, ( )( )( )( )2 2z m c s,1 s,2d v dx 2 g g T T x L\u3b2\u3c1 µ= \u2212 < (e) Integrating the foregiong expression, we obtain ( )( )( )( )2z m c s,1 s,2 1dv dx 4 g g T T x L C\u3b2\u3c1 µ= \u2212 + ( )( )( )( )3z m c s,1 s,2 1 2v 12 g g T T x L C x C\u3b2\u3c1 µ= \u2212 + + Applying the boundary conditions vz(-L) = vz(L) = 0, it follows that C1 = ( )( )( )m c s,1 s,212 g g T T L\u3b2\u3c1 µ\u2212 \u2212 and C2 = 0. Hence, ( )( )( ) ( ) ( )2 3 3z m c s,1 s,2v L 12 g g T T x L x L\u3b2\u3c1 µ \uf8ee \uf8f9= \u2212 \u2212\uf8ef \uf8fa\uf8f0 \uf8fb < COMMENTS: The validity of assuming fully-developed conditions improves with increasing plate length and would be satisfied precisely for infinite plates. PROBLEM 9.82 KNOWN: Dimensions of vertical rectangular fins. Temperature of fins and quiescent air. FIND: Optimum fin spacing and corresponding fin heat transfer rate. SCHEMATIC: ASSUMPTIONS: (1) Isothermal fins, (2) Negligible radiation, (3) Quiescent air, (4) Negligible heat transfer from fin tips, (5) Negligible radiation. PROPERTIES: Table A-4, Air (Tf = 325K, 1 atm): n = 18.41 ´ 10 -6 m 2 /s, k = 0.0282 W/m×K, a = 26.1 ´ 10 -6 m 2 /s, Pr = 0.703. ANALYSIS: From Table 9.3 ( ) ( ) 1/41/4 s3 opt S g T T S 2.71 Ra / S H 2.71 H b an -- ¥é ù-= = ê ú ë û ( ) ( ) 1/412 opt 6 2 6 2 9.8m/s 325K 50K S 2.71 7.12mm 26.1 10 m / s 18.4 10 m / s 0.15m -- - - é ù ê ú= = ê ú´ ´ ´ ´ë û < From Eq. 9.45 and Table 9.3 ( ) ( )s 1/2 2 1/2 S S 576 2.87 Nu Ra S / L Ra S / L -é ù ê ú= + ê úë û ( ) ( ) ( ) ( ) ( )412 34s S 6 2 6 2 9.8m/s 325K 50K 7.12 10 mg T T S Ra S / L H 25.4 10 m / s 18.4 10 m /s 0.15m b an - - ¥ - - ´- = = ´ ´ ´ ´ ( )SRa S / L 53.2= ( ) ( ) [ ]S 1/2 1/2 2 1 / 2 576 2.87 Nu 0.204 0.393 1.29 53.2 53.2 - - é ù ê ú= + = + = ê úë û ( )S 2h Nu k / S 1.29 0.0282W/m K/0.00712m 5.13W/m K.= = × = × With N = W/(t + S) = (355 mm)/(8.62 ´ 10-3 m) = 41.2 » 41, ( ) ( ) ( )( )2sq 2Nh L H T T 82 5.13W/m K 0.02m 0.15m 50K¥= ´ - = × ´ q 63.1W.= < COMMENTS: Sopt = 7.12 mm is considerably less than the value of 34 mm predicted from previous considerations. Hence, the corresponding value of q = 63.1 W is considerably larger than that of the previous predication. PROBLEM 9.83 KNOWN: Length, width and spacing of vertical circuit boards. Maximum allowable board temperature. FIND: Maximum allowable power dissipation per board. SCHEMATIC: ASSUMPTIONS: (1) Circuit boards are flat with uniform heat flux at each surface, (2) Negligible radiation. PROPERTIES: Table A-4, Air ( )T 320K,1atm := n = 17.9 ´ 10-6 m2/s, k = 0.0278 W/m×K, a = 25.5 ´ 10-6 m2/s. ANALYSIS: From Eqs. 9.41 and 9.46 and Table 9.3, ( ) 1/2 s * 2/5*s,L S S q S 48 2.51 T T k Ra S / L Ra S / L - ¥ é ù ê ú¢¢ = +ê ú - ê ú ê úë û where ( ) ( ) ( )( ) 1 525 s* s S 6 2 6 2 9.8m/s 320K 0.025m qg q SS Ra L k L 0.0278W/m K 25.5 10 m / s 17.9 10 m / s 0.4m b an - - - ¢¢¢¢ = = × ´ ´ * S s S Ra 58.9q L ¢¢= and ( ) s s s s,L q 0.025m qS 0.015q . T T k 60 K 0.0278W/m K¥ ¢¢ ¢¢× ¢¢= = - × Hence, ( ) 1/2 s 0.4s s 0.815 0.492 0.015q . q q -é ù ê ú¢¢ = + ¢¢ê ú¢¢ë û A trial-and-error solution yields 2 sq 287 W / m .¢¢ = Hence, ( ) ( )2 2s sq 2A q 2 0.4m 287W/m 91.8 W.¢¢= = = < COMMENTS: Larger heat rates may be achieved by using a fan to superimpose a forced flow on the buoyancy driven flow. PROBLEM 9.84 KNOWN: Array of isothermal vertical fins attached to heat sink at 42°C with ambient air temperature at 27°C. FIND: (a) Heat removal rate for 24 pairs of fins and (b) Optimum fin spacing for maximizing heat removal rate if overall size of sink is to remain unchanged. SCHEMATIC: ASSUMPTIONS: (1) Fins form vertical, symmetrically heated, isothermal plates, (2) Negligible radiation effects, (3) Ambient air is quiescent. PROPERTIES: Table A-4, Air ( ) ( )( )f sT T T / 2 42 27 / 2 C 308K,1atm :¥= + = + ° = 6 2 6 2 316.69 10 m /s, 23.5 10 m / s , k 26.9 10 W / m K.n a- - -= ´ = ´ = ´ × ANALYSIS: Considering the fins as vertical isothermal plates, the heat rate can be determined from Eq. 9.37