PROBLEM 4.78 KNOWN: Nodal network and boundary conditions for a water-cooled cold plate. FIND: (a) Steady-state temperature distribution for prescribed conditions, (b) Means by which operation may be extended to larger heat fluxes. ASSUMPTIONS: (1) Steady-state conditions, () Two-dimensional conduction, (3) Constant properties. ANALYSIS: Finite-difference equations must be obtained for each of the 8 nodes. Applying the energy balance method to regions 1 and 5, which are similar, it follows that Node 1: ( ) ( ) ( ) ( ) Node 5: ( ) ( ) ( ) ( ) y x T + x y T6 y x + x y T1= 0 y x T4 + x y T10 y x + x y T5 = 0 Nodal regions, 3 and 4 are similar, and the energy balance method yields a finite-difference equation of the form Nodes,3,4: ( y x)( Tm 1,n + Tm+ 1,n ) + ( x y) Tm,n 1 ( y x) + ( x y) Tm,n = 0 Energy balances applied to the remaining combinations of similar nodes yield the following finitedifference equations. Continued...
PROBLEM 4.78 (Cont.) Nodes 6, 14: ( x y) T1 + ( y x) T7 [( x y) + ( y x) + ( h x k) ] T6 = ( h x k) T ( x y) T19 + ( y x) T15 [( x y) + ( y x) + ( h x k) ] T14 = ( h x k) T Nodes 7, 15: ( )( + ) + ( ) [( ) + ( ) + ( )] = ( ) y x T6 T8 x y T y x x y h x k T7 h x k T y x T14 + T16 + x y T0 [ y x + x y + h x k ] T15 = ( h x k) T ( )( ) ( ) ( ) ( ) ( ) y x T + y x T + x y T + x y T 3 y x + 3 x y Nodes 8, 16: ( ) 7 ( ) 9 ( ) 11 ( ) 3 [ ( ) ( ) + ( hk)( x+ y)] T8 = ( hk)( x+ yt ) ( y x) T15 + ( y x) T17 + ( x y) T11 + ( x y) T1 [ 3( y x) + 3( x y) + ( )( + )] = ( )( + ) hk x y T16 hk x yt Node 11: ( x y) T8 + ( x y) T16 + ( y x) T1 [ ( x y) + ( y x) + ( h y k) ] T11 = ( h y k) T Nodes 9, 1, 17, 0, 1, : ( y x) Tm 1,n + ( y x) Tm+ 1,n + ( x y) Tm,n+ 1 + ( x y) Tm,n 1 [ ( x y) + ( y x) ] Tm,n = 0 Nodes 10, 13, 18, 3: ( x y) Tn + 1,m + ( x y) Tn 1,m + ( y x) Tm 1,n [ ( x y) + ( y x) ] Tm,n = 0 Node 19: ( x y) T14 + ( x y) T4 + ( y x) T0 [ ( x y) + ( y x) ] T19 = 0 Nodes 4, 8: ( x y) T19 + ( y x) T5 [ ( x y) + ( y x) ] T4 = ( q o x k) ( x y) T + ( y x) T [ ( x y) + ( y x) ] T = ( q x k) 3 7 8 o Nodes 5, 6, 7: ( y x) T + ( y x) T + ( x y) T [ ( x y) + ( y x) ] T = ( q x k) m 1,n m+ 1,n m,n+ 1 m,n o Evaluating the coefficients and solving the equations simultaneously, the steady-state temperature distribution ( C), tabulated according to the node locations, is: 3.77 3.91 4.7 4.61 4.74 3.41 3.6 4.31 4.89 5.07 5.70 6.18 6.33 8.90 8.76 8.6 8.3 8.35 30.7 30.67 30.57 30.53 30.5 3.77 3.74 3.69 3.66 3.65 Alternatively, the foregoing results may readily be obtained by accessing the IHT Tools pat and using the -D, SS, Finite-Difference Equations options (model equations are appended). Maximum and minimum cold plate temperatures are at the bottom (T 4 ) and top center (T 1 ) locations respectively. (b) For the prescribed conditions, the maximum allowable temperature (T 4 = 40 C) is reached when q o = 1.407 10 5 W/m (14.07 W/cm ). Options for extending this limit could include use of a copper cold plate (k 400 W/m K) and/or increasing the convection coefficient associated with the coolant. With k = 400 W/m K, a value of q o = 17.37 W/cm may be maintained. With k = 400 W/m K and h = 10,000 W/m K (a practical upper limit), q o = 8.65 W/cm. Additional, albeit small, improvements may be realized by relocating the coolant channels closer to the base of the cold plate. COMMENTS: The accuracy of the solution may be confirmed by verifying that the results satisfy the overall energy balance q o( 4 x) = h[ ( x ) ( T6 T ) + x( T7 T ) + ( x+ y)( T8 T ) ( ) ( )( ) ( ) ( )( )] + y T11 T + x+ y T16 T + x T15 T + x T14 T.
PROBLEM 5.109 KNOWN: Thick slab of copper as treated in Example 5.9, initially at a uniform temperature, is suddenly exposed to large surroundings at 1000 C (instead of a net radiant flux). FIND: (a) The temperatures T(0, 10 s) and T(0.15 m, 10s) using the finite-element software FEHT for a surface emissivity of 0.94 and (b) Plot the temperature histories for x = 0, 150 and 600 mm, and explain key features of your results. ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, () Slab of thickness 600 mm approximates a semi-infinite medium, (3) Slab is small object in large, isothermal surroundings. ANALYSIS: (a) Using FEHT, an outline of the slab is drawn of thickness 600 mm in the x-direction and arbitrary length in the y-direction. Click on Setup Temperatures in K, to enter all temperatures in kelvins. The boundary conditions are specified as follows: on the y-planes and the x = 600 mm plane, treat as adiabatic; on the surface (0,y), select the convection coefficient option, enter the linearized radiation coefficient after Eq. 1.9 written as 0.94 * 5.67e-8 * (T + 173) * (T^ + 173^) and enter the surroundings temperature, 173 K, in the fluid temperature box. See the Comments for a view of the input screen. From View Temperatures, find the results: T(0, 10 s) = 339 K = 66 C T(150 mm, 10 s) = 305K = 3 C (b) Using the View Temperatures command, the temperature histories for x = 0, 150 and 600 mm (10 mm mesh, Nodes 18, 3 and 15, respectively) are plotted. As expected, the surface temperature increases markedly at early times. As thermal penetration increases with increasing time, the temperature at the location x = 150 mm begins to increase after about 30 s. Note, however, that the temperature at the location x = 600 mm does not change significantly within the 150 s exposure to the hot surroundings. Our assumption of treating the boundary at the x = 600 mm plane as adiabatic is justified. A copper plate of 600 mm is a good approximation to a semi-infinite medium at times less than 150 s. Continued..
PROBLEM 5.109 (Cont.) COMMENTS: The annotated Input screen shows the outline of the slab, the boundary conditions, and the triangular mesh before using the Reduce-mesh option.
PROBLEM 6.40 KNOWN: Drag force and air flow conditions associated with a flat plate. FIND: Rate of heat transfer from the plate. ASSUMPTIONS: (1) Chilton-Colburn analogy is applicable. PROPERTIES: Table A-4, Air (70 C,1 atm): ρ = 1.018 kg/m 3, c p = 1009 J/kg K, Pr = 0.70, ν = 0. 10-6 m /s. ANALYSIS: The rate of heat transfer from the plate is ( ) ( s ) q = h L T T where h may be obtained from the Chilton-Colburn analogy, Cf /3 h j /3 H = = St Pr = Pr ρ u cp Cf 1 τs 1 ( 0.075 N/ )/ ( 0.m) 5.76 10 4 = = =. ρ u / 3 1.018 kg/m ( 40 m/s ) / Hence, C h = f ρ u -/3 c p Pr -4 3 /3 h = 5.76 10 ( 1.018kg/m ) 40m/s ( 1009J/kg K ) ( 0.70) h = 30 W/m K. The heat rate is $ q = 30 W/m K 0.m 10 0 C ( ) ( ) ( ) q = 40 W. COMMENTS: Although the flow is laminar over the entire surface (Re L = u L/ν = 40 m/s 0.m/0. 10-6 m /s = 4.0 10 5 ), the pressure gradient is zero and the Chilton-Colburn analogy is applicable to average, as well as local, surface conditions. Note that the only contribution to the drag force is made by the surface shear stress.
PROBLEM 7.10 KNOWN: Speed and temperature of atmospheric air flowing over a flat plate of prescribed length and temperature. FIND: Rate of heat transfer corresponding to Re x,c = 10 5, 5 10 5 and 10 6. ASSUMPTIONS: (1) Flow over top and bottom surfaces. PROPERTIES: Table A-4, Air (T f = 348K, 1 atm): ρ = 1.00 kg/m 3, ν = 0.7 10-6 m /s, k = 0.099 W/m K, Pr = 0.700. ANALYSIS: With u L 5 m/s 1m Re 6 L = = = 1.1 10 ν 0.7 10-6 m /s the flow becomes turbulent for each of the three values of Re x,c. Hence, ( L ) Nu 4/5 1/3 L = 0.037 Re A Pr A = 0.037 Re 4/5 1/ x,c 0.664 Re x,c Re x,c 10 5 5 10 5 10 6 A 160 871 1671 Nu L 7 1641 931 h L W/m K 67.9 49.1 7.8 ( ) where ( ) q ( W/m) 13,580 980 5560 q = hll Ts T is the total heat loss per unit width of plate. COMMENTS: Note that L h decreases with increasing Re x,c, as more of the surface becomes covered with a laminar boundary layer.
PROBLEM 7.0 KNOWN: Material properties, inner surface temperature and dimensions of roof of refrigerated truck compartment. Truck speed and ambient temperature. Solar irradiation. FIND: (a) Outer surface temperature of roof and rate of heat transfer to compartment, (b) Effect of changing radiative properties of outer surface, (c) Effect of eliminating insulation. ASSUMPTIONS: (1) Negligible irradiation from the sky, () Turbulent flow over entire outer surface, (3) Average convection coefficient may be used to estimate average surface temperature, (4) Constant properties. PROPERTIES: Table A-4, air (p = 1 atm, T f 300K): ν = 15.89 10-6 m /s, k = 0.063 W/m K, Pr = 0.707. ANALYSIS: (a) From an energy balance for the outer surface, Ts,o Ts,i αsgs + q conv E= q cond = R tot 4 Ts,o Ts,i αsgs+ h( T Ts,o) εσts,o = R p + Ri 5 where ( ) = = ( ) Rp t 1/ kp.78 10 m K / W, 6 7 = 9. m / s 10m /15.89 10 m / s = 1.84 10, Ri t / ki 1.93m K / W, = = and with Re L = u L /ν Hence, 4/5 ( ) ( ) k 4/5 1/3 0.063 W / m K 7 1/3 h = 0.037 ReL Pr = 0.037 1.84 10 0.707 = 56. W / m K L 10m ( ) ( s,o ) 8 4 4 0.5 750 W / m K + 56. W / m K 305 T 0.5 5.67 10 W / m K Ts,o = Solving, we obtain Ts,o 63K ( ) 5 5.56 10 + 1.93 m K / W Ts,o = 306.8K = 33.8 C Hence, the heat load is 33.8 + 10 C q = W L qcnd = 3.5m 10m = 797 W 1.93m K / W ( ) ( ) ( ) (b) With the special surface finish ( α = ε = ) S 0.15, 0.8, Continued..
PROBLEM 7.0 (Cont.) Ts,o = 301.1K = 7.1 C q = 675.3W (c) Without the insulation (t = 0) and with α S = ε = 0.5, Ts,o = 63.1K = 9.9 C q = 90, 630W COMMENTS: (1) Use of the special surface finish reduces the solar input, while increasing radiation emission from the surface. The cumulative effect is to reduce the heat load by 15%. () The thermal resistance of the aluminum panels is negligible, and without the insulation, the heat load is enormous.
PROBLEM 7.43 KNOWN: Initial temperature, power dissipation, diameter, and properties of heating element. Velocity and temperature of air in cross flow. FIND: (a) Steady-state temperature, (b) Time to come within 10 C of steady-state temperature. ASSUMPTIONS: (1) Uniform heater temperature, () Negligible radiation. PROPERTIES: Table A.4, air (assume T f 450 K): ν = 3.39 10-6 m /s, k = 0.0373 W/m K, Pr = 0.686. ANALYSIS: (a) Performing an energy balance for steady-state conditions, we obtain q conv = h ( π D)( T T ) = Pelec = 1000 W m With VD ( 10 m s) 0.01m ReD = = = 3,087 ν 3.39 10 6 m s the Churchill and Bernstein correlation, Eq. 7.57, yields 4/5 0.6 Re 1/ 1/3 5/8 D Pr ReD NuD = 0.3 + 1 + /3 1/4 8, 000 1+ ( 0.4 Pr) 1/ 1/3 5/8 4/5 0.6( 3087) ( 0.686) 3087 NuD = 0.3 + 1 + 8. /3 1/4 8, 000 = 1+ ( 0.4 0.686) k 0.0373W m K h = Nu D = 8. = 105. W m K D 0.010 m Hence, the steady-state temperature is Pelec 1000 W m T = T + = 300 K + = 603K π Dh π 0.01m 105. W m K ( ) (b) With Bi = hro k = 105. W/m K(0.005 m)/40 W/m K = 0.00, a lumped capacitance analysis may be performed. The time response of the heater is given by Eq. 5.5, which, for T i = T, reduces to ( ) ( ) T= T + b a 1 exp at Continued...
PROBLEM 7.43 (Cont.) 7 7 = 0.0173 s -1 and b/a = where a = 4 hdρ cp = 4 105. W m K 0. 01m 700 kg m 3 900 J kg K π Dh = 1000 W m ( 0.01m 105. W m K) Pelec π = 30.6 K. Hence, ( ) 593 300 K 1 exp( 0.0173t) = = 0.968 30.6 K t 00s COMMENTS: (1) For T = 603 K and a representative emissivity of ε = 0.8, net radiation exchange between the heater and surroundings at T sur = T = 300 K would be ( )( q 4 4 ) rad = εσ π D T Tsur = 0.8 5.67 10-8 W/m K 4 (π 0.01 m)(603 4-300 4 )K 4 = 177 W/m. Hence, although small, radiation exchange is not negligible. The effects of radiation are considered in Problem 7.46. () The assumed value of T f is very close to the actual value, rendering the selected air properties accurate.