PROBLEM Node 5: ( ) ( ) ( ) ( )

Transcription

1 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...

2 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) 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: 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 = 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 = 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.

3 PROBLEM 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..

4 PROBLEM (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.

5 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 = kg/m 3, c p = 1009 J/kg K, Pr = 0.70, ν = 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 ( N/ )/ ( 0.m) = = =. ρ u / kg/m ( 40 m/s ) / Hence, C h = f ρ u -/3 c p Pr -4 3 /3 h = ( 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/ m /s = ), 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.

6 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, and 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 = m /s, k = W/m K, Pr = ANALYSIS: With u L 5 m/s 1m Re 6 L = = = ν m /s the flow becomes turbulent for each of the three values of Re x,c. Hence, ( L ) Nu 4/5 1/3 L = Re A Pr A = Re 4/5 1/ x,c Re x,c Re x,c A Nu L h L W/m K ( ) where ( ) q ( W/m) 13, 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.

7 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 = m /s, k = W/m K, Pr = 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 m K / W, 6 7 = 9. m / s 10m / m / s = , Ri t / ki 1.93m K / W, = = and with Re L = u L /ν Hence, 4/5 ( ) ( ) k 4/5 1/ W / m K 7 1/3 h = ReL Pr = = 56. W / m K L 10m ( ) ( s,o ) W / m K W / m K 305 T W / m K Ts,o = Solving, we obtain Ts,o 63K ( ) m K / W Ts,o = 306.8K = 33.8 C Hence, the heat load is 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..

8 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.

9 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 = m /s, k = W/m K, Pr = 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 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 = /3 1/4 8, ( 0.4 Pr) 1/ 1/3 5/8 4/5 0.6( 3087) ( 0.686) 3087 NuD = /3 1/4 8, 000 = 1+ ( ) k W m K h = Nu D = 8. = 105. W m K D 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...

10 PROBLEM 7.43 (Cont.) 7 7 = s -1 and b/a = where a = 4 hdρ cp = W m K 0. 01m 700 kg m J kg K π Dh = 1000 W m ( 0.01m 105. W m K) Pelec π = 30.6 K. Hence, ( ) K 1 exp( t) = = 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 = W/m K 4 (π 0.01 m)( )K 4 = 177 W/m. Hence, although small, radiation exchange is not negligible. The effects of radiation are considered in Problem () The assumed value of T f is very close to the actual value, rendering the selected air properties accurate.