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+ bx + cx2. The right-hand term represents the rate of energy storage per unit volume, [ ] [ ] ( )4 2 5 3stE k 0 b 2cx k 0 2c 5 W / m K 2 2 10 C / m 2 10 W / m x \u2202 \u2032\u2032 = + + = + = \u22c5 × \u2212 × ° = \u2212 × \u2202 \ufffd < (g) With no heat generation, the wall will eventually (t \u2192 \u221e) come to equilibrium with the fluid, T(x,\u221e) = T\u221e = 20°C. To determine the energy that must be removed from the wall to reach this state, apply the conservation of energy requirement over an interval basis, Eq. 1.11b. The \u201cinitial\u201d state is that corresponding to the steady-state temperature distribution, Ti, and the \u201cfinal\u201d state has Tf = 20°C. We\u2019ve used T\u221e as the reference condition for the energy terms. in out st f i inE E E E E with E 0.\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032\u2212 = \u2206 = \u2212 = ( ) ( )Lout p f p iLE c 2L T T c T T dx\u3c1 \u3c1 + \u221e \u221e \u2212 \u2032\u2032 \u2212 = \u2212 \u2212 \u2212\u222b LL 2 2 3 out p pL L E c a bx cx T dx c ax bx / 2 cx / 3 T x\u3c1 \u3c1 ++ \u221e \u221e \u2212 \u2212 \uf8ee \uf8f9 \uf8ee \uf8f9\u2032\u2032 = + + \u2212 = + + \u2212\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\u222b 3 out pE c 2aL 0 2cx / 3 2T L\u3c1 \u221e\uf8ee \uf8f9\u2032\u2032 = + + \u2212\uf8ef \uf8fa\uf8f0 \uf8fb ( )3 4 2CoutE 2600kg / m 800J / kg K 2 82 C 0.020m 2 2 10 / m\uf8ee °\uf8ef\uf8f0\u2032\u2032 = × \u22c5 × ° × + \u2212 × ( ) ( )30.020m / 3 2 20 C 0.020m\uf8f9\uf8fa\uf8fb\u2212 ° 6 2 outE 4.94 10 J / m\u2032\u2032 = × < COMMENTS: (1) In part (a), note that the temperature gradient is larger at x = + L than at x = - L. This is consistent with the results of part (c) in which the conduction heat fluxes are evaluated. Continued \u2026.. PROBLEM 2.25 (Cont.) (2) In evaluating the conduction heat fluxes, ( )xq x ,\u2032\u2032 it is important to recognize that this flux is in the positive x-direction. See how this convention is used in formulating the energy balance in part (c). (3) It is good practice to represent energy balances with a schematic, clearly defining the system or surface, showing the CV or CS with dashed lines, and labeling the processes. Review again the features in the schematics for the energy balances of parts (c & d). (4) Re-writing the heat diffusion equation introduced in part (b) as d dTk q 0 dx dx \uf8eb \uf8f6 \u2212 \u2212 + =\uf8ec \uf8f7\uf8ed \uf8f8 \ufffd recognize that the term in parenthesis is the heat flux. From the differential equation, note that if the differential of this term is a constant ( )q / k ,\ufffd then the term must be a linear function of the x-coordinate. This agrees with the analysis of part (e). (5) In part (f), we evaluated stE ,\ufffd the rate of energy change stored in the wall at the instant the volumetric heat generation was deactivated. Did you notice that 5 3stE 2 10 W / m= \u2212 ×\ufffd is the same value of the deactivated q?\ufffd How do you explain this? PROBLEM 2.26 KNOWN: Steady-state conduction with uniform internal energy generation in a plane wall; temperature distribution has quadratic form. Surface at x=0 is prescribed and boundary at x = L is insulated. FIND: (a) Calculate the internal energy generation rate, q\ufffd , by applying an overall energy balance to the wall, (b) Determine the coefficients a, b, and c, by applying the boundary conditions to the prescribed form of the temperature distribution; plot the temperature distribution and label as Case 1, (c) Determine new values for a, b, and c for conditions when the convection coefficient is halved, and the generation rate remains unchanged; plot the temperature distribution and label as Case 2; (d) Determine new values for a, b, and c for conditions when the generation rate is doubled, and the convection coefficient remains unchanged (h = 500 W/m2\u22c5K); plot the temperature distribution and label as Case 3. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constant properties and uniform internal generation, and (3) Boundary at x = L is adiabatic. ANALYSIS: (a) The internal energy generation rate can be calculated from an overall energy balance on the wall as shown in the schematic below. in out gen in convE E E 0 where E q\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032\u2212 + = =\ufffd \ufffd \ufffd \ufffd ( )oh T T q L 0\u221e \u2212 + =\ufffd (1) ( ) ( )2 6 3oq h T T / L 500 W / m K 20 120 C / 0.050 m 1.0 10 W / m\u221e= \u2212 \u2212 = \u2212 \u22c5 \u2212 ° = ×\ufffd < (b) The coefficients of the temperature distribution, T(x) = a + bx + cx2, can be evaluated by applying the boundary conditions at x = 0 and x = L. See Table 2.1 for representation of the boundary conditions, and the schematic above for the relevant surface energy balances. Boundary condition at x = 0, convection surface condition ( ) ( )in out conv x x x 0 dTE E q q 0 0 where q 0 k dx = \uf8f6 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2212 = \u2212 = = \u2212 \uf8f7\uf8f8\ufffd \ufffd ( ) ( )o x 0h T T k 0 b 2cx 0\u221e =\uf8ee \uf8f9\u2212 \u2212 \u2212 + + =\uf8f0 \uf8fb Continued \u2026.. PROBLEM 2.26 (Cont.) ( ) ( )2 4ob h T T / k 500 W / m K 20 120 C / 5 W / m K 1.0 10 K / m\u221e= \u2212 \u2212 = \u2212 \u22c5 \u2212 ° \u22c5 = × < Boundary condition at x = L, adiabatic or insulated surface ( ) ( )in out x x x L dTE E q L 0 where q L k dx = \uf8f6 \u2032\u2032 \u2032\u2032 \u2212 = \u2212 = = \u2212 \uf8f7\uf8f8\ufffd \ufffd [ ]x Lk 0 b 2cx 0=+ + = (3) ( )4 5 2c b / 2L 1.0 10 K / m / 2 0.050m 1.0 10 K / m= \u2212 = \u2212 × × = \u2212 × < Since the surface temperature at x = 0 is known, T(0) = To = 120°C, find ( )T 0 120 C a b 0 c 0 or a 120 C= ° = + \u22c5 + \u22c5 = ° (4) < Using the foregoing coefficients with the expression for T(x) in the Workspace of IHT, the temperature distribution can be determined and is plotted as Case 1 in the graph below. (c) Consider Case 2 when the convection coefficient is halved, h2 = h/2 = 250 W/m2\u22c5K, 6q 1 10= ×\ufffd W/m3 and other parameters remain unchanged except that oT 120 C.\u2260 ° We can determine a, b, and c for the temperature distribution expression by repeating the analyses of parts (a) and (b). Overall energy balance on the wall, see Eqs. (1,4) 6 3 2 oa T q L / h T 1 10 W / m 0.050m / 250 W / m K 20 C 220 C\u221e= = + = × × \u22c5 + ° = °\ufffd < Surface energy balance at x = 0, see Eq. (2) ( ) ( )2 4ob h T T / k 250 W / m K 20 220 C / 5 W / m K 1.0 10 K / m\u221e= \u2212 \u2212 = \u2212 \u22c5 \u2212 ° \u22c5 = × < Surface energy balance at x = L, see Eq. (3) ( )4 5 2c b / 2L 1.0 10 K / m / 2 0.050m 1.0 10 K / m= \u2212 = \u2212 × × = \u2212 × < The new temperature distribution, T2 (x), is plotted as Case 2 below. (d) Consider Case 3 when the internal energy volumetric generation rate is doubled, 6 3 3q 2q 2 10 W / m ,= = ×\ufffd \ufffd h = 500 W/m 2 \u22c5K, and other parameters remain unchanged except that oT 120 C.\u2260 ° Following the same analysis as part (c), the coefficients for the new temperature distribution, T (x), are 4 5 2a 220 C b 2 10 K / m c 2 10 K / m= ° = × = \u2212 × < and the distribution is plotted as Case 3 below. Continued \u2026.. PROBLEM 2.26 (Cont.) COMMENTS: Note the following features in the family of temperature distributions plotted above. The temperature gradients at x = L are zero since the boundary is insulated (adiabatic) for all cases. The shapes of the distributions are all quadratic, with the maximum temperatures at the insulated boundary. By halving the convection coefficient for Case 2, we expect the surface temperature To to increase relative to the Case 1 value, since the same heat flux is removed from the wall ( )qL\ufffd but the convection resistance has increased. By doubling the generation rate for Case 3, we expect the surface temperature To to increase relative to the Case 1 value, since double the amount of heat flux is removed from the wall ( )2qL .\ufffd Can you explain why To is the same for Cases 2 and 3, yet the insulated boundary temperatures are quite different? Can you explain the relative magnitudes of T(L) for the three cases? 0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0 W a ll p o s i tio n , x (m m ) 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0 Te m pe ra tu re , T (C ) 1 . h = 5 0 0 W /m ^2 .K , q d o t = 1 e 6 W /m ^3 2 . h = 2 5 0 W /m ^2 .K , q d o t = 1 e 6 W /m ^3 3 . h = 5 0 0 W /m ^2 .K , q d o t = 2 e 6 W /m ^3 PROBLEM 2.27 KNOWN: Temperature distribution and distribution of heat generation in central layer of a solar pond. FIND: (a) Heat fluxes at lower and upper surfaces of the central layer, (b)