fundamentals of heat and mass transfer solutions manual phần 3 pdf

PROBLEM 4.33 KNOWN: Dimensions and thermal conductivities of a heater and a finned sleeve. Convection conditions on the sleeve surface. FIND: (a) Heat rate per unit length, (b) Generation rate and centerline temperature of heater, (c) Effect of fin parameters on heat rate. SCHEMATIC: ASSUMPTIONS: (1) Steady state, (2) Constant properties, (3) Negligible contact resistance between heater and sleeve, (4) Uniform convection coefficient at outer surfaces of sleeve, (5) Uniform heat generation, (6) Negligible radiation. ANALYSIS: (a) From the thermal circuit, the desired heat rate is () ss t,o tot cond 2D TT TT q RRR ∞∞ −− ′ == ′′′ + The two-dimensional conduction resistance, may be estimated from Eq. (4.27) and Case 6 of Table 4.2 () ()() () 4 cond 2D ss ln 1.08w / D ln 2.16 1 R 5.11 10 m K/ W S k 2 k 2 240W/m K ππ − ′ == = =× ⋅ ′ ⋅ The thermal resistance of the fin array is given by Eq. (3.103), where η o and A t are given by Eqs. (3.102) and (3.99) and η f is given by Eq. (3.89). With L c = L + t/2 = 0.022 m, m = (2h/k s t) 1/2 = 32.3 m -1 and mL c = 0.710, c f c tanh mL 0.61 0.86 mL 0.71 η === ()( ) tfb A NA A N 2L t 4w Nt 0.704m 0.096m 0.800m ′′′ =+= ++−= + = () () f of t NA 0.704m 1 1 1 0.14 0.88 A 0.800m η η ′ =− − =− = ′ () ( ) 1 1 23 t,o o t R h A 0.88 500W / m K 0.80m 2.84 10 m K / W η − − − ′′ ==×⋅×=×⋅ () ( ) 43 300 50 C q 74,600W /m 5.11 10 2.84 10 m K / W −− −° ′ == ×+× ⋅ < (b) Eq. (3.55) may be used to determine q,  if h is replaced by an overall coefficient based on the surface area of the heater. From Eq. (3.32), PROBLEM 4.33 (Cont.) ()( ) 11 ss s tot U A U D R 3.35m K/ W 298W/m K π −− ′′ == = ⋅ = ⋅ () 2 s U 298W /m K / 0.02m 4750W /m K π =⋅×= ⋅ () ( ) () 283 ss q 4U T T / D 4 4750W / m K 250 C /0.02m 2.38 10 W/ m ∞ =−= ⋅° =×  < From Eq. (3.53) the centerline temperature is () () () () 22 83 s h q D/ 2 2.38 10 W/m 0.01m T 0 T 300 C 315 C 4k 4 400W / m K × =+= +°=° ⋅  < (c) Subject to the prescribed constraints, the following results have been obtained for parameter variations corresponding to 16 ≤ N ≤ 40, 2 ≤ t ≤ 8 mm and 20 ≤ L ≤ 40 mm. N t(mm) L(mm) f η () qW/m ′ 16 4 20 0.86 74,400 16 8 20 0.91 77,000 28 4 20 0.86 107,900 32 3 20 0.83 115,200 40 2 20 0.78 127,800 40 2 40 0.51 151,300 Clearly there is little benefit to simply increasing t, since there is no change in t A ′ and only a marginal increase in f . η However, due to an attendant increase in t A, ′ there is significant benefit to increasing N for fixed t (no change in f η ) and additional benefit in concurrently increasing N while decreasing t. In this case the effect of increasing t A ′ exceeds that of decreasing f . η The same is true for increasing L, although there is an upper limit at which diminishing returns would be reached. The upper limit to L could also be influenced by manufacturing constraints. COMMENTS: Without the sleeve, the heat rate would be () s q Dh T T 7850W /m, π ∞ ′ =−= which is well below that achieved by using the increased surface area afforded by the sleeve. PROBLEM 4.34 KNOWN: Dimensions of chip array. Conductivity of substrate. Convection conditions. Contact resistance. Expression for resistance of spreader plate. Maximum chip temperature. FIND: Maximum chip heat rate. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Constant thermal conductivity, (3) Negligible radiation, (4) All heat transfer is by convection from the chip and the substrate surface (negligible heat transfer from bottom or sides of substrate). ANALYSIS: From the thermal circuit, () hh hsp h,cnv t,c sp,cnv tsp TT TT qq q RRRR ∞∞ −− =+ = + ++ () ( ) () 1 1 1 2 22 h,cnv s,n h R h A hL 100W / m K 0.005m 400K / W − − −  === ⋅ =   () ()() 357 rrrr tsp sub h 1 1.410A 0.344A 0.043A 0.034A 1 0.353 0.005 0 0 R 0.408K / W 4k L 4 80W / m K 0.005m −+ + + −+++ === ⋅ () 42 t,c t,c 22 h R 0.5 10 m K / W R 2.000K / W L 0.005m − ′′ ×⋅ == = () () 1 1 222 sp,cnv sub s,h R h A A 100W / m K 0.010m 0.005m 133.3K / W − − =− = ⋅ − =      () 70 C 70 C q 0.18W 0.52W 0.70W 400K / W 0.408 2 133.3 K / W °° =+ =+= ++ < COMMENTS: (1) The thermal resistances of the substrate and the chip/substrate interface are much less than the substrate convection resistance. Hence, the heat rate is increased almost in proportion to the additional surface area afforded by the substrate. An increase in the spacing between chips (S h ) would increase q correspondingly. (2) In the limit () r tsp A0,R → reduces to 1/2 sub h 2kD π for a circular heat source and sub h 4k L for a square source. PROBLEM 4.35 KNOWN: Internal corner of a two-dimensional system with prescribed convection boundary conditions. FIND: Finite-difference equations for these situations: (a) Horizontal boundary is perfectly insulated and vertical boundary is subjected to a convection process (T ∞ ,h), (b) Both boundaries are perfectly insulated; compare result with Eq. 4.45. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties, (4) No internal generation. ANALYSIS: Consider the nodal network shown above and also as Case 2, Table 4.2. Having defined the control volume – the shaded area of unit thickness normal to the page – next identify the heat transfer processes. Finally, perform an energy balance wherein the processes are expressed using appropriate rate equations. (a) With the horizontal boundary insulated and the vertical boundary subjected to a convection process, the energy balance results in the following finite-difference equation: inout123456 EE0 qqqqqq0 −=+++++= && ( ) ( ) m-1,nm,nm,n-1m,n m,n TTTT xy ky1k1h1TT x2y2 ∞ −− ∆∆  ∆⋅+⋅+⋅−  ∆∆  ( ) m+1,nm,nm,n+1m,n TTTT y 0k1kx10. 2xy −− ∆  ++⋅+∆⋅=  ∆∆  Letting ∆x = ∆y, and regrouping, find ( ) ( ) m-1,nm,n+1m+1,nm,n-1m,n hxhx 2TTTTT6T0. kk ∞ ∆∆  ++++−+=   < (b) With both boundaries insulated, the energy balance would have q 3 = q 4 = 0. The same result would be obtained by letting h = 0 in the previous result. Hence, ( ) ( ) m-1,nm,n+1m+1,nm,n-1m,n 2TTTT6 T0. +++−= < Note that this expression compares exactly with Eq. 4.45 when h = 0, which corresponds to insulated boundaries. PROBLEM 4.36 KNOWN: Plane surface of two-dimensional system. FIND: The finite-difference equation for nodal point on this boundary when (a) insulated; compare result with Eq. 4.46, and when (b) subjected to a constant heat flux. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional, steady-state conduction with no generation, (2) Constant properties, (3) Boundary is adiabatic. ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)⋅∆y, and using the conduction rate equation, it follows that inout123 EE0 qqq0 −=++= && (1,2) ( ) m-1,nm,nm,n-1m,nm,n+1m,n TTTTTT xx ky1k1k10. x2y2y −−− ∆∆  ∆⋅+⋅+⋅=  ∆∆∆  (3) Note that there is no heat rate across the control volume surface at the insulated boundary. Recognizing that ∆x =∆y, the above expression reduces to the form m-1,nm,n-1m,n+1m,n 2TTT4T0. ++−= (4) < The Eq. 4.46 of Table 4.3 considers the same configuration but with the boundary subjected to a convection process. That is, ( ) m-1,nm,n-1m,n+1m,n 2hxhx 2TTTT22T0. kk ∞ ∆∆  +++−+=   (5) Note that, if the boundary is insulated, h = 0 and Eq. 4.46 reduces to Eq. (4). (b) If the surface is exposed to a constant heat flux, o q, ′′ the energy balance has the form 123o qqqqy0 ′′ +++⋅∆= and the finite difference equation becomes o m-1,nm,n-1m,n+1m,n qx 2TTT4T. k ′′ ∆ ++−=− < COMMENTS: Equation (4) can be obtained by using the “interior node” finite-difference equation, Eq. 4.33, where the insulated boundary is treated as a symmetry plane as shown below. PROBLEM 4.37 KNOWN: External corner of a two-dimensional system whose boundaries are subjected to prescribed conditions. FIND: Finite-difference equations for these situations: (a) Upper boundary is perfectly insulated and side boundary is subjected to a convection process, (b) Both boundaries are perfectly insulated; compare result with Eq. 4.47. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties, (4) No internal generation. ANALYSIS: Consider the nodal point configuration shown in the schematic and also as Case 4, Table 4.2. The control volume about the node – shaded area above of unit thickness normal to the page – has dimensions, (∆x/2)(∆y/2)⋅1. The heat transfer processes at the surface of the CV are identified as q 1 , q 2 ⋅⋅⋅. Perform an energy balance wherein the processes are expressed using the appropriate rate equations. (a) With the upper boundary insulated and the side boundary subjected to a convection process, the energy balance has the form inout1234 EE0 qqqq0 −=+++= && (1,2) ( ) m-1,nm,nm,n-1m,n m,n TTTT yxy k1k1h1TT00. 2x2y2 ∞ −− ∆∆∆  ⋅+⋅+⋅−+=  ∆∆  Letting ∆x = ∆y, and regrouping, find m,n-1m-1,nm,n hx1hx TTT21T0. k2k ∞ ∆∆  ++−+=   (3) < (b) With both boundaries insulated, the energy balance of Eq. (2) would have q 3 = q 4 = 0. The same result would be obtained by letting h = 0 in the finite-difference equation, Eq. (3). The result is m,n-1m-1,nm,n TT2T0. +−= < Note that this expression is identical to Eq. 4.47 when h = 0, in which case both boundaries are insulated. COMMENTS: Note the convenience resulting from formulating the energy balance by assuming that all the heat flow is into the node. PROBLEM 4.38 KNOWN: Conduction in a one-dimensional (radial) cylindrical coordinate system with volumetric generation. FIND: Finite-difference equation for (a) Interior node, m, and (b) Surface node, n, with convection. SCHEMATIC: (a) Interior node, m (b) Surface node with convection, n ASSUMPTIONS: (1) Steady-state, one-dimensional (radial) conduction in cylindrical coordinates, (2) Constant properties. ANALYSIS: (a) The network has nodes spaced at equal ∆r increments with m = 0 at the center; hence, r = m∆r (or n∆r). The control volume is ( ) V2 rr2mrr. ππ=⋅∆⋅=∆∆⋅ ll The energy balance is ingab EEqqqV0 +=++= && & ( ) m-1mm+1m rTTrTT k2rk2r+q2mrr0. 2r2r πππ ∆−∆−   −++∆∆=    ∆∆   & lll Recognizing that r = m∆r, canceling like terms, and regrouping find 2 m-1m+1m 11qmr mTm+T2mT0. 22k ∆  −+−+=   & < (b) The control volume for the surface node is ( ) V2 rr/2. π =⋅∆⋅ l The energy balance is ingdconv EEqqqV=0.+=++ && & Use Fourier’s law to express q d and Newton’s law of cooling for q conv to obtain [ ] ( ) ( ) n-1n n rTTr k2rh2 rTTq2nr0. 2r2 πππ ∞  ∆−∆  −+−+∆=   ∆   & lll Let r = n∆r, cancel like terms and regroup to find 2 n-1n 11hnrqnrhnr nTnTT0. 22k2kk ∞ ∆∆∆  −−−+++=     & < COMMENTS: (1) Note that when m or n becomes very large compared to ½, the finite-difference equation becomes independent of m or n. Then the cylindrical system approximates a rectangular one. (2) The finite-difference equation for the center node (m = 0) needs to be treated as a special case. The control volume is ( ) 2 Vr/2π=∆ l and the energy balance is 2 10 inga TT rr EEqqVk2q0. 2r2 ππ − ∆∆ +=+=+= ∆          && && ll Regrouping, the finite-difference equation is 2 o1 qr TT0. 4k ∆ −++= & PROBLEM 4.39 KNOWN: Two-dimensional cylindrical configuration with prescribed radial (∆r) and angular (∆φ) spacings of nodes. FIND: Finite-difference equations for nodes 2, 3 and 1. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in cylindrical coordinates (r,φ), (3) Constant properties. ANALYSIS: The method of solution is to define the appropriate control volume for each node, to identify relevant processes and then to perform an energy balance. (a) Node 2. This is an interior node with control volume as shown above. The energy balance is inabcd Eqqqq0. ′′′′ =+++= & Using Fourier’s law for each process, find ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 5232 i i i212 i i TTTT 3 krrkr 2rrr TTTT 1 krrkr0. 2rrr φ φ φ φ −−   +∆∆+∆+   ∆+∆∆   −−   ++∆∆+∆=   ∆+∆∆   Canceling terms and regrouping yields, ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 22 i2i531ii 22 i i rr 131 2rrTrrTTTrrT0. rr22 rr φφ ∆∆ −+∆+++∆++++∆= +∆ ∆+∆∆        (b) Node 3. The adiabatic surface behaves as a symmetry surface. We can utilize the result of Part (a) to write the finite-difference equation by inspection as ( ) ( ) ( ) ( ) ( ) ( )( ) 22 i3i62ii 22 i i r2r 131 2rrTrrTTrrT0. rr22 rr φφ ∆∆ −+∆+++∆+⋅++∆= +∆ ∆+∆∆        (c) Node 1. The energy balance is abcd qqqq0. ′′′′ +++= Substituting, ( ) ( ) ( ) ( ) 4121 i i TTTT 3 krrkr 22rrr φ φ −− ∆  +∆+∆+   ∆+∆∆   ( ) ( )( ) i1 i1 TT 1 krrhrTT0 22r φ ∞ − ∆  ++∆+∆−=   ∆   < This expression could now be rearranged. PROBLEM 4.40 KNOWN: Heat generation and thermal boundary conditions of bus bar. Finite-difference grid. FIND: Finite-difference equations for selected nodes. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)(∆y/2)⋅1, find the FDE for node 1, ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) o1 u121 t,c 61 t,cou26 ky/21 TT x h1TTTT R/y/212x kx/21 TTqx/2y/210 y x/kRThx/kTTT ∞ ∞ ∆⋅ − ∆  +⋅−+−  ′′ ∆∆  ∆⋅  +−+∆∆=  ∆ ′′ ∆+∆++ & ( ) ( ) ( ) 2 t,cu1 qx/2kx/kRhx/k2T0.  ′′ +∆−∆+∆+=  & < (b) Performing an energy balance on the control volume, (∆x)(∆y/2)⋅1, find the FDE for node 13, ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( )( )( ) ( ) l131213 8131413 hx1TTk/xy/21TT k/yx1TTk/xy/21TTqxy/210 ∞ ∆⋅−+∆∆⋅− +∆∆⋅−+∆∆⋅−+∆⋅∆⋅= & ( ) ( ) ( ) ( ) 2 l12814l13 hx/kT1/2T2TTqx/2khx/k2T0. ∞ ∆++++∆−∆+= & < COMMENTS: For fixed T o and T ∞ , the relative amounts of heat transfer to the air and heat sink are determined by the values of h and t,c R. ′′ [...]... < 1.0 k 0 1 2 3 4 5 6 7 T1 34 0 33 8.9 33 8 .3 338 .8 33 9.4 33 9.8 34 0.1 34 0 .3 T2 33 0 33 6 .3 337 .4 33 8.4 33 8.8 33 9.2 33 9.4 33 9.5 T3 31 5 32 4 .3 328.0 32 8.2 32 8.9 32 9 .3 329.7 32 9.9 T4 250 237 .2 241.4 247.7 251.6 254.0 255.4 256.4 T5 225 232 .1 241.5 245.7 248.7 250.5 251.7 252.5 T6 205 225.4 226.6 230 .6 232 .9 234 .5 235 .7 236 .4 T7 195 175.2 178.6 180.5 182 .3 1 83. 7 184.7 185.5 T8 160 1 63. 1 169.6 175.6 178.7 180.6... × 0.1m/1.5W/m ⋅ K = 3. 33 2 ( 3. 33 + 2 ) T3 = ( 2 × 1 03. 5 + 45.8 + 67.0 ) °C + 2 × 3. 33 × 30 o C 1 ( 31 9.80 + 199.80) °C=48.7°C 10.66 (b) The heat rate per unit thickness from the surface to the fluid is determined from the sum of the convection rates from each control volume surface ′ ′ ′ q′ conv = qa + qb + qc + q ′ d T3 = q i = h∆y i ( T − T ) i ∞ q′ conv = 50  0.1 m 45.8 − 30 .0 °C + ( )  m ⋅K... 7 3 & 2 2 / k = 0 and q∆x = 5 × 10 W/m ( 0.005m ) = 62.5K & Tneighbors − 4Ti + q ∆x ∑ k Node A (to find T2): Node 3 (to find T3): Node 1 (to find T1): 20 W/m ⋅ K & 2T2 + 2TB − 4TA + q∆ x2 / k = 0 1 T2 = ( −2 × 37 4.6 + 4 × 39 8.0 − 62.5) K = 39 0.2K 2 & Tc + T2 + TB + 30 0K − 4T3 + q∆ x2 / k = 0 1 T3 = ( 34 8.5 + 39 0.2 + 37 4.6 + 30 0 + 62.5 ) K = 36 9.0K 4 & 30 0 + 2TC + T2 − 4T1 + q∆ x2 / k = 0 1 T1 = ( 30 0... ) T3 (°C ) 200 1000 2000 5000 49.80 49.02 48.11 45.67 49.79 48.97 48.00 45.44 49.84 49.24 48. 53 46.61 T4 (°C ) T5 (°C ) 49.96 49. 83 49.66 49. 23 49. 93 49.65 49 .33 48.46 T6 (°C ) T7 (°C ) q ′ ( W / m ) 49.91 49.55 49. 13 48.00 49.90 49.52 49.06 47.86 477 232 5 4510 10 ,34 0 There are two resistances to heat transfer between the outer surface of the heat sink and the fluid, that due to conduction in the heat. .. 0.25 ( Ts + T3 + T5 + T1 ) + 15.625 (2) T3 = 0.25 ( Ts + T2 + T6 + T2 ) + 15.625 (3) T4 = 0.25 ( T1 + T5 + T1 + Ts ) + 15.625 (4) T5 = 0.25 ( T2 + T6 + T2 + T4 ) + 15.625 (5) T6 = 0.25 ( T3 + T5 + T3 + T5 ) + 15.625 (6) With Ts = 30 0 K, the set of equations was written directly into the IHT workspace and solved for the nodal temperatures, T1 T2 T3 T4 T5 T6 (K) 34 8.6 36 8.9 37 4.6 36 2.4 39 0.2 39 8.0 <  (b)... in the form required of the Gauss-Seidel method (see Section 4.5.2), and with Bi = h∆x/k = 100 W/m2⋅K × 0. 030 m/1 W/m⋅K = 3, we obtain: Node 1: T1 = 1 1 1 (T2 + Ts + BiT∞ ) = (T2 + 50 + 3 ×100 ) = (T2 + 35 0) 5 5 ( Bi + 2 ) (1) Node 2: T2 = 1 1 1 (T1 + 2Ts + T3 ) = (T1 + T3 + 2 × 50 ) = (T1 + T3 + 100 ) 4 4 4 (2) Node 3: T3 = 1 1 1 (T2 + 3Ts ) = (T2 + 3 × 50 ) = ( T2 + 150 ) 4 4 4 (3) k Denoting each... the unknown temperatures T1, T2, T3 and T4 Recognizing that these nodes may be treated as interior nodes, the nodal equations from Eq 4 .33 are (T2 + 25 + T2 + 35 0) - 4T1 = 0 (T1 + 25 + T3 + 35 0) - 4T2 = 0 (T2 + 25 + T4 + 35 0) - 4T3 = 0 (T3 + 25 + 25 + T3) - 4T4 = 0 The Gauss-Seidel iteration method is convenient for this system of equations and following the procedures of Section 4.5.2, they are rewritten... solution to the system of equations (1, 2, 3) of part (a) is T1 = 81.70°C, T2 = 58.44°C, and T3 = 52.12°C The fact that only 4 iterations were required to obtain agreement within 0.01°C is due to the close initial guesses (2) Note that the rate of heat transfer by convection to the top surface of the rod must balance the rate of heat transfer by conduction to the sides and bottom of the rod NOTE TO INSTRUCTOR:... 182 .3 180.8 180.4 180 .3 180 .3 185 186.6 167.0 1 63. 3 162.5 162 .3 162 .3 162.2 185 105.8 96.0 94.2 93. 8 93. 7 93. 6 93. 6 ← initial estimate ← ε . ) 2 1 T 237 4.6 439 8.062.5K390.2K 2 =−×+×−= < Node 3 (to find T 3 ): 2 c2B3 TTT300K4Tqx/k0 +++−+∆= & ( ) 3 1 T348. 539 0. 237 4. 630 062.5K369.0K 4 =++++= < Node 1 (to find T 1 ): 2 C21 30 02TT4Tqx/k0 ++−+∆= & (. schematic, cv1 234 qqqqq ′′′′′ =+++ ()( ) ( ) ( )()( ) cv 1 2 3 4 q h x2 T T hxT T hxT T h x2 T T ∞∞∞ ∞ ′ =∆ −+∆ −+∆ −+∆ − ()()()() 2 cv q 50W m K 0.1m 430 30 0 2 422 30 0 39 4 30 0 36 3 30 0 2 K ′  =⋅×. Eq. (3. 1 03) , where η o and A t are given by Eqs. (3. 102) and (3. 99) and η f is given by Eq. (3. 89). With L c = L + t/2 = 0.022 m, m = (2h/k s t) 1/2 = 32 .3 m -1 and mL c = 0.710, c f c tanh