chapter2-solutions

chapter2-solutions - PROBLENI 2.7 KNOWN: Irradiation and...

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Unformatted text preview: PROBLENI 2.7 KNOWN: Irradiation and absmptivity of ahuninum. glass and aerogel. FIND: Ability of the protective barrier to withstand the irradiation in terms of the teinperatm‘e gradients that develop in response to the irradiation. SCHEMATIC: —> G = 10x 105W1m2—p —’ |—> X (131 : 0.2 ugl : 0.9 (1a = 0.8 ASSUMPTIONS: (1) One-dimensional conduction in the x-direction. (2) Constant properties. (c) Negligible emission and convection from the exposed surface. PROPERTIES: Table A. 1. pure aluminum (300 K): k3. = 238 W.-"'1n-K. Table A.3. glass (300 K): kgl = 1.4 Wt’in-K. ANALYSIS: From Eqs. 1.6 and 2.30 6T .. 'kg : Cls : Gabs : UG x20 or E 2 E 6x x20 k The temperature gradients at X = 0 for the three materials are: < at r ("K‘sz (Kan) Material '— aluinintun 8.4 x 103 glass 6.4 x 106 aerogel 1.6 x 109 COMMENT: It is unlikely that the aerogel bairier can sustain the thermal stresses associated with the large temperature gradient. Low thermal conductivity solids are prone to large te111perature gradients. and are often brittle. PROBLEM 2.12 KNOWS: Length and thennal conductivity of a shaft. Temperature distribution along shaft. FIND: Temperature and heat rates at ends of shaft. SCHEMATIC': Suppor‘l'mg sfiaff) k=25W/m-K,L=1m, ASSUMPTIONS: [1) Steady-state conditions. (2) One-dimensional conduction in x. (3) Constant properties. ANALYSIS: Temperatures at the top and bottom of the shaft are. respectively. T(0) : 1000C T(L) : 40°C. < Applying Fourier’ s law. Eq. 2.1. (Ix : — " j—T : —25 W3”1n-K(0.005 1112)[—150 + 20x)0 (7.5m X qX :0.125(_150 -20X)W. Hence. (1:40): 18.75 W qx(L) : 16.25 w. < The difference in heat rates. qx(0) Z> qx(L). is due to heat losses (1 {r from the side of the shaft. CON-'IX-IENTS: Heat loss from the side requires the existence of temperature gradients over the shaft cross-section. Hence. specification of T as a function of only X is an approximation. PROBLENI 2.14 KNOWN: Dimensions of and temperature difference across an aircraft window. Window materials and cost of energy. FIND: Heat loss through one window and cost of heating for 180 windows on 8-hour trip. SCHENIATIC: _.| |.¢—L=0.t]1 m ASSUMPTIONS: (1) Steady-state conditions. (2) One-dimensional conduction in the X- direction. (3) Constant properties. PROPERTIES: Table 131.3. soda lime glass (300 K): kgl = 1.4 W.-"'1n-K. ANALYSIS: From Eq. 2.1. qx = -kAd—T =ka b—(Tl '9 dx L For glass. qx = 1.4 W x 0.3111>< 0.31n><' 80 C = 1010 w < ’g 1n-K [0.01111 The cost associated with heat loss tln‘ough N windows at a rate of R = Sl..-"'kW'h over a t = 8 11 flight time is S ><8h>< 1k“: =$1050 < kW .11 1000w Cg. =qu.gRt = 130 >< 1010 w x1 Repeating the calculation for the polycarbonate yields qxp = 151 w. cp = $15.? < while for aerogel, q“ = 10.1 wica =s1o < COMMENT: Polycarbonate provides significant savings relative to glass. It is also lighter (pp = 1200 kgs’mi) relative to glass (pg = 2500 leg-"1113). The aerogel offers the best thermal performance and is very light (pa = 2 leg-"1113) but would be relatively expensive. PROBLEM 2.1 7 KNOWN: Electrical heater sandwiched between two identical cylindrical (30 111111 dia. x 60 mm length) samples whose opposite ends contact plates maintained at TO. FIND: (a) Thermal conductivity of S8316 samples for the prescribed conditions (A) and their average temperature. (b) Thermal conductivity of Annco iron sample for the prescribed conditions (B). (c) Comment on advantages of experimental arrangement. lateral heat losses. and conditions for which ATl iATz. SCHE KIA—[ICE 7;:77DC 7;: 77°C yeah,” A7; = 25. 0 °C Heafer’ 1971—4510? 100V, SS 516 100V; Armco iron 0553A atg=250°c 95"” $45.01: T: ‘1: tit-“MW T=77°C a 77 Case A 0 Case B ASSUMPTIONS: (I) One-dimensional heat transfer in samples. (2) Steady-state conditions. (3) Negligible contact resistance between materials. PROPERTIES: Table/1.2. Stainless steel 316 (T=400 kss : 15.2 W.-’111-K; Armco iron [T=3 80 Kg): 1ng1 = 67".2 W..-"'1n-K. ANALYSIS: (a) For Case A recognize that half the heater power will pass through each of the samples which are presumed identical. Apply Fourier‘s law to a sample AT q c Ax qAx _ 0.5[100Vx0.353A)><0.0151n AcAT 1110.030111')21-"'4><25.0C'C = 15.0 wan-K. < The total temperature drop across the length of the sample is AT1(L..-‘"AX) = 250C [60 min-"'15 mm) = 100°C. Hence. the l1eate1‘te111pe1ature is Th 2 177°C. Thus the average temperature of the sample is T:(r0 + Th j) 2 =127°c:400 K. < We compare the calculated value of k with the tabulated value (see above) at 400 K and note the good agreement. (b) For Case B. we assume that the thermal conductivity of the SS316 sample is the same as that found in Part (a). The heat rate through the Armco iron sample is Continued . . PROBLEN‘I 2.17 (CONTJ :r(0.030 m)2 15.00 C a : ‘— . :IOOVXOfiOlA—ISDW.-"'1n-K>< x— Clrion (Illeater st 4 0.015111 (111-011 : [60.1—10.6]W=49.5 W where (133 = kSSACATZ AXE. Applying l-"ourier= s law to the iron sample. -- A1 49.5 W 0.015 . kiron = quon K2 2 X m = 70.0 Willi-K. < AcATz 1:40.030 m)2 .-‘"4><15.00C The total drop across the iron sample is 15°C(60.-"15) : 600C: the heater temperature is (7? + 60)°C : 13?”C. Hence the average temperature of the iron sample is T: (137 + 77f Cf2=107°C=380 K. < We compare the computed value of k with the tabulated value (see above) at 380 K and note the good agreement. (c) The principal advantage of having two identical samples is the assurance that all the electrical power dissipated in the heater will appear as equivalent heat flows through the samples. With only one sample. heat can flow from the backside of the heater even though insulated. Heat leakage out the lateral surfaces of the cylindric ally shaped samples will become significant when the sample thermal conductivity is comparable to that of the insulating material. Hence. the method is suitable for metallics. but must be used with caution on nonmetallic materials. For any combination of materials in the upper and lower position. we expect All = AT2. However. if the insulation were improperly applied along the lateral surfaces. it is possible that heat leakage will occur. causing ATl 7-: AT} PROBLEM 2.26 KNOWX: 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. :1 . by applying an overall energy balance to the wall. (D) Determine the coefficients a._ b. and c. by applying the boundary conditions to the prescribed form of the temperature distribution: plot the telnperatin‘e 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 temperatm‘e distribution and label as Case 2'. ((1) Determine new values for a._ b. and c for conditions when the generation rate is doubled. and the convection coefficient remains unchanged (11 = 500 V\-’..-"'1112-K): plot the temperature distribution and label as Case 3. SCHENIATIC: T(x) = a + bx + 0142 V k = 5 Wim-K, d g Tlol = To : 12000 % Insulated ._.\ g boundary : o -—\ Ten 20 C {Emide I h = 500 WimE-K “x N- L) x L = 50 mm ASSUMPTIONS: (l) 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 _ Ebut + Egen = 0 Where Bin 2 qgonv thw—TO)+€1L:0 (1) q = —h(Tx. —T0).-"L = —500W'i1112‘K(:20—120]°Ci0.050 1n =1.0><106 Win? < To I222 TO I I ., I la; q’conv : qCOI‘W : qxw) GIRL) : —-» .222 —' —>. .2; I —’| /,'/ I I , _ rig; I I , Egen L‘) | 1—)x —I—) x=L x L (a) Overall energy balance (b) Surface energy balances (b) The coefficients of the temperature distribution. T(X) = a + bx + cX‘. can be evaluated by applying the boundary conditions at x = 0 and X = L. See Table 2.2 for representation of the boundary conditions. and the schematic above for the relevant surface energy balances. Bonndma‘ condition (tin: = 0, convection snifoce condiiion Bin ’ but 2 (liconv * (13’; (0) Z 0 Where q; (0) 2 7k— lilTw —rO )—[—k(0+b+2cx)x=0] :0 Continued . . PROBLEM 2.26 (Cont) b = —11[T.I —ro )rk 2—500VV351112 .Kt20—120)°c,=‘5w 5111- K =l.0><104K.-"III] t2)< Boundary condition or x = L, adiabrtrfr or insulated surface - - " ' n \ dT Ein _ Eout : _qx : 0 where qx{‘]_r}: \ x=L l:[0+b+2cx]sz =0 (3) c=—b.-*'2L=—1.0x104Kr‘mt(2x0.0501n}=_1_0x105K;mE < Since the surface temperature at x = 0 is lcnown. T(O) = T0 = 1200C: find T(0]=1200C=a7b-0+c-0 or a=1200C (4)< Using the foregoing coefficients with the expression for T(x) in the Workspace of IHTE 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. h; = 11/2 = 250 Whirl-K, q = 1><106 - 3 . t Wr'ni and other parameters remain unchanged except that To 9: 120°C. We can detemnne a. b._ and c for the temperature distribution expression by repeating the analyses of parts (a) and (13). Overall energt' baiance on the it'd”, see 593. (2,4) a = To = (tr/11+ rm 2 1 x106W..-"1113 x 0.050111r250w.s'm3-K + zocc = 220°C < Surface energr betcha? on = 0, .seeEq. (3) b : —11[r.x —To)t-"'k:—2501V.s"1n2-K(20— 220)°c..-‘5wrm. K : 1.0><104K.-"1n < Surface energy {minute on I L, see Eq. (3) c = —b.-*' 2L 2 _1.0 ><104K.-"'111f(2X0.050111:} = —l_0>< 105K511]: < The new temperature distribution. T2 (x). is plotted as Case 2 below. (d) Consider Case 3 when the intemal energy volumetric generation rate is doubled. , ('13 2 2q 2 2 >< 106 W I 1113. h = 500 waif-K. and other parameters remain michanged except that TO 1*: 120°C Following the same analysis as part (c). the coefficients for the new temperature distribution. T (x). are a2220°C b=2><104K.-"1nc=—2><105K..-"'1112 < and the distribution is plotted as Case 3 below Continued . . PROBLEM 2.26 (Cont) 300 700 60E] 500 400 Temperature. T (C) 300 200 'lOD O 5 10 'l 5 20 25 30 35 4-0 45 50 Wall posttion,x(mm) 500 WImAZ K, qdot= 166 WIm’G 250 WIm"2.K, qdot= 166 WIm’G — l h —'— 2. h + 3. h 500 WIm"2.K, qd0t=2e6 WIm’x'B COMMEXTS: 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 temperattu‘es at the insulated boundaiy. 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} but the convection resistance has increased. By doubling the generation rate for Case 3. we expect the surface temperatm‘e T0 to increase relative to the Case 1 value. since double the amount of heat flux is removed from the wall (qu). 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 111a gnitudes of T(L) for the three cases? PROBLEM 2.47 KNOWN: Plane wall. initially at a uniform temperature Ti. is suddenly exposed to convection with a fluid at To,3 at one surface. while the other surface is exposed to a constant heat flux qg. FIND: (a) Temperature distributions, T(X.t). for initial. steady-state and two intennediate times. (b) Coiresponding heat fluxes on q; — x coordinates. (c) Heat flux at locations X = 0 and x = L as a function of time. (d) Expression for the steady-state temperature of the heater. T(0.os-)._ in terms of (13. T93, k. 11 and L. SCHEMATIC: # 9=Qk Heater, 9o J7 Insulahon PX L 76,0):7’: ASSUMPTIONS: (l) One-dimensional conduction. (2) No heat generation. [3) Constant properties. ANALYSIS: (a) For Ti < Tm, the teniperatiu‘e distributions are Note the constant gradient at X = 0 since q; [0} : (13. (b) The heat flux distribution. q"X (x,t), is determined from lolowledge of the temperature gradients. evident from Part (a). and Pourier‘s law. (c) On (1')} — t coordinates. the heat fluxes at the bomidaries are shown above. (d) Pel‘foiin a siu‘face energy balance at X = L and an energy balance on the wall: qgond : qcoiiv :11[T(L=°'3')—Toc-] (1): (Icond : qflo- (2) | For the wall. under steadV-state conditions. Foul‘ier‘s law gives i i 9" ‘21.: ,, dT T(0___oo) — T(L,o':-) road I v (10 _ k 2 k - (3) xéL dx L Combine Eqs. (1). (2). (3) to find: (1’6 T o ,- 2T ..+—. ( ’0“) 3" 1.r’h+L-..="1-; PROBLEM 2.53 ') KNOW-1V: Thin electrical heater dissipating 4000 W.-"m‘ sandwiched between two 25-min thick plates whose surfaces experience convection. FIND: (a) On T-x coordinates. sketch the steady-state temperature distribution for -L x +L: calculate values for the surfaces x = L and the mid-point. x = 0: label this distribution as Case 1 and explain key features; (b) Case 2: sudden loss of coolant causing existence of adiabatic condition on the x = +L surface: sketch temperature distribution on same T-x coordinates as part (a) and calculate 'alues for x = 0. : L'. explain key features: (c) Case 3: further loss of coolant and existence of adiabatic condition on the X = - L surface: situation goes undetected for 15 minutes at which time power to the heater is deactivated: determine the eventual (t —> so) uniform. steady-state temperature distribution; sketch temperature distribution on same T-X coordinates as parts (ab): and (d) On T-t coordinates. sketch the temperature-time history at the plate locations X = 0. i L during the transient period between the steady-state distributions for Case 2 and Case 3: at what location and when will the teinperatiu‘e in the system achieve a maximum value? SCHEMATIC: Electric heater qL= 4000 me2 Tm = 20°C h = 400 Wim2-K Plates p = 2500 kglm3 cp = 700 Jikg-K “’1 k = 5 WIm-K . it its) '->X +L=25mm it. ASSUMPTIONS: [1) One-dimensional conduction. [2) Constant properties. (3) .\'0 internal volumetric generation in plates. and (3) Negligible thermal resistance between the heater surfaces and the plates. ANALYSIS: (a) Since the system is symmetrical. the heater power results in equal conduction fluxes tln‘ough the plates. By applying a surface energy balance on the surface X = +L as shown m the schematic. determine the temperatures at the mid-point. X = 0. and the exposed surface= X — L. i ' T(+L) I I qiionv —*I I—* q;(+L) . . res“ I I Tm, h ,3 I I X_/‘“‘( Ein _ Eout : 0 q’)’{ (+L) — c13011V : 0 where q’)’{ (+L) 2 (13 r" 2 q; ..--'2711[r[+L)7r.fi] = 0 Ti (+L] : (is i" 211 + T. J : 4000 w 1112 i (2 >< 400 w 1112 -K) + 20°C : 25°C < From Pourier‘s law for the conduction flux through the plate. find T(0). (13;: q; = k[T[0]i T (+L]].-=' L T1(0) = T1 [+L)—qu.-=' 2k = ascc— 4000w..--"1i12 -K >< 0025111212 >< 5w: 1n - K) = 350C < The temperature. distribution islshown on the T—X coordinates below and labeled Case 1. The key features of the distribution are its synnnetiy about the heater plane and its linear dependence With distance. Continued . . PROBLEM 2.53 (Cont) 86.1 Case 33300 | Case 2. T200 I I 40 t r Case 1, Tltx} 30 ' T1(O) = 35°C T = 20 — — — — ' (b) Case 2: sudden loss of coolant with the existence of an adiabatic condition on surface it = —L. For this situation. all the heater power will be conducted to the coolant through the left-hand plate. From a surface energy balance and application of Fourier= s law as done for part (a). find r2 (—L) : qg 11 + r...) : 4000 w r 1112 400 w 1113 - K + 20°C : 30°C < T2 (0) 2 T2 [—L) + qu k = 30°C + 4000 w 1112 X 0025 1115 w 111- K = 50°C < The temperature distribution is shown on the T-x coordinates above and labeled Case 2. The distribution is linear in the left-hand plate. with the maximum valrre at the mid-point. Since no heat flows tln‘ough the right-hand plate. the gradient rrrust zero and this plate is at the maximum temper‘atru‘e as well. The maximum temperature is higher than for Case 1 because the heat flux tlu'ough the left—hand plate has increased two-fold. (c) Case 3: sudden loss of coolant occurs at the X = -L surface also. For this situation. there is no heat transfer out of either plate. so that for a 15-minute period. Ate. the heater dissipates 4000 V‘Vr’nr2 and then is deactivated. To determine the eventual. uniform steady-state temperature distribution. apply the conservation of energy requirement on a tirrre-interval basis. Eq. 1.1]b. The initial condition corresponds to the temperature distribution of Case 2. and the final condition will be a uniform. elevated temperature Tf= T 3 representing Case 3. We have used Tog. as the reference condition for' the energy terms. Bin _ Eout + E’gen : AEgt : Ef ‘Ei (1) Note that 3311 — Egut = 0. and the dissipated electrical energy is E’éen 2 game 2 4000 WT1112[15 X 60) s = 3.600 x 106 J r‘ 1112 (2) For the fural condition. Ef = pc(2L)[Tf — rm] 2 2500kg #1113 X 700i..-"1<g-K[2 X0.025111)[Tf —20]°C B; = 8.75 ><104 [rf — 20] 1112 where Tf = T 3,. the final uniform temperature. Case 3. For the initial condition. , —L . v , 0 y , +L . ‘r E1: chL [r2 (x)—r.,.]ds = pC{J_L [r2 (a)—Tx.]da—J.O [r2 (or—rm]de (4) where T2 (24} is linear for —L s X g 0 and constant at T2 (0) for 0 X s +L. T2(x):T3[0}+[T3[0)—T2[L)]x..-"L —L£Xg0 T2 (x) = 50°C + [50 a 30] ocrxr' 0.025111 T2[x):50°C+800X (5) Substituting for T2 (x }. Eq. (5). into Eq. (4) Continued . . PROBLEM 2.53 (Cont) H Ei’ = pe I:[so+800x —Tc,:.]dx +[r3 [0)—Tx._L% E Ef:peil i i: poiifiom 400x2 —T.73x]: ——[T3 [Oi—TOOJL} i [*SOL + 400L2 + TOOL] W [t2 (0) 7 tech. Ei’ : ch{+50— 400L —TI.. + T2 (0}—t0¢.} E = 2500kgm3 x 700 J r kg’K >< 0.025111{+507 400 x 0.0257 20+ 507 20} K 135* = 2.188xio6J.-‘m2 (6) Returning to the energy balance. Eq. (I). and substituting Eqs. (2). (3) and (6). find Tf= T3,. 3.600><106 J..-"m2 = 8.75 ><104 [T3 7 20] i 2.188><106 1112 T3 :(66.l+20)OC:86.1°C < The temperature distribution is shown on the T-X coordinates above and labeled Case 3. The distribution is uniform. and considerably higher than the maximum value for Case 2. (d) The temperature-time histon at the plate locations X I 0. : L during the transient period between the distributions for Case 2 and Case 3 are shown on the T-t coordinates below. T(x,t) Maximum point D 15 Case 2 Heater deactivated Time (min) Note the temperatures for the locations at time t = 0 corresponding to the instant when the surface x I - L becomes adiabatic. These temperatures correspond to the distribution for Case 2. The heater remains energized for yet another 15 minutes and then is deactivated. The midpoint temperature. T(0.t). is always the hottest location and the maximum value slightly exceeds the final temperature T 3,. ...
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chapter2-solutions - PROBLENI 2.7 KNOWN: Irradiation and...

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