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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) Onedimensional conduction in the xdirection. (2) Constant properties. (c)
Negligible emission and convection from the exposed surface. PROPERTIES: Table A. 1. pure aluminum (300 K): k3. = 238 W."'1nK. Table A.3. glass (300 K):
kgl = 1.4 Wt’inK. 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/mK,L=1m, ASSUMPTIONS: [1) Steadystate conditions. (2) Onedimensional 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”1nK(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'IXIENTS: Heat loss from the side requires the existence of temperature gradients over the shaft
crosssection. 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 8hour trip. SCHENIATIC: _. .¢—L=0.t]1 m ASSUMPTIONS: (1) Steadystate conditions. (2) Onedimensional conduction in the X
direction. (3) Constant properties. PROPERTIES: Table 131.3. soda lime glass (300 K): kgl = 1.4 W."'1nK. 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 1nK [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) Onedimensional heat transfer in samples. (2) Steadystate conditions. (3)
Negligible contact resistance between materials. PROPERTIES: Table/1.2. Stainless steel 316 (T=400 kss : 15.2 W.’111K; Armco iron
[T=3 80 Kg): 1ng1 = 67".2 W.."'1nK. 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 wanK. < 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."'1nK>< x—
Clrion (Illeater st 4 0.015111 (111011 : [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 WilliK. < 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: Steadystate 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\’.."'1112K): plot the temperature distribution and
label as Case 3. SCHENIATIC:
T(x) = a + bx + 0142 V
k = 5 WimK, d g
Tlol = To : 12000 % Insulated
._.\ g boundary
: o —\
Ten 20 C {Emide I
h = 500 WimEK “x N L)
x L = 50 mm ASSUMPTIONS: (l) Steadystate conditions. (2) Onedimensional 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=a7b0+c0 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 WhirlK, 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'm3K + zocc = 220°C <
Surface energr betcha? on = 0, .seeEq. (3)
b : —11[r.x —To)t"'k:—2501V.s"1n2K(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 waifK. 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 40 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. steadystate 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 steadystate 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) Onedimensional 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 steadVstate 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 ')
KNOW1V: Thin electrical heater dissipating 4000 W."m‘ sandwiched between two 25min thick plates
whose surfaces experience convection. FIND: (a) On Tx coordinates. sketch the steadystate temperature distribution for L x +L:
calculate values for the surfaces x = L and the midpoint. 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 Tx 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. steadystate temperature
distribution; sketch temperature distribution on same TX coordinates as parts (ab): and (d) On Tt
coordinates. sketch the temperaturetime history at the plate locations X = 0. i L during the transient
period between the steadystate 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 Wim2K Plates p = 2500 kglm3 cp = 700 JikgK
“’1 k = 5 WImK
. it its) '>X +L=25mm it. ASSUMPTIONS: [1) Onedimensional 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 midpoint. 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 lefthand 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 Tx coordinates above and labeled Case 2. The
distribution is linear in the lefthand plate. with the maximum valrre at the midpoint. Since no heat
ﬂows tln‘ough the righthand 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 twofold. (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 15minute period. Ate. the heater dissipates 4000 V‘Vr’nr2 and
then is deactivated. To determine the eventual. uniform steadystate temperature distribution. apply
the conservation of energy requirement on a tirrreinterval 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<gK[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 TX coordinates above and labeled Case 3. The
distribution is uniform. and considerably higher than the maximum value for Case 2. (d) The temperaturetime 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 Tt 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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