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Unformatted text preview: \“i iii/U £52 gait/{Film} ME {lit TQM A spherical container of inner radius rl , outer radius r; , and thermal conductivity i: is given. The boundary condition on the inner surface of the
container for steady onedimensional conduction is to be expressed for the
following cases: (a) Speciﬁed temperature of 30°C: T01) = 50°C dTO‘l) =30 WI 3
ctr m (b) Speciﬁed heat ﬂux of 30 W/‘rn1 towards the center: it (c) Convection to a medium at Tm with a heat transfer coefﬁcient of in: kg? = h[T(rE)— T5,}
: Heat conduction through the bottom section of a steel pan that is used to boil water on top of an
electric range is considered. Assuming constant thermal conductivity and onedimensional heat transfer,
the mathematical formulation (the differential equation and the boundary conditions) of this heat
conducnon problem is to be obtained for steady operation. Assmnptians 1 Heat transfer is given to be steady and onedimensional. 2 Thermal conductivity is given to
be constant. 3 There is no heat generation in the medium. 4 The top surface atx = L is subjected to
convection and the bottom surface at :c = 0 is subjected to uniform heat ﬂux. Analysis The heat flux at the bottom of the pan is ' E,  a
4“, =gi=i=w=3lsgo W/m' AA. mo /4 7:10.20 m)2 M Then the differential equation and the boundary conditions for this heat conduction problem can be
expressed as 2
dr=0
tit"
k thO) = q, = 33,230 vii/m1
cit
—k “m” = mgr.)  rm] a} .5 A steam pipe is subjected to convection on the inner surface and to speciﬁed temperature on the
outer surface. The mathematical formulation, the variation of temperature in the pipe, and the rate of heat
loss are to be determined for steady one~dimensional heat transfer. Assumptions 1 Heat conduction is steady and onedimensional since the pipe is long relative to its
thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There
is no heat generation in the pipe. Properties The thermal conductivity is given to be k = 7.2 Btu/liﬁ°F. Analysis (0) Noting that heat transfer is onedimensional in the radial r direction, the mathematical
formulation of this problem can be expressed as d dT T=160°F
._..... r— = O A
or [ Cir] Steam dT  and —k ('1) mm, —T(r1)] 250°?
0'" h=12.5 FIFE): T2 =160DF (b) Integrating the differential equation once with respect to r gives i L = 30 R E I f'“"‘_ = C1
dr Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,
dT _ C]
a:  7
TOP) = C1 lnr+C2 where C] and C3 are arbitrary constants. Applying the boundary conditions give r=r;: —kﬂ=h[1'1m—(C'1 lnrl+C2)]
"i
r=r3: T(r3)=Cl lnr2 +C2 =T2
Solving for C1 and C: simultaneously gives
T —T T —T
C1=—3—°“— and C,=T,—C1 lnr7=T1— 1 '1' lnr,
"2 k ' " ' " r1 k '
[n — + — In —' + —
Fl in} r1 [1"] Substituting C1 and G; into the general solution, the variation of temperature is determined to be r —r 
m) =0. Im~+r2 —C1 inr, =c,(1nr—1nr,)+r, =—3—°=—1n'—+T2
_, —  r; k 1'1
ln—+— '
r1 In]
. (1 soatsmn7 r :
=———.——————1 +150°r=—24.741 +160°F
“1&4. 7.2 Btu/h a °1= ” 2.4 in n 2.4 in
2 (12.5 Btu/h e3 ~°F)(2 112 a)
(c) The rate of heat conduction through the pipe is
. r, 4"
dr r r3
ln—+—
1', hr]
nw—nmm
=—2 0 new . .u =33 600Bt /h
”(3 ﬁx? u h n F) lnEi 7.2 Btu/hft“F , u 2 (12.5 Btu/h a1 °F)(2/'12 rt) PROBLEM . a: 4
=KNOWl‘J: Temperature distribution in a composite wall. END: (3) Relative magnitudes of interfacial heat ﬂuxes, (13) Relative magnitudes of thermal
conductivities, and (c) Heat ﬂux as a function of distance x. SCHEMATIC: ASSUMPTIONS: (l) Steadystate conditions, (2) Onedimensional conducrion, (3) Consrant
;_properties, _ ANALYSIS: (3.) For the prescribed conditions (onedimensional, steadystate, constant k), the
parabolic temperamre distribution in C implies the existence of heat generation." Hence, since
1117th increases with decreasing it, the heat ﬂux in C increases with decreasing x. Hence, q; > all However, the linear temperature distributions in A and B indicate no generation, in which case
C12 = C13
(b) Since conservation of energy requires that (11,313 =qglc and dT/d:t)B < dT/dx)c, it follows
from Fourier's law that ' k}; > kc .
Similarly, since qu = c133 and dT/dx)A > dT/dx)E. it follows that
RA < k3 . (c) It follows that the ﬂux disn'ibution appears as shown below. COMMENTS: Note that, with dT/dx)4c m 0, the interface at 4 is adiabatic. PROBLEM ' a: 5’ KNOWN: Plane wall with internal heat generation which is insulated at the inner surface and
subjected to a convection process at the outer surface. . FIND: Maximum temperature in the wall. SCHEMATIC:_  15:25W/mK . .
I T9: 0.5x106W/m3 051 Insu/a'l'ion ""1? =920C
2017,, SHTT h ~K500W/m’
x   . ASSUMPTIONS: (1) Steadystate conditions, (2) Onedimensional conduction with uniform
volumetric heat generation, (3) Inner sm'face is adiabatic. ANALYSIS: From Eq. 3 ”42 the temperature at the' inner surface IS given by Eq 3 43 and 15 the
maximum temperature within the wall Tn = q 1.2/21; + T5. ‘ The outer surface temperature follows from Eq. 3.46.
T5 = Tu. + 6; Mb
W T5 = 92°C + 0.3x105—— x o 1mJSOOW/m K 92°C + 60°C=152°C.
m3 It follows then that
Ta = 0.3m6 W/m3 3 (0.1m)2/2x25W/mK + 152°C T=60°C+152°C=212°C.  ' <1 COMMENTS: The heat ﬂux leaving the wall can be determined from knowledge of h, T5 and
T, using NewtonI 5 law of cooling. . 23m = has—L.) = sonar/m2 K(152—92)°C = 301cm];2 . This same result can be determined from an energy balance on the entire wall which has the
form ' Eg " 1gout = 0
where E3 = dAL and EM = q'c’m A. _
Hence, glam, = 51.1. =—. 0.3x105 W/m3 x 0.1m = 3OltW/m2 . PROBLEM; he (y KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces
of a rear window. FIND: (a) Inner and outer window surface temperatures. T“ and T“. and (b) T“ and Tm as a function of
the outside air temperature Tm“ and for selected values of outer convection coefﬁcient. h“. SCHEMATIC:
Glass TS": h... “'VVV ‘VVVVV El :1
T 31000 Hi Tm m use Uk 1th valmove r. = 0.004m_s—_.[: I hi: 30 Win12 “ K
ASSUMPTIONS: (l) Steadystate conditions. (2) Onedimensional conduction. (3) Negligible radiation
effects. (4) Consrant properties. . 1 PROPERTIES: Table A3. Glass (300 K): k = [.4 WimK.
ANALYSIS: (a) The heat ﬂux may be obtained from Eqs. 3.11 and 3.12. H TM a Tag 40%: — (—10'c) ‘1 =——1 L l =__“1 ' '0'“.00‘4"m""""_‘—“‘"1
—+—+— —_———i——+—+—~—m1—
ha k hi 63W/m'K 1.4W/mK 30W/m'K q” 50 C =9ssw/m1.‘ = (0.0154 + 0.0029 + 0.0333)m3 K/W Hence. with q" = hi('l.‘,ti — T“), the inner surface temperature is H 3
ni=eiﬁ—=4o~c—M=
' ' h. sow/mK l 7.7“C ' <
Similarly for the outer surface temperature with q" = hD(T5_D — Tm) find rr 4
T.D=Tn—3—=—10“C——363i/ET‘—
 h 63W/m'K ﬂ = 4.91: < ( b) Using the same analysis, Tm and T5.El have been computed and plotted as a function of the outside air
temperature, Two, for outer convection coefﬁcients of h‘J = 2. 65, and 100 W/ITIE'K. As expected, T5,. and
T”, are linear with changes in the outside air temperature. The difference between T5; and T“; increases
with increasing convection coefficient, since the heat flux through the window likewise increases. This difference is larger at IDWer outside air temperatures for the same reason Note that with h“ = 2 W/mzK.
Tm  '1'”, is too small to show on the plot. Surface temperatures, Tsl orTsu [Ci Outside air temperature. Ticic (0) —:h— T51: ha 3 “ID Wan‘EM
1b— Tee: he: tut: Wim‘th
E— Tat: ho = 55 Wim*2.K
‘5'" Tan: no =r BE Whn"2.K
Tel when i‘io : 2 Wlm“.K COMMENTS: (1) The largest resistance is that associated with convection at the inner surface. The
values of T“ and T5,” could be increased by increasing the value of hi. (2) The [HT Thermal Resistance Network Made was used to create a model of the window and generate
the above plot. The Workspace is shown below. _ I N am at : '
seesaw W age E a; 0 r I I I’D V‘ct ‘
ll Heat rates Into node Lot]. through thermal resistance Hi]  P g m ‘ I4 #@1 a! q21=(1'2T1)In21
qaa = (Ta  T2) I R32
q43 = (T4  T3) I H43 II Nodal energy balances
qt + q21 = 0 q2q21 4.413220 :13 ~ £132 + Q43 = 0 :14  q43 = 0 l' Assigned variables list: deselect the qi', Fill and Ti whichlare unknowns: set cl = 0 for embedded nodal points
at which there is no external source of heat. 'I ' T1 = Ttnle ll Outside alrtamperaiure. 0 [lot = 1/ Heat rate. W  T2 = Tso // Outer surface temperature; 0 q2 = 0 I! Heat rate. W: node 2. no external heat source T3 2 Tel ll inner surface temperature. 0 q3 = 0 I! Heat rate. W: node 2. no external heat source T4 = ‘l’inll 1/ Inside alriemperature. C l/q4 = I! Heat rate. W II Thermal Hesietances: R21 .= 1 i ( ho ' As ) ll Convection thermal resistance. KlW: outer. suriaca
H32 = Li ( k ‘ As i ll Conduction thermal resistance. KJW; glass H43 = 1 l (hi ' As ) ll Convection thermal resistance. KJW: inner surface
If Other Assigned Variables: Tintc = 10 ll Outside air temperature. 0 ho = 65 ll Convection coefficient. W/mﬁ2.i<: outer surface L = 0.004 ll Thickness, m; glass it = 1.4 II Thermal conductivity. W/m.K: glass Tinll = 40 If Inside air temperature. C bl = 30 'l/ Convection coefﬁcient. Wimr“2.l<; inner surlace A5 = 1 If Crosssectional area. m"2; unit area ...
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 '08
 Rhee,Jinny

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