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Unformatted text preview: nw an: 1 Samuel/x ME we F03 PROBLEM 1 KNOWN: Inner'surface temperature and thermal conductivity of a concrete wall, FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from
15 to 38°C. SCHEMA'I‘IC: A = 20 m2 k = 1, 0.75. or 1.25 WlmK T1 = 25 “0
T2 = 4510 38 DC '
To!)
—>L H. = 0.30 rn
X ASSUMPTIONS: (1) One—dimensional conduction in the xdirection, (2) Steadystate conditions. (3)
Constant properties, (4) Outside wall temperature is that of the ambient air. ANALYSIS: From Fourier's law, it is evident that the gradient, dT/dx = ~—q;'/lt, is a constant, and
hence the temperature distribution is linear, if q’x' and k are each constant. The heat ﬂux must be
constant under onedimensional, steadystate conditions; and k is approximately constant if it depends only weakly on temperature. The heat ﬂux and heat rate when the outside wall temperature is T: = 15°C
are err T —T 25'C—(—15“c) ,
“:4: =1: 1 2=1w vI{—————=133.3W ~. ' 1
q“ dx L m 0.30m /m ( )
q‘=q§xA=133.3W/m3x20m2=2667W. (g)< Combining Eqs. (1) and (2), the heat rate q,‘ can be determined for the range of ambient temperature, 15
S T: S 38°C, with different wall thermal conductivities, it. :15ch
2500
'15
a: 1500
n
.2
a 500 r
I
son
4590 2D 1D D 19 20 30 40
Amblenl air tamporaiurn, T! (C)  I:  Wall thermal cunﬂuctlvlly. k a: 1.25 WimK
——~ I: 1 WimK. concraia wall
4— k u 0.75 Wink For the concrete wall. I: = i W/mK, the heat loss varies linearin from +2667 W to 867 W and is zero when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with
increasing thermal conductivity. COMMENTS: Without steadystate conditions and constant k, the temperature distribution in a plane
wall would not be linear. impingiihi 5.,;..‘.:g::;" I PROBLEM «at 2
KNO‘W 1“ : Chip Width and maximum alIGWable temperature. Coolant conditions. FIND: Maximum allowable chip power. for air and liquid coolents. 5 CHEMIATEC:
#Wﬁ'nrm 'é’
a E . . ..
739:5 C —i‘>
Air, i7=200i/‘£/m3K _ GNP: latex: ‘35 C
Dieisci'rlc fluid, a”: '
I1: 3000 W/mzK I  — 4' ASSUZ‘JIFTEONS: [1) Steadystate conditions, (2) Negligible heat transfer from sides
and button, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat
transfer by radiation in air. AJVALYSIS: All of the electrical power dissipated in the chip is transferred by
convection to. the coolant. Hence, ‘ P = q
and from Newton's law of cooling,
P = new — T9,) = hW2(T — Tm). In oz'r,  PM, = 900 “vii/n12 £01005 mﬂss  15) ° C = 0.35 W. <1 In the dielectric liquid Pm, = 3000 W/m’JKmnns m)2(85 — 15) ° C = 5.25 W. <1 DOWNTS: Relative to liquids, air is a. poor heat transfer ﬂuid. Hence, in air the
chip can dissipate far less energ}r than in the dielectric liquid. 3 .
For a medium in which the heat conduction equation is given in its simplest by i; = ii : ax‘ Ct at (:7) Heat transfer is transient, (b) it is onedimensional, (c) there is no heat generation, and (d) the thermal
conduativity is constant, For a medium in which the heat conduction equation is given by
l a Er?" a BT .
~“(h'T]+T[k*:J+ age“ = 0! r 51' or o: .. (:1) Heat transfer is steady, (b) it is twodimensional, (c) there is heat generation, and (d) the thermal
conductivity is variable. Psoerara 4&3 KNOWN: Area, emissivity and temperature of a surface placed in a large, evacuated chamber
of prescribed temperature. FWD: (:1) Rate of surface radiation emission, ([3) Net rate of radiation exchange between
surface and chamber walls. SCHEMATIC: ASSUMPTIONS: (1) Area of the enclosed surface is much less than that of chamber walls. ANALYSIS: (3.) From Eq. 1.5, the rate at which radiation is emitted by the surface is
qenﬁt=q;nit A=£AUT§ clam = 08(05 1112) 5.67 x 10'8 W/mzK4 [(150 + 273)K]4 gem, = 726 w'. ' <1 (13) From Eq. 1.7, the net rate at which radiation is transferred from the surface to the chamber
walls is q = e A o (Ti—T3... q= 0.3(0.5 mg) 5.57 x 10'“8 W/mZK4 [(423104 — (298194] q=547w. . <1 COMMENTS: The foregoing result gives the net heat loss from the surface which occurs at
the instant the surface is placed in the chamber. The surface would, of course, cool due to this
heat loss and its temperature, as well as the heat loss, would decrease with increasing time. Steadystate conditions would eventually be achieved when the temperature of the surface
reached that of the surroundings. ' PROBLEM r it 4 KNOWN: Onedimensional system with prescribed thermal conductivity and
thiokness.  ' ‘ FIND: Unknowns for various temperature conditions and sketch distribution.  SCHEMATIC: ASSUMIPTEDNS: (1) Steadystate conditions, (2) Onedimensional conduction, (3)
No internal heat generation, (4) Constant properties. ANALYSIS: The rate equation and temperature gradient for this system are
u (11' £1; __ T2 _T1 qx # ——k — and dx dx L ' (1’2) Using Eqs. (1) and (2), the unknown quantities for each case can be determined. dT _ (—20 —— 50) K (a) dx 0.25m —280 /
n W . K 
=—5 — —280——— =. . W 1112.
qx Gm'Kx[ m] 140k / (b) £=W=30Kﬁn dx 0.25m 
H W K
= —50 —— so— = —i 2
qx mDK x[ m] OkW/rn
n W K ‘
= —50 —— 1'60 — = — . 2
(c) qx WK >< [ m] s o kw/m _
T2 =L—dl + T1 = 0.25111><[160E +70 " 0.
dx 111
T2 = 110 " c.
H W K
d =—5o —— ksoh— = . 2
Us); WK x m 40kW/m
T1 =T2 .
dx 111
T1 a C.
Ir W ‘K I
= —50 —. 200—— = — . 2
(e) qx WK x [ m] 10 o kW/nti T1 = T2 —— Lac—13 = 30 ° c — 0.2542005] = —20 "to.
dx m a PROBLEM e5” KNOWN: Temperature distribution in a onedimensional wall with prescribed thickness and
thermal conductivity. FIND: (a) The heat generation rate, q, in the wall, (in) Heat fluxes at the wall faces and relation ASSUMPTIONS: (1) Steadystate conditions, (2) Onedimensional heat ﬂow, (3) Constant
properties. ‘ ' ANALYSES: (a) The appropriate form of the heat equation for steadyState, onedimensional
conditions with constant properties is Eq. 2.15 rewritten as  . d dT
= _k .—
‘1 d3; dx ]
Using the form for the temperature distribution, evaluate the gradient giving,
 _‘ _ d _d_ 2  .. _d_ .— 9
q .. 1: dx dx (a+bx )] — k dx [2bx] — "bit
q = —2(—2000°C/m2) x 50 W/nnK = 2.0x105 W/m3 . . 4
‘ (b) The heat ﬂuxes at the wall faces can be evaluated fromFourier’s'law,
n dT
QXCX)=“k a x Using the temperature distribution T(x) to evaluate the gradient, ﬁnd
_ q;(x) = —k % [a—tbx2] = —2kbx .
The ﬂux at the face x=0, is then . q;(0)=o <1
and at x = L, q;(L) = —2kbL = 2 x SOW/mK (—2000°C/m2) x 0.050m gm): 10,000 W/mz.  < 
; COMMENTS: _ From an overall energy balance on the wall, it follows that '. _ E0111 + 1‘31:} = 0 "' + 611‘ = 0
E q nO 2— ‘ 9
Q=M=W=g_0xloﬁw/m3. _ ...
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This note was uploaded on 09/08/2010 for the course ME 114 at San Jose State.
 '08
 Rhee,Jinny

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