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Unformatted text preview: In order to increase the rate of heat dissipation from one side of an electronic device
1 m wide and 1 m tall the addition of aluminum fins has been suggested. The fins
are to be rectangular in cross section, 2.5 cm long and 0.25 cm thick. There are to be 100 ﬁns per meter. The convective heat transfer coefﬁcient, both for the wall and the
ﬁns, is estimated at 35 Wz’m2 K. With this information determine the percent increase
in the rate of heat transfer of the ﬁnned wall compared to the bare wall. GIVEN  Aluminum fins with a rectangular cross section e Dimensions: 2.5 cm long and 0.25 mm thick  Number of fins per meter = 100 __  The convective heat transfer coefficient (11:) = 35 W/m2 K FIND
The percent increase in the rate of heat transfer of the finned wall compared to the bare
wall. ASSUMPTIONS
 Steady State heat transfer SKETCH PROPERTIES AND CONSTANTS
From Appendix 2, Table 12:
The thermal conductivity of aluminum (k) = 240 W/m K at 127°C SOLUTION
Since the fins are of uniform cross section, Table 2.1 can be used to calculate the heat
transfer rate from a single tin with convection at the tip: M sinh (mL) + (Hg/mk) cosh (mL)
q‘ ' cosh (mu—T { Jka sin—h {mL} where M = ‘Ih‘cPkA' as = ,{ﬁc 2{t+w) k (tw)‘ a,
BS : T u T S u For a l m width [w='l m): M = H35 (1.0025 m)(240 (.0025 m?) 05 = 6.49 0‘ ¥ '5 P E 2[t+w) {35 %)2(1.0025m)
L k‘_A :1. _° =0.025m —'“ mL 1‘ (tw) (240 m_“‘l'<)(.0025m1)
Lm = 0.025m[10.81 %) = 0.270
F 35 4
__°_ = _—M = 0.0135
rnk (10.31 #240 Therefore the rate of heat transfer from one fin, 1 meter wide is:
W sinhf0.27} + 0.0135 cosh(0.27} = 6.493 _ ____._.___._.._.___ qf " K C0511 (0.27} + 0.0135 Sinhw27)
W
= 1.7929 —
qf 5 K In I rn2 of wall area there are 100 fins covering “100 tw = 100 (.0025 m)(1m) z 0.25 rrt2 of
wall area leaving 0.75 rn2 of bare wall. The total rate of heat transfer from the wall with
fins is the sum of the heat transfer from the bare wall and the heat transfer from 100
fins. qlot ; qbare + qun = FAlure as + qfin _ W a w _ W
q“ _ (35ﬁ)(0.75m )0, + 100 (1.792) 0‘?  205.3 0‘? The rate of heat transfer from the wall without fins is: gm = Rae, = [35—1] (1:112) 0' = 35.011;
The percent increase due to the addition of fins is:
205.3 — 35 x 100 = 486%
3:: % increase = COMMENTS This problem illustrates the dramatic increase in the rate of heat transfer that can be
achieved with properly designed fins. The assumption that the convective heat transfer coefficient is the same for the fins and
the wall is an oversimplification of the real situation, but does not affect the ﬁnal results
appreciably. In later chapters we will learn how to evaluate the heat transfer coefficient
from physical parameters and the geometry of the system. The tip of a soldering iron consists of a 0.6 cm 01') copper rod, 7.6 cm long. if the tip
must be 204°C, what is the required minimum temperature of the base and the heat
flow, in Btu’s per hour and in watts, into the base? Assume that h = 22.7 W'sz K
and 13,, = 21°_c. GIVEN  Tip of soldering iron consists of a copper rod
 Outside diameter (D) = 0.6 cm = .006 m  Length (L) = 7.6 cm = ‘076 rn  Temperature of the tip (TL) = 204°C  Heat transfer coefﬁcient (5} x 22.7 W/m2 K  Ambient temperature (T_) = 21°C FIND
(a) Minimum temperature of the base (T)
(b) Heat ﬂow into the base (:1) in Btu/h and W ASSUMPTIONS  The tip is in stead)»r state  The thermal conductivity of copper is uniform and constant, i.e. not a function of
temperature.  The copper tip can be treated as a fin SKETCH T i wait Tw=21°0 PROPERTIES AND CONSTANTS
From Appendix 2, Table 12: The thermal conductivity of copper (K)= 388 W/m K at 227%: SOLUTION (a). From Table 2.1, the temperature distribution for a fin with a uniform cross section
and convection from the tip is: e cosh[m(l,ex)]+(_)sinh[rn[L—x)] _ mk 8’ cosh (m L} + sinh [m L) where 6 = T—T_ and 6s = 8(0) = T'T_ 4 (22.71)
rn’K EP F100 45
Lm=L ..._=L = _.=0.G76m
kA kEDZ kD Lm = 0.076 m[6.25_1.] = 0.475 (388.3%) (.006 m) m a 227—?
_k = _._""i_ = 0.00935
:11 (6.25 Ln] [388.3%]
Evaluating the temperature at x = L:
9L = TLT cosh(0) + .0093651nhw) _  .___...._.—.... = 0.8932 0 T‘——T_ ‘ 005 H.475) + .00936 50010475) Solving for the base temperature:
T T o _ :
T=T+ L '=21°C+M
’ " .8932 .8932 = 226” C (b). To maintain steady state conditions, the rate of heat transfer into the base must
be equal to the rate of heat loss from the rod. From Table 2.1, the rate of heat loss is: sinh(m L) + c05h(m L} where M =‘lﬁ PkA 0‘ =1lE k‘T'EP (TsT) q! = M —..
cosh (m L) + sinh (m L} 2
M: ‘02.? W )(3ss£)“_(.006m)3 (226°C—21°C)=14.045W
m2K mK 4 sinh (.475) + .00936 cosh {.475} = 6 3 W
‘cosh _(.475')—+ .00'936" sinh (.475) ' 3.412 Btu/h c1I = 14.045W 6.3WE ) = 21.513111!!! COMMENTS
A small soldering iron such as this will typically be rated at 30W to allow for radiation
heat losses and more rapid heatup. One end of a 0.3 m iong steel rod is connected to a wall at 204°C. The other end is
connected to a wall which is maintained at 93°C. Air is blown across the rod so that
a heat transfer coefﬁcient of 1‘? W I m2 K is maintained over the entire surface. If the diameter of the rod is 5 cm and the temperature of the air is 38°C, what is the net
rate of heat loss to the air? GIVEN  A steel rod connected to walls at both ends
 Length of rod (L) = 0.3 m  Diameter of the rod (D) = 5 cm = .05m  Wall temperatures: Ts= 204°C TL: 93°C
 Heat transfer coefficient (11—0) = '17 W/m2 K
 Air temperature (T__) = 38°C FIND
The net rate of heat loss to the air (q.) ASSUMPTIONS  The wall temperatures are constant  The system is in steady state  The rod is 1% carbon steel  The thermal conductivity of the rod is uniform and not dependent on temperature
 One dimensional conduction along the rod SKETCH Tl ‘ €300 PROPERTIES AND CONSTANTS
Pram Appendix 2, Table 10.
The thermal conductivity of 1% carbon steel (k) : 43 W/m K [at 20°C) SOLUTION The rod can be idealized as a tin of uniform cross section with fixed temperatures at
both ends. From Table 2.1 the rate of heat loss is: cosh (m L)  (elf 95) = M
q‘ sinh(m L) where: 8L =TL T_ = 93°C  38°C = 55°C and 3' = "I"  T“ = 204°C  38°C =166°C E P ﬁcﬂD
Lm=L 31.4 =
le‘
4
1 2
M=JHPkA es= hﬁfnakef \(17.;K)14.(.05m}3(43% (166°C)=78.82W cosh (1.687) — E
166 = 78.82W __.._ = 74.4 w
q‘ sinh (1.687) COMMENTS
In a real situation the convective heat transfer coefficient will not be uniform over the circumference. It will be higher over the side facing the air stream. But because of the
hi h thermal conductivit the tem erature at any given section will be nearly uniform. Both ends of a 0.6 cm copper Ushaped rod, as shown in the accompanying sketch,
are rigidly affixed to a vertical wall the temperature of which is maintained at 93°C.
The developed length of the rod is 0.6 m and it is exposed to air at 38°C. The
combined radiative and convective heat transfer coefﬁcient for this system is 34 W/m2 K. {a} Calculate the temperature of the midpoint'of rbd. (b) What will
the rate of heat transfer from the rod be? '  .  3 '. . _ GIVEN  Ushaped copper rod rigidly affixed to a wall
 Diameter (D) = 0.6 Cm = 0.006 m  Developed length (L) = 0.6 rn  Wall temperature is constant at (T) = 93 C  Air temperature (T_) = 38°C  Heat transfer coefficient (h) = 34 W/m2 K FIND
(a) Temperature of the midpoint (TU)
(b) Rate of heat transfer from the rod (M) ASSUMPTIONS  The system is in steady state  Variation in the thermal conductivity of copper is negligible  The Ushaped rod can be approximated by a straight rod of equal length
 Uniform temperature across any section of the rod SKETCH \/A .. I’D391 I L L— 0.6m use area
= 0 Ar 05 re: D: 0.6 cm Z N )2 END PROPERTIES AND CONSTANTS From Appendix 2, Table 12, thermal conductivity of copper (k) = 396 W/m2 K at 64 C
SOLUTION By symmetry, the conduction thrctugh the rod at the center must be zero. Therefore, the rod can be thought of as two pin fins with insulated ends as shown in the sketch above. Continued . . . (a). From Table 2.1, the temperature distribution for a tin of uniform cross section
with an adiabatic tip is:
9 coshIm (L.r  x)] _ _ where: 9 :TT_, 8 =T T__ and Lt=1ength ofthe fin
{ia cosh(m L) ‘ ’ W 4(34—2—
m: .11“ = _ﬁ.’iP_= = w—M__=7.57i
1‘" HE D?) (396_}(.006rn) m
4 rnK
Evaluating the temperature of the tip of the pin fin:
BU.) cosh [m (LlLg] 1
3s _ cosh(m L) H eoshim L)
The length of the fin is half of the wire length {L[= 0.3 m):
9(L T L — T
8'): (TﬂT~= 1___=o.2os
8 _  coshi7.57':(0.3m) TH.) = .205 (T5 T_) + T“ = .205 {93°C 38°C) + 38°C 2 492°C The temperature at the tip of the fin is the temperature at the midpoint of the curved
rod (492°C). (b). From Table 2.1, the heat transfer from the fin is: qﬂn =Mtanh(m L) where: M=JHPkA 9.: He: 13114392) (Tl—T.) M = i “T: (34%)(396 TIT—K) (.006 m)3 (93°C 38°C) = 4.653W q“n = 4.553w tanh (7.57%) (03111) = 4.56W The rate of heat transfer from the curved rod is approximately twice the heat transfer
of the pin fin: qmd = 2% = 2 (4.56W) = 9.12 w mmmﬁxssm : A circumferential tin of rectangular cross section, 3.7 cm OD and 0.3 cm thick
surrounds a 2.5 cm diameter tube. The {in is constructed of mild steel. Air blowing
over the fin produces a heat transfer coefﬁcient of 28.4 W/rn2 K. It the temperatures
of the base of the fin and the air are 260”C and 38°C, respectively, calculate the heat
transfer rate from the ﬁn. GIVEN  A mild steel circumferential tin of rectangular cross section on a tube
 Tube diameter (DJ = 2.5 cm = .025 m  Fin outside diameter (DD = 3.7 cm = 0.037 m  Fin thickness (t) = 0.3 cm = 0.003 m  Heat transfer coefficient (hc) = 28.4 W/m’ l<  Fin base temperature (T) : 260°C  Air temperature (T_.) = 38°C FIND
The rate of heat transfer from the fin (qr...) ASSUMPTIONS  The system has reached steady state  The mild steel is 1% carbon steel  The thermal conductivity of the steel is uniform  Radial conduction only (temperature is uniform across the CTOSS section of the fin)  The heat transfer from the end of the fin can be accounted for by increasing the length
by half the thickness and assuming the end is insulated. SKETCH mt
v 0,3.7“. PROPERTIES AND CONSTANTS
Thermal conductivity of 1% carbon steeltk) = 43 W/m K at 20°C SOLUTION
The rate of heat transfer for the fin can be calculated using the fin efficiency determined
from the efﬁciency graph for this geometry, Figure 2.17. Continued . . . The length of a ﬁn (L) = (D,  DINZ = 0.006 m The parameters needed are: ri=,[2_)‘=0_‘[25m r°=%+L=O.125m +.006m=.0185m
3 w \1
a I 1. 2(28.4_. '’
(ra + g — ri)’ 7 =[.0815m + '00?“ .0125m “'K)
* ro " (43%)(003mX0185m —.0125m)
= 0.176
(r hi)” : 0.0185m + .0015m =16
° 2 ' .0125m From Figure 2.17, the fin efficiency for these parameters is: n{=98% The rate of heat transfer from the fin is: 2 qﬁn:n[EcAfin(TsT.) =ntE¢Zﬂ to 4"; ‘riz (T.'T.) qﬁn=(.98) (28.4%) 21t[(0.085m + 0.0015m)2 {0.0125m)’](260°c —38°C) =9.46 w A turbine blade 6.3 cm long, with crosssectional area A = 4.6 x 1134 m3 and
perimeter P 2 0.12 In, is made of stainless steel (k = 18 W/m K). The temperature
of the root, T“ is 482°C. The blade is exposed to a hot gas at 871°C, and the heat
transfer coefﬁcient 11 is 454 W/mz K. Determine the temperature of the blade tip and
the rate of heat ﬂow at the root of the blade. Assume that the tip is insulated. GIVEN  Stainless steel turbine blade  Length of blade (L) = 6.3 cm = 0.063 m » Crosssectional area (A) = 4.6 x 10* in2  Perimeter (P) = 0.12 m  Thermal conductivity (k) = '18 W/m K  Temperature of the root (T) = 482°C  Temperature of the hot gas (T) = B71“C  Heat transfer coefficient (h) = 454 W/rn2 K. FIND
(a) The temperature of the blade tip (T L)
(b) The rate of heat flow (q) at the root of the blade ASSUMPTIONS  Steady state conditions prevail  The thermal conductivity is uniform  The tip is insulated  The crosssection of the blade is uniform
~ One dimensional conduction SKETCH Ta = 51th A224 (A): 4.6.! .6" M’ ?emasrer. (P)= on». 13. 452%
LII 6.3 on Continued... SOLUTION (a) The temperature distribution in a fin of uniform crosssection with an insulated ti p,
from Table 2.1, is: 4541!(o.12m)
3 =_—C°5h [ma‘ "01 where:m = E}: = ________"‘ K =31.1i
a, cosh {m L) A islxmsxiwmz) m
3 =1“  T“ At the blade tip, x = L, therefore: 9:_ = TL“T. = cosh [m(0)l = 1 F; T‘—T_ cosh (m L) cosh (In L} T T a _ c. = , .. = 8710C + 482 C 871 C = 86an L “+ m L)
C05 111. cosh (81.1%)(0063mll (b) The rate of heat transfer from the fin is given by Table 2.1 to be: q = MtanhﬁnL} where M = ‘lhcPkA 8‘ M = 4S4%(0.12m)(18%)[4.6x104m2) (482°C 371°C) = 261W q = (261W)tanh[81.1.;.(0.063m)] = —261W (out of the blade) COMMENTS in a real situation the heat transfer coefficient will vary over the surface with the highest
value near the leading edge. But because of the hibh thermal conductivity of the blade
the temperature at any section will be esentially unifonn. ...
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This note was uploaded on 04/08/2008 for the course ENGR 302 taught by Professor Gadelhak during the Spring '08 term at VCU.
 Spring '08
 GadelHak

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