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Unformatted text preview: PR M KEQWN: The heat flux in a stainless steel (A181 304) pan containing boiling water may be as
high as 5.0 x 105 W/m2 and the temperature at the water/pan interface is 120°C. FIND ; The temperature gradient in the pan at the water/pan interface. CHEMATIC: W Onedimensional heat conduction. BMW From table A14 for stainless steel (M81 304) at 393.15 K:
k stainless steel: 16.48 WIm"C W
ci'=kg1]I=5x105[§]=—16.48[ﬁ5]%
5 105 1V— a g—E: = 30.34 x 103 <=<=<=<=<=<= MMENT DIS CU ION: The wmperature gradient is inﬂuenced by the surface temperature in the stainless Steel since the themal
conductivity is temperature dependenL It the thermal conductivity is evaluate at 400 K, k = 16.6 WIm"C and BTIBT] = {50.12 x 103 °Clm. This Difference would be negligible for most engineering
calculations. PR BLEM  E EEQWN : The heat flux on the 30° diagonal surface of the Bakelite wedge is 680 Btufhrft2 in the
normal direction. FIND; Determine the heat flux and temperature gradient in the x and y directions. SQHEMATIQ; ASS LIMPTIQNS: The heat flux is constant Over the entire wedged surface.
PR! IPERTIES ; From table 85 for Bakelite at 80°F:
k Bakelite = 0.8089 Btu/hrft»°F
9 = 30“
ANALYSIS:
o, the angle between nETA and x axis. is 90° + 6 = 120°
q"; = on (cos ¢) = 680 cos 120" =  340 '='<===
hrft hrft
q} = q'}, (sin 4») = 680 sin 120 ° = 588.9 =<==<=
hrft brft
C] y = 'kBakelite 5; and q x = ‘kBakelite a—x
 340 a ‘31; = M = 420.3 [—F] =‘=‘=‘=
3X13 akelile 03039 ft 
hrftv'F
588.9 ,, Bl = hr t't2 ___ _ 728IO[ F] <=<=<=<=
aYBakettte 413039 ft
hrft‘F C MMENT DISCU I N: The heat flux and the resulting thermal gradient can be broken down into its x and y vector components
with textbook trigonometry. PRQBLEM 82
KNQWN: (a) k] = 14 W/m“C T1 = 100°C Lb) Q” = 0.86 kW/m2 Ta, = 20°C
[(2 = 0.5 W/m’C T2 = 30°C ' k1 = 15 WIm"C h = 10 W/msz
k2 = 60 W/m°C
IND“ Draw the temperature proﬁles in the two composite section shown. Calculate the total
rate of heat transfer per unit area for case (a) SQHEMATIQ; lcm ASS UMPTIQNS; The thermal properties are constant and the tem is steady state.
The twodimensional conducﬁon from the ends is negligible, i.e. the materials are treated as inﬁnite slabs. P ER I ' Stated above in Known section.
QNALYSIs. a) q" ___ _kl (T1 1312) = _k2 (Tl1:210)
__w__ (100  T1—2)['C] _ _ty__ (T1~2  30) ['C]
14 [m'C] 0.01 [m]  0'5 [mt] 0.03 [m] Tm = 99.17 ['C] q" = 1152. cm:
b) q" = h (Tsurface ‘ a.) Tsurl'ace = 106cc :4: = _k 3_T= _k (Tsurrace  T12 ) ___ _15 (106  Ttz )i'C]
3x 3‘1 m‘C 0.02 [In]
0.02 [m] o a T1_2=86 th]. ____— +106 c =107.15 c
(L: J gem: T24 =860[Mg —°——"‘— +107.15°c =107.29°c
m 603%] an: TH —86({l‘i] E%':l + 107.29°c = 108.44°C
:c=<= QQMMENTS DISC Q SSIQN ; Thermal contact resistances between the materials has been neglected. PR BLE 11 KN WN: A Plain carbon steel pipe (5.25 cm inside diameter and 6.03 cm outside diameter) is
covered with 2 cm thick sixply asbestos corrugated paper. The temperature of the
steam inside is 150°C and that of the surrounding air is 25°C. h SW = 1500 Wlmz“C h ah : 5 Wlm2°C
l Estimate the outside surface temperature of the insulation and the rate of heat transfer
per meter of pipe length. '
SQﬂEMATIC:
2 cm thick. 0
sixvply asbestos T” = 25 C r
corrugated paper h m 39 mm
thick pipe
5. .
hls‘ t"
l
l I ASS ﬂMPTIONS ; The thermal properties are constant. The heat conduction from the ends
is negligible, i.e. the materials are treated as inﬁnite cylinders. The heat
l transfer is one dimensional and steady state.
EBQPERTIES: h swam =1500 Wlm2°C hat. =5 Wlm2°C from problem statement.
k carbon steel = 56.7 WIm"C at 400 K from Table Al4
k asbest conugalcd “Um“C, K, A‘15.4.
A N A LYS IS : cm; (Tsteam  T..) 1.;
z RlL ﬁ .
 w I I z RrL = ]  + 1n(routpipe’rinpiE) + ln(routineulJlfroutpiE) I
. 2 . 1; hip: kinsulation Ilitipiptrl'lsteam I'outiusul.'hair z RpL =_L { 1 +ln(3.015/2.625)+1n(5.015/3.015) 1 }= 1.592 I ‘~ 2 n 0026254500 56.7 0035 0.05.0155
1: ,  _ (150  25) _ E ¢=¢=¢=c=¢=¢:
. .. gash—1592 7s.523[m] (Ii/L = 2"it'l'tiutinsulfl‘air(_Toutside surface temp. ' Too) = 2'75'0»050 1 5'S(T0l115ldt: surface temp. ' 25) 5 Toutside surface temp. = 74.84%: ¢=<=<=<:¢=¢= W KNQWN; A plain carbon steel pipe with an inside diameter of 9 cm and a thickness of 0,5 cm
carries hot water which is at a temperature of 150°C. The pipe is insulated with 5 cm thick 85% magnesia. The inside heat transfer coefﬁcient is 7100 Wlm2'C. The
convection heat transfer coefﬁcient on the outside of the insulation is 57 Wlm2'C.
Scale builds up on .the inside of the pipe yielding a fouling factor of 0.0005 m2'CIW. I D ' Determine: (a) The overall heat transfer coefficient based on the outside surface
 area of the insulation, in watts per meter squareddegree
(b) The rate of heat transfer lost from the pipe per meter length if the outside air temperature is 20°C
SQHEMAIIE;
Tw= 150°C
hw==7100 WImZBC
rw= 0.0005 m2'C/W
W
EBQPERTIES; Thermal conductivity of pipe and insulation;
Pipe; kp = 55.83 WIm“C @423 K Table (Al4)
Insulation; 1:: = 0.055 WIm“C @365 K Table (A15.4)
ANALYSIS;
2 R (a) ‘ hwrtdyL marl. ampL 2rrkiL hondgL 2R‘=___l____+_0.9_005_+ 1n 10/9 + ln(20l10 +____1____
7100rr(0.09}l «(0.09M 2n55.33l 2n~0.055l 571r(0.20)—l 2 R‘ = {4.981 x 10" + 1.768 x 10" + 3.004 x 10" + 2.006 + 2.792 x 10'2} 2 R.=2.036[°%] ; UoA3=EL{ ; Uo=—1——— where A3=1td3(l)
l U =0.782[ 2v ] ¢=¢=<=¢==
m 'C 0 = ____l___
n(O.2)(2.036) (b) (1/1. = UJ'A3'AT =(0.782)(1t)(0.2)(150  20): 63.87 since L = 1 meter ¢=¢=<=¢=<= PRQBLEM Blﬁ KN WN: An electrical heating element is shrunk in a hollow cylinder of amorphous carbon, k =
1.6 WIm“C, Fig. P816. The outside of the carbon is in contact with 20 °C air. The convection heat transfer coefﬁcient is 40 Wlm2°C. FIND: Determine the maximum electrical heating (Watts per meter of cylinder length) which
can be applied if the maximum allowable temperature of the carbon is 200°C. The
resistance to the transfer of heat within the heating element may be neglected. SQHEMATIC: ASSUMPTIONS;
PR! QPE RTIES: Carbon kc = 1.6 W/m"C from problem statement. ANALYSIS; T.  To. in [dgidﬂ 1
. Where 2 R. — zﬂ'kc‘L + humdTL . LetL— 1 meter _ In (0.02/0.01) _ _ Z Rt _——————2'n.“ ASH + ———1————40_M0'02H _ 0.0689 + 0.3979 _ 0.4668 w 3 = ' = 385.6 Wlm (since L = 1 meter) <=<=<=<=<=<=<= PR BLE 1 KN! [WE ; The ceiling of a house is composed of 1.5 cm plaster board. From a typical winter day,
the r00m temperature is 21"C and the air temperature in the attic is 10”C. IND ' Determine the amount of heat saved if the attic floor is covered with 1.5 cm plywood
_ and the 9 cm space between this ﬂoor and plaster board is ﬁlled with a glass ﬁber blanket with a density of 16 kg/m3, Fig. P8—19. The convection ﬁlm convection ﬁlm
coefficient are both 40 W/m2'C. . SCHEMATIQ: See below:
ASSUMPIIQES; BRQPERTIES: Thermal Properties:
Plaster board kpa = 0.17 W/m° Table (A~15.2) C @ 300 K
Plywood kpy = 0.12 WIm“C @ 300 K Table (AlS.2)
Fiberglass kf = 0.046 W/m"C @ 300 K Table (A—15.3)  .___...__ ._ __ _p__.—— ANALY§I§:
(a) Plaster board alone; Tn: 40°C h...= 40 Win? .°C 
A" = 15 cm BL in EL
1 pa 0 Ti = 21°C h; = 40 W/m2°C _._A_T__ =_(2_1;(;1_0))__= 31 _ = 11 = .\_V_
i M _1_ L+Q.D_Li+_t_ 0.025 +0.033+0.025 0.138 224‘6Ln2] ‘=‘=‘=‘=
hi +kpa+hm 40 0.17 40 i:
A (b) For insulated ceiling,
1;. = 10"C 11.; 40 WIm2 °C mm w ''V'‘<V='>_
‘t, ::::;'3;g¢3_0W;W9m+ it: ééitéliié'iiiaﬂon 2:1:
. § asseé‘éséﬁgﬂwg o kpa kf kpg It???" “No.63: ' .. (AT J_+A_xt+é£2+é£1+J_
ho kpa ho ‘ .0.
l J_+Q.QL5_+.Q.Q9_+Q.D_L1+_L 2.219 ' m2
40 0.17 0.046 0.12 40 Savings A—q = 224.63  13.97 = 210.7 [31] =<=<=<=¢=<=
A m2 ERQ LEM 323 KJQW_N; Hot water at 98°C ﬂows through commercial bronze tube having an inside diameter of
2 cm. The tube is extruded and has the cross sectional proﬁle shown in Figure P 823.
The outside diameter of the ﬁnned tube is 4.8 cm and the ﬁns are 1 cm long and 2 mm
thick. The convection heat transfer coefﬁcient on the water side of the tube is 1200 W/m2 °C. The ﬁnned tube is surrounded by 15 “C air and the convection heat transfer
coefﬁcient is 5 W/m2 “C. ' E IND ; Determine the rate of heat transfer per meter of tube length. SQEEMATI; 2; ASS UMPTIQNS: The thermal properties are constant and the problem is steady state. EBQPERTIES: k bronze = 52 WIm"C @ 300 K, from table A14.
A NALYS IS ; The problem is very similar to the previous two problems and example
84 in the text. V For 1 meter axial length; L; Ab = it'd — 12t = “It0.028  120.002 = 0.064 m2 _ ln (rout/fin) _ln (2.8/2.0) _ _3 _ l _ I _ E
Rp—z'ﬂ‘kbronZe‘L _ 2.1r52l _l’03X10 w R” Thump!) ‘5.0_064 '3125 [w] A. = L . thickness = 10.002: 0.002 [ml] P = 2 (L) = 2.0 [m] = '—= ——"—=9.806
I.c=L+AC/P=0.01+—03082[m]=0,011[m] m VioAc .V 520.002 R = _ __1_ .._= =0.760
‘ NtlhPkAc tanh [xiLC 121520520002 tanh (9.80.011)
_ RbRf _ 3.1250760 _ .
R“? ‘Rb + R, ‘ 3.125 + 0.760 “0'6” [ CM] =_1__ = 3 = n
2 R MAin + RP +Req momma02 + 0.00103 + 0.611 0.625 [ CIW] =Imter'_Tair. —m = 132.3[31] c=¢=¢=¢=
m .g _
“L 2R 0.625['CIW] h
I .
l i
PRQB LEM 339 I
KNQ E N ; The thermocouple shown in the sketch, Fig. P8—30, is to be used as the sensing _ _
element in a temperature control unit The control unit is set to take corrective action if
the fluid temperature is equal to or greater than 150°C. The normal operating fluid
temperature is 100 “C. A malfunction in the system results in an instantaneous increase
in the fluid temperature to 200 °C. The thennocouple diameter is 0.5 mm.
F D ' How long will it take the thermocouple to sense that the control unit must take i
corrective action if the convective heat transfer coefﬁcient is 500 W/m2°C?
SQHE M ATIQ ; a
d = 0.5 mm ‘
> i
To = a
1;: 200°C 2 i
h = 500 W/m °C 5
ASSQMPTIQNS; 1
i
ERQPE RTIE§: The average thermocouple properties from the problem statement are: k = 23 W/m'"C p = 8920 kglm3 c = 334 Ifkg“C / r
g
N Y I  " 1
Check Bi number: 3‘ ‘ k ‘ L‘ A J
4 n r3 r Q (0.0005) _ I
=————=—= = =8.33X10 m I
3 (4 m2) 3 6 6 S[ 1 1'
_ 5
Bi=w=1311x103 .l 23
Since Bi < 0.10 use the LumpedCapacity method. T‘ To. h»t
To  11,, ‘ exp [pcLc J ; and To = 100°C; T,” = 200°C; T = 150°C j M _ ex _____'10_9L___ —   !
_ P —exp[ 1.752 t] 
100 — 200 [(8920}(384)(8.33 x 105) '.
0.5 = exp [0.693]: 1.752t = 0.693 or t = 0.396 sec. <=<=<=<=<=<= PRQB LEM 845 K EQWE: Most foods are blanched in a water bath before they can be processed for canning. In
the blanching process the minimum temperature of the food must be high encugh to
destroy the undesirable enzymes. The speciﬁc produce of concern is mushrooms
which have an average spherical shape of 2 cm in diameter. The thermodynamic
propenies of the mushrooms are approximately the same as those for water and the convection heat transfer coefﬁcient is 1400 WImZ‘C. EIND ; If the minimum allowable temperature is 75°C. determine the minimum amount of time
that the mushrooms must remain in the water bath. The initial temperature of the
mushroom entering the water is 10°C and the bath temperature is 95°C. HEMATIC: /
Evaluate thermal Tof 10F
. T—%C
properties at: u T": 75.C
(75+95)/2 = 85 C h = 1400 Wimzrc Diameter = 2 cm ASS UMPTIQNS; Treat the mushroom as a sphere with R = 1 cm (0.01 m). Assume center must reach 75°C. PRQPERTIES: From Table A9 @ 85 "C
p = 968 kg/m3; c = 4.201 x 103 JIkg‘C; k = 0.670 WIm“C ANALYSIS; a = JL = _.Mla_._ = 1.655 x 101 [W2]
P'c 9684201 x 103 J . . 7.
Bi :3: 1400 (0'01 = 20.8 ; Fo =H=M1= 1.655x 101:
1: 0.673 R2 (0.01)2 Use Figure 826: BizFo;100 ; (20.8)2(1.655x 103t)=100 t: 139.6 seconds ¢=¢=¢=¢=¢=¢=¢=¢= ...
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