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Unformatted text preview: PROBLEM 3.38 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface. Inner and
outer wall temperatures. Temperature of ﬂuid adjoining outer wall. FIND: Effect of wall thermal conductivity, thermal contact resistance, and convection coefﬁcient on
total heater power and heat rates to outer ﬂuid and inner surface. SCHEMATIC:
T0 = 25 0C Electrical heater, (1'
W 2
ro=75mm 10éhé1000W/m «K
T. : no: ‘1 0 OC ri=25 mm ’ 1éké200 WlmK OéR'tcéDJ mK/W 0 5. 7e
qf M R't,c (1l21t rah) ‘70
275k q" ASSUMPTIONS: (1) Steady—state conditions, (2) One—dimensional conduction, (3} Constant properties,
(4) Negligible temperature drop across heater, (5) Negligible radiation. ANALYSIS: Applying an energy balance to a control surface about the heater, d=d+%
q, T0 —T1 + T0 —Tw
111(r0/r1)+ R, (l/erroh)
Zak t’c Selecting nominal values of k = 10 W/mK, R'LC = 0.01 mKIW and h = 100 W/mZK, the following parametric variations are obtained 350° ' —r l "Ilﬂ—F"‘fg'—1 3000
3000 ~ 7 "I"? v
\ "7 2500
2500 ‘ r  P;
E ' , “ x a ‘“
“g 2000 / r ., .  .< g 0“— +>~ﬁ
v 7, 10.,
u: . ‘ . . . " Tl; ‘
In: 1500 I e’ . E
I 1000 r , _ . _ I ‘
500 7  ‘  , v1 ’ "
/ fAFlr ‘ p it
0 —‘ , —y— "r— a z i l , r‘r—l
0 50 100 150 200 D 0.02 0.04 0.06 0.08 0.1
Thermal conductivity, lefm K) Contact resistance, RlclmK/W)
E— q; —El— q.
ﬂ" 0 H3" 0
, E, qo e9" qo Continued... PROBLEM 3.38 (Cont) 20000 i
I
16000
’E‘
 12000
E
2
S
a 3000
a;
I
4000
o i t t
0 200 400 600 800
Convection coefficient, htW/m"2.K)
—FJ— qi
’5 q
+ W For a prescribed value of h, (12) is ﬁxed, while qfl , and hence q' , increase and decrease, respectively, with increasing k and R'LC. These trends are attributable to the effects of k and R'LC on the total
(conduction plus contact) resistance separating the heater from the inner surface. For ﬁxed k and RLC , qi is ﬁxed, while q'0 , and hence q' , increase with increasing h due to a reduction in the convection
resistance.
COMMENTS: For the prescribed nominal values of k, R'LC and h, the electric power requirement is q' z 2377 W/rn. To maintain the prescribed heater temperature, q' would increase with any changes
which reduce the conduction, contact and/‘or convection resistances. PROBLEM 3.41 KNOWN: Thin electrical heater fitted between two concentric cylinders, the outer surface of which
experiences convection. FIND: (a) Electrical power required to maintain outer surface at a speciﬁed temperature, (b)
Temperature at the center SCHEMATIC: AJkA=O.I5ﬂ%/( r, :20?” m 73in electrical hea‘fcr '3 = 4 0m 7" (r: ) TearI 0C
a B,k3=1.5ﬂ,ﬂ_/K h=505W/m2'K ASSUMPTIONS: (1) Onedimensional, radial conduction, (2) Steadystate conditions, (3) Heater
element has negligible thickness, (4) Negligible contact resistance between cylinders and heater, (5)
Constant properties, (6) No generation. ANALYSIS: (3) Perform an energy balance on the
composite system to determine the power required to maintain T(r2) = TS = 5°C. Ein _ Ebut + Egen : Est +qelec ‘qconv : 0 Using Newton’s law of cooling,
qelec : qconv = h‘ 27: r2 (Ts ‘Too) qelec=50 EV x27z(0.040m) [5—(~15):0C:251W/m. <
m K (b) From a control volume about Cylinder A, we recognize that the cylinder must be isothermal, that
is, T(0) = Tm).
Represent Cylinder B by a thermal circuit:
T0,) 7;
W0 qr:T(rl),_Ts
R 3 RB For the cylinder, from Eq. 3.28,
Rig :lnr2/r1/27r kB giving
T(r1)=TS +q'R'B : 5”C+253.l—“—]——hlﬁO/L— = 235°C
m 27r><l.5 W/mK
Hence, T(0) = T(r1) : 235°C. < Note that k A has no influence on the temperature T(0). ...
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 Fall '08
 Rothamer
 Heat Transfer

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