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Unformatted text preview: PROBLEM 5.7 KNOWN: Solid steel sphere (A181 1010), coated with dielectric layer of prescribed thickness and
thermal conductivity, Coated sphere, initially at uniform temperature) is suddenly quenched in an oil
bath. FIND: Time required for sphere to reach 140°C.
SCH E MAT 1C: Die/cc fric A layer
Sphere, D: 30017:»; k = 0.04W/m-K
AISI 1010 5feel) @ T T "I [sizimm
74‘} T(0)=500°C 7;=100 0C
h = 3300 W/mz-K PROPERTIES: Table A-1,AISI 1010 Steel (T = [500 +140]°C/2 : 320°C 2 600K): p = 7832 kg/m3, c : 559 J/kg-K, k =48.8 W/rn- K. ASSUMPTIONS: (1) Steel sphere is space-wise isothermal, (2) Dielectric layer has negligible
thermal capacitance compared to steel sphere, (3) Layer is thin compared to radius of sphere, (4)
Constant properties. ANALYSIS: The thermal resistance to heat transfer from the sphere is due to the dielectric layer and
the convection coefﬁcient. That is, 2
R” g 1 0002‘“ + 1 —(0.050+0.0003):0.0503m K k ' h 0.04W/rn-K 3300 W/m2~K W ; or in terms of an overall coefﬁcient, U : l/R” : 19.88 W/m2 ~ K. The effective Biot number is
_ ULC # U(Ib/3) #1988 W/m2 -K><(0.300/6)m
k k 48.8 W/m~ K where the characteristic length is LC 2 r0/3 for the sphere, Since Big < 0.1, the lumped capacitance
approach is applicable. Hence, Eq 5.5 is appropriate with h replaced by U, hall} 9i _& my; U AS “793’ U AS T(t)eTw
Substituting numerical values with (V/AS) = r0/3 : D/6,
7 7832 kg/m3x559 J/kgAK [0.300111] 1n(500 400)” c
19.88 W/mZK 6 (140—100)“ C Bie 70.0204 1 t: 25,3583 : 7.04h. < COMMENTS: (1) Note from calculation of R" that the resistance of the dielectric layer dominates
and therefore nearly all the temperature drop occurs across the layer. PROBLEM 5.17 KNOWN: Diameter, resistance and current flow for a wire. Convection coefﬁcient and temperature
of surrounding oil. FIND: Steadynstate temperature of the wire. Time for the wire temperature to come within 1°C of
its steady—state value. SCHEMATIC :
. W“ ,
—>1‘.—.100A ’4 |_> 1ETA/the,.D=1mm 5‘ X ASSUMPTIONS: (1) Constant properties, (2) Wire temperature is independent of x.
i . 3 ,
PROPERTIES: Wire (given): p = 8000 kg/m , cp = 500 J/kg-K, k = 20 W/m-K, R6 = 0.01 Q/m. ANALYSIS: Since 2 ‘ -4
_ hue/2) 500W/m -K(2.5><10 m)
k 20W/m-K the lumped capacitance method can be used. The problem has been analyzed in Example 1.3, and
without radiation the steady-state temperature is given by 71: Dh(T—Too) :12Rg.
Hence 2 I 2 l R 0 100A 0019/ m e = 25 C +——-———————( ) 2
75 Dh 7:(0.001 rn)500 W/rn -K
With no radiation, the transient thermal response of the wire is governed by the expression (Example 1.3) T=T°°+ =88.7°C. < 2 I
dT I R 4h
_=___—32___ (T_']; dt pep (73D /4) :0ch With "f = T1 = 25°C at t = 0, the solution is T—Tm—(izag/noh) 4h
——-=exp H t . Ti—Tm—(izRg/n Dh) 9ch Substituting numerical values, ﬁnd 87.7—25—63.7 4x500 W/m2 -K t
25-25—63-7 8000 kg/m3 x500 J/kg-Kx0.001 m
t= 8.31s. < COMMENTS: The time to reach steady state increases with increasing 0, ep and D and with
decreasing h. ...
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at Wisconsin.
- Fall '08
- Heat Transfer