This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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/mK
AISI 1010 5feel) @ T T "I [sizimm
74‘} T(0)=500°C 7;=100 0C
h = 3300 W/mzK PROPERTIES: Table A1,AISI 1010 Steel (T = [500 +140]°C/2 : 320°C 2 600K): p = 7832 kg/m3, c : 559 J/kgK, k =48.8 W/rn K. ASSUMPTIONS: (1) Steel sphere is spacewise 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/rnK 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 _&[1] 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“ ,
5:500 w/mzK
—>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/kgK, k = 20 W/mK, R6 = 0.01 Q/m. ANALYSIS: Since 2 ‘ 4
_ hue/2) 500W/m K(2.5><10 m)
BIS—=—————:0.006<0.l
k 20W/mK the lumped capacitance method can be used. The problem has been analyzed in Example 1.3, and
without radiation the steadystate 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
2525—637 8000 kg/m3 x500 J/kgKx0.001 m
t= 8.31s. < COMMENTS: The time to reach steady state increases with increasing 0, ep and D and with
decreasing h. ...
View
Full
Document
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
 Rothamer
 Heat Transfer

Click to edit the document details