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# ch05

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```heat fluxes are nearly the same
at t = 60 days? Are these results consistent with the temperature distributions? What happens to the
heat flux values for times much greater than 60 days? Use your IHT model to confirm your
explanation.
(2) Selected portions of the IHT code for the semi-infinite medium model are shown below.
// Models | Transient Conduction | Semi-infinite Solid | Constant temperature Ts
/* Model: Semi-infinite solid, initially with a uniform temperature T(x,0) = Ti, suddenly subjected to
prescribed surface boundary conditions. */
// The temperature distribution (Tx,t) is
T_xt = T_xt_semi_CST(x,alpha,t,Ts,Ti) // Eq 5.55
// The heat flux in the x direction is
q''_xt = qdprime_xt_semi_CST(x,alpha,t,Ts,Ti,k) //Eq 5.56
// Input parameters
/* The independent variables for this system and their assigned numerical values are */
Ti = 20 // initial temperature, C
k = 0.52 // thermal conductivity, W/m.K; base case condition
alpha = 1.38e-7 // thermal diffusivity, m^2/s; base case
//alpha = 1.0e-7
//alpha = 3.0e-7
// Calculating at x-location and time t,
x = 0 // m, surface
// x = 0.68 // m, burial depth
t = t_day * 24 * 3600 // seconds to days time covnersion
//t_day = 60
//t_day = 1
//t_day = 5
//t_day = 10
//t_day = 30
t_day = 20
// Surface condition: constant surface temperature
Ts = -15 // surface temperature, K
PROBLEM 5.74
KNOWN: Tile-iron, 254 mm to a side, at 150°C is suddenly brought into contact with tile over a
subflooring material initially at Ti = 25°C with prescribed thermophysical properties. Tile adhesive
softens in 2 minutes at 50°C, but deteriorates above 120°C.
FIND: (a) Time required to lift a tile after being heated by the tile-iron and whether adhesive
temperature exceeds 120°C, (2) How much energy has been removed from the tile-iron during the time it
has taken to lift the tile.
SCHEMATIC:
ASSUMPTIONS: (1) Tile and subflooring have same thermophysical properties, (2) Thickness of
adhesive is negligible compared to that of tile, (3) Tile-subflooring behaves as semi-infinite solid
experiencing one-dimensional transient conduction.
PROPERTIES: Tile-subflooring (given): k = 0.15 W/m\u22c5K, \u3c1cp = 1.5 × 106 J/m3\u22c5K, \u3b1 = k/\u3c1cp = 1.00 ×
10-7 m2/s.
ANALYSIS: (a) The tile-subflooring can be approximated as a semi-infinite solid, initially at a uniform
temperature Ti = 25°C, experiencing a sudden change in surface temperature Ts = T(0,t) = 150°C. This
corresponds to Case 1, Figure 5.7. The time required to heat the adhesive (xo = 4 mm) to 50°C follows
from Eq. 5.57
( )
( )
o o s o
1/ 2i s o
T x , t T x
erf
T T 2 t\u3b1
\uf8eb \uf8f6
\u2212 \uf8ec \uf8f7= \uf8ec \uf8f7\u2212 \uf8ed \uf8f8
( )1/ 27 2 o
50 150 0.004 m
erf
25 150
2 1.00 10 m s t\u2212
\uf8eb \uf8f6\uf8ec \uf8f7
\u2212
= \uf8ec \uf8f7
\u2212 \uf8ec \uf8f7× ×\uf8ec \uf8f7\uf8ed \uf8f8
( )1/ 2o0.80 erf 6.325t\u2212=
to = 48.7s = 0.81 min
using error function values from Table B.2. Since the softening time, \u2206ts, for the adhesive is 2 minutes,
the time to lift the tile is
( )o st t t 0.81 2.0 min 2.81min= + \u2206 = + =
&quot;
. <
To determine whether the adhesive temperature has exceeded 120°C, calculate its temperature at t
&quot;
=
2.81 min; that is, find T(xo, t
&quot;
)
( )
( )
o
1/ 27 2
T x , t 150 0.004m
erf
25 150
2 1.0 10 m s 2.81 60s\u2212
\uf8eb \uf8f6\uf8ec \uf8f7
\u2212
= \uf8ec \uf8f7
\u2212 \uf8ec \uf8f7× × ×\uf8ec \uf8f7\uf8ed \uf8f8
&quot;
Continued...
PROBLEM 5.74 (Cont.)
( ) ( )oT x , t 150 125erf 0.4880 125 0.5098\u2212 = \u2212 = ×&quot;
( )oT x , t 86 C= \$&quot; <
Since T(xo, t
&quot;
) < 120°C, the adhesive will not deteriorate.
(b) The energy required to heat a tile to the lift-off condition is
( )t x s0Q q 0, t A dt\u2032\u2032= \u22c5\u222b &quot; .
Using Eq. 5.58 for the surface heat flux \u2032\u2032qs (t) = \u2032\u2032qx (0,t), find
( )
( )
( )
( )
t s i s i 1/ 2
s s1/ 2 1/ 2 1/ 20
k T T 2k T TdtQ A A t
t\u3c0\u3b1 \u3c0\u3b1
\u2212 \u2212
= =\u222b &quot;
&quot;
( )
( ) ( ) ( )
2 1/ 2
1/ 27 2
2 0.15W m K 150 25 CQ 0.254 m 2.81 60s 56kJ
1.00 10 m s\u3c0 \u2212
× \u22c5 \u2212
= × × × =
× ×
\$
<
COMMENTS: (1) Increasing the tile-iron temperature would decrease the time required to soften the
(2) From the energy calculation of part (b) we can estimate the size of an electrical heater, if operating
continuously during the 2.81 min period, to maintain the tile-iron at a near constant temperature. The
power required is
P Q t 56kJ 2.81 60s 330 W= = × =
&quot;
.
Of course a much larger electrical heater would be required to initially heat the tile-iron up to the
operating temperature in a reasonable period of time.
PROBLEM 5.75
KNOWN: Heat flux gage of prescribed thickness and thermophysical properties (\u3c1, cp, k)
initially at a uniform temperature, Ti, is exposed to a sudden change in surface temperature
T(0,t) = Ts.
FIND: Relationships for time constant of gage when (a) backside of gage is insulated and (b)
gage is imbedded in semi-infinite solid having the same thermophysical properties. Compare
with equation given by manufacturer, ( )2 2p4d c / k.\u3c4 \u3c1 \u3c0=
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.
ANALYSIS: The time constant \u3c4 is defined as the time required for the gage to indicate,
following a sudden step change, a signal which is 63.2% that of the steady-state value. The
manufacturer\u2019s relationship for the time constant
( )2 2p4d c / k\u3c4 \u3c1 \u3c0=
can be written in terms of the Fourier number as
2 2 2p
k 4Fo 0.4053.
cd d
\u3b1\u3c4 \u3c4
\u3c1 \u3c0
= = \u22c5 = =
The Fourier number can be determined for the two different installations.
(a)

For the gage having its backside insulated, the
surface and backside temperatures are Ts and
T(0,t), respectively. From the sketch it follows
that
( ) s
o
i s
T 0, T
0.368.
T T
\u3c4
\u3b8\u2217 \u2212= =
\u2212
From Eq. 5.41,
( )2o 1 10.368 C exp Fo\u3b8 \u3b6\u2217 = = \u2212
Using Table 5.1 with Bi = 100 (as the best approximation for Bi = hd/k \u2192 \u221e, corresponding
to sudden surface temperature change with h \u2192 \u221e), 1\u3b6 = 1.5552 rad and C1 = 1.2731. Hence,
2
a0.368 1.2731exp( 1.5552 Fo )= \u2212 ×
aFo 0.513.= <
Continued \u2026..
PROBLEM 5.75 (Cont.)
(b) For the gage imbedded in a semi-infinite
medium having the same thermophysical
properties, Table 5.7 (case 1) and Eq. 5.57 yield
( ) ( )
( )
1/ 2s
i s
1/ 2
T x, T
0.368 erf d/2
T T
d/2 0.3972
\u3c4
\u3b1\u3c4
\u3b1\u3c4
\u2212 \uf8ee \uf8f9
= = \uf8ef \uf8fa\uf8f0 \uf8fb\u2212
=
( )b 2 2
1Fo 1.585
d 2 0.3972
\u3b1\u3c4
= = =
×
<
COMMENTS: Both models predict higher values of Fo than that suggested by the
manufacturer. It is understandable why Fob > Foa since for (b) the gage is thermally
connected to an infinite medium, while for (a) it is isolated. From this analysis we conclude
that the gage\u2019s transient response will depend upon the manner in which it is applied to the
surface or object.
PROBLEM 5.76
KNOWN: Procedure for measuring convection heat transfer coefficient, which involves
melting of a surface coating.
FIND: Melting point of coating for prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in solid rod (negligible losses to
insulation), (2) Rod approximated as semi-infinite medium, (3) Negligible surface radiation,
(4) Constant properties, (5) Negligible thermal resistance of coating.
PROPERTIES: Copper rod (Given): k = 400 W/m\u22c5K, \u3b1 = 10-4 m2/s.
ANALYSIS: Problem corresponds to transient conduction in a semi-infinite solid. Themal
response is given by
( )
( ) ( )
( )1/ 22i
1/ 2 2 1/ 2i
T x,t T h tx hx h t x
erfc exp erfc .
T T k kk2 t 2 t
\u3b1\u3b1
\u3b1 \u3b1\u221e
\uf8ee \uf8f9\uf8eb \uf8f6 \uf8eb \uf8f6\uf8ee \uf8f9\uf8eb \uf8f6
\u2212 \uf8ef \uf8fa\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ef \uf8fa\uf8ec \uf8f7= \u2212 + +\uf8ec \uf8f7 \uf8ef \uf8fa\uf8ec \uf8f7 \uf8ec \uf8f7\u2212 \uf8ef \uf8fa\uf8ed \uf8f8\uf8f0 \uf8fb\uf8ed \uf8f8 \uf8ed \uf8f8\uf8f0 \uf8fb
For x = 0, erfc(0) = 1 and T(x,t) = T(0,t) = Ts. Hence
( )1/ 22s i
2i
h tT T h t1 exp erfc
T T kk
\u3b1\u3b1
\u221e
\uf8eb \uf8f6\uf8eb \uf8f6
\u2212 \uf8ec \uf8f7\uf8ec \uf8f7= \u2212 \uf8ec \uf8f7 \uf8ec \uf8f7\u2212 \uf8ed \uf8f8 \uf8ed \uf8f8
with
( ) ( )1/ 22 -4 21/ 2m 200 W/m```