ch04
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ch04


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315 C
4k 4 400 W / m K
×
= + = + ° = °
\u22c5
\ufffd
<
(c) Subject to the prescribed constraints, the following results have been obtained for parameter
variations corresponding to 16 \u2264 N \u2264 40, 2 \u2264 t \u2264 8 mm and 20 \u2264 L \u2264 40 mm.
 N t(mm) L(mm) f\u3b7 ( )q W / m\u2032
16 4 20 0.86 74,400
16 8 20 0.91 77,000
28 4 20 0.86 107,900
32 3 20 0.83 115,200
40 2 20 0.78 127,800
40 2 40 0.51 151,300
Clearly there is little benefit to simply increasing t, since there is no change in tA\u2032 and only a
marginal increase in f .\u3b7 However, due to an attendant increase in tA ,\u2032 there is significant benefit to
increasing N for fixed t (no change in f\u3b7 ) and additional benefit in concurrently increasing N while
decreasing t. In this case the effect of increasing tA\u2032 exceeds that of decreasing f .\u3b7 The same is
true for increasing L, although there is an upper limit at which diminishing returns would be reached.
The upper limit to L could also be influenced by manufacturing constraints.
COMMENTS: Without the sleeve, the heat rate would be ( )sq Dh T T 7850 W / m,\u3c0 \u221e\u2032 = \u2212 =
which is well below that achieved by using the increased surface area afforded by the sleeve.
PROBLEM 4.34
KNOWN: Dimensions of chip array. Conductivity of substrate. Convection conditions. Contact
resistance. Expression for resistance of spreader plate. Maximum chip temperature.
FIND: Maximum chip heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Constant thermal conductivity, (3) Negligible radiation, (4)
All heat transfer is by convection from the chip and the substrate surface (negligible heat transfer
from bottom or sides of substrate).
ANALYSIS: From the thermal circuit,
( )
h h
h sp
h,cnv t,c sp,cnvt sp
T T T Tq q q
R R R R
\u221e \u221e\u2212 \u2212
= + = +
+ +
( ) ( ) ( ) 111 22 2h,cnv s,n hR h A hL 100 W / m K 0.005m 400K / W\u2212\u2212\u2212 \uf8ee \uf8f9= = = \u22c5 =\uf8ef \uf8fa\uf8f0 \uf8fb
( ) ( )( )
3 5 7
r r r r
t sp
sub h
1 1.410 A 0.344 A 0.043A 0.034 A 1 0.353 0.005 0 0
R 0.408K / W
4 k L 4 80 W / m K 0.005m
\u2212 + + + \u2212 + + +
= = =
\u22c5
( )
4 2
t,c
t,c 2 2
h
R 0.5 10 m K / WR 2.000 K / W
L 0.005m
\u2212\u2032\u2032 × \u22c5
= = =
( ) ( ) 11 2 2 2sp,cnv sub s,hR h A A 100 W / m K 0.010m 0.005m 133.3K / W\u2212\u2212= \u2212 = \u22c5 \u2212 =\uf8ee \uf8f9\uf8ee \uf8f9\uf8f0 \uf8fb \uf8ef \uf8fa\uf8f0 \uf8fb
( )
70 C 70 Cq 0.18W 0.52 W 0.70 W
400K / W 0.408 2 133.3 K / W
° °
= + = + =
+ +
<
COMMENTS: (1) The thermal resistances of the substrate and the chip/substrate interface are much
less than the substrate convection resistance. Hence, the heat rate is increased almost in proportion to
the additional surface area afforded by the substrate. An increase in the spacing between chips (Sh)
would increase q correspondingly.
(2) In the limit ( )r t spA 0, R\u2192 reduces to 1/ 2 sub h2 k D\u3c0 for a circular heat source and sub h4k L
for a square source.
PROBLEM 4.35
KNOWN: Internal corner of a two-dimensional system with prescribed convection boundary
conditions.
FIND: Finite-difference equations for these situations: (a) Horizontal boundary is perfectly insulated
and vertical boundary is subjected to a convection process (T¥,h), (b) Both boundaries are perfectly
insulated; compare result with Eq. 4.45.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties, (4) No internal generation.
ANALYSIS: Consider the nodal network shown above and also as Case 2, Table 4.2. Having
defined the control volume \u2013 the shaded area of unit thickness normal to the page \u2013 next identify the
heat transfer processes. Finally, perform an energy balance wherein the processes are expressed
using appropriate rate equations.
(a) With the horizontal boundary insulated and the vertical boundary subjected to a convection process,
the energy balance results in the following finite-difference equation:
in out 1 2 3 4 5 6E E 0 q q q q q q 0- = + + + + + =& &
( ) ( )m-1,n m,n m,n-1 m,n m,nT T T Tx yk y 1 k 1 h 1 T Tx 2 y 2 ¥
- -D Dé ù é ùD × + × + × -ê ú ê úD Dë û ë û
( )m+1,n m,n m,n+1 m,nT T T Ty 0 k 1 k x 1 0.
2 x y
- -Dé ù+ + × + D × =ê ú D Dë û
Letting Dx = Dy, and regrouping, find
( ) ( )m-1,n m,n+1 m+1,n m,n-1 m,nh x h x2 T T T T T 6 T 0.k k¥
D Dé ù+ + + + - + =ê úë û
<
(b) With both boundaries insulated, the energy balance would have q3 = q4 = 0. The same result would
be obtained by letting h = 0 in the previous result. Hence,
( ) ( )m-1,n m,n+1 m+1,n m,n-1 m,n2 T T T T 6 T 0.+ + + - = <
Note that this expression compares exactly with Eq. 4.45 when h = 0, which corresponds to insulated
boundaries.
PROBLEM 4.36
KNOWN: Plane surface of two-dimensional system.
FIND: The finite-difference equation for nodal point on this boundary when (a) insulated; compare
result with Eq. 4.46, and when (b) subjected to a constant heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction with no generation, (2) Constant
properties, (3) Boundary is adiabatic.
ANALYSIS: (a) Performing an energy balance on the control volume, (Dx/2)×Dy, and using the
conduction rate equation, it follows that
in out 1 2 3E E 0 q q q 0- = + + =& & (1,2)
( ) m-1,n m,n m,n-1 m,n m,n+1 m,nT T T T T Tx xk y 1 k 1 k 1 0.
x 2 y 2 y
- - -D Dé ù é ùD × + × + × =ê ú ê úD D Dë û ë û
(3)
Note that there is no heat rate across the control volume surface at the insulated boundary.
Recognizing that Dx =Dy, the above expression reduces to the form
m-1,n m,n-1 m,n+1 m,n2T T T 4T 0.+ + - = (4) <
The Eq. 4.46 of Table 4.3 considers the same configuration but with the boundary subjected to a
convection process. That is,
( )m-1,n m,n-1 m,n+1 m,n2h x h x2T T T T 2 2 T 0.k k¥
D Dé ù+ + + - + =ê úë û
(5)
Note that, if the boundary is insulated, h = 0 and Eq. 4.46 reduces to Eq. (4).
(b) If the surface is exposed to a constant heat flux, oq ,¢¢ the energy balance has the form
1 2 3 oq q q q y 0¢¢+ + + ×D = and the finite difference equation becomes
o
m-1,n m,n-1 m,n+1 m,n
q x
2T T T 4T .
k
¢¢ D+ + - = - <
COMMENTS: Equation (4) can be obtained by using the \u201cinterior node\u201d finite-difference equation,
Eq. 4.33, where the insulated boundary is treated as a symmetry plane as shown below.
PROBLEM 4.37
KNOWN: External corner of a two-dimensional system whose boundaries are subjected to
prescribed conditions.
FIND: Finite-difference equations for these situations: (a) Upper boundary is perfectly insulated and
side boundary is subjected to a convection process, (b) Both boundaries are perfectly insulated;
compare result with Eq. 4.47.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties, (4) No internal generation.
ANALYSIS: Consider the nodal point configuration shown in the schematic and also as Case 4, Table
4.2. The control volume about the node \u2013 shaded area above of unit thickness normal to the page \u2013
has dimensions, (Dx/2)(Dy/2)×1. The heat transfer processes at the surface of the CV are identified as
q1, q2 ×××. Perform an energy balance wherein the processes are expressed using the appropriate rate
equations.
(a) With the upper boundary insulated and the side boundary subjected to a convection process, the
energy balance has the form
in out 1 2 3 4E E 0 q q q q 0- = + + + =& & (1,2)
( )m-1,n m,n m,n-1 m,n m,nT T T Ty x yk 1 k 1 h 1 T T 0 0.2 x 2 y 2 ¥
- -D D Dé ù é ù é ù× + × + × - + =ê ú ê ú ê úD Dë û ë û ë û
Letting Dx = Dy, and regrouping, find
m,n-1 m-1,n m,n
h x 1 h x
T T T 2 1 T 0.
k 2 k¥
D Dé ù+ + - + =ê úë û
 (3) <
(b) With both boundaries insulated, the energy balance of Eq. (2) would have q3 = q4 = 0. The same
result would be obtained by letting h = 0 in the finite-difference equation, Eq. (3). The result is
m,n-1 m-1,n m,nT T 2T 0.+ - = <
Note that this expression is identical to Eq. 4.47 when