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ch09


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Numerical coefficient 4600 49
COMMENTS: (1) For these condition, the convection coefficient for the water is nearly two orders
of magnitude higher than for air.
(2) Using the average-coefficient approach, the time-to-cool, to, values for both fluids is 15-20%
faster than the more accurate numerical integration approach. Evaluating the average coefficient at
sT results in systematically over estimating the coefficient.
(3) The IHT code used for numerical integration of the energy balance equation and the correlations is
shown below for the fluid water.
// LCM energy balance
- hDbar * As * (Ts - Tinf) = M * cps * der(Ts,t)
As = pi * D^2
M = rhos * Vs
Vs = pi * D^3 / 6
// Input variables
D = 0.025
// Ts = 85 + 273 // Initial condition, Ts
Tinf_C = 25
rhos = 8933 // Table A.1, copper, pure
cps = 382
ks = 399
/* Correlation description: Free convection (FC), sphere (S), RaD<=10^11, Pr >=0.7, Churchill
correlation, Eqs 9.25 and 9.35 . See Table 9.2 . */
NuDbar = NuD_bar_FC_S(RaD,Pr) // Eq 9.35
NuDbar = hDbar * D / k
RaD = g * beta * deltaT * D^3 / (nu * alpha) //Eq 9.25
deltaT = abs(Ts - Tinf)
g = 9.8 // gravitational constant, m/s^2
// Evaluate properties at the film temperature, Tf.
Tf = Tfluid_avg(Tinf,Ts)
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
x = 0 // Quality (0=sat liquid or 1=sat vapor)
nu = nu_Tx(&quot;Water&quot;,Tf,x) // Kinematic viscosity, m^2/s
k = k_Tx(&quot;Water&quot;,Tf,x) // Thermal conductivity, W/m·K
Pr = Pr_Tx(&quot;Water&quot;,Tf,x) // Prandtl number
beta = beta_T(&quot;Water&quot;,Tf) // Volumetric coefficient of expansion, K^(-1) (f, liquid, x = 0)
alpha = k / (rho * cp) // Thermal diffusivity, m^2/s
// Conversions
Ts_C = Ts - 273
 Tinf_C = Tinf - 273
PROBLEM 9.81
KNOWN: Temperatures and spacing of vertical, isothermal plates.
FIND: (a) Shape of velocity distribution, (b) Forms of mass, momentum and energy equations for
laminar flow, (c) Expression for the temperature distribution, (d) Vertical pressure gradient, (e)
Expression for the velocity distribution.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar, incompressible, fully-developed flow, (2) Constant properties, (3)
Negligible viscous dissipation, (4) Boussinesq approximation.
ANALYSIS: (a) For the prescribed conditions, there must be buoyancy driven ascending and
descending flows along the surfaces corresponding to Ts,1 and Ts,2, respectively (see schematic).
However, conservation of mass dictates equivalent rates of upflow and downflow and, assuming constant
properties, inverse symmetry of the velocity distribution about the midplane.
(b) For fully-developed flow, which is achieved for long plates, vx = 0 and the continuity equation yields
zv z 0\u2202 \u2202 = <
Hence, there is no net transfer of momentum or energy by advection, and the corresponding equations
are, respectively,
( ) ( ) ( )2 2z c0 dp dz d v dx g gµ \u3c1= \u2212 + \u2212 <
0 = (dT2/dx2) <
(c) Integrating the energy equation twice, we obtain
T = C1x + C2
and applying the boundary conditions, T(-L) = Ts,1 and T(L) = Ts,2, it follows that C1 = -(Ts,1 - Ts,2)/2L
and C2 = (Ts,1 + Ts,2)/2 \u2261 Tm, in which case,
m
s,1 s,2
T T x
T T 2L
\u2212
= \u2212
\u2212
<
Continued...
PROBLEM 9.81 (Cont.)
(d) From hydrostatic considerations and the assumption of a constant density \u3c1m, the balance between the
gravitational and net pressure forces may be expressed as dp/dz = -\u3c1m(g/gc). The momentum equation is
then of the form
( ) ( )( )2 2z m c0 d v dx g gµ \u3c1 \u3c1= \u2212 \u2212
or, invoking the Boussinesq approximation, ( )m m mT T\u3c1 \u3c1 \u3b2\u3c1\u2212 \u2248 \u2212 \u2212 ,
( )( )( )2 2z m c md v dx g g T T\u3b2\u3c1 µ= \u2212 \u2212
or, from the known temperature distribution,
( )( )( )( )2 2z m c s,1 s,2d v dx 2 g g T T x L\u3b2\u3c1 µ= \u2212 <
(e) Integrating the foregiong expression, we obtain
( )( )( )( )2z m c s,1 s,2 1dv dx 4 g g T T x L C\u3b2\u3c1 µ= \u2212 +
( )( )( )( )3z m c s,1 s,2 1 2v 12 g g T T x L C x C\u3b2\u3c1 µ= \u2212 + +
Applying the boundary conditions vz(-L) = vz(L) = 0, it follows that C1 =
( )( )( )m c s,1 s,212 g g T T L\u3b2\u3c1 µ\u2212 \u2212 and C2 = 0. Hence,
( )( )( ) ( ) ( )2 3 3z m c s,1 s,2v L 12 g g T T x L x L\u3b2\u3c1 µ \uf8ee \uf8f9= \u2212 \u2212\uf8ef \uf8fa\uf8f0 \uf8fb <
COMMENTS: The validity of assuming fully-developed conditions improves with increasing plate
length and would be satisfied precisely for infinite plates.
PROBLEM 9.82
KNOWN: Dimensions of vertical rectangular fins. Temperature of fins and quiescent air.
FIND: Optimum fin spacing and corresponding fin heat transfer rate.
SCHEMATIC:
ASSUMPTIONS: (1) Isothermal fins, (2) Negligible radiation, (3) Quiescent air, (4) Negligible heat
transfer from fin tips, (5) Negligible radiation.
PROPERTIES: Table A-4, Air (Tf = 325K, 1 atm): n = 18.41 ´ 10
-6
 m
2
/s, k = 0.0282 W/m×K, a =
26.1 ´ 10
-6
 m
2
/s, Pr = 0.703.
ANALYSIS: From Table 9.3
( ) ( )
1/41/4 s3
opt S
g T T
S 2.71 Ra / S H 2.71
H
b
an
-- ¥é ù-= = ê ú
ë û
( ) ( )
1/412
opt 6 2 6 2
9.8m/s 325K 50K
S 2.71 7.12mm
26.1 10 m / s 18.4 10 m / s 0.15m
--
- -
é ù
ê ú= =
ê ú´ ´ ´ ´ë û
<
From Eq. 9.45 and Table 9.3
( ) ( )s
1/2
2 1/2
S S
576 2.87
Nu
Ra S / L Ra S / L
-é ù
ê ú= +
ê úë û
( ) ( )
( ) ( ) ( )412 34s
S 6 2 6 2
9.8m/s 325K 50K 7.12 10 mg T T S
Ra S / L
H 25.4 10 m / s 18.4 10 m /s 0.15m
b
an
- -
¥
- -
´-
= =
´ ´ ´ ´
( )SRa S / L 53.2=
( ) ( )
[ ]S
1/2
1/2
2 1 / 2
576 2.87
Nu 0.204 0.393 1.29
53.2 53.2
-
-
é ù
ê ú= + = + =
ê úë û
( )S
2h Nu k / S 1.29 0.0282W/m K/0.00712m 5.13W/m K.= = × = ×
With N = W/(t + S) = (355 mm)/(8.62 ´ 10-3 m) = 41.2 » 41,
( ) ( ) ( )( )2sq 2Nh L H T T 82 5.13W/m K 0.02m 0.15m 50K¥= ´ - = × ´
q 63.1W.= <
COMMENTS: Sopt = 7.12 mm is considerably less than the value of 34 mm predicted from previous
considerations. Hence, the corresponding value of q = 63.1 W is considerably larger than that of the
previous predication.
PROBLEM 9.83
KNOWN: Length, width and spacing of vertical circuit boards. Maximum allowable board
temperature.
FIND: Maximum allowable power dissipation per board.
SCHEMATIC:
ASSUMPTIONS: (1) Circuit boards are flat with uniform heat flux at each surface, (2) Negligible
radiation.
PROPERTIES: Table A-4, Air ( )T 320K,1atm := n = 17.9 ´ 10-6 m2/s, k = 0.0278 W/m×K, a =
25.5 ´ 10-6 m2/s.
ANALYSIS: From Eqs. 9.41 and 9.46 and Table 9.3,
( )
1/2
s
* 2/5*s,L S S
q S 48 2.51
T T k Ra S / L Ra S / L
-
¥
é ù
ê ú¢¢
= +ê ú
- ê ú
ê úë û
where
( ) ( )
( )( )
1 525
s* s
S 6 2 6 2
9.8m/s 320K 0.025m qg q SS
Ra
L k L 0.0278W/m K 25.5 10 m / s 17.9 10 m / s 0.4m
b
an
-
- -
¢¢¢¢
= =
× ´ ´
*
S s
S
Ra 58.9q
L
¢¢=
and
( )
s s
s
s,L
q 0.025m qS
0.015q .
T T k 60 K 0.0278W/m K¥
¢¢ ¢¢× ¢¢= =
- ×
Hence,
( )
1/2
s 0.4s s
0.815 0.492
0.015q .
q q
-é ù
ê ú¢¢ = +
¢¢ê ú¢¢ë û
A trial-and-error solution yields
2
sq 287 W / m .¢¢ =
Hence, ( ) ( )2 2s sq 2A q 2 0.4m 287W/m 91.8 W.¢¢= = = <
COMMENTS: Larger heat rates may be achieved by using a fan to superimpose a forced flow on
the buoyancy driven flow.
PROBLEM 9.84
KNOWN: Array of isothermal vertical fins attached to heat sink at 42°C with ambient air
temperature at 27°C.
FIND: (a) Heat removal rate for 24 pairs of fins and (b) Optimum fin spacing for maximizing heat
removal rate if overall size of sink is to remain unchanged.
SCHEMATIC:
ASSUMPTIONS: (1) Fins form vertical, symmetrically heated, isothermal plates, (2) Negligible
radiation effects, (3) Ambient air is quiescent.
PROPERTIES: Table A-4, Air ( ) ( )( )f sT T T / 2 42 27 / 2 C 308K,1atm :¥= + = + ° =
6 2 6 2 316.69 10 m /s, 23.5 10 m / s , k 26.9 10 W / m K.n a- - -= ´ = ´ = ´ ×
ANALYSIS: Considering the fins as vertical isothermal plates, the heat rate can be determined
from Eq. 9.37