PROBLEM 9.1
PROBLEM 9.2
PROBLEM 9.3
PROBLEM 9.3 (Cont.)
PROBLEM 9.4
PROBLEM 9.5
PROBLEM 9.6
PROBLEM 9.7
PROBLEM 9.8
PROBLEM 9.9
PROBLEM 9.10
PROBLEM 9.11
PROBLEM 9.12
PROBLEM 9.13
KNOWN: Aluminum plate (alloy 2024) at an initial uniform temperature of 227ºC is suspended in
a room where the ambient air and surroundings are at 27ºC.
Instantaneous temperature and time rate of temperature change of a vertical plate cooling in a
room.
PROBLEM 9.14
PROBLEM 9.15
PROBLEM 9.15 (Cont.)
PROBLEM 9.16
PROBLEM 9.17
PROBLEM 9.17 (Cont.)
PROBLEM 9.18
PROBLEM 9.19
PROBLEM 9.20
PROBLEM 9.21
PROBLEM 9.22
PROBLEM 9.23
PROBLEM 9.23 (Cont.)
PROBLEM 9.24
PROBLEM 9.24 (Cont.)
PROBLEM 9.25
PROBLEM 9.25 (Cont.)
PROBLEM 9.25 (Cont.)
PROBLEM 9.26
PROBLEM 9.26 (Cont.)
PROBLEM 9.27
PROBLEM 9.27 (Cont.)
PROBLEM 9.28
PROBLEM 9.28 (Cont.)
PROBLEM 9.28 (Cont.)
PROBLEM 9.29
PROBLEM 9.30
PROBLEM 9.31
PROBLEM 9.31 (Cont.)
PROBLEM 9.32
PROBLEM 9.33
PROBLEM 9.33 (Cont.)
PROBLEM 9.34
PROBLEM 9.34 (Cont.)
PROBLEM 9.35
PROBLEM 9.36
PROBLEM 9.36 (Cont.)
PROBLEM 9.36 (Cont.)
PROBLEM 9.37
PROBLEM 9.38
PROBLEM 9.39
PROBLEM 9.39 (Cont.)
PROBLEM 9.40
PROBLEM 9.40 (Cont.)
PROBLEM 9.41
PROBLEM 9.41 (Cont.)
PROBLEM 9.41 (Cont.)
PROBLEM 9.42
PROBLEM 9.42 (Cont.)
PROBLEM 9.43
PROBLEM 9.44
PROBLEM 9.45
PROBLEM 9.46
PROBLEM 9.46 (Cont.)
PROBLEM 9.46 (Cont.)
PROBLEM 9.47
PROBLEM 9.48
PROBLEM 9.48 (Cont.)
PROBLEM 9.49
PROBLEM 9.49 (Cont.)
PROBLEM 9.50
PROBLEM 9.50 (Cont.)
PROBLEM 9.50 (Cont.)
PROBLEM 9.51
PROBLEM 9.51 (Cont.)
PROBLEM 9.51 (Cont.)
PROBLEM 9.52
PROBLEM 9.52 (Cont.)
PROBLEM 9.52 (Cont.)
PROBLEM 9.52 (Cont.)
PROBLEM 9.53
PROBLEM 9.54
PROBLEM 9.54 (Cont.)
PROBLEM 9.55
PROBLEM 9.56
PROBLEM 9.56 (Cont.)
PROBLEM 9.57
PROBLEM 9.57 (Cont.)
PROBLEM 9.58
PROBLEM 9.58 (Cont.)
PROBLEM 9.59
PROBLEM 9.60
PROBLEM 9.60 (Cont.)
PROBLEM 9.61
PROBLEM 9.61 (Cont.)
PROBLEM 9.62
PROBLEM 9.62 (Cont.)
PROBLEM 9.63
PROBLEM 9.64
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 = 323K, 1 atm): v = 18.2 × 10-6 m2/s, k = 0.028 W/m ⋅ K,
α = 25.9 × 10-6 m2/s.
ANALYSIS: From Eqs. 9.41 and 9.46 and Table 9.4,


q′′S
S  48
2.51

=
+
Ts,L − T∞ k  Ra *S S / L ( Ra * S / L )2/5 
S


−1/ 2
9.8m/s 2 (323K) −1 (0.025m)5 q′′S
S gβ q′′SS5
where =
Ra *S
=
L ka vL 0.028 W/m ⋅ K ( 25.9 × 10−6 m 2 / s )(18.2 × 10−6 m 2 / s ) 0.4 m
S
= 56.12q′′S
L
q′′S
0.025 m ⋅ q′′S
S
=
= 0.0148q′′S .
Ts,L − T∞ k (60 K) 0.028 W/m ⋅ k
Ra *S
and
Hence,
 0.855 0.501 
′′S 
0.0148q
=
+

0.4
 q′′S
( q′′S ) 
−1/ 2
.
A trial-and-error solution yields
q′′S = 288 W/m 2 .
Hence, =
q 2=
ASq′′S 2(0.4 m) 2 ( 288 W=
/ m 2 ) 92.16 W.
<
COMMENTS: Larger heat rates may be achieved by using a fan to superimpose a forced flow on the
buoyancy driven flow.
PROBLEM 9.65
PROBLEM 9.65 (Cont.)
PROBLEM 9.66
KNOWN: Vertical air vent in front door of dishwasher with prescribed width and height. Spacing
between isothermal and insulated surface of 20 mm.
FIND: (a) Heat loss from the tub surface and (b) Effect on heat rate of changing spacing by ±
10 mm.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Vent forms vertical parallel isothermal/adiabatic
plates, (3) Ambient air is quiescent.
PROPERTIES: Table A-4, ( T
=
(TS + T∞ ) /2 = 315.5 K, 1 atm): v = 17.45 × 10-6 m2/s,
f
α = 24.8 × 10-6 m2/s, k = 27.4 × 10-3 W/m ⋅ K, β = 1/Tf.
ANALYSIS: The vent arrangement forms two vertical plates, one is isothermal, Ts, and the other is
adiabatic ( q′′ = 0). The heat loss can be estimated from Eq. 9.37 with the correlation of Eq. 9.45 using
C1 = 144 and C2 = 2.87 from Table 9.4:
=
Ra S
gβ (TS − T∞ )S3 9.8m/s 2 (1 / 315.5K) (55 − 30) K (0.020 m)3
=
= 14,355
νa
17.45 × 10−6 m 2 / s × 24.8 × 10−6 m 2 / s

C1
C2
k
+
q=
AS (TS − T∞ ) 
2
1/ 2 
S  (Ra SS/L)
(Ra SS/L) 
−1/ 2
=
(0.500 × 0.580) m 2 ×

C1
C2
0.0274 W / m ⋅ K 
+
(55 − 30) K

2
1/ 2 
0.020 m
(Ra SS / L) 
 (Ra SS / L)
−1/ 2
= 28.6 W.
<
(b) To determine the effect of the spacing at S = 30 and 10 mm, we need only repeat the above
calculations with these results
S (mm)
RaS
q (W)
10
1794
25.8
<
30
48,448
286.9
<
Since it would be desirable to minimize heat losses from the tub, based upon these calculations you
would recommend a decrease in the spacing.
COMMENTS: For this situation, according to Table 9.3, the spacing corresponding to the maximum
heat transfer rate is Smax = (Smax/Sopt) × 2.15 (RaS/S3L)−1/4 = 15 mm. Find qmax = 28.4 W. Note that the
heat rate is not very sensitive to spacing for these conditions.
PROBLEM 9.67
KNOWN: Dimensions, spacing and temperature of plates in a vertical array. Ambient air temperature.
Total width of the array.
FIND: Optimal plate spacing for maximum heat transfer from the array and corresponding number of
plates and heat transfer.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible plate thickness, (3) Constant properties.
−6
v 18.5 ×10 m2/s, k = 0.0282 W/m∙K,
PROPERTIES: Table A-4, air (p = 1atm, T = 325.5K ):=
=
α 26.3 ×10−6 m 2 s , Pr = 0.703, β = 0.00307 k-1.
=
L gβ (Ts − T∞ ) / aν L = (9.8 m/s2 × 0.00307 K-1 × 55°C)/(26.3 × 18.5
ANALYSIS: With Ra S /S
3
× 10-12m4/s2 × 0.3m) = 1.134 × 1010 m4, from Table 9.4, the spacing which maximizes heat
transfer for the array is
2.71
2.71
Sopt =
8.30 × 10−3 m =
8.30 mm
=
=
<
1/4
1/4
−
3
10
4
RaS / S L
1.134 × 10 m
(
)
)
(
With the requirement that (N − 1) Sopt ≤ War, it follows that N ≤ 1 + 150 mm/8.30 mm = 19.1, in
which case
N = 19
The corresponding heat rate is q = N (2WL) h (Ts − T∞ ), where, from Eq. 9.45 and Table 9.4,
1/2
k
k  576
2.87 
h = Nu S = 
+

2
S
S  (Ra SS/L) (Ra SS/L)1/2 
With Ra S S/L = (Ra S /S3 L)S4 = 1.134 × 1010 m -4 × ( 0.00830 m ) = 53.8,
4
h=
0.0282 W/m ⋅ k  576
2.87 
+
= 3.4 (0.199 + 0.391) = 2.01 W/m 2 ⋅ K

2
0.00830 m  (53.8) (53.8)1/2 
q = 19 ( 2 × 0.3m × 0.3m ) 2.01W/m 2 ⋅ K × 55°C = 378 W
COMMENTS: It would be difficult to fabricate heater plates of thickness δ << Sopt . Hence, subject to
the constraint imposed on War, N would be reduced, where N ≤ 1 + War /(Sopt + δ ).
PROBLEM 9.68
PROBLEM 9.68 (Cont.)
PROBLEM 9.69
PROBLEM 9.70
PROBLEM 9.71
PROBLEM 9.72
KNOWN: Dimensions of horizontal rectangular duct and radiation shield. Temperatures of duct
and shield walls.
FIND: Convection heat loss per unit length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Uniform duct and shield wall temperatures, (3) Constant
properties, (4) Convection occurs in four distinct rectangular regions, (4) Convection is twodimensional.
=
α 23.68 ×10 m /s,
PROPERTIES: Table A-4, Air (T = 308 K): v = 16.69 × 10-6 m2/s,
−6
2
26.89 ×10−3 W/m ⋅ K, Pr = 0.706, β = 1/T = 0.0032 K-1.
k=
ANALYSIS: Within the vertical portions of the enclosure, motion will occur in the vertical
direction and will likely extend from the bottom to the top of the shield wall. Thus, the aspect
ratio for the vertical regions is (H + 2t)/t = 0.42/0.06 = 7. The Rayleigh number based on the
enclosure width, t, is
g β (T1 − T2 )t 3 9.8 m/s 2 × 0.0032 K −1 × (45 − 25)° C × (0.06 m)3
Ra
=
=
= 3.48 ×105
t
−6
2
−6
2
νa
16.69 ×10 m /s × 23.68 ×10 m /s
Therefore, Eq. 9.50 holds, and yields
 Pr

=
Nu t 0.22 
Rat 
 0.2 + Pr

0.28
−1/4
 H + 2t 
 0.706

=
0.22 
3.48 × 105 


 t 
 0.2 + 0.706

0.28
−1/4
(7)=
4.49
and
h vert =Nu t k / t =4.49 × 26.89 × 10−3 W/m ⋅ K/0.06 m = 2.01W/m 2 ⋅ K
The two horizontal regions differ from one another. The bottom region is heated from above. There is
therefore no air motion and Nu t = 1, which yields
h bot =
Nu t ,bot k / t =
1 × 26.89 × 10−3 W/m ⋅ K/0.06 m = 0.448 W/m 2 ⋅ K
The top region is heated from below and the Rayleigh number exceeds the critical value for convection
to occur, Rat,crit = 1708. From Eq. 9.49,
Continued...
PROBLEM 9.72 (Cont.)
=
Nu t ,top 0.069
=
Rat1/3 Pr 0.074 4.73
and
h top =Nu t ,top k / t =4.73 × 26.89 × 10−3 W/m ⋅ K/0.06 m = 2.12 W/m 2 ⋅ K
Finally, the convection heat loss per unit length is
′
=
qconv
[2h vert ( H + 2t ) + ( h bot + h top )W ](T1 − T2 )
= [2 × 2.01 W/m 2 ⋅ K(0.42 m) + (0.448 + 2.12)W/m 2 ⋅ K × 0.8 m](45 − 25)°C = 74.9 W/m
<
COMMENTS: (1) The identified rectangular regions do not satisfy the adiabatic end condition.
(2) Presumably the shield would have a small emissivity to reduce radiation. However, if both
surfaces have an emissivity of one, radiation across the enclosure is given by
′ =
qrad
σ [2W + 2 H ](T14 − T24 )= 5.67 ×10−8 W/m 2 ⋅ K 4 × [2 × 0.8 m + 2 × 0.3 m]
× (3184 − 2984 ) K 4 = 290 W/m. Radiation can be more significant than free convection.
PROBLEM 9.73
PROBLEM 9.74
PROBLEM 9.74 (Cont.)
PROBLEM 9.75
PROBLEM 9.76
PROBLEM 9.77
PROBLEM 9.77 (Cont.)
PROBLEM 9.78
PROBLEM 9.78 (Cont.)
PROBLEM 9.79
PROBLEM 9.79 (Cont.)
PROBLEM 9.80
PROBLEM 9.80 (Cont.)
PROBLEM 9.81
PROBLEM 9.82
PROBLEM 9.82 (Cont.)
PROBLEM 9.83
PROBLEM 9.83 (Cont.)
PROBLEM 9.83 (Cont.)
PROBLEM 9.83 (Cont.)
PROBLEM 9.84
PROBLEM 9.85
PROBLEM 9.86
PROBLEM 9.87
PROBLEM 9.87 (Cont.)
PROBLEM 9.88
PROBLEM 9.89
PROBLEM 9.89 (Cont.)
PROBLEM 9.90
PROBLEM 9.90 (Cont.)
PROBLEM 9.91
PROBLEM 9.91 (Cont.)
PROBLEM 9.92
PROBLEM 9.93
PROBLEM 9.94
PROBLEM 9.95
PROBLEM 9.95 (Cont.)
PROBLEM 9.96
PROBLEM 9.96 (Cont.)
PROBLEM 9.96 (Cont.)
PROBLEM 9.97
PROBLEM 9.97 (Cont.)
PROBLEM 9.97 (Cont.)