PROBLEM 5.7
KNOWN: Diameter and initial temperature of steel balls cooling in air.
FIND: Time required to cool to a prescribed temperature.
SCHEMATIC:
D = 0. 01 m
h = 25 W/m2• K
ASSUMPTIONS: (1) Negligible radiation effects, (2) Constant properties.
ANALYSIS: Applying Eq. 5.10 to a sphere (Lc = r o /3),
=
Bi
2
hLc h ( ro / 3) 25 W/m ⋅ K ( 0.005 m/3)
=
=
= 0.001.
k
k
40 W/m ⋅ K
Hence, the temperature of the steel remains approximately uniform during the cooling
process, and the lumped capacitance method may be used. From Eqs. 5.4 and 5.5,
t
=
t=
(
)
3
Ti − T∞ ρ p D / 6 cp Ti − T∞
ln
ln
=
hAs
T − T∞
T − T∞
hp D 2
ρ Vcp
7800 kg/m3 ( 0.01 m ) 600 J/kg ⋅ K
6 × 25 W/m 2 ⋅ K
ln
1150 − 325
450 − 325
=
t 589
=
s 0.164 h
COMMENTS: Due to the large value of T i , radiation effects are likely to be significant
during the early portion of the transient. The effect is to shorten the cooling time.
<
PROBLEM 5.9
KNOWN: The temperature-time history of a pure copper sphere in a hydrogen stream.
FIND: The heat transfer coefficient between the sphere and the hydrogen stream.
SCHEMATIC:
T(0) = 75°C
T(97s) = 55°C
D = 20 mm
ASSUMPTIONS: (1) Temperature of sphere is spatially uniform, (2) Negligible radiation
exchange, (3) Constant properties.
3
PROPERTIES: Table A-1, Pure copper (338 K): r = 8933 kg/m , cp = 384 J/kg⋅K, k = 399
W/m⋅K.
ANALYSIS: The time-temperature history is given by Eq. 5.6 with Eq. 5.7.
θ (t)

t 
exp  −
where
=

θi
 R t Ct 
1
p D2
Rt =
As =
hAs
p D3
C t r=
Vcp
V
=
6
θ= T − T∞ .
Recognize that when t = 97 s,
θ (t)
=
θi
( 55 − 27 ) C =
( 75 − 27 ) C
 t
0.583 = exp  −
 tt

 97 s 
 = exp  −


 tt 
and solving for t t find
t t = 180 s.
Hence,
=
h
ρ Vcp
=
Ast t
(
)
8933 kg/m3 p 0.023 m3 / 6 384 J/kg ⋅ K
p 0.022 m 2 ×180 s
=
h 63.5 W/m 2 ⋅ K.
COMMENTS: Note that with Lc = Do/6,
Bi =
hLc
0.02
= 63.5 W/m 2 ⋅ K ×
m/399 W/m ⋅ K = 5.3 ×10-4 .
k
6
Hence, Bi < 0.1 and the spatially isothermal assumption is reasonable.
<
PROBLEM 6.8
KNOWN: Smooth flat plate with either laminar or turbulent conditions starting at the leading edge.
Corresponding heat transfer coefficient correlations.
FIND: Average heat transfer coefficients for both conditions for plates of length L = 0.1 m and 1 m.
SCHEMATIC:
Laminar: hx = 1.74 W/m1.5·K x—0.5
Turbulent: hx = 3.98 W/m1.8·K x-0.2
T∞, u∞
Laminar or turbulent conditions
L= 0.1 m or 1 m
x
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional flow and heat transfer.
ANALYSIS: The average heat transfer coefficient is defined as:
hL =
1 L
hx dx
L ∫0
For laminar flow, hx = Alam x −0.5 where
Alam 1.74 W/m1.5 ⋅ K. Thus,
=
h=
L ,lam
Alam
L
∫
L
0
0.5
x −=
dx
2 Alam 0.5 L 2 Alam 3.48 W/m1.5 ⋅ K
x= =
0
L
L0.5
L0.5
The results for the two plate lengths are:
11.0 W/m 2 ⋅ K, L =
0.1 m
hL ,lam = 
2
1m
3.48 W/m ⋅ K, L =
<
For turbulent flow, hx = Aturb x −0.2 where
=
A
3.98 W/m1.8 ⋅ K. Thus,
turb
hL ,turb
=
Aturb
L
∫
L
0
−0.2
x=
dx
Aturb 0.8 L
Aturb
4.98 W/m1.8 ⋅ K
x
=
=
0
L0.2
0.8 L
0.8 L0.2
The results for the two plate lengths are:
7.88 W/m 2 ⋅ K, L =
0.1 m
hL ,turb = 
2
1m
4.98 W/m ⋅ K, L =
<
COMMENTS: For L = 1 m, the turbulent heat transfer coefficient is larger than the laminar one, as is
usually the case. However, for L =0.1 m, the situation is reversed.
PROBLEM 6.24
KNOWN: Heat transfer rate from a turbine blade for prescribed operating conditions.
FIND: Heat transfer rate from a larger blade operating under different conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Surface area A is
directly proportional to characteristic length L, (4) Negligible radiation, (5) Blade shapes are
geometrically similar.
ANALYSIS: For a prescribed geometry,
Nu
=
hL
= f ( ReL , Pr ) .
k
The Reynolds numbers for the blades are
=
ReL,1
V1L1 / ν )
(=
15 / ν
=
ReL,2
( V=
2L2 /ν )
15 / ν .
Hence, with constant properties, ReL,1 = ReL,2 . Also, Pr1 = Pr2 . Therefore,
Nu 2 = Nu 1
( h 2L2 / k ) = ( h1L1 / k )
L1
L1
q1
=
h2 =
h1
.
L2
L 2 A1 Ts,1 − T∞
(
)
Hence, the heat rate for the second blade is
(
(
)
)
L1 A 2 Ts,2 − T∞
q1
L 2 A1 Ts,1 − T∞
Ts,2 − T∞
( 400 − 35) 1500 W
q2 =
q1
=
(
)
Ts,1 − T∞
( 300 − 35)
(
)
q 2 h 2 A 2 Ts,2 −=
T∞
=
q 2 = 2066 W.
<
COMMENTS: The slight variation of ν from Case 1 to Case 2 would cause Re L,2 to differ
from Re L,1 . However, for the prescribed conditions, this non-constant property effect is
small.