PROBLEM 1.1
PROBLEM 1.2
PROBLEM 1.3
PROBLEM 1.4
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.
FIND: Heat loss by conduction through the wall as a function of outer surface temperatures ranging from -15 to
38°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant
properties.
ANALYSIS: From Fourier’s law, if q′′x and k are each constant it is evident that the gradient, dT dx = − q′′x k , is a
constant, and hence the temperature distribution is linear. The heat flux must be constant under one-dimensional,
steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and
heat rate when the outside wall temperature is T2 = -15°C are
25 C − −15 C
dT
T1 − T2
(1)
−k
=
=
=
q′′x =
k
1W m ⋅ K
133.3 W m 2 .
)
(
dx
L
0.30 m
q x = q′′x × A = 133.3 W m 2 × 20 m 2 = 2667 W .
(2)
<
Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of outer surface temperature, -15 ≤ T2 ≤
38°C, with different wall thermal conductivities, k.
3500
Heat loss, qx (W)
2500
1500
500
-500
-1500
-20
-10
0
10
20
30
40
Ambient
air temperature, T2 (C)
Outside
surface
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K
For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearly from +2667 W to -867 W and is zero when the inside
and outer surface temperatures are the same. The magnitude of the heat rate increases with increasing thermal
conductivity.
COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not
be linear.
PROBLEM 1.5
PROBLEM 1.6
PROBLEM 1.7
PROBLEM 1.8
PROBLEM 1.9
PROBLEM 1.9 (Cont.)
PROBLEM 1.10
PROBLEM 1.11
PROBLEM 1.12
PROBLEM 1.13
PROBLEM 1.14
PROBLEM 1.15
KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder with an imbedded
electrical heater for different air velocities.
FIND: (a) Determine the convection coefficient for each of the air velocity conditions and display the results
graphically, and (b) Assuming that the convection coefficient depends upon air velocity as h = CVn, determine the
parameters C and n.
SCHEMATIC:
V(m/s)
Pe¢(W/m)
h (W/m2⋅K)
1
450
22.0
2
658
32.2
4
983
48.1
8
1507
73.8
12
1963
96.1
ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiation exchange between
the cylinder surface and the surroundings, (3) Steady-state conditions.
ANALYSIS: (a) From an overall energy balance on the cylinder, the power dissipated by the electrical heater is
transferred by convection to the air stream. Using Newton’s law of cooling on a per unit length basis,
=
Pe′ h (π D )( Ts − T∞ )
where Pe′ is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/s condition, using the data
from the table above, find

=
h 450 W m π × 0.025 m 300 − 40 =
C 22.0 W m 2⋅K
(
)
<
Repeating the calculations, find the convection coefficients for the remaining conditions which are tabulated above
and plotted below. Note that h is not linear with respect to the air velocity.
(b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C = 22.12 W/m2⋅K(s/m)n,
assuring a match at V = 1, we can readily find the exponent n from the slope of the h vs. V curve. From the trials with
100
Coefficient, h (W/m^2.K)
Coefficient, h (W/m^2.K)
n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonable choice. Hence, C = 22.12 and n = 0.6.
80
60
40
20
0
2
4
6
8
10
12
Air velocity, V (m/s)
Data, smooth curve, 5-points
100
80
60
40
20
10
1
2
4
Air velocity, V (m/s)
Data , smooth curve, 5 points
h = C * V^n, C = 22.1, n = 0.5
n = 0.6
n = 0.8
COMMENTS: Radiation may not be negligible, depending on surface emissivity.
6
8
10
<
PROBLEM 1.16
PROBLEM 1.17
PROBLEM 1.18
PROBLEM 1.19
KNOWN: Width, input power and efficiency of a transmission. Temperature and convection coefficient associated
with air flow over the casing.
FIND: Surface temperature of casing. Thermal convection resistance.
SCHEMATIC:
ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3) Negligible
radiation.
ANALYSIS: From Newton’s law of cooling,
=
q hAs ( Ts − T∞=
) 6 hW 2 ( Ts − T∞ )
where the output power is hPi and the heat rate is
q = Pi − Po = Pi (1 − h ) =150 hp × 746 W / hp × 0.07 = 7833 W
Hence,
Ts = T∞ +
q
6 hW 2
= 30°C +
7833 W
2
6 × 200 W / m 2 ⋅ K × ( 0.3m )
= 102.5°C
<
From Eq. 1.11, the thermal resistance due to convection is
R t,conv =
∆T / q x = ( Ts − T∞ ) / q x =
0.00926 K/W
(102.5 − 30 ) K / 7833 W =
<
COMMENTS: (1) There will, in fact, be considerable variability of the local convection coefficient over the
transmission case and the prescribed value represents an average over the surface. (2) The convection thermal
resistance could equivalently be calculated from Rt,conv = 1/hA.
PROBLEM 1.20
PROBLEM 1.21
KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of air stream.
Current, voltage drop and surface temperature of wire for a particular application.
FIND: Air velocity
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from the wire by natural
convection or radiation.
ANALYSIS: If all of the electric energy is transferred by convection to the air, the following equality must
be satisfied
Pelec
= EI
= hA ( Ts − T∞ )
where A =
π DL =
π ( 0.0005m × 0.02m ) =
3.14 ×10−5 m 2 .
Hence,
=
h
EI
5V × 0.1A
=
= 318 W/m 2 ⋅ K
A ( Ts − T∞ ) 3.14 ×10−5m 2 50 C
(
(
)
V = 6.25 ×10−5 h 2 = 6.25 ×10−5 318 W/m 2 ⋅ K
)
2
= 6.3 m/s
<
COMMENTS: The convection coefficient is sufficiently large to render buoyancy (natural convection) and
radiation effects negligible.
PROBLEM 1.22
KNOWN: Chip width and maximum allowable temperature. Coolant conditions.
FIND: Maximum allowable chip power for air and liquid coolants.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and bottom, (3) Chip
is at a uniform temperature (isothermal), (4) Negligible heat transfer by radiation in air.
ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to the coolant.
Hence,
P=q
and from Newton’s law of cooling,
2
P = hA(T - T∞) = h W (T - T∞).
In air,
2
2
Pmax = 250 W/m ⋅K(0.005 m) (85 - 15) ° C = 0.44 W.
<
In the dielectric liquid
2
2
Pmax = 3000 W/m ⋅K(0.005 m) (85-15) ° C = 5.25 W.
<
COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip can dissipate far
less energy than in the dielectric liquid.
PROBLEM 1.23
PROBLEM 1.24
PROBLEM 1.25
PROBLEM 1.26
KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient and
emissivity of a person in the room.
FIND: Basis for difference in comfort level between summer and winter. Ratio of thermal convection
resistance to thermal radiation resistance in summer and winter.
SCHEMATIC:
Air
q“rad
“
qconv
o
= 20 C
h = 2 W/m2-K
Tsur = 27oC (summer)
Tsur = -14oC (winter)
TS = 32oC
ε = 0.9
ASSUMPTIONS: (1) Person may be approximated as a small object in a large enclosure.
ANALYSIS: Thermal comfort is linked to heat loss from the human body, and a chilled feeling is
associated with excessive heat loss. Because the temperature of the room air is fixed, the different summer
and winter comfort levels cannot be attributed to convection heat transfer from the body. In both cases, the
convection heat flux is
Summer and Winter: q′′conv
= h ( Ts − T∞=
C 24 W/m 2
) 2 W/m2 ⋅ K ×12 =
However, the heat flux due to radiation will differ, with values of
(
)
(
)
(
)
(
)
Summer:
−8
4
4
2 4
4
4
4
2
q ′′rad =εs Ts − Tsur =0.9 × 5.67 × 10 W/m ⋅ K 305 − 300 K =28.3 W/m
Winter:
−8
4
4
2 4
4
4
4
2
q ′′rad =εs Ts − Tsur =0.9 × 5.67 × 10 W/m ⋅ K 305 − 287 K =95.4 W/m
There is a significant difference between winter and summer radiation fluxes, and the chilled condition is
attributable to the effect of the colder walls on radiation.
<
From Eq. 1.11, the thermal resistance due to convection is
R t,conv =
∆T / q conv =
( Ts − T∞ ) / q′′conv A
and the thermal resistance due to radiation is
R t,rad =
∆T / q rad =
( Ts − Tsur ) / q′′rad A
Continued…
PROBLEM 1.26 (Cont.)
Thus the ratio of resistances is
R t,conv
R t,rad
=
( Ts − T∞ ) / q′′conv
( Ts − Tsur ) / q′′rad
R t,conv
Summer:
=
R t,rad
( 32 − 20 ) K / 24 W/m 2 2.83
=
( 32 − 27 ) K / 28.3 W/m 2
<
R t,conv
Winter:
=
R t,rad
( 32 − 20 ) K / 24 W/m 2 2.65
=
( 32 − 14 ) K / 95.4 W/m 2
<
2
COMMENTS: (1) For a representative surface area of A = 1.5 m , the heat losses are qconv = 36 W,
qrad(summer) = 42.5 W and qrad(winter) = 143.1 W. The winter time radiation loss is significant and if
maintained over a 24 h period would amount to 2,950 kcal. (2) The convection resistance is larger than the
radiation resistance but they are the same order of magnitude. There isn’t much difference in the resistances
between summer and winter conditions; the main difference is the larger temperature difference through
which radiation occurs in the winter as compared to summer.
PROBLEM 1.27
PROBLEM 1.28
KNOWN: Spherical shaped instrumentation package with prescribed surface emissivity within a large spacesimulation chamber having walls at 77 K.
FIND: Acceptable power dissipation for operating the package surface temperature in the range Ts = 40 to 85°C.
Show graphically the effect of emissivity variations for 0.2 and 0.3.
SCHEMATIC:
ASSUMPTIONS: (1) Uniform surface temperature, (2) Chamber walls are large compared to the spherical package,
and (3) Steady-state conditions.
ANALYSIS: From an overall energy balance on the package, the internal power dissipation Pe will be transferred by
radiation exchange between the package and the chamber walls. From Eq. 1.7,
(
4
q rad = Pe = εAs σ Ts4 - Tsur
)
For the condition when Ts = 40°C, with As = πD2 the power dissipation will be
(
)
4
Pe = 0.25 π × 0.102 m 2 × 5.67 ×10-8 W m 2 ⋅ K 4 × ( 40 + 273) - 77 4  K 4 = 4.3 W


Repeating this calculation for the range 40 ≤ Ts ≤ 85°C, we can obtain the power dissipation as a function of surface
temperature for the ε = 0.25 condition. Similarly, with 0.2 or 0.3, the family of curves shown below has been
obtained.
<
Power dissipation, Pe (W)
10
8
6
4
2
40
50
60
70
80
90
Surface temperature, Ts (C)
Surface emissivity, eps = 0.3
eps = 0.25
eps = 0.2
COMMENTS: (1) As expected, the internal power dissipation increases with increasing emissivity and surface
temperature. Because the radiation rate equation is non-linear with respect to temperature, the power dissipation will
likewise not be linear with surface temperature.
(2) What is the maximum power dissipation that is possible if the surface temperature is not to exceed 85°C? What
kind of a coating should be applied to the instrument package in order to approach this limiting condition?
PROBLEM 1.29
PROBLEM 1.30
PROBLEM 1.31
PROBLEM 1.31 (Cont.)
PROBLEM 1.32
PROBLEM 1.32 (Cont.)
PROBLEM 1.33
PROBLEM 1.33 (Cont.)
PROBLEM 1.34
PROBLEM 1.34 (Cont.)
PROBLEM 1.35
KNOWN: Hot and cold reservoir temperatures of an internally reversible refrigerator. Thermal resistances between
refrigerator and hot and cold reservoirs.
FIND: Expressions for modified Coefficient of Performance and power input of refrigerator.
SCHEMATIC:
qout
∙
Rt,h
Rt,c
qin
ASSUMPTIONS: (1) Refrigerator is internally reversible, (2) Steady-state operation.
ANALYSIS: Heat is transferred from the low temperature reservoir (the refrigerated space) at Tc to the refrigerator
unit, through the resistance Rt,c, with Tc > Tc,i. Heat is rejected from the refrigerator unit to the higher temperature
reservoir (the surroundings), through the resistance Rt,h, with Th,i > Th. The heat input and output rates can be
expressed in a manner analogous to Equations 1.18a and 1.18b.
q=
(Tc − Tc ,i ) / Rt ,c
in
(1)
q=
(Th ,i − Th ) / Rt ,h
out
(2)
Equations (1) and (2) can be solved for the internal temperatures, to yield
 1 + COPm 
Th ,i =
Th + qout Rt ,h =
Th + qin Rt ,h 

 COPm 
(3)
Tc ,=
Tc − qin Rt ,c
i
(4)
 1 + COPm 
qout = qin 

 COPm 
(5)
In Equation (3), qout has been expressed as
using the definition of COPm given in the problem statement. The modified Coefficient of Performance can then be
expressed as
Continued…
PROBLEM 1.35 (Cont.)
COPm
=
Tc ,i
=
Th ,i − Tc ,i
Tc − qin Rt ,c
 1 + COPm 
Th + qin Rt ,h 
 − Tc + qin Rt ,c
 COPm 
Manipulating this expression,
(T
h
− Tc + qin Rt ,c ) COPm + qin Rt ,h (1 + COPm ) = Tc − qin Rt ,c
Solving for COPm results in
COPm =
Tc − qin Rtot
Th − Tc + qin Rtot
<
From the definition of COPm, the power input can be determined:
=
W
T −T + q R
qin
=
qin h c in tot
COPm
Tc − qin Rtot
COMMENTS: As qin or Rtot goes to zero, the Coefficient of Performance approaches the maximum Carnot value,
COPm = COPC = Tc /(Th - Tc).
<
PROBLEM 1.36
KNOWN: Hot and cold reservoir temperatures of an internally reversible refrigerator. Thermal resistances between
refrigerator and hot and cold reservoirs under clean and dusty conditions. Desired cooling rate.
FIND: Modified Coefficient of Performance and power input of refrigerator under clean and dusty conditions.
SCHEMATIC:
qout
= 25°C
Rh,n = 0.04 K/W
Rh,d = 0.1 K/W
∙
Rc,n = 0.05 K/W
qin
qin = 750 W
= 5°C
ASSUMPTIONS: (1) Refrigerator is internally reversible, (2) Steady-state operation, (3) Cold side thermal resistance
does not degrade over time.
ANALYSIS: According to Problem 1.35, the modified Coefficient of Performance and power input are given by
COPm =
Tc − qin Rtot
Th − Tc + qin Rtot
T −T + q R
W = qin h c in tot
Tc − qin Rtot
(1)
(2)
Under new, clean conditions, with Rtot,n = Rh,n + Rc,n = 0.09 K/W, we find
278 K − 750 W × 0.09 K/W
= 2.41
298 K − 278 K + 750 W × 0.09 K/W
<
298 K − 278 K + 750 W × 0.09 K/W
750
W
312 W
=
278 K − 750 W × 0.09 K/W
<
=
COPm ,n
Wn
Under dusty, conditions, with Rtot,d = Rh,d + Rc,n = 0.15 K/W, we find
=
COPm ,d
278 K − 750 W × 0.15 K/W
= 1.25
298 K − 278 K + 750 W × 0.15 K/W
<
Continued...
PROBLEM 1.36 (Cont.)
Wd
298 K − 278 K + 750 W × 0.15 K/W
=
750
W
600 W
278 K − 750 W × 0.15 K/W
<
COMMENTS: (1) The cooling rates and power input values are time-averaged quantities. Since the refrigerator
does not run constantly, the instantaneous power requirements would be higher than calculated. (2) In practice, when
the condenser coils become dusty the power input does not adjust to maintain the cooling rate. Rather, the
refrigerator’s duty cycle would increase. (3) The ideal Carnot Coefficient of Performance is COPC = Tc/(Th – Tc) = 14
and the corresponding power input is 54 W. (4) This refrigerator’s energy efficiency is poor. Less power would be
consumed by more thoroughly insulating the refrigerator, and designing the refrigerator to minimize heat gain upon
opening its door, in order to reduce the cooling rate, qin.
PROBLEM 1.37
KNOWN: Width, surface emissivity and maximum allowable temperature of an electronic chip. Temperature of air
and surroundings. Convection coefficient.
2
1/4
FIND: (a) Maximum power dissipation for free convection with h(W/m ⋅K) = 4.2(T - T∞) , (b) Maximum power
2
dissipation for forced convection with h = 250 W/m ⋅K.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between a small surface and a large enclosure,
(3) Negligible heat transfer from sides of chip or from back of chip by conduction through the substrate.
ANALYSIS: Subject to the foregoing assumptions, electric power dissipation by the chip must be balanced by
convection and radiation heat transfer from the chip. Hence, from Eq. (1.10),
where
(a) If heat transfer is by natural convection,
<
(b) If heat transfer is by forced convection,
<
COMMENTS: Clearly, radiation and natural convection are inefficient mechanisms for transferring heat from the
2
chip. For Ts = 85°C and T∞ = 25°C, the natural convection coefficient is 11.7 W/m ⋅K. Even for forced convection
2
with h = 250 W/m ⋅K, the power dissipation is well below that associated with many of today’s processors. To
provide acceptable cooling, it is often necessary to attach the chip to a highly conducting substrate and to thereby
provide an additional heat transfer mechanism due to conduction from the back surface.
PROBLEM 1.38
KNOWN: Width, input power, and efficiency of a transmission. Temperature and convection coefficient for air flow
over the casing. Emissivity of casing and temperature of surroundings.
FIND: Surface temperature of casing. Resistances due to convection and radiation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3) Radiation
exchange with large surroundings.
ANALYSIS: Heat transfer from the case must balance heat dissipation in the transmission, which may be expressed
as q = Pi – Po = Pi (1 - h) = 150 hp × 746 W/hp × 0.07 = 7833 W. Heat transfer from the case is by convection and
radiation, in which case
(
)
4 
q As  h ( Ts − T∞ ) + εs Ts4 − Tsur
=


2
where As = 6 W . Hence,
7833 W
(
)
2
6 ( 0.30 m )  200 W / m 2 ⋅ K ( Ts − 303K ) + 0.8 × 5.67 × 10−8 W / m 2 ⋅ K 4 Ts4 − 3034 K 4 


A trial-and-error solution yields
Ts ≈ 373 K =
100°C
<
The thermal resistances can be found from Eq. 1.11, that is, Rt = ∆T/q. For convection, the relevant temperature
difference is Ts − T∞ .
R t,conv =
( Ts − T∞ ) / qconv =( Ts − T∞ ) /  hAs ( Ts − T∞ ) =1/ hAs
2
=
R t,conv 1/ 6 ( 0.30 =
m ) 200 W / m 2 ⋅ K  0.00926 K / W


<
For radiation, the relevant temperature difference is Ts − Tsur and
Continued …
PROBLEM 1.38 (Cont.)
(
)
4 A 
R t,rad =
( Ts − Tsur ) / q rad =
( Ts − Tsur ) / εs Ts4 − Tsur
s

R t,rad = ( 373 − 303) K / 0.8 × 5.67 ×10−8 W / m 2 ⋅ K 4 3734 − 3034 K 4 6 ( 0.30 m )2 


= 0.262 K / W
(
)
<
COMMENTS: (1) For Ts ≈ 373 K, qconv ≈ 7,560 W and qrad ≈ 270 W, in which case heat transfer is dominated by
convection. This can also be seen by the fact that the resistance to radiation heat transfer is much larger than the
resistance to convection heat transfer. (2) If radiation is neglected, the corresponding surface temperature is Ts =
102.5°C.
PROBLEM 1.39
PROBLEM 1.39 (Cont.)
PROBLEM 1.40
KNOWN: Surface areas, convection heat transfer coefficient, surface emissivity of gear box and generator.
Temperature of nacelle. Electric power generated by the wind turbine and generator efficiency.
FIND: Gear box and generator surface temperatures.
∙
W∙ gen,in
Wgb,in
SCHEMATIC:
Nacelle, Ts = 416 K
qconv,gb
qconv,gen
qrad,gen
qrad,gb
Generator, Tgen, hgen = 0.95, Agen = 4m2
Gear box, Tgb, hgb = 0.93, Agb = 6m2
ASSUMPTIONS: (1) Steady-state conditions, (2) Interior of nacelle can be treated as large surroundings, (3)
Negligible heat transfer between the gear box and the generator.
ANALYSIS: Heat is generated within both the gear box and the generator. The mechanical work into the generator
can be determined from the electrical power, P = 2.5 × 106 W, and the efficiency of the generator as
Wgen,in ==
P / ηgen 2.5 × 106 W / 0.95 =
2.63 × 106 W
Therefore, the heat transfer from the generator is
qgen= Wgen − P= 2.63 × 106 W − 2.5 × 106 W= 0.13 × 106 W
The heat transfer is composed of convection and radiation components. Hence,
4
=
− Ts4 ) 
qgen Agen  h (Tgen − Ts ) + es ( Tgen
W
W
4 

4
− ( 416K ) 
Tgen − 416K ) + 0.90 × 5.67 × 10−8 2 4 Tgen
= 4m 2 ×  40 2
(
m ⋅K
 m ⋅K

6
= 0.13 × 10 W
(
)
The generator surface temperature may be found by using a numerical solver, or by trial-and-error, yielding
Tgen = 785 K = 512°C
<
Continued...
PROBLEM 1.40 (Cont.)
Heat is also generated by the gear box. The heat generated in the gear box may be determined from knowledge of the
heat generated cumulatively by the gear box and the generator, which is provided in Example 3.1 and is q = qgen + qgb
= 0.33 × 106 W. Hence, qgb = q - qgen = 0.33 × 106W - 0.13 × 106W = 0.20 × 106W and
=
qgb Agb  h (Tgb − Ts ) + εs ( Tgb4 − Ts4 ) 
W
W
4 

Tgb − 416K ) + 0.90 × 5.67 × 10−8 2 4 Tgb4 − ( 416K ) 
= 6m 2 ×  40 2
(
m ⋅K
 m ⋅K

6
= 0.20 × 10 W
(
)
which may be solved by trial-and-error or with a numerical solver to find
Tgb = 791 K = 518°C
<
COMMENTS: (1) The gear box and generator temperatures are unacceptably high. Thermal management must be
employed in order to generate power from the wind turbine. (2) The gear box and generator temperatures are of
similar value. Hence, the assumption that heat transfer between the two mechanical devices is small is valid. (3) The
radiation and convection heat transfer rates are of similar value. For the generator, convection and radiation heat
transfer rates are qconv,gen = 5.9 × 104 W and qrad,gen = 7.1 × 104 W, respectively. The convection and radiation heat
transfer rates are qconv,gb = 9.0 × 104 W and qrad,gb = 11.0 × 104 W, respectively, for the gear box. It would be a poor
assumption to neglect either convection or radiation in the analysis.
PROBLEM 1.41
KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive wastes. Surface
convection conditions.
FIND: Total energy generation rate and surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thin container wall.
ANALYSIS: The rate of energy generation is
r
o
2

o
E g = qdV=q
1- ( r/ro )  2π rLdr
0 

=
E g 2π Lq o ro2 / 2 - ro2 / 4
∫
(
∫
)
or per unit length,
πq r
E ′g = o o .
2
2
<
Performing an energy balance for a control surface about the container yields, at an instant,
E ′g − E ′out = 0
and substituting for the convection heat rate per unit length,
π q o ro2
=
h ( 2π ro )( Ts − T∞ )
2
Ts = T∞ +
q o ro
.
4h
<
COMMENTS: The temperature within the radioactive wastes increases with decreasing r from Ts at ro to a
maximum value at the centerline.
PROBLEM 1.42
KNOWN: Thickness and initial temperature of an aluminum plate whose thermal environment is changed.
FIND: (a) Initial rate of temperature change, (b) Steady-state temperature of plate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible end effects, (2) Uniform plate temperature at any instant, (3) Constant properties,
(4) Adiabatic bottom surface, (5) Negligible radiation from surroundings, (6) No internal heat generation.
ANALYSIS: (a) Applying an energy balance, Eq. 1.12c, at an instant of time to a control volume about the plate,
E in − E out =
E st , it follows for a unit surface area.
( ) ( )
( )
(
)
αSGS 1m 2 − E 1m 2 − q′′conv 1m 2 =( d dt )( McT ) =ρ 1m 2 × L c ( dT dt ) .
Rearranging and substituting from Eqs. 1.3 and 1.5, we obtain
=
dT dt
(1 ρ Lc ) αSGS − εσ Ti4 − h ( Ti − T∞ ) .
=
dT dt
( 2700 kg m3 × 0.004 m × 900 J kg ⋅ K )
−1
×
 0.8 × 900 W m 2 − 0.25 × 5.67 × 10−8 W m 2 ⋅ K 4 ( 308 K )4 − 20 W m 2 ⋅ K ( 35 − 30 ) C 


dT dt = .063 C s .
 = 0, and the energy balance reduces to
(b) Under steady-state conditions, E
st
αSGS = εσ T 4 + h ( T − T∞ )
<
(2)
0.8 × 900 W m 2 = 0.25 × 5.67 × 10−8 W m 2 ⋅ K 4 × T 4 + 20 W m 2 ⋅ K ( T − 303 K )
The solution yields T ≈ 331 K = 57.80°C.
<
COMMENTS: The surface radiative properties have a significant effect on the plate temperature, which would
decrease with increasing ε and decreasing αS. If a low temperature is desired, the plate coating should be
characterized by a large value of ε/αS. The temperature would also decrease with increasing h.
PROBLEM 1.43
KNOWN: Blood inlet and outlet temperatures and flow rate. Dimensions of tubing.
FIND: Required rate of heat addition and estimate of kinetic and potential energy changes.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible liquid with negligible kinetic and potential energy
changes, (3) Blood has properties of water.
PROPERTIES: Table A.6, Water ( T ≈ 300 K): cp,f = 4179 J/ kg∙ K, ρf = 1/vf = 997 kg/m3.
ANALYSIS: From an overall energy balance, Equation 1.12e,
 p (Tout - Tin )
q = mc
where
∀ = 997 kg/m3 × 200 m/min × 10-6 m3 /m
∀ = ρf ∀
m
60 s/min = 3.32 × 10-3 kg/s
Thus
q = 3.32 × 10-3 kg/s × 4179 J/kg ⋅ K × (37o C - 10o C) = 375 W
<
The velocity in the tube is given by
∀ /A = 200 m/min × 10-6 m3 /m
V=∀
c
(60 s/min × 6.4 × 10-3 m × 1.6 × 10-3 m) = 0.33 m/s
The change in kinetic energy is
1
1
2
2
 V 2 - 0) = 3.32 × 10-3 kg/s ×
m(
× (0.33 m/s)2 = 1.8 × 10-4 W
<
The change in potential energy is
 = 3.32 × 10-3 kg/s × 9.8 m/s 2 × 3 m = 0.097 W
mgz
<
COMMENT: The kinetic and potential energy changes are both negligible relative to the thermal energy change.
PROBLEM 1.44
PROBLEM 1.45
PROBLEM 1.46
PROBLEM 1.47
PROBLEM 1.47 (Cont.)
PROBLEM 1.48
KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot and cool zones,
respectively.
FIND: (a) Initial rate of change of the wafer temperature corresponding to the wafer temperature Tw,i = 300 K, and
(b) Steady-state temperature reached if the wafer remains in this position. How significant is convection for this
situation? Sketch how you’d expect the wafer temperature to vary as a function of vertical distance.
SCHEMATIC:
Too = 700 K
hu = 8 W/m2-K
′′q“
qconv,u
cv,u
qrad
“ ,h
Too = 700 K
hl = 4 W/m2-K
“
′′qcv,l
qconv,l
“ ,c
qrad
Tsur,h = 1500 K
d = 0.78 mm
Wafer
Tw,i = 300 K, or Tw,ss
ρ = 2700 kg/m3
c = 875 J/kg-K
ε = 0.65
Tsur,c = 330 K
ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Transient conditions when wafer is initially positioned, (3)
Hot and cool zones have uniform temperatures, (3) Radiation exchange is between small surface (wafer) and large
enclosure (chamber, hot or cold zone), and (4) Negligible heat losses from wafer to mounting pin holder.
ANALYSIS: The energy balance on the wafer illustrated in the schematic above includes convection from the upper
(u) and lower (l) surfaces with the ambient gas, radiation exchange with the hot- and cool-zone (chamber)
surroundings, and the rate of energy storage term for the transient condition.
E ′′in − E ′′out =
E ′′st
dT
q′′rad,h + q′′rad,c − q′′conv,u − q′′conv,l =
r cd w
dt
(
)
(
)
4
4
εs Tsur,h
− Tw4 + εs Tsur,c
− Tw4 − h u ( Tw − T∞ ) − h l ( Tw − T∞ ) =
r cd
d Tw
dt
(a) For the initial condition, the time rate of temperature change of the wafer is determined using the energy balance
Tw T=
above with =
w,i 300 K,
(
)
(
)
0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 1500 4 − 3004 K 4 + 0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 3304 − 3004 K 4
−8 W / m 2 ⋅ K ( 300 − 700 ) K − 4 W / m 2 ⋅ K ( 300 − 700 ) K =
2700 kg / m3 × 875J / kg ⋅ K ×0.00078 m ( d Tw / dt )i
( d Tw / dt )i = 104 K / s
<
(b) For the steady-state condition, the energy storage term is zero, and the energy balance can be solved for the steadystate wafer temperature, Tw = Tw,ss .
Continued …..
PROBLEM 1.48 (Cont.)
(
)
(
)
4
4
0.65s 15004 − Tw,ss
K 4 + 0.65s 3304 − Tw,ss
K4
−8 W / m 2 ⋅ K ( Tw,ss − 700 ) K − 4 W / m 2 ⋅ K ( Tw,ss − 700 ) K =
0
Tw,ss = 1251 K
<
To determine the relative importance of the convection processes, re-solve the energy balance above ignoring those
K / s and Tw,ss 1262 K. We conclude that the radiation exchange processes
=
processes to=
find ( d Tw / dt )i 101
control the initial time rate of temperature change and the steady-state temperature.
If the wafer were elevated above the present operating position, its temperature would increase, since the lower
surface would begin to experience radiant exchange with progressively more of the hot zone chamber. Conversely, by
lowering the wafer, the upper surface would experience less radiant exchange with the hot zone chamber, and its
temperature would decrease. The temperature-distance trend might appear as shown in the sketch.
PROBLEM 1.49
PROBLEM 1.49 (Cont.)
PROBLEM 1.50
PROBLEM 1.51
PROBLEM 1.52
KNOWN: Inner surface heating and new environmental conditions associated with a spherical shell of prescribed
dimensions and material.
FIND: (a) Governing equation for variation of wall temperature with time. Initial rate of temperature change, (b)
Steady-state wall temperature, (c) Effect of convection coefficient on canister temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible temperature gradients in wall, (2) Constant properties, (3) Uniform, timeindependent heat flux at inner surface.
PROPERTIES: Table A.1, Stainless Steel, AISI 302: ρ = 8055 kg/m3, c p = 535 J/kg⋅K.
 − E
E st . Identifying
ANALYSIS: (a) Performing an energy balance on the shell at an instant of time, E
in
out =
relevant processes and solving for dT/dt,
4
dT
q′′i 4pprp
ri2 − h 4 ro2 ( T − T=
ro3 − ri3 cp
∞)
3
dt
dT
3
q′′i ri2 − hro2 ( T − T∞ )  .
=

dt r c r 3 − r 3 
(
) (
p
(
o
)
i
(
)
)
Substituting numerical values for the initial condition, find
W
W
2

3 105 2 ( 0.5m ) − 500 2
( 0.6m )2 ( 500 − 300 ) K 
dT 
⋅
m
m
K


 =
kg
J 
3
3 3
dt i
8055 3 510
( 0.6 ) − ( 0.5)  m
kg ⋅ K 
m
dT 
= −0.084 K/s .
dt i
<
 = 0, it follows that
(b) Under steady-state conditions with E
st
(
)
) (
=
q′′i 4π ri2 h 4π ro2 ( T − T∞ )
2
2
q′′  r 
105 W/m 2  0.5m 
T=
T∞ + i  i  =
300K +
439K

 =
h  ro 
500W/m 2 ⋅ K  0.6m 
<
Continued …..
PROBLEM 1.52 (Cont. )
(c) Parametric calculations were performed using the IHT First Law Model for an Isothermal Hollow Sphere. As
shown below, there is a sharp increase in temperature with decreasing values of h < 1000 W/m2⋅K. For T > 380 K,
boiling will occur at the canister surface, and for T > 410 K a condition known as film boiling (Chapter 10) will occur.
The condition corresponds to a precipitous reduction in h and increase in T.
1000
900
Temperature, T(K)
800
700
600
500
400
300
100
400
800
2000
6000 10000
Convection coefficient, h(W/m^2.K)
Although the canister remains well below the melting point of stainless steel for h = 100 W/m2⋅K, boiling should be
avoided, in which case the convection coefficient should be maintained at h > 1000 W/m2⋅K.
COMMENTS: The governing equation of part (a) is a first order, nonhomogenous differential equation with constant
coefficients. Its solution is θ=
R
(
)
(S/R ) (1 − e− Rt ) + θi e− Rt , where θ ≡ T − T∞ , S ≡ 3q′′i ri2 / r cp ( ro3 − ri3 ) ,
3hro2 /r c p ro3 − ri3 . Note results for t → ∞ and for S = 0.
PROBLEM 1.53
PROBLEM 1.54
PROBLEM 1.54 (Cont.)
PROBLEM 1.55
PROBLEM 1.56
PROBLEM 1.57
KNOWN: Daily thermal energy generation, surface area, temperature of the environment, and heat transfer
coefficient.
FIND: (a) Skin temperature when the temperature of the environment is 20ºC, and (b) Rate of perspiration to maintain
skin temperature of 33ºC when the temperature of the environment is 33ºC.
SCHEMATIC:
T∞
h = 3 W/m2⋅K
E out
Air
E g
ASSUMPTIONS: (1) Steady-state conditions, (2) Thermal energy is generated at a constant rate throughout the day,
(3) Air and surrounding walls are at same temperature, (4) Skin temperature is uniform, (5) Bathing suit has no effect
on heat loss from body, (6) Heat loss is by convection and radiation to the environment, and by perspiration in Part 2.
(Heat loss due to respiration, excretion of waste, etc., is negligible.), (7) Large surroundings.
PROPERTIES: Table A.11, skin: ε = 0.95, Table A.6, water (306 K): ρ = 994 kg/m3, hfg = 2421 kJ/kg.
ANALYSIS:
(a) The rate of energy generation is:
E g =
2000 × 103 cal/day (0.239 cal/J × 86,400 s/day ) = 96.9 W
Under steady-state conditions, an energy balance on the human body yields:
E g - E out = 0
 = q = 96.9 W. Energy outflow is due to convection and net radiation from the surface to the environment,
Thus E
out
Equations 1.3a and 1.7, respectively.
4
E out = hA(Ts - T∞ ) + εσA(Ts4 - Tsur
)
Substituting numerical values
Continued…
PROBLEM 1.57 (Cont.)
96.9 W = 3 W/m 2 ⋅ K×1.8 m 2 × (Ts - 293 K)
+ 0.95 × 5.67 × 10-8 W/m 2 ⋅ K 4 × 1.8 m 2 × (Ts4 - (293 K) 4 )
and solving either by trial-and-error or using IHT or other equation solver, we obtain
Ts = 299 K = 26ºC
<
Since the comfortable range of skin temperature is typically 32 – 35ºC, we usually wear clothing warmer than a
bathing suit when the temperature of the environment is 20ºC.
(b) If the skin temperature is 33ºC when the temperature of the environment is 33ºC, there will be no heat loss due to
convection or radiation. Thus, all the energy generated must be removed due to perspiration:
 fg
E out = mh
We find:
 = E out /h fg = 96.9 W/2421 kJ/kg = 4.0×10-5 kg/s
m
This is the perspiration rate in mass per unit time. The volumetric rate is:
∀
=
∀
∀ = 4.0 × 10−5 kg/s / 994 kg/m3 = 4.0 × 10−8 m3 /s = 4.0 × 10−5 /s
m/ρ
<
COMMENTS: (1) In Part 1, heat losses due to convection and radiation are 32.4 W and
60.4 W, respectively. Thus, it would not have been reasonable to neglect radiation. Care must be taken to include
radiation when the heat transfer coefficient is small, even if the problem statement does not give any indication of its
importance. (2) The rate of thermal energy generation is not constant throughout the day; it adjusts to maintain a
constant core temperature. Thus, the energy generation rate may decrease when the temperature of the environment
goes up, or increase (for example, by shivering) when the temperature of the environment is low. (3) The skin
temperature is not uniform over the entire body. For example, the extremities are usually cooler. Skin temperature
also adjusts in response to changes in the environment. As the temperature of the environment increases, more blood
flow will be directed near the surface of the skin to increase its temperature, thereby increasing heat loss. (4) If the
perspiration rate found in Part 2 was maintained for eight hours, the person would lose 1.2 liters of liquid. This
demonstrates the importance of consuming sufficient amounts of liquid in warm weather.
PROBLEM 1.58
KNOWN: Thermal conductivity, thickness and temperature difference across a sheet of rigid extruded insulation.
Cold wall temperature, surroundings temperature, ambient temperature and emissivity.
FIND: (a) The value of the convection heat transfer coefficient on the cold wall side in units of W/m2⋅°C or W/m2⋅K,
and, (b) The cold wall surface temperature for emissivities over the range 0.05 ≤ ε ≤ 0.95 for a hot wall temperature of
T1 = 30 °C.
SCHEMATIC:
T2 = 20°C = 293 K
32°C
TT11 ==30°C
L=25 mm
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (c) Constant
properties, (4) Large surroundings.
ANALYSIS:
(a) An energy balance on the control surface shown in the schematic yields
E in = E out or
q cond = q conv + q rad
Substituting from Fourier’s law, Newton’s law of cooling, and Eq. 1.7 yields
k
or
T1 - T2
4
= h(T2 - T∞ ) + εσ(T24 - Tsur
)
L
h=
(1)
T -T
1
4
)]
[k 1 2 - εσ(T24 - Tsur
(T2 - T∞ )
L
Substituting values,
h=
h = 12.2
1
(20 - 5)o C
[0.029
W
(30 - 20)o C
W
×
- 0.95 × 5.67 × 10-8 2 4 (2934 - 3204 ) K 4 ]
m⋅K
0.02 m
m ⋅K
W
m2 ⋅ K
Since temperature differences of 1 K and 1°C are equal, h = 12.2
<
W
.
m ⋅°C
2
<
Continued...
PROBLEM 1.58 (Cont.)
(b) Equation (1) may be solved iteratively to find T2 for any emissivity ε. IHT was used for this purpose, yielding the
following.
Surface Temperature vs. Wall Emissivity
Surface Temperature (K)
295
290
285
280
0
0.2
0.4
0.6
0.8
1
Emissivity
COMMENTS: (1) Note that as the wall emissivity increases, the surface temperature increases since the surroundings
temperature is relatively hot. (2) The IHT code used in part (b) is shown below. (3) It is a good habit to work in
temperature units of kelvins when radiation heat transfer is included in the solution of the problem.
//Problem 1.56
h = 12.2 //W/m^2∙K (convection coefficient)
L = 0.02 //m (sheet thickness)
k = 0.029 //W/m∙K (thermal conductivity)
T1 = 30 + 273 //K (hot wall temperature)
Tsur = 320 //K (surroundings temperature)
sigma = 5.67*10^-8 //W/m^2∙K^4 (Stefan -Boltzmann constant)
Tinf = 5 + 273 //K (ambient temperature)
e = 0.95 //emissivity
//Equation (1) is
k*(T1-T2)/L = h*(T2-Tinf) + e*sigma*(T2^4 - Tsur^4)
PROBLEM 1.59
PROBLEM 1.59 (Cont.)
PROBLEM 1.60
PROBLEM 1.61
PROBLEM 1.62
PROBLEM 1.62 (Cont.)
PROBLEM 1.63
PROBLEM 1.64 (a)
PROBLEM 1.64 (b)
PROBLEM 1.64 (c)
PROBLEM 1.64 (c) (Cont.)
PROBLEM 1.64 (d)
PROBLEM 1.64 (e)
PROBLEM 1.64 (f)
PROBLEM 1.64 (g)
PROBLEM 1.65 (a)
PROBLEM 1.65 (b)
PROBLEM 1.65 (c)
PROBLEM 1.65 (d)
PROBLEM 2.1
PROBLEM 2.2
KNOWN: Hot water pipe covered with thick layer of insulation.
FIND: Sketch temperature distribution and give brief explanation to justify shape.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No
internal heat generation, (4) Insulation has uniform properties independent of temperature and
position.
ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the form
dT
dT
qr =
−kA r
=
−k ( 2π r )
dr
dr
where A r = 2πr and  is the axial length of the pipe-insulation system. Recognize that for steadystate conditions with no internal heat generation, an energy balance on the system requires
E in = E out since E g = E st = 0. Hence
qr = Constant.
That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows that
 dT 
r   = Constant.
 dr 
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r,
remains constant throughout the insulation. For our situation, the temperature distribution must appear
as shown in the sketch.
<
COMMENTS: (1) Note that, while qr is a constant and independent of r, q ′′r is not a constant. How
bg
does q ′′r r vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with
increasing radius.
PROBLEM 2.3
PROBLEM 2.4
KNOWN: Temperature dependence of the thermal conductivity, k(T), for heat transfer through a
plane wall.
FIND: Effect of k(T) on temperature distribution, T(x).
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) No internal heat
generation.
ANALYSIS: From Fourier’s law and the form of k(T),
−k
q′′x =
dT
dT
=
− ( k o + aT ) .
dx
dx
(1)
2
2
The shape of the temperature distribution may be inferred from knowledge of d T/dx = d(dT/dx)/dx.
Since q ′′x is independent of x for the prescribed conditions,
dq′′x
d 
dT 
= - ( k o + aT )  = 0
dx
dx 
dx 
2
2
d T
 dT 
0.
- ( k o + aT )
-a  =
 dx 
dx 2
Hence,
d 2T
-a  dT 
=
 
dx 2 k o + aT  dx 
2
k o + aT=k>0

where   dT  2
  dx  > 0

from which it follows that for
a > 0: d 2 T / dx 2 < 0
a = 0: d 2 T / dx 2 = 0
a < 0: d 2 T / dx 2 > 0.
<
COMMENTS: The shape of the distribution could also be inferred from Eq. (1). Since T decreases
with increasing x,
a > 0: k decreases with increasing x = > | dT/dx | increases with increasing x
a = 0: k = ko = > dT/dx is constant
a < 0: k increases with increasing x = > | dT/dx | decreases with increasing x.
PROBLEM 2.5
KNOWN: Irradiation and absorptivity of aluminum, glass and aerogel.
FIND: Ability of the protective barrier to withstand the irradiation in terms of the temperature
gradients that develop in response to the irradiation.
SCHEMATIC:
G = 10 x 106 W/m2
α al =
0.2
x
α gl =
0.9
α a = 0.8
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Constant properties, (c)
Negligible emission and convection from the exposed surface.
PROPERTIES: Table A.1, pure aluminum (300 K): kal = 238 W/m⋅K. Table A.3, glass (300 K): kgl =
1.4 W/m⋅K.
ANALYSIS: From Eqs. 1.6 and 2.32
-k
∂T
∂x
= q′′s = G abs = αG
x=0
or
∂T
∂x
=x=0
αG
k
The temperature gradients at x = 0 for the three materials are:
Material
aluminum
glass
aerogel
<
∂T / ∂x x=0 (K/m)
8.4 x 103
6.4 x 106
1.6 x 109
COMMENT: It is unlikely that the aerogel barrier can sustain the thermal stresses associated with the
large temperature gradient. Low thermal conductivity solids are prone to large temperature gradients,
and are often brittle.
PROBLEM 2.6
PROBLEM 2.7
PROBLEM 2.8
PROBLEM 2.9
PROBLEM 2.9 (Cont.)
PROBLEM 2.10
PROBLEM 2.10 (Cont.)
PROBLEM 2.11
PROBLEM 2.11 (Cont.)
PROBLEM 2.12
PROBLEM 2.13
KNOWN: Temperatures of various materials.
FIND: (a) Graph of thermal conductivity, k, versus temperature, T, for pure copper, 2024 aluminum
and AISI 302 stainless steel for 300 ≤ T ≤ 600 K, (b) Graph of thermal conductivity, k, for helium and
air over the range 300 ≤ T ≤ 800 K, (c) Graph of kinematic viscosity, ν, for engine oil, ethylene glycol,
and liquid water for 300 ≤ T ≤ 360 K, (d) Graph of thermal conductivity, k, versus volume fraction, ϕ,
of a water-Al2O3 nanofluid for 0 ≤ ϕ ≤ 0.08 and T = 300 K. Comment on the trends for each case.
ASSUMPTION: (1) Constant nanoparticle properties.
ANALYSIS: (a) Using the IHT workspace of Comment 1 yields
Thermal Conductivity of Cu, 2024 Al, and 302 ss
500
Copper
k, W/m^2
400
300
2024 Aluminum
200
100
302 Stainless steel
0
400
300
600
500
T, Celsius
Note the large difference between the thermal conductivities of these metals. Copper conducts thermal
energy effectively, while stainless steels are relatively poor thermal conductors. Also note that,
depending on the metal, the thermal conductivity increases (2024 Aluminum and 302 Stainless Steel)
or decreases (Copper) with temperature.
(b) Using the IHT workspace of Comment 2 yields
Thermal Conductivity of Helium and Air
0.4
k, W/m^2
0.3
Helium
0.2
Air
0.1
0
300
400
500
600
700
800
T, Celsius
Note the high thermal conductivity of helium relative to that of air. As such, He is sometimes used as a
coolant. The thermal conductivity of both gases increases with temperature, as expected from
inspection of Figure 2.8.
Continued…
PROBLEM 2.13 (Cont.)
(c) Using the IHT workspace of Comment 3 yields
Kinematic Viscosity of Engine Oil, Ethylene Glycol and H2O
0.01
Engine Oil
nu, m^2/s
0.001
0.0001
Ethylene Glycol
1E-5
1E-6
H2O
1E-7
300
320
340
360
T, Celsius
The kinematic viscosities vary by three orders of magnitude between the various liquids. For each case
the kinematic viscosity decreases with temperature.
(d) Using the IHT workspace of Comment 4 yields
Thermal Conductivity of Nanofluid and Base Fluid
0.8
k, W/m^2
Nanofluid
0.7
0.6
Base fluid (H2O)
0.5
0
0.02
0.04
0.06
0.08
Volume fraction, j
Note the increase in the thermal conductivity of the nanofluid with addition of more nanoparticles. The
solid phase usually has a higher thermal conductivity than the liquid phase, as noted in Figures 2.5 and
2.9, respectively.
COMMENTS: (1) The IHT workspace for part (a) is as follows.
// Copper (pure) property functions : From Table A.1
// Units: T(K)
kCu = k_T("Copper",T)
// Thermal conductivity,W/m·K
// Aluminum 2024 property functions : From Table A.1
// Units: T(K)
kAl = k_T("Aluminum 2024",T)
// Thermal conductivity,W/m·K
// Stainless steel-AISI 302 property functions : From Table A.1
// Units: T(K)
kss = k_T("Stainless Steel-AISI 302",T)
// Thermal conductivity,W/m·K
T = 300
// Temperature, K
Continued…
PROBLEM 2.13 (Cont.)
(2) The IHT workspace for part (b) follows.
// Helium property functions : From Table A.4
// Units: T(K)
kHe = k_T("Helium",T)
// Thermal conductivity, W/m·K
// Air property functions : From Table A.4
// Units: T(K); 1 atm pressure
kAir = k_T("Air",T)
// Thermal conductivity, W/m·K
T = 300
// Temperature, K
(3) The IHT workspace for part (c) follows.
// Engine Oil property functions : From Table A.5
// Units: T(K)
nuOil = nu_T("Engine Oil",T)
// Kinematic viscosity, m^2/s
// Ethylene glycol property functions : From Table A.5
// Units: T(K)
nuEG = nu_T("Ethylene Glycol",T)
// Kinematic viscosity, m^2/s
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xH2O =0
// Quality (0=sat liquid or 1=sat vapor)
nuH2O = nu_Tx("Water",T,xH2O)
// Kinematic viscosity, m^2/s
T = 300
// Temperature, K
(4) The IHT workspace for part (d) follows.
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xH2O =0
// Quality (0=sat liquid or 1=sat vapor)
kH2O = k_Tx("Water",T,xH2O)
// Thermal conductivity, W/m·K
kbf = kH2O
T = 300
j = 0.01
// Volume fraction of nanoparticles
//Particle Properties
kp = 36
knf = (num/den)*kbf
num = kp + 2*kbf-2*j*(kbf - kp)
den = kp + 2*kbf + j*(kbf - kp)
// Thermal conductivity, W/mK
PROBLEM 2.14
PROBLEM 2.15
KNOWN: Thermal conductivity of helium.
FIND: The helium temperature. Compare to value from Table A.4.
ASSUMPTIONS: (1) Ideal gas behavior.
PROPERTIES: Table A.4: Helium; M = 4.003 kg/kmol, cp = 5.193 kJ/kg⋅K (independent of
temperature). Figure 2.8: Helium, d = 0.219 nm.
ANALYSIS: For helium, the gas constant, specific heat at constant volume, and ratio of specific heats
are:
=
R
R 8.315 kJ/kmol ⋅ K
kJ
=
= 2.077
;
M
4.003 kg/kmol
kg ⋅ K
cv = c p − R = 5.193
c p 5.193
kJ
kJ
kJ
− 2.077
= 3.116
; g =
=
= 1.667
kg ⋅ K
kg ⋅ K
kg ⋅ K
cv 3.166
From Equation 2.12
k=
9g - 5 cv
4 πd2
Nπ
T=
M kB
M k BT
Nπ
 4 kπ d 2 


 (9g - 5)cv 
2
(
-9

6.024 × 10 mol × 1000 mol/kmol × π  4 × 0.15 W/m ⋅ K × π × 0.219 × 10 m
=
×

4.003 kg/kmol × 1.381× 10-23 J/K
(9 × 1.667 - 5) × 3166 J/kg ⋅ K

= 288 K
23
-1
From Table A.4, the thermal conductivity of helium is 0.15 W/m⋅K when T = 294 K.
The computed value of 288 K is within 2% of the reported value.
)
2




2
<
<
COMMENTS: The preceding analysis may be used to estimate the thermal conductivity at various
temperatures. However, the analysis is not valid for extreme temperatures or pressures. For example,
(1) the thermal conductivity is predicted to be independent of the pressure of the gas. As pure vacuum
conditions are approached, the thermal conductivity will suddenly drop to zero, and the preceding
analysis is no longer valid. Also, (2) for temperatures considerably higher or lower than normallyencountered room temperatures, the agreement between the predicted and actual thermal
conductivities can be poor. For example, for carbon dioxide at T = 600 K, the predicted thermal
conductivity is k = 0.0223 W/m∙K, while the actual (tabular) value is k = 0.0407 W/m∙K. For extreme
temperatures, thermal correction factors must be included in the predictions of the thermal
conductivity.
PROBLEM 2.16
PROBLEM 2.16 (Cont.)
PROBLEM 2.17
KNOWN: Diameter, length, and mass of stainless steel rod, insulated on its exterior surface other
than ends. Temperature distribution.
FIND: Heat flux.
SCHEMATIC:
Stainless steel
D = 20 mm
T(x) = 305 K – 10 K (x/L)
x
L = 100 mm
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3)
Constant properties.
ANALYSIS: The heat flux can be found from Fourier’s law,
q′′x = −k
dT
dx
Table A.1 gives values for the thermal conductivity of stainless steels, however we are not told which
type of stainless steel the rod is made of, and the thermal conductivity varies between them. We do
know the mass of the rod, and can use this to calculate its density:
ρ
=
0.248 kg
M
M
=
=
= 7894 kg/m3
2
V π D L / 4 π × (0.02 m) 2 × 0.1 m/4
From Table A.1, it appears that the material is AISI 304 stainless steel. The temperature of the rod
varies from 295 K to 305 K. Evaluating the thermal conductivity at 300 K, k = 14.9 W/m⋅K. Thus,
q′′x =−k
dT
=−k (−b / L) =
14.9 W/m ⋅ K ×10 K / 0.1 m =
1490 W/m 2
dx
COMMENTS: If the temperature of the rod varies significantly along its length, the thermal
conductivity will vary along the rod as much or more than the variation in thermal conductivities
between the different stainless steels.
<
PROBLEM 2.18
PROBLEM 2.19
KNOWN: Temperature distribution in a plane wall. Whether conditions are steady-state or transient.
FIND: (a) Whether thermal energy is being generated within the wall, and if so, whether it is positive
or negative. (b) Whether the volumetric generation rate is positive or negative. (c) and (d) Whether the
temperature is increasing or decreasing with time.
SCHEMATIC:
“
qin
“
qin
T(x,t)
“
qout
T(x,∞)
“
qout
“
qout
T(x,∞)
dx
dx
x
(a)
“
qout
T(x,t)
dx
x
dx
x
(b)
x
(d)
(c)
ASSUMPTIONS: (1) One-dimensional conduction in x-direction, (2) Constant properties.
ANALYSIS: An energy balance on the differential control volume can be expressed as
dEst 
=Ein − E out + E g
dt
(1)
The heat flux is given by Fourier’s law,
q′′x = −k
dT
dx
(2)
Assuming constant thermal conductivity, the slope of the temperature distribution indicates
the magnitude and direction of the heat flux according to Eq. (2). A positive slope means heat
is flowing from right to left and vice versa. The magnitudes and directions of the heat fluxes
are illustrated in the schematic above. With this background we can consider each scenario in
turn.
(a) Conditions are steady-state, therefore dEst/dt = 0 in Eq. (1). Since the slope of the
temperature distribution is positive, heat is flowing from right to left in the schematic.
With the slope higher at the right than the left, more heat is entering at the right than
leaving at the left. Therefore heat generation must exist and must be negative.
<
(b) Conditions are steady-state, therefore dEst/dt = 0. The slope of the temperature distribution
is positive and is smaller at the right than the left, therefore less heat is entering at the right
than leaving at the left. Therefore heat generation must exist and must be positive.
Continued …
<
PROBLEM 2.19 (Cont.)
(c) Conditions are transient. There is no heat generation, therefore E g = 0 in Eq. (1). Since the
slope of the temperature distribution is negative, heat is flowing from left to right in the
schematic. With the slope steeper at the left than the right, more heat is entering at the left
than leaving at the right. Therefore, there is net heat transfer into the control volume and
dEst/dt > 0. Thus the temperature is increasing with time.
<
(d) Conditions are transient. There is no heat generation. The slope of the temperature
distribution is negative and is smaller at the left than the right, therefore less heat is
entering at the left than leaving at the right. Therefore, there is net heat transfer out of the
control volume and dEst/dt < 0. Thus the temperature is decreasing with time.
COMMENTS: If the thermal conductivity is not constant, it is not possible to tell whether the heat
flux is higher or lower at the two sides of the control volume.
<
PROBLEM 2.20
KNOWN: Diameter D, thickness L and initial temperature Ti of pan. Heat rate from stove to bottom
of pan. Convection coefficient h and variation of water temperature T∞(t) during Stage 1.
Temperature TL of pan surface in contact with water during Stage 2.
FIND: Form of heat equation and boundary conditions associated with the two stages.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in pan bottom, (2) Heat transfer from stove is
uniformly distributed over surface of pan in contact with the stove, (3) Constant properties.
ANALYSIS:
Stage 1
∂ 2T
Heat Equation:
∂x 2
Boundary Conditions:
Initial Condition:
=
1 ∂T
α ∂t
−k
q
∂T
=
q′′o =o
∂x x = 0
π D2 / 4
−k
∂T
= h T ( L, t ) − T∞ ( t ) 
∂x x = L
(
)
T ( x, 0 ) = Ti
Stage 2
Heat Equation:
Boundary Conditions:
d 2T
dx 2
−k
=0
dT
=
q′′o
dx x = 0
T ( L ) = TL
COMMENTS: Stage 1 is a transient process for which T∞(t) must be determined separately. As a
first approximation, it could be estimated by neglecting changes in thermal energy storage by the pan
bottom and assuming that all of the heat transferred from the stove acted to increase thermal energy
storage within the water. Hence, with q ≈ mcp dT∞/dt, where m and cp are the mass and specific heat
of the water in the pan, T∞(t) ≈ (q/mcp) t.
PROBLEM 2.21
PROBLEM 2.21 (Cont.)
PROBLEM 2.22
PROBLEM 2.23
PROBLEM 2.23 (Cont.)
PROBLEM 2.23 (Cont.)
PROBLEM 2.23 (Cont.)
PROBLEM 2.24
KNOWN: Transient temperature distributions in a plane wall.
FIND: Appropriate forms of heat equation, initial condition, and boundary conditions.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible radiation.
ANALYSIS: The general form of the heat equation in Cartesian coordinates for constant k is
Equation 2.21. For one-dimensional conduction it reduces to
∂ 2T q 1 ∂T
+ =
∂x 2 k α ∂t
At steady state this becomes
d 2T q
+ =
0
dx 2 k
If there is no thermal energy generation the steady-state temperature distribution is linear (or could be
constant). If there is uniform thermal energy generation the steady-state temperature distribution must
be parabolic.
Continued…
PROBLEM 2.24 (Cont.)
In case (a), the steady-state temperature distribution is constant, therefore there must not be any
thermal energy generation. The heat equation is
∂ 2T 1 ∂T
=
∂x 2 α ∂t
<
The initial temperature is uniform throughout the solid, thus the initial condition is
<
T ( x, 0) = Ti
At x = 0, the slope of the temperature distribution is zero at all times, therefore the heat flux is zero
(insulated condition). The boundary condition is
∂T
∂x
<
=0
x =0
At x = L, the temperature is the same for all t > 0. Therefore the surface temperature is constant:
<
T ( L, t ) = Ts
For case (b), the steady-state temperature distribution is not linear and appears to be parabolic,
therefore there is thermal energy generation. The heat equation is
∂ 2T q 1 ∂T
+ =
∂x 2 k α ∂t
<
The initial temperature is uniform, the temperature gradient at x = 0 is zero, and the temperature at x =
L is equal to the initial temperature for all t > 0, therefore the initial and boundary conditions are
T ( x, 0) = Ti ,
∂T
∂x
= 0,
<
T ( L, t ) = Ti
x =0
With the left side insulated and the right side maintained at the initial temperature, the cause of the
decreasing temperature must be a negative value of thermal energy generation.
In case (c), the steady-state temperature distribution is constant, therefore there is no thermal energy
generation. The heat equation is
∂ 2T 1 ∂T
=
∂x 2 α ∂t
<
Continued…
PROBLEM 2.24 (Cont.)
The initial temperature is uniform throughout the solid. At x = 0, the slope of the temperature
distribution is zero at all times. Therefore the initial condition and boundary condition at x = 0 are
T ( x, 0) = Ti ,
∂T
∂x
=0
<
x =0
At x = L, neither the temperature nor the temperature gradient are constant for all time. Instead, the
temperature gradient is decreasing with time as the temperature approaches the steady-state
temperature. This corresponds to a convection heat transfer boundary condition. As the surface
temperature approaches the fluid temperature, the heat flux at the surface decreases. The boundary
condition is:
−k
∂T
∂x
= h [T ( L, t ) − T∞ ]
<
x=L
The fluid temperature, T∞, must be higher than the initial solid temperature to cause the solid
temperature to increase.
For case (d), the steady-state temperature distribution is not linear and appears to be parabolic,
therefore there is thermal energy generation. The heat equation is
∂ 2T q 1 ∂T
+ =
∂x 2 k α ∂t
<
Since the temperature is increasing with time and it is not due to heat conduction due to a high surface
temperature, the energy generation must be positive.
The initial temperature is uniform and the temperature gradient at x = 0 is zero. The boundary
condition at x = L is convection. The temperature gradient and heat flux at the surface are increasing
with time as the thermal energy generation causes the temperature to rise further and further above the
fluid temperature. The initial and boundary conditions are:
T ( x, 0) = Ti ,
∂T
∂x
= 0,
x =0
−k
∂T
∂x
= h [T ( L, t ) − T∞ ]
x=L
COMMENTS: 1. You will learn to solve for the temperature distribution in transient conduction in
Chapter 5. 2. Case (b) might correspond to a situation involving a spatially-uniform endothermic
chemical reaction. Such situations, although they can occur, are not common.
<
PROBLEM 2.25
KNOWN: Rod consisting of two materials with same lengths. Ratio of thermal conductivities.
FIND: Sketch temperature and heat flux distributions.
SCHEMATIC:
T1
T2
T1 < T2
A
x
0.5 L
B
L
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties, (3) No internal generation.
ANALYSIS: From Equation 2.19 for steady-state, one-dimensional conduction with constant
properties and no internal heat generation,
∂  ∂T
k
∂x  ∂x
∂q′′x

=0
 = 0 or
∂x

From these equations we know that heat flux is constant and the temperature gradient is inversely
proportional to k. Thus, with kA = 0.5kB, we can sketch the temperature and heat flux distributions as
shown below:
<
COMMENTS: (1) Note the discontinuity in the slope of the temperature distribution at x/L = 0.5.
The constant heat flux is in the negative x-direction. (2) A discontinuity in the temperature distribution
may occur at x/L = 0.5 due the joining of dissimilar materials. We shall address thermal contact
resistances in Chapter 3.
PROBLEM 2.26
PROBLEM 2.26 (Cont.)
PROBLEM 2.26 (Cont.)
PROBLEM 2.27
KNOWN: Three-dimensional system – described by cylindrical coordinates (r,φ,z) –
experiences transient conduction and internal heat generation.
FIND: Heat diffusion equation.
SCHEMATIC: See also Fig. 2.12.
ASSUMPTIONS: (1) Homogeneous medium.
ANALYSIS: Consider the differential control volume identified above having a volume
given as V = dr⋅rdφ⋅dz. From the conservation of energy requirement,
(1)
+ E = E .
+q −q
+q −q
q −q
r
r +dr
φ
φ +dφ
z
g
z+dz
st
The generation and storage terms, both representing volumetric phenomena, are
 = q ( dr ⋅ rdφ ⋅ dz ) E = r Vc∂ T/∂ t = r ( dr ⋅ rdφ ⋅ dz ) c ∂ T/∂ t.
(2,3)
E = qV
g
g
Using a Taylor series expansion, we can write
∂
∂
( )
q r+dr =
qr +
qφ +
qφ dφ ,
( q r ) dr, qφ +dφ =
∂ r
∂ φ
∂
q z+dz =
qz +
( q z ) dz. (4,5,6)
∂ z
Using Fourier’s law, the expressions for the conduction heat rates are
−kA r∂ T/∂ r =
−k ( rdφ ⋅ dz ) ∂ T/∂ r
qr =
(7)
qφ =
−kAφ ∂ T/r∂φ =
−k ( dr ⋅ dz ) ∂ T/r∂φ
(8)
qz =
−kA z∂ T/∂ z =
−k ( dr ⋅ rdφ ) ∂ T/∂ z.
(9)
Note from the above, right schematic that the gradient in the φ-direction is ∂T/r∂φ and not
∂T/∂φ. Substituting Eqs. (2), (3) and (4), (5), (6) into Eq. (1),
−
∂
∂
∂
∂ T
( q r ) dr − qφ dφ − ( q z ) dz + q dr ⋅ rdφ ⋅ dz = r ( dr ⋅ rdφ ⋅ dz ) c .
∂ r
∂φ
∂ z
∂ t
( )
(10)
Substituting Eqs. (7), (8) and (9) for the conduction rates, find
−
∂ 
∂ T
∂
−
⋅
−
φ
k
rd
dz
dr
(
)
∂ r 
∂ r 
∂ φ

∂ T
∂ 
∂ T
 −k ( drdz ) r∂φ  dφ − ∂ z  −k ( dr ⋅ rdφ ) ∂ z  dz




+q dr ⋅ rdφ ⋅ dz = r ( dr ⋅ rdφ ⋅ dz ) c
∂ T
.
∂ t
(11)
Dividing Eq. (11) by the volume of the CV, Eq. 2.26 is obtained.
1 ∂  ∂ T 1 ∂  ∂ T ∂  ∂ T
∂ T
kr
+
k
+
k
+ q =
rc






2
r ∂ r  ∂ r  r ∂φ  ∂φ  ∂ z  ∂ z 
∂ t
<
PROBLEM 2.28
KNOWN: Three-dimensional system – described by spherical coordinates (r,φ,q) – experiences
transient conduction and internal heat generation.
FIND: Heat diffusion equation.
SCHEMATIC: See Figure 2.13.
ASSUMPTIONS: (1) Homogeneous medium.
ANALYSIS: The differential control volume is V = dr⋅rsinqdφ⋅rdq, and the conduction terms are
identified in Figure 2.13. Conservation of energy requires
q r − q r +dr + q φ − q φ +dφ + qq − qq +dq + E g = E st .
(1)
The generation and storage terms, both representing volumetric phenomena, are
 =⋅
E g =
qV
q [ dr r sinq dφ ⋅ rdq ]
∂ T
∂ T
E st =
rr
Vc
=
[dr ⋅ r sinq dφ ⋅ rdq ] c .
∂ t
∂ t
(2,3)
Using a Taylor series expansion, we can write
∂
∂
( )
∂
q r+dr =
qr +
qφ +
qφ dφ , qq +dq =
qq +
( qq ) dq .
( q r ) dr, qφ +dφ =
∂ r
∂φ
∂q
(4,5,6)
From Fourier’s law, the conduction heat rates have the following forms.
qr =
−kA r∂ T/∂ r =
−k [ r sinq dφ ⋅ rdq ] ∂ T/∂ r
(7)
qφ =
−kAφ ∂ T/r sinq∂φ =
−k [ dr ⋅ rdq ] ∂ T/r sinq∂φ
(8)
−kAq ∂ T/r∂q =
−k [ dr ⋅ r sinq dφ ] ∂ T/r∂q .
qq =
(9)
Substituting Eqs. (2), (3) and (4), (5), (6) into Eq. (1), the energy balance becomes
−
∂
∂
∂
∂ T
r [ dr ⋅ r sinq dφ ⋅ rdq ] c
( q r ) dr − qφ dφ − ( qq ) dq +q [ dr ⋅ r sinq dφ ⋅ rdq ] =
∂ r
∂φ
∂q
∂ t
( )
(10)
Substituting Eqs. (7), (8) and (9) for the conduction rates, find
−
−
∂
∂q
∂ 
∂ T
∂ 
∂ T 
−k [ r sinθ dφ ⋅ rdθθ
dφ
]  dr −  −k [dr ⋅ rd ]

r sinθ∂φ 
∂r 
∂ r
∂φ 
∂ T
∂ T

r [ dr ⋅ r sinq dφ ⋅ rdq ] c
 −k [ dr ⋅ r sinq dφ ] r∂q  dq + q [ dr ⋅ r sinq dφ ⋅ rdq ] =
∂ t
(11)
Dividing Eq. (11) by the volume of the control volume, V, Eq. 2.29 is obtained.
∂  ∂ T
∂ 
∂ T
∂ T
1 ∂  2 ∂ T
1
1

+
+
q
+
=
r
kr
k
k
sin
q
c
.



∂ r  r 2sin 2q ∂φ  ∂ φ  r 2 sinq ∂q 
∂ q 
∂ t
r2 ∂ r 
<
COMMENTS: Note how the temperature gradients in Eqs. (7) - (9) are formulated. The numerator
is always ∂T while the denominator is the dimension of the control volume in the specified coordinate
direction.
PROBLEM 2.29
KNOWN: Temperature distribution in a semi-transparent medium subjected to radiative flux.
FIND: (a) Expressions for the heat flux at the front and rear surfaces, (b) Heat generation rate q ( x ) ,
(c) Expression for absorbed radiation per unit surface area in terms of A, a, B, C, L, and k.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in medium, (3)
Constant properties, (4) All laser irradiation is absorbed and can be characterized by an internal
volumetric heat generation term q ( x ) .
ANALYSIS: (a) Knowing the temperature distribution, the surface heat fluxes are found using
Fourier’s law,
 A

 dT 
-k   =
-q′′x =
k a ) e-ax + B
(
 dx 
 ka 2

Front Surface, x=0:
 A

A

q′′x ( 0 ) =−k  + ⋅1 + B =−  + kB
 ka

a

Rear Surface, x=L:
 A

A

- k  + e-aL + B =
-  e-aL + kB .
q′′x ( L ) =
 ka

a

<
(b) The heat diffusion equation for the medium is
d  dT  q
0 or

+ =
dx  dx  k
-k
q ( x ) =

q=-k
d  dT 


dx  dx 
d  A -ax

+ e + B =
Ae-ax .

dx  ka

<
(c) Performing an energy balance on the medium,
E in − E out + E g = 0
 g represents the absorbed irradiation. On a unit area basis
recognize that E
)
(
A
-E ′′in + E ′′out =
-q′′x ( 0 ) + q′′x ( L ) =
+ 1 - e-aL .
E ′′g =
a
 ′′ by integration over the volume of the medium,
Alternatively, evaluate E
g
=
E ′′g
(
)
L A
L
L
A
-aL .

q ( x )dx= Ae-ax dx=- e-ax=
 0 a 1 - e
0
0
a 
∫
∫
<
<
PROBLEM 2.30
PROBLEM 2.31
2
KNOWN: Steady-state temperature distribution in a one-dimensional wall is T(x) = Ax +
Bx + C, thermal conductivity, thickness.
FIND: Expressions for the heat fluxes at the two wall faces (x = 0,L) and the heat generation
rate in the wall per unit area.
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3)
Homogeneous medium.
ANALYSIS: The appropriate form of the heat diffusion equation for these conditions is
d 2T
q
+
=0
dx 2 k
or
q = -k
d 2T
dx 2
.
Hence, the generation rate is

q=-k
d  dT 
d
=
-k [ 2Ax + B + 0]


dx  dx 
dx

q=-k
[ 2A ]
<
which is constant. The heat fluxes at the wall faces can be evaluated from Fourier’s law,
dT
q′′x =
−k
=
−k [ 2Ax + B]
dx
using the expression for the temperature gradient derived above. Hence, the heat fluxes are:
Surface x=0:
q′′x ( 0 ) = −kB
<
Surface x=L:
q′′x ( L ) = −k [ 2AL +B] .
COMMENTS: (1) From an overall energy balance on the wall, find
E ′′in − E ′′out + E ′′g =
0
q′′x ( 0 ) − q′′x ( L ) + E ′′g =
0
( −kB ) − ( −k ) [ 2AL + B] + E ′′g =
E ′′g = −2AkL.
From integration of the volumetric heat rate, we can also find E ′′g as
L
L
E ′′g = ∫ q ( x )dx= ∫ -k [ 2A ]dx=-k [ 2AL ]
0
0
which agrees with the above, as it should.
<
PROBLEM 2.32
KNOWN: Plane wall with no internal energy generation.
FIND: Determine whether the prescribed temperature distribution is possible; explain your
reasoning. With the temperatures T(0) = 0°C and T∞ = 20°C fixed, compute and plot the temperature
T(L) as a function of the convection coefficient for the range 10 ≤ h ≤ 100 W/m2⋅K.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation, (3) Constant
properties, (4) No radiation exchange at the surface x = L, and (5) Steady-state conditions.
ANALYSIS: (a) Is the prescribed temperature distribution possible? If so, the energy balance at the
surface x = L as shown above in the Schematic, must be satisfied.
(1,2)
=
E in − E out 0
=
q′′x ( L ) − q′′cv 0
where the conduction and convection heat fluxes are, respectively,
T ( L) − T ( 0)
dT 

−k
=
−k
=
−4.5 W m ⋅ K × (120 − 0 ) C 0.18 m =
−3000 W m 2
q′′x ( L ) =

dx  x = L
L
=
q′′cv h [ T ( L ) −=
T∞ ] 30 W m 2 ⋅ K × (120 − 20=
) C 3000 W m 2
Substituting the heat flux values into Eq. (2), find (-3000) - (3000) ≠ 0 and therefore, the temperature
distribution is not possible.
(b) With T(0) = 0°C and T∞ = 20°C, the temperature at the surface x = L, T(L), can be determined
from an overall energy balance on the wall as shown above in the schematic,
T ( L) − T ( 0)
′′cv 0
= 0
−k
− h [ T ( L ) − T=
E in − E out
q′′x (0) − q=
∞] 0
L
−4.5 W m ⋅ K  T ( L ) − 0 C  0.18 m − 30 W m 2 ⋅ K  T ( L ) − 20 C  =
0




<
T(L) = 10.9°C
20
Surface temperature, T(L) (C)
Using this same analysis, T(L) as a function of
the convection coefficient can be determined
and plotted. We don’t expect T(L) to be
linearly dependent upon h. Note that as h
increases to larger values, T(L) approaches T∞ .
To what value will T(L) approach as h
decreases?
16
12
8
4
0
0
20
40
60
Convection cofficient, h (W/m^2.K)
80
100
PROBLEM 2.33
PROBLEM 2.33 (Cont.)
PROBLEM 2.34
KNOWN: Cylindrical system with negligible temperature variation in the r,z directions.
FIND: (a) Heat equation beginning with a properly defined control volume, (b) Temperature
distribution T(φ) for steady-state conditions with no internal heat generation and constant
properties, (c) Heat rate for Part (b) conditions.
SCHEMATIC:
ASSUMPTIONS: (1) T is independent of r,z, (2) ∆r = (ro - ri) << ri.
ANALYSIS: (a) Define the control volume as V = ridφ⋅∆r⋅L where L is length normal to page.
Apply the conservation of energy requirement, Eq. 1.12c,
E in − E out + E g = E st
 = ρVc
q φ − q φ +dφ + qV
qφ =−k ( ∆r ⋅ L )
where
∂ T
ri∂ φ
∂T
∂t
qφ +dφ =qφ +
(1,2)
∂
qφ dφ .
∂ φ
( )
(3,4)
Eqs. (3) and (4) follow from Fourier’s law, Eq. 2.1, and from Eq. 2.25, respectively. Combining
Eqs. (3) and (4) with Eq. (2) and canceling like terms, find
1 ∂  ∂ T
∂ T
 rc
k
.
+ q=


∂ t
ri2 ∂ φ  ∂ φ 
(5) <
Since temperature is independent of r and z, this form agrees with Eq. 2.26.
(b) For steady-state conditions with q = 0, the heat equation, (5), becomes
d
dφ
 dT 
 k dφ  = 0.


(6)
With constant properties, it follows that dT/dφ is constant which implies T(φ) is linear in φ. That
is,
dT T2 − T1
1
= =
+ ( T2 − T1 )
dφ φ2 − φ1
π
or
1
T (φ ) =
T1 + ( T2 − T1 ) φ .
π
(7,8) <
(c) The heat rate for the conditions of Part (b) follows from Fourier’s law, Eq. (3), using the
temperature gradient of Eq. (7). That is,
qφ =−k ( ∆r ⋅ L )
 ro − ri 
1 1

T
T
k
+
−
=−
(
)
2
1

 L ( T2 − T1 ) .

ri  π
r
π
i 

(9) <
COMMENTS: Note the expression for the temperature gradient in Fourier’s law, Eq. (3),
is ∂T/ri∂φ not ∂T/∂φ. For the conditions of Parts (b) and (c), note that qφ is independent of
φ; this is first indicated by Eq. (6) and confirmed by Eq. (9).
PROBLEM 2.35
KNOWN: Heat diffusion with internal heat generation for one-dimensional cylindrical,
radial coordinate system.
FIND: Heat diffusion equation.
SCHEMATIC:
ASSUMPTIONS: (1) Homogeneous medium.
ANALYSIS: Control volume has volume, V = A r ⋅ dr = 2πr ⋅ dr ⋅ 1, with unit thickness
normal to page. Using the conservation of energy requirement, Eq. 1.12c,
E in − E out + E gen = E st
 = rVc p
q r − q r +dr + qV
∂T
.
∂t
Fourier’s law, Eq. 2.1, for this one-dimensional coordinate system is
q r = − kA r
∂T
∂T
= − k × 2πr ⋅ 1 ×
.
∂r
∂r
At the outer surface, r + dr, the conduction rate is
q r+dr = q r +
∂
∂
∂ T
( q r ) dr=q r +  −k ⋅ 2π r ⋅  dr.
∂ r
∂ r
∂ r
Hence, the energy balance becomes

∂ 
∂ T 
∂ T
−k2pprp
q r − q r +
r
dr  + q ⋅ 2 rdr= ⋅ 2 rdr ⋅ cp


∂ r
∂ r 
∂ t

Dividing by the factor 2πr dr, we obtain
1 ∂  ∂ T
∂ T
 r cp
kr
.
+ q=


r∂ r ∂ r
∂ t
<
COMMENTS: (1) Note how the result compares with Eq. 2.26 when the terms for the
φ,z coordinates are eliminated. (2) Recognize that we did not require q and k to be
independent of r.
PROBLEM 2.36
KNOWN: Heat diffusion with internal heat generation for one-dimensional spherical,
radial coordinate system.
FIND: Heat diffusion equation.
SCHEMATIC:
ASSUMPTIONS: (1) Homogeneous medium.
2
ANALYSIS: Control volume has the volume, V = Ar ⋅ dr = 4πr dr. Using the
conservation of energy requirement, Eq. 1.12c,
E in − E out + E gen = E st
 = rVc p
q r − q r +dr + qV
∂T
.
∂t
Fourier’s law, Eq. 2.1, for this coordinate system has the form
q r = − kA r
∂T
∂T
= − k ⋅ 4πr 2 ⋅
.
∂r
∂r
At the outer surface, r + dr, the conduction rate is
q r+dr = q r +
∂
∂
∂ T
( q r ) dr = q r +  −k ⋅ 4π r 2 ⋅  dr.
∂ r
∂ r
∂ r
Hence, the energy balance becomes

∂ 
∂ T 
∂ T
−k ⋅ 4pprp
q r − q r +
r2 ⋅
dr  + q ⋅ 4 r 2dr= ⋅ 4 r 2dr ⋅ cp
.


∂ r
∂ r 
∂ t

Dividing by the factor 4πr 2 dr, we obtain
1 ∂  2 ∂ T
∂ T
 r cp
+ q=
kr
.


∂ r
∂ t
r2 ∂ r 
<
COMMENTS: (1) Note how the result compares with Eq. 2.29 when the terms for the
q,φ directions are eliminated.
(2) Recognize that we did not require q and k to be independent of the coordinate r.
PROBLEM 2.37
PROBLEM 2.38
KNOWN: Radii and thermal conductivity of conducting rod and cladding material. Volumetric
rate of thermal energy generation in the rod. Convection conditions at outer surface.
FIND: Heat equations and boundary conditions for rod and cladding.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3) Constant
properties.
ANALYSIS: From Equation 2.26, the appropriate forms of the heat equation are
Conducting Rod:
k r d  dTr
r
r dr  dr
 
0
+q =

<
Cladding:
d  dTc 
r
= 0.
dr  dr 
<
Appropriate boundary conditions are:
(a)
dTr / dr|r=0 = 0
<
(b)
Tr ( ri ) = Tc ( ri )
<
(c)
kr
(d)
dTc
- k=
|r h Tc ( ro ) - T∞ 
c
dr o
dT
dTr
|ri =k c c |ri
dr
dr
<
<
COMMENTS: Condition (a) corresponds to symmetry at the centerline, while the interface
conditions at r = ri (b,c) correspond to requirements of thermal equilibrium and conservation of
energy. Condition (d) results from conservation of energy at the outer surface. Note that contact
resistance at the interface between the rod and cladding has been neglected.
PROBLEM 2.39
PROBLEM 2.39 (Cont.)
PROBLEM 2.39 (Cont.)
PROBLEM 2.40
PROBLEM 2.40 (Cont.)
PROBLEM 2.40 (Cont.)
PROBLEM 2.40 (Cont.)
PROBLEM 2.41
KNOWN: Temperature distribution in a spherical shell.
FIND: Whether conditions are steady-state or transient. Manner in which heat flux and heat rate
vary with radius.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties.
ANALYSIS: From Equation 2.29, the heat equation reduces to
1 ∂  2∂ T 1 ∂ T
.
r
=
r2 ∂ r  ∂ r  α ∂ t
Substituting for T(r),
1∂ T
1 ∂  2 C1 
0.
=
−
r
=
α ∂ t
r2 ∂ r  r2 
<
Hence, steady-state conditions exist.
From Equation 2.28, the radial component of the heat flux is
C
∂ T
=
k 1.
∂ r
r2
−k
q ′′r =
(
)
Hence, q ′′r decreases with increasing r 2 q′′r ∝ 1/ r 2 .
<
At any radial location, the heat rate is
q r = 4πr 2 q ′′r = 4πkC1.
Hence, qr is independent of r.
<
COMMENTS: The fact that qr is independent of r is consistent with the energy conservation
requirement. If qr is constant, the flux must vary inversely with the area perpendicular to the
2
direction of heat flow. Hence, q ′′r varies inversely with r .
PROBLEM 2.42
PROBLEM 2.43
PROBLEM 2.43 (Cont.)
PROBLEM 2.44
KNOWN: Plane wall, initially at a uniform temperature, is suddenly exposed to convective
heating.
FIND: (a) Differential equation and initial and boundary conditions which may be used to find the
temperature distribution, T(x,t); (b) Sketch T(x,t) for these conditions: initial (t ≤ 0), steady-state, t
→ ∞, and two intermediate times; (c) Sketch heat fluxes as a function of time for surface locations;
3
(d) Expression for total energy transferred to wall per unit volume (J/m ).
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal heat
generation.
ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has
the form,
∂ 2T 1 ∂ T
=
∂ x2 α ∂ t
and the
conditions are:
Initial, t ≤ 0 : T ( x,0 ) =
Ti

∂
∂
Boundaries:
x=0
T/
x)0 = 0


x=L − k∂ T/∂ x)L = h T ( L,t ) − T∞ 

uniform
adiabatic
convection
(b) The temperature distributions are shown on the sketch.
Note that the gradient at x = 0 is always zero, since this boundary is adiabatic. Note also that the
gradient at x = L decreases with time.
(c) The heat flux, q′′x ( x,t ) , as a function of time, is shown on the sketch for the surfaces x = 0
and x = L.
Continued …
PROBLEM 2.44 (Cont.)
For the surface
at x 0,=
=
q′′x ( 0, t ) 0 since it is adiabatic. At x = L and t = 0, q′′x ( L,0 ) is a
maximum (in magnitude)
′′x ( L,0 ) h T ( L,0 ) − T∞
q=
where T(L,0) = Ti. The temperature difference, and hence the flux, decreases with time.
(d) The total energy transferred to the wall may be expressed as
∞
0
Ein = ∫ q′′conv As dt
∞
T∞ − T ( L,t ) dt
0
=
Ein hAs ∫
(
)
Dividing both sides by AsL, the energy transferred per unit volume is
Ein h ∞
T∞ − T ( L,t ) dt
=
V
L ∫0 
 J/m3 


COMMENTS: Note that the heat flux at x = L is into the wall and is hence in the negative x
direction.
PROBLEM 2.45
KNOWN: Qualitative temperature distributions in two cases.
FIND: For each of two cases, determine which material (A or B) has the higher thermal
conductivity, how the thermal conductivity varies with temperature, description of the heat flux
distribution through the composite wall, effect of simultaneously doubling the wall thickness and
thermal conductivity.
SCHEMATIC:
T(x)
LA
LB
kB
kA
T(x)
LA
kA
x
Case 1.
LB
kB
x
Case 2.
ASSUMPTIONS: (1) Steady-state, one-dimensional conditions, (2) Negligible contact
resistances, (3) No internal energy generation.
ANALYSIS: Under steady-state conditions with no internal generation, the conservation of
energy requirement dictates that the heat flux through the wall must be constant.
<
dT
dT
"
=
−kA A =
q"B =
−k B B . Therefore,
For Materials A and B, Fourier’s law is written qA
dx
dx
kA dTB / dx
=
> 1 and kB < kA for both cases.
k B dTA / dx
<
Since the heat flux through the wall is constant, Fourier’s law dictates that lower thermal
conductivity material must exist where temperature gradients are larger. For Case 1, the
temperature distributions are linear. Therefore, the temperature gradient is constant in each
material, and the thermal conductivity of each material must not vary significantly with
temperature. For Case 2, Material A, the temperature gradient is larger at lower temperatures.
Hence, for Material A the thermal conductivity increases with increasing material temperature. For
Case 2, Material B, the temperature gradient is smaller at lower temperatures. Hence, for Material
B the thermal conductivity decreases with increases in material temperature.
<
COMMENTS: If you were given information regarding the relative values of the thermal
conductivities and how the thermal conductivities vary with temperature in each material, you
should be able to sketch the temperature distributions provided in the problem statement.
PROBLEM 2.46
PROBLEM 2.47
KNOWN: Qualitative temperature distribution in a composite wall with one material
experiencing uniform volumetric energy generation.
FIND: Which material experiences uniform volumetric generation. The boundary condition at
x = -LA. Temperature distribution if the thermal conductivity of Material A is doubled.
Temperature distribution if the thermal conductivity of Material B is doubled. Sketch the heat flux
distribution q"x ( x ) through the composite wall.
SCHEMATIC:
T(x)
qx″
LA
LB
kA
kB
x
ASSUMPTIONS: (1) Steady-state, one-dimensional conditions, (2) Constant properties.
ANALYSIS: Consider a control volume with the LHS control surface at the interface between
the two materials and the RHS control surface located at an arbitrary location within Material B, as
shown in the schematic. For this control volume, conservation of energy and Fourier’s law may be
combined to yield, for uniform volumetric generation in Material B,
q" ( ) = q"x = −k
dT
dx
or
x =
dT
dx
x =
∝ 
(1)
The temperature distribution of the problem reflects the preceding proportionality between the
temperature gradient and the distance  , and it is appropriate to assume that uniform volumetric
<
generation occurs in Material B but not in Material A.
The boundary condition at x = -LA is associated with perfectly insulated conditions,
dT
0=
− LA ) =
−k
q"x ( x =
dx
or
x =− LA
dT
dx
=0
<
x =− LA
The temperature distribution in Material A corresponds to q"x ,A = 0 , and is independent of its thermal
<
conductivity.
Continued...
PROBLEM 2.47 (Cont.)
If the volumetric energy generation rate, q , is unchanged, Equation (1) requires that the
temperature gradient everywhere in Material B will be reduced by half if the thermal conductivity
of Material B is doubled. Hence, the difference between the minimum and maximum temperatures
<
in the composite wall would be reduced by half.
Considering Eq. 1, it follows that the heat flux distribution throughout the composite wall is as
shown in the sketch below.
qx″
0
<
-LA
0
x
LB
COMMENTS: If you were given information regarding which material experiences internal
energy generation, the boundary condition at x = -LA, and the thermal conductivities of both
materials, you should be able to sketch the temperature and heat flux distributions.
PROBLEM 2.48
KNOWN: Size and thermal conductivities of a spherical particle encased by a spherical shell.
FIND: (a) Relationship between dT/dr and r for 0 ≤ r ≤ r1, (b) Relationship between dT/dr and r
for r1 ≤ r ≤ r2, (c) Sketch of T(r) over the range 0 ≤ r ≤ r2.
SCHEMATIC:
Chemical reaction
.
q
r
r1
Ambient air
T∞ , h
r2
Control volume B
Control volume A
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat
transfer.
ANALYSIS:
(a) The conservation of energy principle, applied to control volume A, results in
E in + E g - E out = E st
4
∀ = q πr 3
where E∀ g = q∀∀
3
since
(1)
(2)
E st = 0
dT
E in - E out = q′′r A = - (-k1
)(4πr 2 )
dr
(3)
Substituting Eqs. (2) and (3) in Eq. (1) yields
4
dT
q πr 3 + k1
(4πr 2 ) = 0
3
dr
or
dT
q r
=dr
3 k1
<
Continued…
PROBLEM 2.48 (Cont.)
(b) For r > r1, the radial heat rate is constant and is
4
E∀ g = q r = q∀∀
∀1 = q πr13
3
(4)
dT
E in - E out = q′′r A = - (-k 2
)4πr 2
dr
(5)
Substituting Eqs. (4) and (5) into Eq. (1) yields
k2
dT
4
4πr 2 + q π r13 = 0
dr
3
or
 13
qr
dT
=dr
3k 2 r 2
<
(c) The temperature distribution on T-r coordinates is
COMMENTS: (1) Note the non-linear temperature distributions in both the particle and the shell.
(2) The temperature gradient at r = 0 is zero. (3) The discontinuous slope of T(r) at r1/r2 = 0.5 is a
result of k1 = 2k2.
PROBLEM 2.49
KNOWN: Temperature distribution in a plane wall of thickness L experiencing uniform
volumetric heating having one surface (x = 0) insulated and the other exposed to a convection
process characterized by T∞ and h. Suddenly the volumetric heat generation is deactivated while
convection continues to occur.
FIND: (a) Determine the magnitude of the volumetric energy generation rate associated with the
initial condition, (b) On T-x coordinates, sketch the temperature distributions for the initial
condition (T ≤ 0), the steady-state condition (t → ∞), and two intermediate times; (c) On
-t
coordinates, sketch the variation with time of the heat flux at the boundary exposed to the
calculate the corresponding value of the heat flux at t = 0; and (d)
convection process,
2
Determine the amount of energy removed from the wall per unit area (J/m ) by the fluid stream as
the wall cools from its initial to steady-state condition.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, and (3) Uniform
internal volumetric heat generation for t < 0.
ANALYSIS: (a) The volumetric heating rate can be determined by substituting the temperature
distribution for the initial condition into the appropriate form of the heat diffusion equation.
<
(b) The temperature distributions are shown in the sketch below.
Continued …
PROBLEM 2.49 (Cont.)
(c) The heat flux at the exposed surface x = L, q ′′x ( L, 0 ) , is initially a maximum value and
decreases with increasing time as shown in the sketch above. The heat flux at t = 0 is equal to the
convection heat flux with the surface temperature T(L,0). See the surface energy balance
represented in the schematic.
q′′x ( L, 0 ) = q′′conv ( t = 0 ) = h ( T ( L, 0 ) − T∞ ) = 1000 W / m 2 ⋅ K ( 200 − 20 ) °C = 1.80 × 105 W / m 2
<
2
where T ( L, 0 ) = a + bL2 = 300°C − 1.0 × 104°C / m 2 ( 0.1m ) = 200°C.
(d) The energy removed from the wall to the fluid as it cools from its initial to steady-state
condition can be determined from an energy balance on a time interval basis, Eq. 1.12b. For the
2
initial state, the wall has the temperature distribution T(x,0) = a + bx ; for the final state, the wall
is at the temperature of the fluid, Tf = T∞. We have used T∞ as the reference condition for the
energy terms.
E′′in − E′′out =
∆E′′st =
E′′f − E′′i
with
E′′in =
0
x =L
T ( x, 0 ) − T∞ dx
E′′out ρ cp ∫
=
x =0 
x =L 
= ρ cp ∫
= ρ cp ax + bx 3 / 3 − T∞ x 
E′′out
a + bx 2 − T∞ dx
x = 0 


0
L
3
=
E′′out 7000 kg / m3 × 450 J / kg ⋅ K 300 × 0.1 − 1.0 ×104 ( 0.1) / 3 − 20 × 0.1 K ⋅ m


E′′out
= 7.77 ×107 J / m 2
<
COMMENTS: (1) In the temperature distributions of part (a), note these features: initial
condition has quadratic form with zero gradient at the adiabatic boundary; for the steady-state
condition, the wall has reached the temperature of the fluid; for all distributions, the gradient at the
adiabatic boundary is zero; and, the gradient at the exposed boundary decreases with increasing
time.
(2) In this thermodynamic analysis, we were able to determine the energy transferred during the
cooling process. However, we cannot determine the rate at which cooling of the wall occurs
without solving the heat diffusion equation.
PROBLEM 2.50
PROBLEM 2.50 (Cont.)
PROBLEM 2.51
KNOWN: Thickness of composite plane wall consisting of material A in left half and material B
in right half. Exothermic reaction in material A and endothermic reaction in material B, with equal
and opposite heat generation rates. External surfaces are insulated.
FIND: Sketch temperature and heat flux distributions for three thermal conductivity ratios, kA/kB.
SCHEMATIC:
Material A
Material B
q∙A
q∙B = - q∙A
-L
L
x
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties.
ANALYSIS: From Equation 2.19 for steady-state, one-dimensional conduction, we find
∂  ∂T
k
∂x  ∂x
∂q′′x

= q
 = −q or
∂x

From the second equation, with uniform heat generation rate, we see that q′′x varies linearly with x,
and its slope is + qA in material A and - qA in material B. Furthermore, since the wall is insulated
on both exterior surfaces, the heat flux must be zero at x = ±L. Thus, the heat flux is as shown in
the graph below and does not depend on the thermal conductivities. The heat generated in the left
half is conducting to the right and accumulating as it goes. Once it reaches the centerline, it begins
to be consumed by the exothermic reaction and drops to zero at x = L.
kA = 0.5kB
kA = kB
kA = 2kB
T
qx"
0
-1
-0.5
0
0.5
1
x/L
Continued…
PROBLEM 2.51 (Cont.)
Since q′′x = − k
∂T
, the temperature gradient is negative everywhere, and its magnitude is greatest
∂x
where the heat flux is greatest. Thus the slope of the temperature distribution is zero at x = -L, it
becomes more negative as it reaches the center, and then becomes flatter again until it reaches a
slope of zero at x = L. When kA = kB, the temperature distribution has equal and opposite slopes on
either side of the centerline. If kB is held fixed and kA is varied, the results are as shown in the plot
above. Since the temperature gradient is inversely proportional to the thermal conductivity, it is
steeper in the region that has the smaller thermal conductivity. Physically, when thermal
conductivity is larger, heat conducts more readily and causes the temperature to become more
uniform.
If qB = −2qA , an energy balance on the wall gives:
dEst 
=Ein − E out + E g
dt
dEst 
=
(qA + qB )V =
− qAV
Eg =
dt
where V is the volume. Since dEst/dt is non-zero, the wall cannot be at steady-state. With the
exothermic reaction greater than the endothermic reaction, the wall will continuously decrease in
temperature.
<
COMMENTS: (1) Given the information in the problem statement, it is not possible to calculate
actual temperatures. There are an infinite number of correct solutions regarding temperature
values, but only one correct solution regarding the shape of the temperature distribution. (2)
Chemical reactions would cease if the temperature became too small. It would not be possible to
continually cool the wall for the case when, initially, qB = −2qA .
PROBLEM 2.52
KNOWN: Radius and length of coiled wire in hair dryer. Electric power dissipation in the wire,
and temperature and convection coefficient associated with air flow over the wire.
FIND: (a) Form of heat equation and conditions governing transient, thermal behavior of wire
during start-up, (b) Volumetric rate of thermal energy generation in the wire, (c) Sketch of
temperature distribution at selected times during start-up, (d) Variation with time of heat flux at r =
0 and r = ro.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties, (3) Uniform
volumetric heating, (4) Negligible radiation from surface of wire.
ANALYSIS: (a) The general form of the heat equation for cylindrical coordinates is given by Eq.
2.26. For one-dimensional, radial conduction and constant properties, the equation reduces to
1 ∂  ∂T  q r c p ∂T 1 ∂T
+
 r =
r ∂r  ∂r  k
k
The initial condition is
T ( r, 0 ) = Ti
The boundary conditions are:
∂T / ∂r r = 0 =
0
−k
∂T
∂r r = r
o
<
=
∂t α ∂t
<
<
<
= h [ T ( ro , t ) − T∞ ]
(b) The volumetric rate of thermal energy generation is
q∀
=
E∀ g
P
500 W
8
3
= elec
=
= 3.18 × 10 W / m
2
2
∀
π ro L π ( 0.001m ) ( 0.5m )
<
Under steady-state conditions, all of the thermal energy generated within the wire is transferred to
the air by convection. Performing an energy balance for a control surface about the wire,
− E out + E g =
0, it follows that −2π ro L q ′′ ( ro , t → ∞ ) + Pelec =0. Hence,
q ′′ ( ro , t → ∞
=
)
Pelec
500 W
5
2
=
= 1.59 × 10 W / m
2π ro L 2π ( 0.001m ) 0.5m
<
COMMENTS: The symmetry condition at r = 0 imposes the requirement that ∂T / ∂r r = 0 =
0, and
hence q ′′ ( 0, t ) = 0 throughout the process. The temperature at ro, and hence the convection heat
flux, increases steadily during the start-up, and since conduction to the surface must be balanced
by convection from the surface at all times, ∂T / ∂r r = r also increases during the start-up.
o
PROBLEM 2.53
KNOWN: Temperature distribution in a composite wall.
FIND: (a) Relative magnitudes of interfacial heat fluxes, (b) Relative magnitudes of
thermal conductivities, and (c) Heat flux as a function of distance x.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3)
Constant properties.
ANALYSIS: (a) For the prescribed conditions (one-dimensional, steady-state, constant
k), the parabolic temperature distribution in C implies the existence of heat generation.
Hence, since dT/dx increases with decreasing x, the heat flux in C increases with
decreasing x. Hence,
q′′3 > q′′4
However, the linear temperature distributions in A and B indicate no generation, in which
case
q′′2 = q′′3
(b) Since conservation of energy requires
that q′′3,B q′′3,C and dT/dx)B < dT/dx)C , it
=
follows from Fourier’s law that
k B > kC.
Similarly,
since q′′2,A q′′2,B and dT/dx)A > dT/dx)B , it follows that
=
k A < k B.
(c) It follows that the flux distribution appears as shown below.
COMMENTS: Note that, with dT/dx)4,C = 0, the interface at 4 is adiabatic.
PROBLEM 3.1
KNOWN: One-dimensional, plane wall separating hot and cold fluids at T∞ ,1 and T∞ ,2 ,
respectively.
FIND: Temperature distribution, T(x), and heat flux, q ′′x , in terms of T∞ ,1 , T∞ ,2 , h1 , h 2 , k
and L.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant
properties, (4) Negligible radiation, (5) No generation.
ANALYSIS: For the foregoing conditions, the general solution to the heat diffusion equation is
of the form, Equation 3.2,
T (=
x ) C1x + C2 .
(1)
The constants of integration, C1 and C2, are determined by using surface energy balance
conditions at x = 0 and x = L, Equation 2.34, and as illustrated above,
dT 
dT 
−k 
=h1 T∞,1 − T ( 0 ) 
−k 
=h 2 T ( L ) − T∞,2  .
(2,3)
dt  x=0
dx  x=L
For the boundary condition at x = 0, Equation (2), use Equation (1) to find
−k ( C=
1 + 0 ) h1  T∞,1 − ( C1 ⋅ 0 + C2 ) 
(4)
and for the boundary condition at x = L to find
−k ( C1=
+ 0 ) h 2 ( C1L + C2 ) − T∞,2  .
(5)
Multiply Eq. (4) by h2 and Eq. (5) by h1, and add the equations to obtain C1. Then substitute C1
into Eq. (4) to obtain C2. The results are
T∞,1 − T∞,2
T∞,1 − T∞,2
C1 =
C2 =
−
−
+ T∞,1
1
1
1 L
1 L
k +
h1  +
+ 
+ 
 h1 h 2 k 
 h1 h 2 k 
T∞,1 − T∞,2  x 1 
<
T(x) =
−
 +  + T∞,1.
1
1 L   k h1 
h + h + k 
2
 1

(
)
(
(
)
)
From Fourier’s law, the heat flux is a constant and of the form
T∞,1 − T∞,2
dT
q′′x =
−k
=
−k C1 =
+
.
dx
1
1 L
h + h + k 
2
 1

(
)
<
PROBLEM 3.2
PROBLEM 3.3
PROBLEM 3.4
PROBLEM 3.5
KNOWN: Curing of a transparent film by radiant heating with substrate and film surface subjected to
known thermal conditions.
FIND: (a) Thermal circuit for this situation, (b) Radiant heat flux, q′′o (W/m2), to maintain bond at
curing temperature, To, (c) Compute and plot q′′o as a function of the film thickness for 0 ≤ Lf ≤ 1 mm,
and (d) If the film is not transparent, determine q′′o required to achieve bonding; plot results as a function
of Lf.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) All the radiant heat
flux q′′o is absorbed at the bond, (4) Negligible contact resistance.
ANALYSIS: (a) The thermal circuit for
this situation is shown at the right. Note
that terms are written on a per unit area
basis.
(b) Using this circuit and performing an energy balance on the film-substrate interface,
q′′o= q1′′ + q′′2
=
q′′o
To − T∞
T −T
+ o 1
R ′′cv + R ′′f
R ′′s
where the thermal resistances are
R ′′cv = 1 h = 1 50 W m 2 ⋅ K = 0.020 m 2 ⋅ K W
=
⋅ K 0.010 m 2 ⋅ K W
R ′′f Lf =
k f 0.00025 m 0.025 W m=
′′s Ls =
R
=
k s 0.001m 0.05 W m =
⋅ K 0.020 m 2 ⋅ K W


( 60 − 20 ) C
( 60 − 30 ) C
+
=
q′′o =
2833 W m 2
(1333 + 1500 ) W m 2 =
2
2
[0.020 + 0.010] m ⋅ K W 0.020 m ⋅ K W
<
(c) For the transparent film, the radiant flux required to achieve bonding as a function of film thickness Lf
is shown in the plot below.
(d) If the film is opaque (not transparent), the thermal circuit is shown below. In order to find q′′o , it is
necessary to write two energy balances, one around the Ts node and the second about the To node.
.
The results of the analysis are plotted below.
Continued...
PROBLEM 3.5 (Cont.)
COMMENTS: (1) When the film is transparent, the radiant flux is absorbed on the bond. The flux
required decreases with increasing film thickness. Physically, how do you explain this? Why is the
relationship not linear?
(2) When the film is opaque, the radiant flux is absorbed on the surface, and the flux required increases
with increasing thickness of the film. Physically, how do you explain this? Why is the relationship
linear?
(3) The IHT Thermal Resistance Network Model was used to create a model of the film-substrate system
and generate the above plot. The Workspace is shown below.
// Thermal Resistance Network Model:
// The Network:
// Heat rates into node j,qij, through thermal resistance
Rij
q21 = (T2 - T1) / R21
q32 = (T3 - T2) / R32
q43 = (T4 - T3) / R43
// Nodal energy balances
q1 + q21 = 0
q2 - q21 + q32 = 0
q3 - q32 + q43 = 0
q4 - q43 = 0
/* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points
at which there is no external source of heat. */
T1 = Tinf
// Ambient air temperature, C
//q1 =
// Heat rate, W; film side
T2 = Ts
// Film surface temperature, C
q2 = 0
// Radiant flux, W/m^2; zero for part (a)
T3 = To
// Bond temperature, C
q3 = qo
// Radiant flux, W/m^2; part (a)
T4 = Tsub
// Substrate temperature, C
//q4 =
// Heat rate, W; substrate side
// Thermal Resistances:
R21 = 1 / ( h * As )
// Convection resistance, K/W
R32 = Lf / (kf * As)
// Conduction resistance, K/W; film
R43 = Ls / (ks * As) // Conduction resistance, K/W; substrate
// Other Assigned Variables:
Tinf = 20
// Ambient air temperature, C
h = 50
// Convection coefficient, W/m^2.K
Lf = 0.00025
// Thickness, m; film
kf = 0.025
// Thermal conductivity, W/m.K; film
To = 60
// Cure temperature, C
Ls = 0.001
// Thickness, m; substrate
ks = 0.05
// Thermal conductivity, W/m.K; substrate
Tsub = 30// Substrate temperature, C
As = 1
// Cross-sectional area, m^2; unit area
PROBLEM 3.6
PROBLEM 3.7
KNOWN: Composite wall with known properties and prescribed convection processes at inner
and outer surfaces.
FIND: Required insulation thickness to limit heat flux to 60 W/m2.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Negligible
contact resistance.
PROPERTIES: Given, kins = 1.5 W/m∙K, kss = 15 W/m∙K.
ANALYSIS: The expression for the total thermal resistance for a unit area of the composite
wall follows from Eq. 3.18.
L
L
1 L
1
R ′′tot = + ss + ins + ss +
.
h1 k ss k ins k ins h 2
The heat flux through the wall is
(
)
∆T/R ′′tot =−
q′′ =
T∞,1 T∞,2 / R ′′tot .
The requirement that q′′ = 60 W/m2 determines the required R ′′tot and insulation thickness:
′′tot
=
R
( T∞,1 − T∞,2 ) / q′′
(
)
 T∞,1 − T∞,2

Lss 1 
Lss 1 
1
1


′′
=
−=
− −2
−
Lins k ins  R tot − − 2
 k ins
′′


h
k
h
q
h
k
h
1
ss
2
1
ss
2


 ( 50 − 25 ) °C 1
0.001 m
1 
L=
− −2
−

ins 1.5 W/m ⋅ K 
2
15 W/m ⋅ K h 2 
 60 W/m ⋅ K h1
Continued…
PROBLEM 3.7 (Cont.)
For Case 1, h1 = h2 = 5 W/m2⋅K, and the insulation thickness is:
 ( 50 − 25 ) °C

1
0.001 m
1
Lins =
1.5 W/m ⋅ K 
−
−2
−
0.0248 m
=
2
2
15 W/m ⋅ K 5 W/m 2 ⋅ K 
 60 W/m ⋅ K 5 W/m ⋅ K
<
The other cases can be calculated similarly and the results are shown in the table.
______________________________________________
Case
h1 (W/m2∙K) h2 (W/m2∙K) Lins (m)
______________________________________________
1
5
5
0.0248
2
50
5
0.295
3
50
500
0.592
4
5000
5000
0.624
______________________________________________
<
COMMENTS: As the heat transfer coefficients increase, the required insulation thickness
increases. As the heat transfer coefficients approach infinity, the maximum required insulation
thickness is 0.625 m.
PROBLEM 3.8
PROBLEM 3.9
KNOWN: Design and operating conditions of a heat flux gage.
FIND: (a) Convection coefficient for water flow (Ts = 27°C) and error associated with neglecting
conduction in the insulation, (b) Convection coefficient for air flow (Ts = 125°C) and error associated
with neglecting conduction and radiation, (c) Effect of convection coefficient on error associated with
neglecting conduction for Ts = 27°C.
SCHEMATIC:
e ε= =0.15
0.15
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant k.
ANALYSIS: (a) The electric power dissipation is balanced by convection to the water and conduction
through the insulation. An energy balance applied to a control surface about the foil therefore yields
′′ = q′′conv + q′′cond = h ( Ts − T∞ ) + k ( Ts − Tb ) L
Pelec
Hence,
′′ − k ( Ts − Tb ) L 2000 W m 2 − 0.04 W m ⋅ K ( 2 K ) 0.01m
Pelec
=
Ts − T∞
2K
h
=
( 2000 − 8 ) W
m2
= 996 W m 2 ⋅ K
2K
h
=
<
If conduction is neglected, a value of h = 1000 W/m2⋅K is obtained, with an attendant error of (1000 996)/996 = 0.40%
(b) In air, energy may also be transferred from the foil surface by radiation, and the energy balance yields
)
(
4
′′ = q′′conv + q′′rad + q′′cond = h ( Ts − T∞ ) + es Ts4 − Tsur
+ k ( Ts − Tb ) L
Pelec
Hence,
h=
)
(
4
′′ − es Ts4 − Tsur
Pelec
− k ( Ts − T∞ ) L
Ts − T∞
2
2000 W m − 0.15 × 5.67 × 10
=
−8
2
W m ⋅K
4
( 398
4
− 298
4
)K
4
− 0.04 W m ⋅ K (100 K) / 0.01m
100 K
=
( 2000 − 146 − 400 ) W
m2
= 14.5 W m 2 ⋅ K
100 K
<
Continued...
PROBLEM 3.9 (Cont.)
If conduction, radiation, or conduction and radiation are neglected, the corresponding values of h and the
percentage errors are 18.5 W/m2⋅K (27.6%), 16 W/m2⋅K (10.3%), and 20 W/m2⋅K (37.9%).
(c) For a fixed value of Ts = 27°C, the conduction loss remains at q′′cond = 8 W/m2, which is also the
′′ and q′′conv . Although this difference is not clearly shown in the plot for
fixed difference between Pelec
10 ≤ h ≤ 1000 W/m2⋅K, it is revealed in the subplot for 10 ≤ 100 W/m2⋅K.
200
Power dissipation, P''elec(W/m^2)
Power dissipation, P''elec(W/m^2)
2000
1600
1200
800
400
0
160
120
80
40
0
0
200
400
600
800
Convection coefficient, h(W/m^2.K)
No conduction
With conduction
1000
0
20
40
60
80
100
Convection coefficient, h(W/m^2.K)
No conduction
With conduction
Errors associated with neglecting conduction decrease with increasing h from values which are significant
for small h (h < 100 W/m2⋅K) to values which are negligible for large h.
COMMENTS: In liquids (large h), it is an excellent approximation to neglect conduction and assume
that all of the dissipated power is transferred to the fluid.
PROBLEM 3.10
KNOWN: A layer of fatty tissue with fixed inside temperature can experience different outside
convection conditions.
FIND: (a) Ratio of heat loss for different convection conditions, (b) Outer surface temperature
for different convection conditions, and (c) Temperature of still air which achieves same cooling
as moving air (wind chill effect).
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-state
conditions, (3) Homogeneous medium with constant properties, (4) No internal heat generation
(metabolic effects are negligible), (5) Negligible radiation effects.
PROPERTIES: Table A-3, Tissue, fat layer: k = 0.2 W/m⋅K.
ANALYSIS: The thermal circuit for this situation is
Hence, the heat rate is
=
q
Ts,1 − T∞
Ts,1 − T∞
=
.
R tot
L/kA + 1/ hA
Therefore,
L 1
 k + h 
windy
q′′calm
=
.
q′′windy
L 1
 k + h 
calm
Applying a surface energy balance to the outer surface, it also follows that
q′′cond = q′′conv .
Continued …
PROBLEM 3.10 (Cont.)
Hence,
)
(
(
k
Ts,1 − Ts,2 = h Ts,2 − T∞
L
k
T∞ +
Ts,1
hL
Ts,2 =
.
k
1+
hL
)
To determine the wind chill effect, we must determine the heat loss for the windy day and use it
to evaluate the hypothetical ambient air temperature, T∞′ , which would provide the same heat
loss on a calm day, Hence,
=
q′′
′
Ts,1 − T∞
Ts,1 − T∞
=
L 1
L 1
 k + h 
 k + h 
windy
calm
From these relations, we can now find the results sought:
0.003 m
1
+
q′′calm
0.2 W/m ⋅ K 65 W/m 2 ⋅ K 0.015 + 0.0154
(a)
=
=
0.003 m
1
q′′windy
0.015 + 0.04
+
0.2 W/m ⋅ K 25 W/m 2 ⋅ K
q′′calm
= 0.553
q′′windy
(b)
Ts,2 
calm
<
0.2 W/m ⋅ K
−15aa
C+
40 C
25 W/m 2 ⋅ K ( 0.003 m )
= 24.97a C
0.2 W/m ⋅ K
1+
25 W/m 2 ⋅ K ( 0.003 m )
(
)
(
)
−15 C +
Ts,2 
(c)
windy
(
0.2 W/m ⋅ K
)
65 W/m 2 ⋅ K ( 0.003m )
40 C
= 12.88 C
0.2 W/m ⋅ K
1+
65 W/m 2 ⋅ K ( 0.003m )

<
)
(
′ =
T∞
40 C − ( 40 + 15 ) C
<
( 0.003/0.2 + 1/ 25) =
−59 C
0.003
/
0.2
+
1/
65
(
)
<
COMMENTS: The wind chill effect is equivalent to a decrease of Ts,2 by 12.09°C and increase
-1
in the heat loss by a factor of (0.553) = 1.81.
PROBLEM 3.11
PROBLEM 3.11 (Cont.)
PROBLEM 3.12
KNOWN: Composite wall of a house with prescribed convection processes at inner and outer
surfaces.
FIND: (a) Expression for thermal resistance of house wall, Rtot; (b) Total heat loss, q(W); (c)
Effect on heat loss due to increase in outside heat transfer convection coefficient, ho; and (d)
Controlling resistance for heat loss from house.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Negligible
contact resistance.
(
)

PROPERTIES: Table A-3, T =
( Ti + To ) / 2 =
( 20 − 15) C/2=2.5C ≈ 300K : Fiberglass
3
blanket, 28 kg/m , kb = 0.038 W/m⋅K; Plywood siding, ks = 0.12 W/m⋅K; Plasterboard, kp =
0.17 W/m⋅K.
ANALYSIS: (a) The expression for the total thermal resistance of the house wall follows from
Eq. 3.18.
Lp
L
L
1
1
<
R tot =
+
+ b + s +
.
hi A k p A k b A ks A h o A
(b) The total heat loss through the house wall is
q=
∆T/R tot =
( Ti − To ) / R tot .
Substituting numerical values, find
R tot =
1
+
0.01m
+
0.10m
30W/m 2 ⋅ K × 350m 2 0.17W/m ⋅ K × 350m 2 0.038W/m ⋅ K × 350m 2
0.02m
1
+
+
2
2
0.12W/m ⋅ K × 350m
60W/m ⋅ K × 350m 2
R tot = [9.52 + 16.8 + 752 + 47.6 + 4.76] ×10−5 C/W = 831×10−5 C/W
The heat loss is then,

q=  20- ( -15 )  C/831×10-5 C/W=4.21 kW.
2
<
-5
(c) If ho changes from 60 to 300 W/m ⋅K, Ro = 1/hoA changes from 4.76 × 10 °C/W to 0.95 ×
-5
-5
10 °C/W. This reduces Rtot to 826 × 10 °C/W, which is a 0.6% decrease and hence a 0.6%
increase in q.
(d) From the expression for Rtot in part (b), note that the insulation resistance, Lb/kbA, is
752/830 ≈ 90% of the total resistance. Hence, this material layer controls the resistance of the
wall. From part (c) note that a 5-fold decrease in the outer convection resistance due to an
increase in the wind velocity has a negligible effect on the heat loss.
PROBLEM 3.13
KNOWN: Composite wall of a house with prescribed convection processes at inner and outer
surfaces.
FIND: Daily heat loss for prescribed diurnal variation in ambient air temperature.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall
thermal energy storage over 24 h periods), (2) Negligible contact resistance.
3
PROPERTIES: Table A-3, T ≈ 300 K: Fiberglas blanket (28 kg/m ), kb = 0.038 W/m⋅K;
Plywood, ks = 0.12 W/m⋅K; Plasterboard, kp = 0.17 W/m⋅K.
ANALYSIS: The heat loss may be approximated as Q =
24h T − T
∞,i
∞,o
∫
0
R tot
dt where
L p L b Ls
1 
+
+
+
 +

A  h i k p k b k s h o 

1 
1
0.01m
0.1m
0.02m
1
=
+
+
+
+
R tot


200m 2  30 W/m 2 ⋅ K 0.17 W/m ⋅ K 0.038 W/m ⋅ K 0.12 W/m ⋅ K 60 W/m 2 ⋅ K 
R tot=
11
R tot = 0.01454 K/W.
Hence the heat rate is
Q
=
12h
1  
2π

 ∫  293 −  273 + 5 sin
R tot  
24

 0
=
Q 68.8
24h

t   dt + ∫

12

2π

 293 −  273 + 11 sin 24



 
t   dt 
 

W  
2π t  12 
2π t  24 
 24 
 24 
+
20t+11
cos
  20t+5   cos
K ⋅h
 2π 
K  
24  0 
24  12 
 2π 


60
132
 
=
Q 68.8   240 + ( −1 − 1)  +  480 − 240 +
(1 + 1)  W ⋅ h
π
π
 


Q 68.8 {480-38.2+84.03} W ⋅ h
Q=36.18 kW ⋅ h=1.302 ×108J.
COMMENTS: From knowledge of the fuel cost, the total daily heating bill could be
determined. For example, at a cost of 0.18$/kW⋅h, the heating bill would be $6.52/day.
<
PROBLEM 3.14
PROBLEM 3.15
KNOWN: What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having
10 studs, each 2.5 m high)? Assume surfaces parallel to the x-direction are adiabatic.
FIND: Wall thermal resistance.
SCHEMATIC:
0.65 m
LA = 8 mm
Hardwood siding (A)
x
LB = LD = 130 mm
Hardwood (B)
40 mm
Gypsum (C)
Insulation
Glass fiber,
paper faced (D)
(28 kg/m3)
LC = 12 mm
ASSUMPTIONS: (1) Steady-state, one-dimensional conditions, (2) Planes parallel to x are adiabatic, (3)
Constant properties, (4) Negligible contact resistance.
PROPERTIES: Table A-3 (T ≈ 300 K): Hardwood siding, kA = 0.094 W/m⋅K; Hardwood, kB = 0.16
W/m⋅K; Gypsum, kC = 0.17 W/m⋅K; Insulation (glass fiber paper faced, 28 kg/m3), kD = 0.038 W/m⋅K.
ANALYSIS: Using the adiabatic
surface assumption, the thermal
circuit associated with a single unit
(enclosed by dashed lines) of the wall
is as shown to the right. The various
resistances are
RA,2 = LA/(kAAD)
RA,1 = LA/(kAAB)
RB= LB/(kBAB)
RD = LD/(kDAD)
RC,1 = LC/(kCAB)
RC,2 = LC/(kCAD)
0.008 m
= 0.8511 K/W
0.094 W/m ⋅ K (0.04 m × 2.5 m)
0.13 m
RB = (LB/kBAB) =
= 8.125 K/W
0.16 W/m ⋅ K (0.04 m × 2.5 m)
0.012 m
RC,1 = (LC/kCAB) =
= 0.7059 K/W
0.17 W/m ⋅ K (0.04 m × 2.5 m)
0.008 m
RA,2 = (LA/kAAD) =
= 0.0558 K/W
0.094 W/m ⋅ K (0.61 m × 2.5 m)
0.13 m
RD = (LD/kDAD) =
= 2.243 K/W
0.038 W/m ⋅ K (0.61 m × 2.5 m)
0.012 m
RC,2 = (LC/kCAD) =
= 0.0463 K/W
0.17 W/m ⋅ K (0.61 m × 2.5 m)
RA,1 = (LA/kAAB) =
Continued…
PROBLEM 3.15 (Cont.)
The total unit resistance is


1
1
+
Rtot,1 = 
 R A,1 +R B +R C,1 R A,2 +R D +R C,2 


−1
1
1


=
+

 0.8511+8.125+0.7059 0.0558+2.243+0.0463 
−1
= 1.888 K/W
With 10 such units in parallel, the total wall resistance is Rtot = (10 × 1/Rtot,1)-1 = 0.1888 K/W.
<
COMMENTS: (1) Contact resistance will increase the overall wall resistance relative to that calculated
here. (2) The total wall resistance assuming isothermal surfaces normal to the x direction is 0.1854 K/W,
which is within 2 % of the value found in this solution.
PROBLEM 3.16
PROBLEM 3.16 (Cont.)
PROBLEM 3.17
PROBLEM 3.18
PROBLEM 3.19
PROBLEM 3.20
PROBLEM 3.20 (Cont.)
PROBLEM 3.21
KNOWN: Dimensions of a container of water. Fusion temperature and outer surface
temperature. Initial solidification rate.
FIND: Interfacial thermal resistance shortly after the onset of freezing.
SCHEMATIC:
ASSUMPTIONS: (1) Linear temperature distribution through container walls, (2) Onedimensional heat transfer (corners can be ignored).
PROPERTIES: Given in Example 1.5, hsf =334 kJ/kg.
ANALYSIS: The heat transfer rate through the container walls can be related to the total
resistance between the outer wall surface and the ice surface. The total thermal resistance
includes the thermal resistance due to conduction through the wall and the interfacial thermal
resistance at the wall/ice interface.
=
q
Tm − T1
Tm − T1
=
R tot
L / k w A + R ′′t,c / A
 is given by
where Tm is the fusion or melting temperature. The solidification rate, m,
 q=
=
m
/ h sf
( Tm − T1 ) / hsf
L / k w A + R ′′t,c / A
This can be solved for the interfacial thermal resistance:
=
R ′′t,c
( Tm − T1 ) A −
 sf
mh
L
kw
( 0 + 30 ) °C × 6 × (0.1 m)2
0.005 m
=
−
=0.00778 m 2 ⋅ K/W
3
3
−
0.05 ×10 kg/s × 334 ×10 J/kg 0.05 W/m ⋅ K
<
COMMENTS: This analysis applies at the initial moment of solidification, when the ice layer
has negligible thickness and thermal resistance. As more ice solidifies, the overall thermal
resistance increases due to the thermal resistance of the thickening layer of ice. The rate of
solidification will therefore decrease with time.
PROBLEM 3.22
PROBLEM 3.22 (Cont.)
PROBLEM 3.23
PROBLEM 3.24
PROBLEM 3.25
PROBLEM 3.25 (Cont.)
PROBLEM 3.26
PROBLEM 3.27
KNOWN: Operating conditions, measured temperatures and heat input, and theoretical thermal
conductivity of a carbon nanotube.
FIND: (a) Thermal contact resistance between the carbon nanotube and the heating and sensing islands,
(b) Fraction of total thermal resistance between the heating and sensing islands due to thermal contact
resistance for 5 µm ≤ s ≤ 20 µm.
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat transfer,
(4) Isothermal heating and sensing islands, (5) Negligible radiation and convection heat transfer.
PROPERTIES: kcn,T = 5000 W/m∙K
ANALYSIS:
(a) The total thermal resistance between the heated and sensing island is
R t,tot =
s
k cn,T A cn
+ 2R t,c
The value of this total resistance is the same as the one posed in Example 3.4 with
kcn = 3113 W/m∙K and Rt,c = 0 or
s
k cn,T A cn
+ 2R t,c =
s
k cn A cn
for which
R t,c =
s
2A cn
 1
1 
5 × 10-6 m
1
1


=
× 



-16
2
 3113 W/m ⋅ K 5000 W/m ⋅ K 
 k cn k cn,T  2 × 1.54 × 10 m
<
= 1.97 × 106 K/W
(b) The fraction of the total resistance due to the thermal contact resistance is
f=
2R t,c


s
2R t,c + 

 k cn,T A cn 
=
2 × 1.97 × 106 K/W
s


2 × 1.97 × 106 K/W + 
-16
2
 5000 W/m ⋅ K × 1.54 × 10 m 
As evident in the plot below, the fraction of the total thermal resistance due to thermal contact decreases
from 0.38 at s = 5 µm to 0.13 at s = 20 µm.
Continued…
PROBLEM 3.27 (Cont.)
Fraction of Thermal Resistance due to Contact
0.4
Fraction, f
0.3
0.2
0.1
5
10
15
20
s, micrometers
COMMENT: To desensitize the experiment to uncertainty due to the unknown thermal contact
resistance values, a large separation distance between the islands is desired. As the separation distance
becomes large, however, the surface area of the carbon nanotube increases and surface heat losses by
radiation may invalidate the assumption of a linear temperature distribution along the length of the
nanotube. An optimal separation distance exists that will minimize the undesirable effects of the thermal
contact resistances and radiation loss from the surface of the nanotube.
PROBLEM 3.28
PROBLEM 3.28 (Cont.)
PROBLEM 3.29
KNOWN: Oak wood with a grain structure. Grains are highly porous and the wood is dry.
FIND: Fraction of oak cross-section that appears as being grained.
SCHEMATIC:
Radial
Cross grain
T1
T1
T2
T2
x
x
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) Thermal
conductivity of highly porous grain is that of air.
PROPERTIES: Table A.3, Oak, cross grain (300 K): kcross = 0.17 W/m⋅K; Oak, radial (300 K): krad =
0.19 W/m∙K. Table A.4, Air (300 K): kair = 0.0263 W/m∙K.
ANALYSIS: The cross grain condition is characterized by the lowest effective thermal conductivity.
Therefore,
kcross = k min = 0.17W/m ⋅ K=
1
1
=
(1)
(1 − εεεε
) / k s + / k f (1 − ) / k s + / 0.0263 W/m ⋅ K
Likewise, the radial condition exhibits the highest effective thermal conductivity. Hence,
k rad = k max = 0.19W/m ⋅ K =εεεε
k f + (1 − )k s = × 0.0263 W/m ⋅ K + (1 − )k s
(2)
The preceding two equations may be solved simultaneously to determine the two unknowns, ks and e,
yielding
e = 0.022
<
COMMENTS: (1) The predicted value of the solid thermal conductivity is ks = 0.1937 W/m∙K. (2) The
predicted value of e is rather low. In reality, the thermal conductivity of the grain is greater than that of
air. Doubling the value of kf to 0.0526 W/m∙K yields e = 0.061, which is consistent with estimates of e
for oak.
PROBLEM 3.30
KNOWN: Density of glass fiber insulation.
FIND: Maximum and minimum possible values of the effective thermal conductivity of the insulation at
T = 300 K, and comparison with the value listed in Table A.3.
SCHEMATIC:
T1
T2
x
ASSUMPTIONS: (1) Constant properties, (2) Negligible radiation, (3) Atmospheric pressure.
PROPERTIES: Table A.3, Glass (plate): kgl = 1.4 W/m∙K, rgl = 2500 kg/m3.Glass fiber batt (paper
faced, rins = 28 kg/m3) kins = 0.038 W/m∙K. Table A.4, Air (300 K): kair = 0.0263 W/m∙K, rair = 1.1614
kg/m3. Given: rins = 28 kg/m3.
ANALYSIS: The density of the glass fiber insulation may be related to the density of the air and glass
phases, and the volume fraction, e, as follows.
rins= εrair + (1 − ε ) r gl
Therefore,
=
ε
r gl − rins
2500 − 28
=
= 0.989
r gl − rair 2500 − 1.164
The minimum effective thermal conductivity is
keff,min
=
1
=
(1 − e ) / kglass + e / kair
1
(1 − 0.989 ) /1.4 W/m ⋅ K + 0.989/0.0263 W/m ⋅ K
<
= 0.0265 W/m ⋅ K
The maximum effective thermal conductivity is
keff,max = e k f + (1 − e ) k s = 0.989 × 0.0263 W/m ⋅ K + (1 − 0.989 ) × 1.4 W/m ⋅ K
= 0.0414 W/m ⋅ K
<
As expected, the predicted minimum and maximum effective thermal conductivities bracket the actual
effective thermal conductivity of kins = 0.038 W/m∙K.
<
COMMENT: Radiation internal to the glass fiber batting may be significant. If so, this will reduce the
insulating capability of the matt.
PROBLEM 3.31
PROBLEM 3.32
KNOWN: Density of stone mix concrete slab, within which are small spherical pockets of air.
FIND: Thermal conductivity of the aggregate.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties.
PROPERTIES: Table A.3 (T = 300 K): Stone mix concrete; ks = 1.4 W/m⋅K, rs= 2300 kg/m3, cp,s = 880
J/kg⋅K. Table A.4 (T = 300 K): Air; kf = 0.0263 W/m⋅K, rf = 1.1614 kg/m3, cp,f = 1007 J/kg⋅K.
ANALYSIS: The density of the aggregate, ra, is equal to its mass divided by its volume, where
the mass is the sum of the concrete and air masses. Thus the porosity, ε = V f / V , can be found
from the following equality:
ms + m f ρρ
sVs +
fVf
ρρερε
=
= s (1 − ) + f
a=
V
V
ρρ
1500 − 2300
a − s
=
= 0.348
ε =
1.1614 − 2300
ρρ
f − s
Once the porosity is known, the effective thermal conductivity can be found from Maxwell’s expression,
Eq. 3.25. Hence,
 k f + 2ks − 2e ( ks − k f ) 
 ks
keff ,a = 
 k f + 2k s + e ( k s − k f ) 
 0.0263 W/m ⋅ K + 2 ×1.4 W/m ⋅ K − 2 × 0.348 × (1.4 W/m ⋅ K − 0.0263 W/m ⋅ K ) 

 × 1.4 W/m ⋅ K
 0.0263 W/m ⋅ K + 2 ×1.4 W/m ⋅ K + 0.348 × (1.4 W/m ⋅ K − 0.0263 W/m ⋅ K ) 
= 0.79 W/m∙K
<
COMMENT: The thermal conductivity and density are reduced significantly relative to the stone mix
concrete values.
PROBLEM 3.33
KNOWN: Porosity distribution in a one-dimensional plane wall, e(x), thermal conductivities of solid and
fluid, wall thickness, temperature at x = 0, and heat flux.
FIND: Plot of the temperature distribution using the expressions for the maximum and minimum
effective thermal conductivities, Maxwell’s expression, and for the constant property case; keff(x) = ks.
SCHEMATIC:
e(x)=emax(x/L)
emax= 0.25
qx" =100 W/m2
ks = 10 W/m∙K
kf = 0.1 W/m∙K
T(x = 0) = 30°C
x
L=1m
ASSUMPTIONS: (1) Constant properties within the two phases, (2) Negligible radiation, (3) No
thermal energy generation, (4) Steady state, one-dimensional heat transfer.
PROPERTIES: Given: ks = 10 W/m∙K, kf = 0.1 W/m∙K.
ANALYSIS: Fourier’s law may be expressed as
dT
q"
= − x
dx
keff
(1)
We note that the heat flux is constant. The effective thermal conductivity may be evaluated from the
various formulae as follows.
Maximum effective thermal conductivity:
keff = keff,max = e k f + (1 − e ) k s
(2)
Minimum effective thermal conductivity:
=
keff k=
eff,min
1
(1 − e ) / ks + e / k f
(3)
Maxwell’s expression:
 k f + 2k s − 2e ( k s − k f ) 
keff k=
=

 ks
eff,Max
 k f + 2k s + e ( k s − k f ) 
(4)
Continued...
PROBLEM 3.33 (Cont.)
No dispersed phase:
keff = k s
(5)
Equations 2, 3, 4, or 5 may be substituted into Equation 1 and the expression may be integrated
numerically using a commercial code. IHT was used to generate the following temperature distributions.
Temperature Distribution T(x)
35
T (C)
30
keff, cp
25
keff, min
20
keff, Max
keff, max
15
0
0.2
0.6
0.4
0.8
1
x (m)
Constant Property (keff = ks)
Maxwell's Expression
Expression for keff, max
Expression for keff, min
The constant property solution exhibits a linear temperature distribution.
The introduction of the low thermal conductivity matter within the medium decreases its effective thermal
conductivity. From Equation (1), the temperature gradient must become larger in order to sustain the
imposed heat flux as the effective thermal conductivity decreases. The increased temperature gradients
are evident in the plot.
Predictions using Maxwell’s expression, and the expression for the maximum effective thermal
conductivity are in relatively good agreement. This is because Maxwell’s expression describes conduction
within a porous medium that is characterized by a contiguous solid phase; thermal energy can be
transferred across the entire thickness of the plane wall within the solid phase only. The concept of a
contiguous solid phase is also embedded in the assumptions that were made in deriving the expression for
keff,max. In contrast, the predictions using the expression for the minimum effective thermal conductivity
are not consistent with the other predictions. In deriving the expression for keff,min, it is assumed that
thermal energy must be transferred through the low thermal conductivity fluid phase as it propagates
through the porous wall. The non-contiguous solid phase associated with the minimum thermal
conductivity expression manifests itself as very large temperature gradients through the plane wall.
Continued...
PROBLEM 3.33 (Cont.)
COMMENTS: (1) It is important to be cognizant of the morphology of the porous medium before
selecting the appropriate expression for the effective thermal conductivity. (2) The IHT code is shown
below.
//Input phase properties, dimensions, thermal boundary condition, and porosity parameter.
ks = 10
kf = 0.1
L=1
emax = 0.25
qflux = 100
T1 = 30
eps = emax*(x/L)
//W/mK
//W/mK
//m
//dimensionless
//W/m^2
//Degrees C
//Constant Property Solution
Der(Tcp,x) = -qflux/ks
//Minimum Effective Thermal Conductivity Solution
Der(Tmin,x) = -qflux/keffmin
keffmin = 1/dena
dena=(1-eps)/ks + eps/kf
//Maximum Effective Thermal Conductivity Solution
Der(Tmax,x) = -qflux/keffmax
keffmax = eps*kf + (1 - eps)*ks
//Maxwell’s Expression
Der(TMax,x) = -qflux/keffMax
keffMax = ks*num/denb
num = kf+2*ks-2*eps*(ks-kf)
denb = kf+2*ks+eps*(ks-kf)
PROBLEM 3.34
PROBLEM 3.35
KNOWN: Conduction in a hollow cylinder with known dimensions and properties.
FIND: Expression for the radial heat rate and its value for given conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r-direction,
(3) No internal heat generation, (4) Constant properties.
PROPERTIES: Given, k= 2.5 W/m⋅K.
ANALYSIS: Based upon the assumptions, and following similar methodology to Example 3.5,
qr is a constant independent of r and the area normal to the direction of heat flow is A = 2π rL .
Accordingly,
dT
dT
qr =
−kA
=
−k2π rL
dr
dr
(1)
Separating variables and integrating from r1 to r2 to find the heat rate,
r2 dr
T
qr
= − ∫ s,2 dT.
∫
Ts,1
2π Lk r1 r
(2)
Integrating and solving for qr,
qr
ln(r2 / r1) =
Ts,1 − Ts,2 ,
2π Lk
Ts,1 − Ts,2
qr =
2π Lk
ln(r2 / r1)
<
Evaluating the expression for qr under the given conditions yields
q r = 2π × 2 m × 2.5 W/m ⋅ K ×
(100 − 67 ) °C
ln(75 mm / 50 mm)
= 1280 W
<
COMMENTS: (a) The expression for the heat rate agrees with Equation 3.32, as it must. (b) If
the temperature distribution were desired, the integration in Eq. (2) could now be repeated using
integration to arbitrary upper limits r and T(r).
PROBLEM 3.36
PROBLEM 3.37
KNOWN: Temperature dependence of the thermal conductivity, k.
FIND: Heat flux and form of temperature distribution for a plane wall.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-state
conditions, (3) No internal heat generation.
ANALYSIS: For the assumed conditions, qx and A(x) are constant and Eq. 3.26 gives
L
0
T
To
q′′x ∫ dx =
− ∫ 1 ( k o + aT )dT
=
q′′x
(
1
a
k o ( To − T1 ) + To2 − T12

L
2
) .
From Fourier’s law,
q′′x =
− ( k o + aT ) dT/dx.
Hence, since the product of (ko+aT) and dT/dx) is constant, decreasing T with increasing x
implies,
a > 0: decreasing (ko+aT) and increasing |dT/dx| with increasing x
a = 0: k = ko => constant (dT/dx)
a < 0: increasing (ko+aT) and decreasing |dT/dx| with increasing x.
The temperature distributions appear as shown in the above sketch.
PROBLEM 3.38
KNOWN: Temperature dependence of tube wall thermal conductivity.
FIND: Expressions for heat transfer per unit length and tube wall thermal (conduction)
resistance.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) No
internal heat generation.
ANALYSIS: From Eq. 3.29, the appropriate form of Fourier’s law is
dT
dT
qr =
−kA r
=
−k ( 2π rL )
dr
dr
dT
q′r = −2π kr
dr
dT
q′r =
−2π rk o (1 + aT ) .
dr
Separating variables,
q′ dr
− r
=k o (1 + aT ) dT
2π r
and integrating across the wall, find
q′
− r
2π
q′
− r
2π
q′
− r
2π
ro dr
∫ri
r
T
Ti
=
k o ∫ o (1+aT ) dT

r
aT 2  To
ln o =
k o T +

ri
2  Ti


ro
a


ln= k o ( To − Ti ) + To2 − Ti2 
ri
2


(
)
 a
 ( T − Ti )
−2π k o 1 + ( To + Ti )  o
q′r =
.
 2
 ln ( ro / ri )
It follows that the overall thermal resistance per unit length is
ln ( ro / ri )
Ti − To
R ′t =
.
=
q′r
 a

2π k o 1 + ( To + Ti ) 
 2

COMMENT: Note the necessity of the stated assumptions to treating q′r as independent of r.
<
<
PROBLEM 3.39
KNOWN: Steady-state temperature distribution of convex shape for material with k = ko(1 +
αT) where α is a constant and the mid-point temperature is ∆To higher than expected for a linear
temperature distribution.
FIND: Relationship to evaluate α in terms of ∆To and T1, T2 (the temperatures at the
boundaries).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal
heat generation, (4) α is positive and constant.
ANALYSIS: At any location in the wall, Fourier’s law has the form
dT
(1)
−k o (1 + α T ) .
q′′x =
dx
Since q′′x is a constant, we can separate Eq. (1), identify appropriate integration limits, and
integrate to obtain
L
T2
− ∫ k o (1 + α T )dT
∫0 q′′x dx =
T1
(2)

α T22  
α T12  
k o 


 .
′′
(3)
−
− T1 +
qx =
T2 +
L 
2  
2 
 


We could perform the same integration, but with the upper limits at x = L/2, to obtain

α T2  
α T12  
2k 

(4)
q′′x =
− o  TL/2 + L/2  −  T1 +
L 
2  
2 

 

where
T +T
(5)
=
TL/2 T=
( L/2 ) 1 2 + ∆To .
2
Setting Eq. (3) equal to Eq. (4), substituting from Eq. (5) for TL/2, and solving for α, it follows
that
2∆To
<
.
α=
2
T22 + T12 / 2 − ( T1 + T2 ) / 2 + ∆To 
(
)
PROBLEM 3.40
KNOWN: Construction and dimensions of a device to measure the temperature of a surface. Ambient
and sensing temperatures, and thermal resistance between the sensing element and the pivot point.
FIND: (a) Thermal resistance between the surface temperature and the sensing temperature, (b) Surface
temperature for Tsen = 28.5°C.
SCHEMATIC:
q
Tsen = 28.5°C
D = 100 nm
Silicon
dioxide
r(x)
x
L = 50 nm
Air
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Negligible
nanoscale effects, (4) Constant properties.
PROPERTIES: Table A.2, polycrystalline silicon dioxide (300 K): k = 1.38 W/m⋅K. Table A.4, air (300
K): k = 0.0263 W/m⋅K.
ANALYSIS:
(a) At any x location, heat transfer in the x-direction occurs by conduction in the air as well as
conduction in the probe. Applying Fourier’s law,
q x = - k a Aa
dT
dT
- k p Ap
dx
dx
(1)
Since the probe radius is r = Dx/2L, the probe area is
Ap =
πD 2
4L2
x 2 and A a =
πD 2
πD 2 
x2 
- Ap =
1 - 2 
4
4 
L 
(2a, 2b)
Substituting Eqs. (2a) and (2b) into Eq. (1) yields
qx = -
πD 2 
dT
k (L2 - x 2 ) + k p x 2 
2  a

dx
4L
Separating variables and integrating,
Continued…
PROBLEM 3.40 (Cont.)
L
qx
dx
∫
k (L2 - x 2 ) + k p x 2
x=0 a
πD 2
=-
4L2
Tsen
∫
dT = -
T=Tsurf
πD 2
4L2
( Tsen
- Tsurf )
Therefore, the thermal resistance associated with the probe is
R sen
(T
- Tsen )
4L2
= surf
=
qx
πD 2
L
∫
x=0
dx
k a L + (k p - k a )x 2
2
Carrying out the integration yields
R sen =
4L
πD
2
kp - ka
1
tan -1
ka
k a (k p - k a )
Substituting values gives
R sen =
4 × 50 × 10-9 m
π × (100 × 10 m)
-9
× tan -1
2
×
1
0.0263 W/m ⋅ K × (1.38 - 0.0263) W/m ⋅ K
(1.38 - 0.0263) W/m ⋅ K
= 48.3 × 106 K/W
0.0263 W/m ⋅ K
<
(b) The thermal circuit is
Tsen = 28.5°C
Tsurf
RT
Rsen
= 48.3
T∞ = 25°C
×106
K/W
= 5×106 K/W
Hence,
(Tsurf - Tsen )
(T - T∞ )
= sen
R sen
RT
Tsurf = (Tsen - T∞ )
Tsurf = 62.3°C
R sen
48.3 × 106 K/W
+ Tsen = (28.5 - 25)°C ×
+ 28.5°C
RT
5 × 106 K/W
<
COMMENT: Heat transfer within the probe region will not be one-dimensional and modification of heat
transfer due to nanoscale effects may be important. However, the probe may be calibrated by measuring
the surface temperature of a large isothermal object.
PROBLEM 3.41
KNOWN: Thickness and inner surface temperature of calcium silicate insulation on a steam pipe.
Convection and radiation conditions at outer surface.
FIND: (a) Heat loss per unit pipe length for prescribed insulation thickness and outer surface
temperature. (b) Heat loss and radial temperature distribution as a function of insulation thickness.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties.
PROPERTIES: Table A-3, Calcium Silicate (T = 645 K): k = 0.089 W/m⋅K.
ANALYSIS: (a) From Eq. 3.32 with Ts,2 = 490 K, the heat rate per unit length is
=
q′ q=
r L
q′ =
(
2π k Ts,1 − Ts,2
ln ( r2 r1 )
)
2π ( 0.089 W m ⋅ K )( 800 − 490 ) K
ln ( 0.08 m 0.06 m )
<
q′ = 603 W m .
(b) Performing an energy for a control surface around the outer surface of the insulation, it follows that
′cond q′conv + q′rad
q=
Ts,1 − Ts,2
Ts,2 − T∞ Ts,2 − Tsur
=
+
ln ( r2 r1 ) 2π k 1 ( 2π r2 h ) 1 ( 2π r2 h r )
(
where h r =
εs Ts,2 + Tsur
2
2
+ Tsur
) ( Ts,2
).
Solving this equation for Ts,2, the heat rate may be determined
from
(
)
(
)
=
q′ 2π r2  h Ts,2 − T∞ + h r Ts,2 − Tsur 
Continued...
PROBLEM 3.41 (Cont.)
and from Eq. 3.31 the temperature distribution is
=
T(r)
Ts,1 − Ts,2
ln ( r1 r2 )
 r 
 + Ts,2
 r2 
ln 
As shown below, the outer surface temperature of the insulation Ts,2 and the heat loss q′ decay
precipitously with increasing insulation thickness from values of Ts,2 = Ts,1 = 800 K and q′ = 11,600
W/m, respectively, at r2 = r1 (no insulation).
800
10000
Heat loss, qprime(W/m)
Temperature, Ts2(K)
700
600
500
400
300
1000
100
0
0.12
0.08
0.04
0
0.04
Insulation thickness, (r2-r1) (m)
0.08
0.12
Insulation thickness, (r2-r1) (m)
Outer surface temperature
Heat loss, qprime
When plotted as a function of a dimensionless radius, (r - r1)/(r2 - r1), the temperature decay becomes
more pronounced with increasing r2.
Temperature, T(r) (K)
800
700
600
500
400
300
0
0.4
0.2
0.6
0.8
1
Dimensionless radius, (r-r1)/(r2-r1)
r2 = 0.20m
r2 = 0.14m
r2= 0.10m
Note that T(r2) = Ts,2 increases with decreasing r2 and a linear temperature distribution is approached as r2
approaches r1.
COMMENTS: An insulation layer thickness of 20 mm is sufficient to maintain the outer surface
temperature and heat rate below 350 K and 1000 W/m, respectively.
PROBLEM 3.42
PROBLEM 3.42 (Cont.)
PROBLEM 3.43
PROBLEM 3.44
PROBLEM 3.44 (Cont.)
PROBLEM 3.45
KNOWN: Diameter, wall thickness and thermal conductivity of steel tubes. Temperature of steam
flowing through the tubes. Thermal conductivity of insulation and emissivity of aluminum sheath.
Temperature of ambient air and surroundings. Convection coefficient at outer surface and maximum
allowable surface temperature.
FIND: (a) Minimum required insulation thickness (r3 – r2) and corresponding heat loss per unit length,
(b) Effect of insulation thickness on outer surface temperature and heat loss.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction, (3) Negligible contact
resistances at the material interfaces, (4) Negligible steam side convection resistance (T∞,i = Ts,i), (5)
Negligible conduction resistance for aluminum sheath, (6) Constant properties, (7) Large surroundings.
ANALYSIS: (a) To determine the insulation thickness, an energy balance must be performed at the outer
surface, where
With
and
it follows that
A trial-and-error solution yields r3 = 0.394 m = 394 mm, in which case the insulation thickness is
<
The heat rate is then
<
(b) The effects of r3 on Ts,o and
have been computed and are shown below.
Continued …
PROBLEM 3.45 (Cont.)
Outer surface temperature, C
240
200
160
120
80
40
0.2
0.26
0.32
0.38
0.44
0.5
Outer radius of insulation, m
Ts,o
Heat rates, W/m
2500
2000
1500
1000
500
0
0.2
0.26
0.32
0.38
0.44
0.5
Outer radius of insulation, m
Total heat rate
Convection heat rate
Radiation heat rate
Beyond r3 ≈ 0.40 m, there are rapidly diminishing benefits associated with increasing the insulation
thickness.
COMMENTS: Note that the thermal resistance of the insulation is much larger than that for the tube
wall. For the conditions of Part (a), the radiation coefficient is hr = 1.37 W/m, and the heat loss by
radiation is less than 25% of that due to natural convection ( q ′rad = 78 W / m, q ′conv,o = 342 W / m ) .
PROBLEM 3.46
PROBLEM 3.47
PROBLEM 3.48
PROBLEM 3.48 (Cont.)
PROBLEM 3.49
KNOWN: Tube diameter and refrigerant temperature for evaporator of a refrigerant system. Convection
coefficient and temperature of outside air.
FIND: (a) Rate of heat extraction without frost formation, (b) Effect of frost formation on heat rate, (c)
Time required for a 2 mm thick frost layer to melt in ambient air for which h = 2 W/m2⋅K and T¥ = 20°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible convection resistance for
refrigerant flow T∞,i = Ts,1 , (3) Negligible tube wall conduction resistance, (4) Negligible radiation
(
)
exchange at outer surface.
ANALYSIS: (a) The cooling capacity in the defrosted condition (δ = 0) corresponds to the rate of heat
extraction from the airflow. Hence,
(
)

q′ h2π r1 T∞,o=
=
− Ts,1 100 W m 2 ⋅ K ( 2π × 0.005 m )( −3 + 18 ) C
<
q′ = 47.1W m
(b) With the frost layer, there is an additional (conduction) resistance to heat transfer, and the extraction
rate is
T∞,o − Ts,1
T∞,o − Ts,1
=
q′ =
R ′conv + R ′cond 1 ( h2π r2 ) + ln ( r2 r1 ) 2π k
For 5 ≤ r2 ≤ 9 mm and k = 0.4 W/m⋅K, this expression yields
Thermal resistance, Rt(m.K/W)
Heat extraction, qprime(W/m)
50
45
40
35
0
0.001
0.002
0.003
Frost layer thickness, delta(m)
Heat extraction, qprime(W/m)
0.004
0.4
0.3
0.2
0.1
0
0
0.001
0.002
0.003
0.004
Frost layer thickness, delta(m)
Conduction resistance, Rtcond(m.K/W)
Convection resistance, Rtconv(m.K/W)
Continued...
PROBLEM 3.49 (Cont.)
The heat extraction, and hence the performance of the evaporator coil, decreases with increasing frost
layer thickness due to an increase in the total resistance to heat transfer. Although the convection
resistance decreases with increasing δ, the reduction is exceeded by the increase in the conduction
resistance.
(c) The time tm required to melt a 2 mm thick frost layer may be determined by applying an energy
balance, Eq. 1.12c, over the differential time interval dt and to a differential control volume extending
inward from the surface of the layer.
E in
=
dt dE
=
st dU lat
(
)
h ( 2π rL ) T∞,o − Tf dt = − h sf r d∀ = − h sf r ( 2π rL ) dr
(
h T∞,o − Tf
=
tm
r
− r h sf ∫ 1 dr
) ∫0t m dt =
r2
r h sf ( r2 − r1 )
=
h T∞,o − Tf
(
)
=
=
t m 11,
690 s 3.25 h
(
)
700 kg m3 3.34 × 105 J kg ( 0.002 m )

2 W m 2 ⋅ K ( 20 − 0 ) C
<
COMMENTS: The tube radius r1 exceeds the critical radius rcr = k/h = 0.4 W/m⋅K/100 W/m2⋅K = 0.004
m, in which case any frost formation will reduce the performance of the coil.
PROBLEM 3.50
KNOWN: Conditions associated with a composite wall and a thin electric heater.
FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and inner
heat flows and conditions for which ratio is minimized.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties, (3) Isothermal
heater, (4) Negligible contact resistance(s).
ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as
shown in the schematic.
 = E
(b) Performing an energy balance for the heater, E
in
out , it follows that
q′′h ( 2π r2 ) =q′i + q′o =
Th − T∞,i
Th − T∞,o
+
ln ( r r )
ln ( r r )
( hi 2π r1 )−1 + 2 1 ( h o 2π r3 )−1 + 3 2
2π k B
2π k A
<
(c) From the circuit,
q′o
=
q′i
(
(
ln ( r2 r1 )
2π k B
×
ln ( r r )
Th − T∞,i
( h o 2π r3 )−1 + 3 2
2π k A
Th − T∞,o
)
)
( hi 2π r1 )−1 +
To reduce q′o q′i , one could increase kB, hi, and r3/r2, while reducing kA, ho and r2/r1.
COMMENTS: Contact resistances between the heater and materials A and B could be important.
<
PROBLEM 3.51
PROBLEM 3.52
PROBLEM 3.53
KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipe
and ice layer formation on the inner surface.
FIND: Ice layer thickness δ.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermal
resistance, (3) negligible ice/wall contact resistance, (4) Constant k.
PROPERTIES: Table A.3, Ice (T = 265 K): k ≈ 1.94 W/m⋅K.
ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it follows
that, for a unit length of pipe,
q′conv = q′cond
Ts,i − Ts,o
h i ( 2π r1 ) T∞,i − Ts,i =
ln ( r2 r1 ) 2π k
(
)
Dividing both sides of the equation by r2,
ln ( r2 r1 )
( r2 r1 )
Ts,i − Ts,o
k
1.94 W m ⋅ K
15 C
=
×
=
×
=0.097
C
2
h i r2 T∞,i − Ts,i
3
2000 W m ⋅ K ( 0.05 m )
(
)
The equation is satisfied by r2/r1 = 1.114, in which case r1 = 0.050 m/1.114 = 0.045 m, and the ice layer
thickness is
δ = r2 − r1 = 0.005 m = 5 mm
<
COMMENTS: With no flow, hi → 0, in which case r1 → 0 and complete blockage could occur. The
pipe should be insulated.
PROBLEM 3.54
PROBLEM 3.55
KNOWN: Sphere of radius ri, covered with insulation whose outer surface is exposed to a
convection process.
FIND: Critical insulation radius, rcr.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (spherical)
conduction, (3) Constant properties, (4) Negligible radiation at surface.
ANALYSIS: The heat rate follows from the thermal circuit shown in the schematic,
q= ( Ti − T∞ ) / R tot
R tot R t,conv + R t,cond and
where=
R t,conv
=
1
1
=
hAs 4π hr 2
(3.9)
=
R
t,cond
1  1 1
 − 
4π k  ri r 
(3.36)
If q is a maximum or minimum, we need to find the condition for which
d R tot
= 0.
dr
It follows that
d  1  1 1
1  
1 1
1 1
=+
−


 − +

 =0
dr  4π k  ri r  4π hr 2   4π k r 2 2π h r 3 
giving
k
h
The second derivative, evaluated at r = rcr, is
d  dR tot 
1
1
3
1
=
−
+

dr  dr 
2π k r 3 2π h r 4  r=r
rcr = 2
cr
=
1

−
1

( 2k/h )3  2π k
+
1 
=

2π h 2k/h 
3
1
1 

( 2k/h )3 2π k 
3
−1 +  > 0
2
Hence, it follows no optimum Rtot exists. We refer to this condition as the critical insulation
radius. See Example 3.6 which considers this situation for a cylindrical system.
PROBLEM 3.56
PROBLEM 3.57
PROBLEM 3.58
PROBLEM 3.59
PROBLEM 3.60
PROBLEM 3.61
PROBLEM 3.62
KNOWN: Volumetric heat generation occurring within the cavity of a spherical shell of
prescribed dimensions. Convection conditions at outer surface.
FIND: Expression for steady-state temperature distribution in shell.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Steady-state conditions, (3)
Constant properties, (4) Uniform generation within the shell cavity, (5) Negligible radiation.
ANALYSIS: For the prescribed conditions, the appropriate form of the heat equation is
d  2 dT 
=0
r
dr  dr 
Integrate twice to obtain,
dT
r2
C1
and
=
dr
C
T=
− 1 + C2 .
r
(1,2)
The boundary conditions may be obtained from energy balances at the inner and outer surfaces.
At the inner surface (ri),
 / 3k.
(3)
=
−k 4π r 2 dT/dr)
−qr
E =
q 4/3π r 3 =
q
dT/dr =
g
(
i
)
cond,i
(
i
)
ri
ri
i
At the outer surface (ro),
− k4π ro2 dT/dr)ro =
q cond,o =
q conv =
h4π ro2 T ( ro ) − T∞ 
dT/dr r =
− ( h/k ) T ( ro ) − T∞  .
o
(4)
 3 / 3k. From Eqs. (1), (2) and (4)
From Eqs. (1) and (3), C1 = −qr
i
3
3
 


qr
 h  qr
− i =
−    i + C2 − T∞ 

 k   3ro k
3kro2


3
3


qr
i − qri + T .
C2 =
∞
3hro2 3ro k
Hence, the temperature distribution is
 3  1 1  qr
 3
qr
i
i +T .
T=
∞
 − +
3k  r ro  3hro2
COMMENTS: Note that E g = q cond,i = q cond,o = q conv .
<
PROBLEM 3.63
PROBLEM 3.64
PROBLEM 3.65
KNOWN: Spherical tank of liquid nitrogen insulated with layer of known thermal conductivity and
outer radius. Inner and outer temperatures.
FIND: Radius of tank that minimizes heat transfer rate and value of minimum heat transfer rate per unit
nitrogen volume.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conduction
through the insulation, and (3) Inner surface of insulation at T = 77 K.
PROPERTIES: Given, k = 0.15 W/m⋅K.
ANALYSIS: The heat gain to the tank is
Ts,2 − Ts,1
Ts,2 − Ts,1
=
R ins
 1/ r1 − 1/ r2 


4π k


=
q
(1)
The heat rate per unit volume of nitrogen is
(
3k Ts,2 − Ts,1
Ts,2 − Ts,1
q
= =
V  1/ r1 − 1/ r2  4 3
r12 − r13 / r2
π r1


4π k

3
(
)
)
(2)
The heat transfer rate will be minimum when the denominator is maximum. The critical radius r1,c
corresponding to the minimum heat transfer rate can therefore be found by differentiating the
denominator with respect to r1 and setting the derivative to zero:
)
(
d 2 3
r1 − r1 / r2 =
2r1 − 3r12 / r2 =
0, r1,c =
2r2 / 3 =
2 × 0.5 m / 3 =
0.33 m
dr
<
The corresponding minimum heat transfer rate per volume of nitrogen can be found from Eq. (2):
(
)
3k Ts,2 − Ts,1
3 × 0.15 W/m ⋅ K × ( 20 + 273 − 77 ) K
q
=
=
= 2620 W/m3
 
2
3
2
V
 min
r1,c − r1c / r2
(0.33 m) − (0.33 m)3 / 0.5 m
(
)
(
)
<
COMMENTS: The existence of a minimum heat transfer rate per nitrogen volume occurs because as r1 is
reduced (insulation thickness is increased), the heat transfer rate decreases, but so does the nitrogen
volume.
PROBLEM 3.66
KNOWN: Dimensions and resistivity of current-carrying cable. Temperature and heat transfer
coefficient of environment.
FIND: Maximum operating current and corresponding minimum cable temperature for copper,
aluminum, and tin cables.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Constant properties, (3) Negligible radiation exchange with
surroundings.
PROPERTIES: Given, Cu: ρe = 10 × 10-8 W⋅m, Al: ρe = 10 × 10-8 W⋅m, Sn: ρe = 20 × 10-8 W⋅m. Table
A.1, Cu: Tm = 1358 K, k = 339 W/m⋅K; Al: Tm = 933 K, k = 218 W/m⋅K; Sn: Tm = 505 K, k = 62.2
W/m⋅K. Thermal conductivities have been evaluated at highest available temperature.
ANALYSIS: The maximum temperature should not exceed the melting temperature of the material. The
maximum temperature occurs at the centerline and can be found from Equation 3.58 evaluated at r = 0.
The surface temperature can be related to the known environment temperature through Equation 3.60.
Thus,
 2 qr

qr
Tmax =T(r =0) = o + o + T∞
(1)
4k
2h
Or solving for q ,
Tmax − T∞
2
ro / 4k + ro / 2h
The heat generation rate depends on the current and resistivity according to:
q max =
2
I 2 R e I ( re L / A c ) I 2 re
=
q∀
=
= =
∀
LA c
A c2
I 2 re
( )
2
π ro2
(2)
(3)
Thus the maximum allowable current is given by:
Continued...
PROBLEM 3.66 (Cont.)

 q max 
Tmax − T∞
2 
Imax =
=
 π ro 
2
 re 
 re ro / 4k + ro / 2h

1/2
1/2
(
)





π ro2
(4)
The minimum temperature occurs at the surface of the cable, so from Equation 3.60,
Tmin =
o
qr
I2 r r
+ T∞ = max e o + T∞
2
2h
2h π ro2
(5)
( )
Evaluating Eqs.(4) and (5) for the properties of copper yields,
1/2
Imax






− 293) K
(1358

 π × ( 0.02 m )2 9610 A
=
2



0.02 m
 10 ×10−8 W ⋅ m  ( 0.02 m )

+

 4 × 339 W/m ⋅ K 2 × 55 W/m 2 ⋅ K  



Tmin
=
(9610 A)2 ×10 ×10−8 W ⋅ m × 0.02 m
=
+ 293 K 1356 K
2
2
2
2 × 55 W/m ⋅ K × π (0.02 m)
(
)
<
<
Similarly for aluminum,
=
Imax 7450
=
A, Tmin 931 K
<
And for tin,
=
Imax 3020
=
A, Tmin 503 K
<
COMMENTS: (1) The minimum (surface) temperature is almost equal to the maximum (melting)
temperature; the cable is nearly isothermal. (2) The neglect of radiation is a poor assumption for such
large temperatures. Assuming the surroundings are also at 293 K and the emissivity is one, the ratio of
4 − T 4 ) / [h(T
σ (Tmin
radiation to convection heat transfer is given by q rad / q conv =
∞
min − T∞ )] = 3.3,
1.2, 0.28 for Cu, Al, and Sn, respectively. If radiation were included in the analysis, it would enhance heat
transfer and increase the current-carrying capacity.
PROBLEM 3.67
KNOWN: Wall of thermal conductivity k and thickness L with uniform generation q ; strip heater with
uniform heat flux q′′o ; prescribed inside and outside air conditions (hi, T∞,i, ho, T∞,o).
FIND: (a) Sketch temperature distribution in wall if none of the heat generated within the wall is lost to
the outside air, (b) Temperatures at the wall boundaries T(0) and T(L) for the prescribed condition, (c)
Value of q ′′o required to maintain this condition, (d) Temperature of the outer surface, T(L), if
 but q′′o corresponds to the value calculated in (c).
q=0
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetric
generation, (4) Constant properties.
ANALYSIS: (a) If none of the heat generated within the wall is
lost to the outside of the chamber, the gradient at x = 0 must be zero.
Since q is uniform, the temperature distribution is parabolic, with T(L) >
T∞,i.
(b) To find temperatures at the boundaries of wall, begin with the general
solution to the appropriate form of the heat equation (Eq.3.40).
q 2
T(x) =
−
x + C1x+C2
2k
(1)
From the first boundary condition,
dT
=0
dx x=o
→
C1 =0.
(2)
Two approaches are possible using different forms for the second boundary condition.
T1
Approach No. 1: With boundary condition → T ( 0 ) =
q 2
T(x) =
−
x + T1
2k
(3)
To find T1, perform an overall energy balance on the wall
E in − E out + E g =
0

− h T ( L ) − T∞,i  + qL=0
T (L) =
T2 =
T∞,i +

qL
h
(4)
Continued …
PROBLEM 3.67 (Cont.)
and from Eq. (3) with x = L and T(L) = T2,
q
T (L) =
− L2 + T1
2k
or
T1 =+
T2

 2
q 2
qL
qL
L =
T∞,i +
+
2k
h
2k
(5,6)
Substituting numerical values into Eqs. (4) and (6), find
T2 = 50 C+1000 W/m3 × 0.200 m/20 W/m 2 ⋅ K=50C+10C=60C
=
T1 60 C+1000 W/m3 × ( 0.200 m ) / 2 × 4 W/m ⋅ K=65C.
2
<
<
Approach No. 2: Using the boundary condition
−k
dT
=h T ( L ) − T∞,i 
dx x=L
yields the following temperature distribution which can be evaluated at x = 0,L for the required
temperatures,
)
(

q
qL
T(x) =
x 2 − L2 +
−
+ T∞,i .
2k
h
(c) The value of q′′o when T(0) = T1 = 65°C
follows from the circuit
q′′o =
T1 − T∞,o
1/ h o

=
q′′o 5 W/m 2 ⋅ K ( 65-25 ) C=200 W/m 2 .
<

(d) With q=0,
the situation is represented
by the thermal circuit shown. Hence,
′′o q′′a + q′′b
q=
q′′o
=
T1 − T∞,o
1/ h o
+
T1 − T∞,i
L/k+1/h i
which yields
T1 = 55 C.
<
PROBLEM 3.68
PROBLEM 3.69
KNOWN: Diameter, thermal conductivity and microbial energy generation rate in cylindrical hay bales.
Ambient conditions.
FIND: The maximum hay temperature for q = 1, 10, and 100 W/m3.
SCHEMATIC:
Air
T∞ = 0°C, h = 25 W/m2·K
.
Ts
q = 1, 10 or 100 W/m3
D=2m
k = 0.04 W/m∙K
r
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat transfer
(4) Uniform volumetric generation, (5) Negligible radiation, (6) Negligible conduction to or from the
ground.
PROPERTIES: k = 0.04 W/m⋅K (given).
ANALYSIS: The surface temperature of the dry hay is (Eq. 3.60)
Ts =
T∞ +
o
qr
1W/m3 × 1m
=
=
0°C+
0.02°C
2h
2 × 25W/m 2 ⋅ K
whereas Ts = 0.2°C and 2.0°C for the moist and wet hay, respectively.
<
<
The maximum hay temperature occurs at the centerline, r = 0. From Eq. 3.58, for the dry hay,
Tmax=
 o2
qr
1W/m3 × (1m)2
+ Ts=
+ 0.02°C= 6.27°C
4k
4 × 0.04 W/m ⋅ K
whereas Tmax = 62.7°C and 627°C for the moist and wet hay, respectively.
<
<
COMMENTS: (1) The hay begins to lose its nutritional value at temperatures exceeding 50°C.
Therefore the center of the moist hay bale will lose some of its nutritional value. (2) The center of the wet
hay bale can experience very high temperatures without combusting due to lack of oxygen internal to the
hay bale. However, when the farmer breaks the bale apart for feeding, oxygen is suddenly supplied to the
hot hay and combustion may occur. (3) The outer surface of the hay bale differs by only 2°C from the dry
to the wet condition, while the centerline temperature differs by over 600 degrees. The farmer cannot
anticipate the potential for starting a fire by touching the outer surface of the hay bale. (4) See Opuku,
Tabil, Crerar and Shaw, “Thermal Conductivity and Thermal Diffusivity of Timothy Hay,” Canadian
Biosystems Engineering, Vol. 48, pp. 3.1 - 3.6, 2006 for hay property information.
PROBLEM 3.70
KNOWN: Diameter, thermal conductivity and microbial energy generation rate in cylindrical hay bales.
Thin-walled tube diameter and insertion location. Temperature of flowing water and convective heat
transfer coefficient inside the tube. Ambient conditions.
FIND: (a) Steady-state heat transfer to the water per unit length of tube, (b) Plot of the radial
temperature distribution, T(r), in the hay (c) Plot of the heat transfer to the water per unit length of tube
for bale diameters of 0.2 m ≤ D ≤ 2 m for q = 100 W/m3.
SCHEMATIC:
Air
T∞,o = 0°C, ho = 25 W/m2·K
.
q = 100 W/m3
Ts,2
T∞,i = 20°C, hi = 200 W/m2·K
r
Ts,1
D =2r2 = 2 m
2r1 = 30 mm
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat transfer
(4) Uniform volumetric generation, (5) Negligible radiation, (6) Negligible conduction to or from the
ground.
PROPERTIES: k = 0.04 W/m⋅K (given).
ANALYSIS: (a) The temperature distribution is found by utilizing the general solution given by Eq.
3.56 with mixed boundary conditions applied at r1 and r2. Specifically,
at r1:
−k
dT
dr
r =r 1
=hi T∞,i − Ts ,1 


at r2: −k
dT
dr
r =r 2
=ho Ts ,2 − T∞,o 
The solutions are given by Eqs. C.16, C.17, and C.2.
From Eq. C.16,
2
h∞,i (T∞,=
i - Ts ,1 ) 200 W/m ⋅ K × (20°C - Ts,1 )
 qr

 2  r2 
k  2  1 - 12  + (Ts ,2 - Ts ,1 ) 



qr
 4k  r2 

= 1- 
2
r1 ln( r2 / r1 )
 100W/m3 (1m)2  (15 × 10-3 m)2 
 (1)
0.04W/m ⋅ K 
1T
T
+

(
)


s
s
,2
,1


(1m)2
100W/m3 × 15 × 10-3m
 4 × 0.04W/m ⋅ K 



=
-3
2
15 × 10 m × ln(1000 /15)
Continued…
PROBLEM 3.70 (Cont.)
From Eq. C.17,
h∞,o (Ts ,2 −=
T∞,o ) 25 W/m 2 ⋅ K × (Ts ,2 − 0°C )
 qr

 2  r2 
k  2  1 − 12  + (Ts ,2 − Ts ,1 ) 



qr
 4k  r2 

= 2− 
2
r 2ln( r2 / r1 )
 100W/m3 (1m)2  (15 × 10−3 m)2 
 (2)
0.04W/m ⋅ K 
1−
+ (Ts ,2 − Ts ,1 ) 


2


(1m)
100W/m3 × 1m

 4 × 0.04W/m ⋅ K 

=
−
2
1m × ln(1000 /15)
Equations (1) and (2) may be solved simultaneously to yield Ts,1 = 21.54°C, Ts,2 = 1.75°C. The heat
transfer to the cold fluid per unit length is
2
−3
=
q ' hi (2π ri )(Ts ,=
=
20)°C 38.7W/m
i − T∞ ,i ) 200W / m ⋅ K × 2 × π × 15 × 10 m × (21.54 −
<
(b) The radial temperature distribution is evaluated from Eq. C.2 and is shown below.
T ( r ) = Ts ,2 +
 ln( r / r )
 22  r 2   qr
 22  r12 
qr
2
 1 − 2  − 
 1 − 2  + (Ts ,2 − Ts ,1 ) 
4k  r2   4k  r2 
ln(
r
2 / r1 )


r 2   100W/m3 × (1m)2  (15 × 10−3 m)2 
100W/m3 × (1m)2 
1 −
−
 1 −
 + (1.75°C − 21.54°C )
2
2
4 × 0.04W/m ⋅ K  (1m )   4 × 0.04W/m ⋅ K 
(1m)




ln(1m / r )
×
ln(1m /15 × 10−3 m)
= 1.75°C +
400
T (C)
300
200
100
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
r (m)
1
Continued…
PROBLEM 3.70 (Cont.)
Note that the maximum temperature occurs at r ≈ 0.35 m.
(W/m)
(c) The rate of heat transfer to the cool fluid, per unit length, is shown versus the bale diameter in the plot
below.
Note that at very small bale diameters, the heat transfer to the inner tube will become negative. That is,
the energy generation in the bale is not sufficient to offset conduction losses from the relatively warm
tube liquid to the relatively cold outside air.
COMMENTS: (1) The energy generated in the bale per unit length is
2
E g' = q × π × ( r22 − r12 ) = 100W/m3 × π × (1m 2 − ( 0.015m ) ) = 314 W/m. Hence, the heat transfer to the
inner tube represents (38.7/314)×100 = 12.3% of the total generated. The remaining 87.6% is lost to the
ambient air. (2) The performance could be improved by inserting more tubes, or by stacking the bales in
adjacent rows so that heat losses from the exterior surface would be minimized. (3) Evaluation of the two
constants appearing in the analytical solution (Eq. 3.56) using the two mixed boundary conditions is very
tedious, resulting in a cumbersome expression. Utilization of the results of Appendix C saves
considerable time.
PROBLEM 3.71
PROBLEM 3.71 (Cont.)
PROBLEM 3.71 (Cont.)
PROBLEM 3.72
PROBLEM 3.73
PROBLEM 3.74
KNOWN: Cylindrical shell with uniform heat generation and surface temperatures.
FIND: Radial distributions of temperature, heat flux, and heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Uniform heat generation, (3)
Constant k.
ANALYSIS: For the cylindrical shell, the appropriate form of the heat equation is
1 d  dT  q
0
r + =
r dr  dr  k
The general solution is
q 2
T (r) =
−
r + C1 ln r + C2
4k
Applying the boundary conditions, it follows that
q 2
T ( r1 ) =
Ts,1 =
−
r1 + C1 ln r1 + C2
4k
q 2
T ( r2 ) =
Ts,2 =
−
r2 + C1 ln r2 + C2
4k
which may be solved for
(
)
(
)
 ) r22 − r12 + Ts,2 − Ts,1  ln ( r2 /r1 )
=
C1 ( q/4k


C2 =+
Ts,2 ( q 4k ) r22 − C1 ln r2
Hence,
ln ( r/r2 )
T ( r ) = Ts,2 + ( q 4k ) r22 − r 2 + ( q 4k ) r22 − r12 + Ts,2 − Ts,1 

 ln ( r /r )
2 1
)
(
)
(
(
)
<
With q′′ = − k dT/dr , the heat flux distribution is
q′′ ( =
r)
q
2
r−
(
)
(
)
k ( q 4k ) r22 − r12 + Ts,2 − Ts,1 


r ln ( r2 /r1 )
<
Continued...
PROBLEM 3.74 (Cont.)
Similarly, with q = q′′ A(r) = q′′ (2πrL), the heat rate distribution is
 2−
=
q ( r ) π Lqr
(
)
(
)
2π Lk ( q 4k ) r22 − r12 + Ts,2 − Ts,1 


ln ( r2 /r1 )
<
COMMENT: An overall energy balance on the cylindrical shell can be expressed as q ( r2 ) − q(r1) =
 where V is the volume of the shell. Thus, the energy balance expresses that the net
π Lq r22 − r12 =
qV
rate at which energy leaves the shell is equal to the rate of energy generated within the shell.
(
)
PROBLEM 3.75
KNOWN: Spherical shell with uniform heat generation and surface temperatures.
FIND: Radial distributions of temperature, heat flux and heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Uniform heat generation, (3)
Constant k.
ANALYSIS: For the spherical shell, the heat equation and general solution are
1 d  2 dT  q
0
r
+ =
r 2 dr  dr  k
T(r) =
− ( q 6k ) r 2 − C1/r + C 2
Applying the boundary conditions, it follows that
T ( r1 ) ==
Ts,1 − ( q 6k ) r12 − C1/r1 + C2
T ( r2 ) ==
Ts,2 − ( q 6k ) r22 − C1/r2 + C2
Hence,
)
(
(
) [(1 r1 ) − (1 r2 )]
=
C1 ( q 6k ) r22 − r12 + Ts,2 − Ts,1 


C2 =+
Ts,2 ( q 6k ) r22 + C1/r2
and
(1 r ) − (1 r2 )
T ( r ) = Ts,2 + ( q 6k ) r22 − r 2 − ( q 6k ) r22 − r12 + Ts,2 − Ts,1 

 (1 r1 ) − (1 r2 )
(
)
(
)
(
)
<
With q′′ (r) = - k dT/dr, the heat flux distribution is
q′′ ( =
r)
q
3
(
)
( q 6 ) r 2 − r 2 + k ( T − T ) 
2 1
s,2
s,1  1

r− 
(1 r1 ) − (1 r2 )
r2
<
Continued...
PROBLEM 3.75 (Cont.)
(
)
and, with q = q′′ 4π r 2 , the heat rate distribution is
4π q 3
=
q (r)
r −
3
(
)
(
)
4π ( q 6 ) r22 − r12 + k Ts,2 − Ts,1 


(1 r1 ) − (1 r2 )
<
COMMENT: An overall energy balance on the sphere can be expressed as q ( r2 ) − q(r1) =
4π q 3 3
 where V is the volume of the spherical shell. Thus, the energy balance expresses that
r2 − r1 =
qV
3
the net rate at which energy leaves the shell is equal to the rate of energy generated within the shell.
(
)
PROBLEM 3.76
KNOWN: Plane wall of thickness 2L, thermal conductivity k with uniform energy generation
For
case 1, boundary at x = -L is perfectly insulated, while boundary at x = +L is maintained at To = 50°C.
For case 2, the boundary conditions are the same, but a thin dielectric strip with thermal resistance
is inserted at the mid-plane.
FIND: (a) Sketch the temperature distribution for case 1 on T-x coordinates and describe key features;
identify and calculate the maximum temperature in the wall, (b) Sketch the temperature distribution for
case 2 on the same T-x coordinates and describe the key features; (c) What is the temperature difference
between the two walls at x = 0 for case 2? And (d) What is the location of the maximum temperature of
the composite wall in case 2; calculate this temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the plane and
composite walls, and (3) Constant properties.
ANALYSIS: (a) For case 1, the temperature distribution, T1(x) vs. x, is parabolic as shown in the
schematic below and the gradient is zero at the insulated boundary, x = -L. From Eq. 3.48 (see discussion
after Eq. 3.49),
and since T1(+L) = To = 50°C, the maximum temperature occurs at x = -L,
<
(b) For case 2, the temperature distribution, T2(x) vs. x, is piece-wise parabolic, with zero gradient at x = L and a drop across the dielectric strip, ∆TAB. The temperature gradients at either side of the dielectric
strip are equal.
(c) For case 2, the temperature drop across the thin dielectric strip follows from the surface energy
balance shown above.
<
(d) For case 2, the maximum temperature in the composite wall occurs at x = -L, with the value,
<
PROBLEM 3.77
KNOWN: Geometry and boundary conditions of a nuclear fuel element.
FIND: (a) Expression for the temperature distribution in the fuel, (b) Form of temperature
distribution for the entire system.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Uniform
generation, (4) Constant properties, (5) Negligible contact resistance between fuel and cladding.
ANALYSIS: (a) The general solution to the heat equation, Eq. 3.44,
d 2T
q
=
+
0
dx 2 k f
is
( −L ≤ x ≤ +L )
q 2
T=
−
x + C1x+C2 .
2k f
The insulated wall at x = - (L+b) dictates that the heat flux at x = - L is zero (for an energy
balance applied to a control volume about the wall, =
E in E=
out 0). Hence
dT 
q
=
− ( −L ) + C1 =
0

dx  x = − L
kf
−
C1 =
or

qL
kf

q 2 qL
−
T=
x −
x+C2 .
2k f
kf
(1)
The value of Ts,1 may be determined from the energy conservation requirement that
=
E g q=
cond q conv , or on a unit area basis.
q ( 2L ) =
Hence,
(
)
(
)
ks
Ts,1 − Ts,2 = h Ts,2 − T∞ .
b
q ( 2 Lb )
Ts,1 =
+ Ts,2
where
ks
q ( 2 Lb ) q ( 2L )
Ts,1=
+
+ T∞ .
ks
h
q ( 2L )
Ts,2 =
+ T∞
h
Continued …
PROBLEM 3.77 (Cont.)
Hence from Eq. (1),
( )
 2
q ( 2 Lb ) q ( 2 L )
3 q L
T (L) =
Ts,1 =
+
+ T∞ =
−
+ C2
ks
h
2 kf
which yields
 2b 2 3 L 
  + +
C2 = T∞ + qL

 ks h 2 k f 
Hence, the temperature distribution for ( −L ≤ x ≤ +L ) is
T=
−
 2b 2 3 L 

q 2 qL
  + +
x −
x+qL
 + T∞
2k f
kf
 ks h 2 k f 
(b) For the temperature distribution shown below,
( −L − b ) ≤ x ≤ −L:
− L ≤ x ≤ +L:
+L ≤ x ≤ L+b:
dT/dx=0, T=Tmax
| dT/dx | ↑ with ↑ x
( dT/dx ) is const.
<
PROBLEM 3.78
PROBLEM 3.78 (Cont.)
PROBLEM 3.79
KNOWN: Three-layer composite cylinder with specified dimensions, thermal conductivities, and
contact resistances, with uniform energy generation q in one layer. Radiation loss at the surface.
FIND: (a) Heat equation in each material, (b) Temperature at surface of Material C; (c) Sketch of
temperature distribution, and (d) Sketch of heat flux distribution.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the radial direction,
and (3) Constant properties.
ANALYSIS: (a) The heat equation in cylindrical radial coordinates with heat generation and constant
thermal conductivity is given in Equation 3.54. There is heat generation in Material B, but not in
Materials A and C. Therefore the heat equations can be written:
1 d  dTA 
=
r
0,
r dr  dr 
1 d  dTB  q B
=
r
+
0,
r dr  dr  k B
1 d  dTC 
=
r
0
r dr  dr 
<
(b) The temperature at the surface, T(rC), can be found by recognizing that the heat generated in Material
B must leave the composite cylinder by radiation at the surface. The rate of heat generation in Material B
2 )L, where L is the length of the cylinder. Thus,
∀B q Bπ (rB2 − rA
is q∀∀
B=
(
)
4 − T 4 = q π (r 2 − r 2 )L
εs 2π rC L TC,2
sur
B B A
which yields:
1/4
 q (r 2 − r 2 )

B B A
4 
TC,2 
=
+ Tsur
 εs 2rC



<
(c,d) The sketches are shown below. These sketches can be developed through the following
considerations:
Continued …
PROBLEM 3.79 (Cont.)
•
•
•
•
For Material A, the differential equation, Eq. (1a), can be multiplied by r and integrated once to give
dTA / dr = C1. Recognizing that the point r = 0 is a symmetry point, the temperature gradient at r =
0 must be zero, therefore C1 = 0. Thus TA is a constant. From a more physical perspective, there can
be no heat transfer within Material A because it only has one surface; if heat enters at that surface, it
has nowhere to go. If heat leaves at that surface, it has nowhere to originate.
For Material B, Eq. (1b) can be integrated to find that the temperature derivative varies as –r. The
slope must be zero at rA since the heat flux is zero coming from Material A. Even though there is a
contact resistance, there is no temperature difference at the A/B interface because the heat flux is
zero.
For Material C, Eq. (1c) can be integrated to find that the temperature derivative varies as 1/r. Since
the heat flux must match across the B/C interface, and the thermal conductivity of Material C is half
that of Material B, the slope of temperature is twice as high for Material C. There is a temperature
jump at the B/C interface because of the contact resistance.
The heat flux is proportional to the (negative) temperature gradient and it must be continuous at both
interfaces.
T(r)
dT/dr
∝ - r’
dTC/dr |r = rB
dTB/dr |r = rB
=
kB
kC
=2
dT/dr ∝ 1/r
r’
r
0
rA
rB
rC
<
q” (r)
q’’ ∝ 1/r
q’’ ∝ r’
0
r’
r
0
rA
rB
rC
<
COMMENTS: (1) The heat transfer rate is zero in Material A. The heat transfer rate increases as radius
squared in Material B as more generated heat must exit toward Material C. The heat transfer rate is
constant in Material C as all the heat generated in Material B is transferred through Material C to the
environment. (2) Solving for the temperature distributions in all three materials can be facilitated by use
of Appendix C.
PROBLEM 3.80
KNOWN: Plane wall with prescribed nonuniform volumetric generation having one boundary
insulated and the other isothermal.
FIND: Temperature distribution, T(x), in terms of x, L, k, q o and To .
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction,
(3) Constant properties.
ANALYSIS: The appropriate form the heat diffusion equation is
d  dT  q
+ =
0.
dx  dx  k
=
q q=
Noting that
( x ) q o (1 − x/L ) , substitute for q ( x ) into the above equation, separate
variables and then integrate,
q 
dT
x2 
=
− o x −
 + C1.
dx
k 
2L 


Separate variables and integrate again to obtain the general form of the temperature distribution
in the wall,
q 
q  x 2 x 3 
x2 
dT =
− o x −
T(x) =
− o
−
 dx+C1dx
 + C1x+C2 .
k 
2L 
k  2 6L 




q
 dT 
− o
d  =
k
 dx 
 x
1 − L  dx
Identify the boundary conditions at x = 0 and x = L to evaluate C1 and C2. At x = 0,
q
T ( 0 ) =To =− o ( 0 − 0 ) + C1 ⋅ 0 + C2
hence, C2 =To
k
At x = L,
q o 
q o L
dT 
L2 
=
0
=
−
L
−
hence, C1 =

 + C1

dx  x=L
k 
2L 
2k


The temperature distribution is
q  x 2 x 3  q o L
<
T(x) =
− o
−
x+To .
+
k  2 6L  2k


COMMENTS: It is good practice to test the final result for satisfying BCs. The heat flux at x =
0 can be found using Fourier’s law or from an overall energy balance
E out
= E=
g
L

∫0 qdV
to obtain
q′′out
= q o L/2.
PROBLEM 3.81
KNOWN: Distribution of volumetric heating and surface conditions associated with a quartz
window.
FIND: Temperature distribution in the quartz.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible
radiation emission and convection at inner surface (x = 0) and negligible emission from outer
surface, (4) Constant properties.
ANALYSIS: The appropriate form of the heat equation for the quartz is obtained by
substituting the prescribed form of q into Eq. 3.44.
d 2T α (1 - β ) q′′o -α x
+
=
e
0
k
dx 2
Integrating,
β ) q′′o -α x
(1 -(1 β ) q′′ e-α x + C x+C
dT
e
T=
=
+
+ C1
o
1
2
dx
k
kα
−k dT/dx) x=o =
β q′′o
−k dT/dx) x=L =
h T ( L ) − T∞ 
Boundary Conditions:
 (1-β )

q′′o + C1  =
β q′′o
-k 
 k

C1 = -q′′o / k
Hence, at x = 0:
At x = L:
 (1-β )

 (1-β )

q′′o=
e-α L + C1  h q′′o e-α L + C1L+C2 - T∞ 
-k 
 k

 kα

Substituting for C1 and solving for C2,
C=
2
q′′
q′′o 
-α L  + q′′o L + o(1-β ) e-α L + T .
-1
1
e
β
(
)
∞

h 
k
kα
Hence, =
T(x)
(1 - β ) q′′o e-α L - e-α x  + q′′o
kα


k
(L - x) +
q′′o
h
-α L  + T . <
1 - (1 β ) e
 ∞
COMMENTS: The temperature distribution depends strongly on the radiative coefficients, α
and β. For α → ∞ or β = 1, the heating occurs entirely at x = 0 (no volumetric heating).
PROBLEM 3.82
KNOWN: Cylindrical shell with uniform volumetric generation is insulated at inner surface and
exposed to convection on the outer surface.
 h, T∞ and k, (b)
FIND: (a) Temperature distribution in the shell in terms of ri , ro , q,
Expression for the heat rate per unit length at the outer radius, q′ ( ro ) .
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (cylindrical)
conduction in shell, (3) Uniform generation, (4) Constant properties.
ANALYSIS: (a) The general form of the temperature distribution and boundary conditions are
q
T (r) =
− r 2 + C1 ln r+C2
4k
dT 
q
1
− ri + C1 + 0
0=
 =
dr r
2k
ri
at r = ri:
q
C1 =ri2
2k
i
−k
at r = ro:
dT 
 = h T ( ro ) − T∞ 
dr r
surface energy balance
o
 q
 q
 q

1 
 q

-k  - ro +  ri2 ⋅   =h  - ro2 +  ri2  ln ro + C2 - T∞ 
ro  
 2k 
 2k
 4k

 2k
  r 
qr
C2 =
− o 1 −  i 
2h   ro 

2
 o2
 + qr
 2k

2


 ri 
1
 −   ln ro  + T∞
 2  ro 



Hence,
(
)
 2  r  qr

q 2 2 qr
T ( r=
ro − r + i ln   − o
)
4k
2k  ro  2h
  2 
1 −  ri   + T∞ .
  ro  


<
(b) From an overall energy balance on the shell,
′ q π r 2 − r 2 .
q′ ( r=
) E=
r o
g
(o i )
<
Alternatively, the heat rate may be found using Fourier’s law and the temperature distribution,
q′r ( r ) =−k ( 2π ro )
 q

 2 1
qr
dT 
i

+ 0 + 0  =q π ro2 − ri2
 =−2π kro − ro +
dr r
2k ro
 2k

o


(
)
PROBLEM 3.83
PROBLEM 3.83 (Cont.)
PROBLEM 3.83 (Cont.)
PROBLEM 3.84
PROBLEM 3.84 (Cont.)
PROBLEM 3.84 (Cont.)
PROBLEM 3.85
KNOWN: Materials, dimensions, properties and operating conditions of a gas-cooled nuclear reactor.
FIND: (a) Inner and outer surface temperatures of fuel element, (b) Temperature distributions for
different heat generation rates and maximum allowable generation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,
(4) Negligible contact resistance, (5) Negligible radiation.
PROPERTIES: Table A.1, Thorium: Tmp ≈ 2000 K; Table A.2, Graphite: Tmp ≈ 2300 K.
ANALYSIS: (a) The outer surface temperature of the fuel, T2, may be determined from the rate equation
T − T∞
q′ = 2
R ′tot
where
R ′tot=
ln ( r3 r2 )
2π k g
+
ln (14 11)
1
=
+
= 0.0185 m ⋅ K W
2π r3h 2π ( 3 W m ⋅ K ) 2π ( 0.014 m ) 2000 W m 2 ⋅ K
1
(
)
and the heat rate per unit length may be determined by applying an energy balance to a control surface
about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that
)
(
)
(
q=′ q π r22 − r12= 108 W m3 × π 0.0112 − 0.0082 m 2= 17, 907 W m
Hence,
T2= q′R ′tot + T∞= 17, 907 W m ( 0.0185 m ⋅ K W ) + 600 K= 931K
<
With zero heat flux at the inner surface of the fuel element, Eq. C.14 yields
T1 =T2 +
 22 
qr
 2 r 
r 2  qr
 1 − 1  − 1 ln  2 
4k t  r 2  2k t  r1 

2 
T1 =
931K +
2
2
108 W m3 ( 0.011m )   0.008 2  108 W m3 ( 0.008 m )
 0.011 
1
−
−
ln 
 
 

4 × 57 W m ⋅ K
2 × 57 W m ⋅ K
 0.008 
  0.011  


Continued...
PROBLEM 3.85 (Cont.)
<
T1 = 931K + 25 K − 18 K = 938 K
(b) The temperature distributions may be obtained by using the IHT model for one-dimensional, steadystate conduction in a hollow tube. For the fuel element ( q > 0), an adiabatic surface condition is
prescribed at r1, while heat transfer from the outer surface at r2 to the coolant is governed by the thermal
resistance R ′′tot,2 = 2π r2 R ′tot = 2π(0.011 m)0.0185 m⋅K/W = 0.00128 m2⋅K/W. For the graphite ( q = 0),
2500
2500
2100
2100
Temperature, T(K)
Temperature, T(K)
the value of T2 obtained from the foregoing solution is prescribed as an inner boundary condition at r2,
while a convection condition is prescribed at the outer surface (r3). For 1 × 108 ≤ q ≤ 5 × 108 W/m3, the
following distributions are obtained.
1700
1300
900
500
0.008
0.009
0.01
Radial location in fuel, r(m)
qdot = 5E8
qdot = 3E8
qdot = 1E8
0.011
1700
1300
900
500
0.011
0.012
0.013
0.014
Radial location in graphite, r(m)
qdot = 5E8
qdot = 3E8
qdot = 1E8
The comparatively large value of kt yields small temperature variations across the fuel element,
while the small value of kg results in large temperature variations across the graphite. Operation
at q = 5 × 108 W/m3 is clearly unacceptable, since the melting points of thorium and graphite are
exceeded and approached, respectively. To prevent softening of the materials, which would occur below
their melting points, the reactor should not be operated much above q = 3 × 108 W/m3.
COMMENTS: A contact resistance at the thorium/graphite interface would increase temperatures in the
fuel element, thereby reducing the maximum allowable value of q .
PROBLEM 3.86
PROBLEM 3.87
KNOWN: Radius, thermal conductivity, heat generation and convection conditions associated
with a solid sphere.
FIND: Temperature distribution.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)
Constant properties, (4) Uniform heat generation.
ANALYSIS: Integrating the appropriate form of the heat diffusion equation,
d  2 dT 

 kr dr  + q=0

r 2 dr 
1
r2
 3
dT
qr
=
−
+ C1
dr
3k
or
 2
d  2 dT 
qr
r
=
−
dr  dr 
k

dT
qr
C
=
− + 1
dr
3k r 2
 2 C1
qr
T (r) =
−
−
+ C2 .
6k
r
dT 
0=
hence
C1 0, and
The boundary conditions=
are:
dr  r=0
dT 
−k  = h T ( ro ) − T∞  .
dr  r
o
Substituting into the second boundary condition (r = ro), find
 qr

o
 2

 2
qr
qr
qr
= h - o + C2 - T∞ 
C2 = o + o + T∞ .
3
3h 6k
 6k

The temperature distribution has the form

q 2 2 qr
T ( r=
ro − r + o + T∞ .
)
6k
3h
COMMENTS: To verify the above result, obtain T(ro) = Ts,
o
qr
Ts
=
+ T∞
3h
Applying energy balance to the control volume about the sphere,
(
)
4

q  π ro3  =
h4π ro2 ( Ts − T∞ )
3


find
o
qr
Ts =
+ T∞ .
3h
<
PROBLEM 3.88
KNOWN: Dimensions and thermal conductivity of a spherical container. Thermal conductivity and
volumetric energy generation within the container. Outer convection conditions.
FIND: (a) Outer surface temperature, (b) Container inner surface temperature, (c) Temperature
distribution within and center temperature of the wastes, (d) Feasibility of operating at twice the energy
generation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional radial
conduction.
ANALYSIS: (a) For a control volume which includes the container, conservation of energy yields
 − q conv =
E g − E out =
0 , or qV
0 . Hence
( )
(
q (=
4 3) π ri3
h4π ro2 Ts,o − T∞
)
and with q = 105 W/m3,
Ts,o =
T∞ +
 i3
qr
3hro2
105 W m 2 ( 0.5 m )
3

25 C +
=
3000 W m 2 ⋅ K ( 0.6 m )
2
36.6 C .
=
<
(b) Performing a surface energy balance at the outer surface, E in − E out =
0 or q cond − q conv =
0.
Hence
4π k ss Ts,i − Ts,o
= h4π ro2 Ts,o − T∞
(1 ri ) − (1 ro )
(
Ts,i = Ts,o +
)
(
)
(
)
h  ro
1000 W m ⋅ K


− 1  ro Ts,o − T∞ = 36.6 C +
( 0.2 ) 0.6 m 11.6 C = 129.4 C .

k ss  ri
15
W
m
K
⋅

(
2
)
<
(c) The heat equation in spherical coordinates is
d  dT 
 2=
k rw  r 2
0.
 + qr
dr  dr 
Solving,
2 dT
 3
qr
=
−
+ C1
dr
3k rw
Applying the boundary conditions,
dT
=0
and
dr r = 0
r
C1 = 0
and
and
 2
qr
C
T (r) =
−
− 1 + C2
6k rw
r
T ( ri ) = Ts,i
 i2 6k rw .
C
=
2 Ts,i + qr
Continued...
PROBLEM 3.88 (Cont.)
Hence
T (r) =
Ts,i +
( ri2 − r2 )
q
6k rw
<
At r = 0,
T (0) =
Ts,i +
 i2
qr
6k rw
=
129.4 C +
105 W m3 ( 0.5 m )
2
6 ( 20 W m ⋅ K )
=
337.7 C
<
(d) The feasibility assessment may be performed by using the IHT model for one-dimensional, steadystate conduction in a solid sphere, with the surface boundary condition prescribed in terms of the total
thermal resistance
[
]
ri2 (1 ri ) − (1 ro ) 1  ri
R ′′tot,i = 4π ri2 R tot = R ′′cnd,i + R ′′cnv,i =
+ 
k ss
h  ro
( )



2
where, for ro = 0.6 m and h = 1000 W/m2⋅K, R ′′cnd,i = 5.56 × 10-3 m2⋅K/W, R ′′cnv,i = 6.94 × 10-4
m2⋅K/W, and R ′′tot,i = 6.25 × 10-3 m2⋅K/W. Results for the center temperature are shown below.
Center temperature, T(0) (C)
675
625
575
525
475
0
2000
4000
6000
8000
10000
Convection coefficient, h(W/m^2.K)
ro = 0.54 m
ro = 0.60 m
Clearly, even with ro = 0.54 m = ro,min and h = 10,000 W/m2⋅K (a practical upper limit), T(0) > 475°C
and the desired condition cannot be met. The corresponding resistances are R ′′cnd,i = 2.47 × 10-3
m2⋅K/W, R ′′cnv,i = 8.57 × 10-5 m2⋅K/W, and R ′′tot,i = 2.56 × 10-3 m2⋅K/W. The conduction resistance
remains dominant, and the effect of reducing R ′′cnv,i by increasing h is small. The proposed extension
is not feasible.
COMMENTS: A value of q = 1.79 × 105 W/m3 would allow for operation at T(0) = 475°C with ro =
0.54 m and h = 10,000 W/m2⋅K.
PROBLEM 3.89
PROBLEM 3.89 (Cont.)
PROBLEM 3.90
KNOWN: Plane wall, long cylinder and sphere, each with characteristic length a, thermal

conductivity k and uniform volumetric energy generation rate q.
FIND: (a) On the same graph, plot the dimensionless temperature, [ T ( x or r ) − T ( a ) ]/[ q a /2k], vs.
the dimensionless characteristic length, x/a or r/a, for each shape; (b) Which shape has the smallest
temperature difference between the center and the surface? Explain this behavior by comparing the
ratio of the volume-to-surface area; and (c) Which shape would be preferred for use as a nuclear fuel
element? Explain why?
2
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties and (4) Uniform volumetric generation.
ANALYSIS: (a) For each of the shapes, with T(a) = Ts, the dimensionless temperature distributions
can be written by inspection from results in Appendix C.3.
2
T ( x ) − Ts
x
Plane wall, Eq. C.22
= 1−  
a
 2 / 2k
qa
Long cylinder, Eq. C.23
Sphere, Eq. C.24
T ( r ) − Ts 1   r 2 
1 −   
=
 2 / 2k 2   a  
qa
T ( r ) − Ts 1   r 2 
1 −   
=
 2 / 2k 3   a  
qa
The dimensionless temperature distributions using the foregoing expressions are shown in the graph
below.
Dimensionless temperature distribution
(T_x,r-Ts) / (qdot*a^2/2*k)
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
Dimensionless length, x/a or r/a
Plane wall, 2a
Long cylinder, a
Sphere, a
Continued …
PROBLEM 3.90 (Cont.)
(b) The sphere shape has the smallest temperature difference between the center and surface, T(0) –
T(a). The ratio of volume-to-surface-area, ∀/As, for each of the shapes is
Plane wall
∀ a (1×1)
= = a
As
(1×1)
Long cylinder
∀ π a 2 ×1 a
= =
As 2π a ×1 2
Sphere
∀ 4π a 3 / 3 a
= =
As
3
4π a 2
The smaller the ∀/As ratio, the smaller the temperature difference, T(0) – T(a).
(c) The sphere would be the preferred element shape since, for a given ∀/As ratio, which controls the
generation and transfer rates, the sphere will operate at the lowest temperature.
PROBLEM 3.91
KNOWN: Radius, thickness, and incident flux for a radiation heat gauge.
FIND: Expression relating incident flux to temperature difference between center and edge
of gauge.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible
temperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5)
Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contact
resistance between foil and heat sink.
ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr,
−k ( 2π rt )
q r + q′′i ( 2π rdr ) =
q r+dr ,
qr =
dT
,
dr
dq
q r+dr =
q r + r dr.
dr
Rearranging, find that
=
q′′i ( 2π rdr
)
d 
dT
( −k2π rt )  dr

dr 
dr 
d  dT 
q′′
r
= − i r.


dr  dr 
kt
Integrating,
dT
q′′i r 2
r
=
−
+ C1
dr
2kt
and
q′′i r 2
T (r) =
−
+ C1lnr+C2 .
4kt
With dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R),
q′′R 2
T (R ) =
− i
+ C2
4kt
or
q′′R 2
C2 =
T (R ) + i .
4kt
Hence, the temperature distribution is
T=
(r)
(
)
q′′i
R 2 − r2 + T ( R ).
4kt
Applying this result at r = 0, it follows that
4kt
4kt
q′′i = T ( 0 ) − T ( R )  = ∆T.
2
R
R2
COMMENTS: This technique allows for determination of a radiation flux from
measurement of a temperature difference. It becomes inaccurate if emission from the foil
becomes significant.
<
PROBLEM 3.92
PROBLEM 3.92 (Cont.)
PROBLEM 3.93
KNOWN: Diameter and base temperature of a silicon carbide nanowire, required temperature of the
catalyst tip.
FIND: Maximum length of a nanowire that may be grown under specified conditions.
SCHEMATIC:
2400 K ≤ Tc ≤ 3000 K
T∞ = 8000 K
h = 105 W/m2∙K
L
D = 15 × 10-9 m
x
Tb = 2400 K
ASSUMPTIONS: (1) Nanowire stops growing when Tc = T(x = L) = 3000 K, (2) Constant properties,
(3) One-dimensional heat transfer, (4) Convection from the tip of the nanowire, (5) Nanowire grows
very slowly, (6) Negligible impact of nanoscale heat transfer effects.
PROPERTIES: Table A.2, silicon carbide (1500 K): k = 30 W/m⋅K.
ANALYSIS: The tip of the nanowire is initially at T = 2400 K, and increases in temperature as the
nanowire becomes longer. At steady-state, the tip reaches T = 3000 K. The temperature distribution
at steady-state is given by Eq. 3.75:
θ
cosh m (L - x) + (h / mk) sinh m (L - x)
=
θb
cosh mL + (h / mk) sinh mL
(1)
where
1/2
 hP 
m= 

 kA c 
1/2
 4h 
=

 kD 
1/2


4 × 105 W/m 2 ⋅ K
=
 30 W/m ⋅ K × 15 × 10-9 m 


= 943 × 103 m-1
and
h
105 W/m 2 ⋅ K
=
= 3.53 × 10-3
3
-1
mk
943 × 10 m × 30 W/m ⋅ K
Equation 1, evaluated at x = L, is
θ
(3000 - 8000) K
1
=
= 0.893 =
3
θb
(2400 - 8000) K
cosh (943 × 10 × L) + 3.53 × 10-3sinh (943 × 103 × L)
A trial-and-error solution yields L = 510 × 10-9 m = 510 nm
<
Continued…
PROBLEM 3.93 (Cont.)
COMMENTS: (1) The importance of radiation heat transfer may be ascertained by evaluating Eq.
1.9. Assuming large surroundings at a temperature of Tsur = 8000 K and an emissivity of unity, the
radiation heat transfer coefficient at the fin tip is
2 
h r = εσ(T(x = L) + Tsur ) T 2 (x=L) + Tsur


= 5.67 × 10-8 W/m 2 ⋅ K 4 × (3000 K + 8000 K) × (3000 K) 2 + (8000 K) 2  = 4.5 × 104 W/m 2 ⋅ K


We see that hr < h, but radiation may be important. (2) The thermal conductivity has been evaluated at
1500 K and extrapolated to a much higher temperature. More accurate values of the thermal
conductivity, accounting for the high temperature and possible nanoscale heat transfer effects, are
desirable. (3) If the nanowire were to grow rapidly, the transient temperature distribution within the
nanowire would need to be evaluated.
PROBLEM 3.94
PROBLEM 3.94 (Cont.)
PROBLEM 3.95
PROBLEM 3.96
PROBLEM 3.96 (Cont.)
PROBLEM 3.97
PROBLEM 3.97 (Cont.)
PROBLEM 3.98
PROBLEM 3.98 (Cont.)
PROBLEM 3.99
KNOWN: Trench length and nanotube diameter. Laser irradiation of known power at two distinct
axial locations. Measured nanotube temperatures at the trench half-width. Nanotube thermal
conductivity. Island temperature.
FIND: Thermal contact resistances at the left and right ends of the nanotube.
SCHEMATIC:
Temperature measurement
Laser irradiation
q = 10 µW
s/2
= 2.5 µm
x
T∞ = Tsur = 300 K
= 1.5 or 3.5 µm
s = 5 µm
Rt,c,L
Carbon nanotube
D = 14 nm
Rt,c,R
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction. (2) Constant properties. (3)
Negligible radiation and convection losses.
ANALYSIS: Thermal circuits may be drawn for the two laser irradiation locations as follows. The
top circuit corresponds to irradiation on the left half of the nanotube. The bottom circuit corresponds
to irradiation of the right half of the nanotube.
q = 10 µW
T1 = 324.5 K
ql,1
T∞
x1
s/2 - x1
kcnAcn
kcnAcn
Rt,c,l
s/2
Rt,c,r
T∞ qr,1
kcnAcn
Tmax,1
x2- s/2
q = 10 µW
kcnAcn
ql,2
T∞
Rt,c,l
s/2
kcnAcn
Tmax,2
s - x2
q
Rt,c,r T∞ r,2
kcnAcn
T2 = 326.4 K
The following equations may be written for irradiation of the left side of the nanotube (top circuit).
=
q ql ,1 + qr ,1
ql ,1 =
(1)
Tmax,1 − T∞
Rt ,c,l +
x1
(2)
kcn Acn
Continued…
PROBLEM 3.99 (Cont.)
qr ,1 =
qr ,1 =
Tmax,1 − T∞
s / 2 − x1
s/2
+
+ Rt ,c,r
kcn Acn
kcn Acn
T1 − T∞
s/2
+ Rt ,c,r
kcn Acn
(3)
(4)
For irradiation of the right side of the nanotube (bottom circuit),
=
q ql ,2 +qr ,2
ql ,2 =
ql ,2 =
qr ,2 =
Tmax,2 − T∞
x − s/2
s/2
+ 2
+ Rt ,c,l
kcn Acn
kcn Acn
T2 − T∞
s/2
+ Rt ,c,l
kcn Acn
Tmax,2 − T∞
s − x2
+ Rt ,c,r
kcn Acn
(5)
(6)
(7)
(8)
With kcn = 3100 W/m⋅K, Acn = 1.54 × 10-16 m2, x1 = 1.5 µm, T1 = 324.5 K, x2 = 3.5 µm, and T2 = 326.4
K, Equations (1) through (8) may be solved simultaneously to yield
Tmax,1 = 331.0 K, ql,1 = 6.896 × 10-6 W, qr,1 = 3.104 × 10-6 W
Tmax,2 = 334.8 K, ql,2 = 4.00 × 10-6 W, qr,2 = 6.00 × 10-6 W
and
Rt,c,l = 1.35 × 106 K/W ; Rt,c,r = 2.65 × 106 K/W
<
COMMENTS: (1) Assuming large surroundings, the maximum possible radiation loss is associated
4
4
- Tsur
) = 5.67 × 10-8
with blackbody behavior and Tmax,1. For this situation, qrad,max = sπDs( Tmax,2
W/m2⋅K4 × π × 14 × 10-9 m × 5 × 10-6 m × (334.84 – 3004)K4 = 55 × 10-10 W. This is much less than the
laser irradiation. Therefore, radiation heat transfer is negligible. (2) The carbon nanotube is not placed
symmetrically between the two islands. It is difficult to place a carbon nanotube with such accuracy.
PROBLEM 3.100
PROBLEM 3.100 (Cont.)
PROBLEM 3.101
PROBLEM 3.101 (Cont.)
PROBLEM 3.102
KNOWN: Dimensions and thermal conductivity of a gas turbine blade. Temperature and convection
coefficient of gas stream. Temperature of blade base and maximum allowable blade temperature.
FIND: (a) Whether blade operating conditions are acceptable, (b) Heat transfer to blade coolant.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3)
Adiabatic blade tip, (4) Negligible radiation.
ANALYSIS: Conditions in the blade are determined by Case B of Table 3.4.
(a) With the maximum temperature existing at x = L, Eq. 3.80 yields
T ( L ) − T∞
Tb − T∞
=
m
=
1
cosh mL
1/ 2
=
( hP/kA
c)
( 250W/m2 ⋅ K × 0.11m/20W/m ⋅ K × 6 ×10−4 m2 )
1/ 2
m = 47.87 m-1 and mL = 47.87 m-1 × 0.05 m = 2.39
From Table B.1, cosh mL = 5.51. Hence,
T ( L=
) 1200 C + (300 − 1200) C/5.51= 1037 C
<
and the operating conditions are acceptable.
(
(b) With M =( hPkA c )1/ 2 Θ b =250W/m 2 ⋅ K × 0.11m × 20W/m ⋅ K × 6 × 10 −4 m 2
) ( −900 C ) =−517W ,
1/ 2

Eq. 3.81 and Table B.1 yield
qf =
M tanh mL =
−517W ( 0.983) =
−508W
Hence, q b =
−q f =
508W
<
COMMENTS: Radiation losses from the blade surface and convection from the tip will contribute to
reducing the blade temperatures.
PROBLEM 3.103
KNOWN: Dimensions of disc/shaft assembly. Applied angular velocity, force, and torque. Thermal
conductivity and inner temperature of disc.
FIND: (a) Expression for the friction coefficient µ, (b) Radial temperature distribution in disc, (c)
Value of µ for prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant
k, (4) Uniform disc contact pressure p, (5) All frictional heat dissipation is transferred to shaft from
base of disc.
ANALYSIS: (a) The normal force acting on a differential ring extending from r to r+dr on the contact
surface of the disc may be expressed as dFn = p2p rdr . Hence, the tangential force is
dFt = µ p2p rdr , in which case the torque may be expressed as
dτ = 2pµ pr 2dr
For the entire disc, it follows that
r
2p
=
τ 2=
pµ p 2 r 2dr
µ pr23
o
3
∫
where p = F p r22 . Hence,
µ=
3 τ
2 Fr2
<
(b) Performing an energy balance on a differential control volume in the disc, it follows that
q cond,r + dq fric − q cond,r + dr =
0
(
)
(
)
With dq=
=
dτ 2 µ Fω r 2 r22 dr , q cond,r
+ dr q cond,r + dq cond,r dr dr , and
fric ω=
q cond,r = −k ( 2π rt ) dT dr , it follows that
d ( rdT dr )
2 µ Fω r 2 r22 dr + 2π kt
dr =
0
dr
(
or
d ( rdT dr )
dr
)
= −
µ Fω 2
r
π ktr22
Integrating twice,
Continued...
PROBLEM 3.103 (Cont.)
dT
µ Fω 2 C1
r +
=
−
2
dr
r
3π ktr2
µ Fω 3
T=
−
r + C1nr + C2
9π ktr22
Since the disc is well insulated
at r r2=
=
, dT dr r
0 and
2
C1 =
µ Fω r2
3π kt
With T ( r1 ) = T1 , it also follows that
µ Fω 3
C2 =
T1 +
r1 − C1nr1
9π ktr22
Hence,
T ( r ) =T1 −
r
µ Fω r2
r 3 − r13 ) +
n
(
3π kt
r1
9π ktr 2
µ Fω
<
2
(c) For the prescribed conditions,
=
m
3
8N ⋅ m
= 0.333
2 200N ( 0.18m )
<
Since the maximum temperature occurs at r = r2,
m Fω r2
Tmax =
T ( r2 ) =
T1 −
9π kt
=
With ( m Fω r2 3π kt
)
  3 
 
1 −  r1   + m Fω r2 n  r2 
  r2   3π kt
 r1 


( 0.333 × 200N × 40rad/s × 0.18m
=
3π ×15W/m ⋅ K × 0.012m
) 282.7a C ,
3
282.7a C   0.02  
 0.18 
1 − 
Tmax =
80aa
C−
  + 282.7 Cn 

3
 0.02 
  0.18  
Tmax =80aaaa
C − 94.1 C + 621.1 C =607 C
COMMENTS: The maximum temperature is excessive, and the disks should be actively cooled (by
convection) at their outer surfaces.
<
PROBLEM 3.104
KNOWN: Dimensions of disc/shaft assembly. Applied angular velocity, force, and torque. Thermal
conductivity and inner temperature of disc.
FIND: (a) Expression for the friction coefficient µ, (b) Radial temperature distribution in disc, (c)
Value of µ for prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant
k, (4) Uniform disc contact pressure p, (5) All frictional heat dissipation is transferred to shaft from
base of disc.
ANALYSIS: (a) The normal force acting on a differential ring extending from r to r+dr on the contact
surface of the disc may be expressed as dFn = p2p rdr . Hence, the tangential force is
dFt = µ p2p rdr , in which case the torque may be expressed as
dτ = 2pµ pr 2dr
For the entire disc, it follows that
r
2p
=
τ 2=
pµ p 2 r 2dr
µ pr23
o
3
∫
where p = F p r22 . Hence,
µ=
3 τ
2 Fr2
<
(b) Performing an energy balance on a differential control volume in the disc, it follows that
q cond,r + dq fric − q cond,r + dr =
0
(
)
(
)
With dq=
=
dτ 2 µ Fω r 2 r22 dr , q cond,r
+ dr q cond,r + dq cond,r dr dr , and
fric ω=
q cond,r = −k ( 2π rt ) dT dr , it follows that
d ( rdT dr )
2 µ Fω r 2 r22 dr + 2π kt
dr =
0
dr
(
or
d ( rdT dr )
dr
)
= −
µ Fω 2
r
π ktr22
Integrating twice,
Continued...
PROBLEM 3.104 (Cont.)
dT
µ Fω 2 C1
r +
=
−
2
dr
r
3π ktr2
µ Fω 3
T=
−
r + C1nr + C2
9π ktr22
Since the disc is well insulated
at r r2=
=
, dT dr r
0 and
2
C1 =
µ Fω r2
3π kt
With T ( r1 ) = T1 , it also follows that
µ Fω 3
C2 =
T1 +
r1 − C1nr1
9π ktr22
Hence,
T ( r ) =T1 −
r
µ Fω r2
r 3 − r13 ) +
n
(
3π kt
r1
9π ktr 2
µ Fω
<
2
(c) For the prescribed conditions,
=
m
3
8N ⋅ m
= 0.333
2 200N ( 0.18m )
<
Since the maximum temperature occurs at r = r2,
m Fω r2
Tmax =
T ( r2 ) =
T1 −
9π kt
=
With ( m Fω r2 3π kt
)
  3 
 
1 −  r1   + m Fω r2 n  r2 
  r2   3π kt
 r1 


( 0.333 × 200N × 40rad/s × 0.18m
=
3π ×15W/m ⋅ K × 0.012m
) 282.7a C ,
3
282.7a C   0.02  
 0.18 
1 − 
Tmax =
80aa
C−
  + 282.7 Cn 

3
 0.02 
  0.18  
Tmax =80aaaa
C − 94.1 C + 621.1 C =607 C
COMMENTS: The maximum temperature is excessive, and the disks should be actively cooled (by
convection) at their outer surfaces.
<
PROBLEM 3.105
KNOWN: Extended surface of rectangular cross-section with heat flow in the longitudinal direction.
FIND: Determine the conditions for which the transverse (y-direction) temperature difference is
negligible compared to the temperature difference between the surface and the environment, such that
the 1-D analysis of Section 3.6.1 is valid by finding: (a) An expression for the conduction heat flux at
the surface, q′′y ( t ) , in terms of Ts and To, assuming the transverse temperature distribution is
parabolic, (b) An expression for the convection heat flux at the surface for the x-location; equate the
two expressions, and identify the parameter that determines the ratio (To – Ts)/(Ts - T∞); and (c)
Developing a criterion for the validity of the 1-D assumption used to model an extended surface.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform convection coefficient and (3) Constant
properties.
ANALYSIS: (a) Referring to the schematics above, the conduction heat flux at the surface y = t at
any x-location follows from Fourier’s law using the parabolic transverse temperature distribution.

∂T 
2y 
2k
Ts ( x ) − To ( x ) 
q′′y ( t ) =
−k
=
−k  Ts ( x ) − To ( x )  
=
−

2
t
∂y  y = t
t y=t

(1)
(b) The convection heat flux at the surface of any x-location follows from the rate equation
=
q′′cv h Ts ( x ) − T∞ 
(2)
Performing a surface energy balance as represented schematically above, equating Eqs. (1) and (2)
provides
q′′y ( t ) = q′′cv
2k
Ts ( x ) − To ( x )  = h Ts ( x ) − T∞ 
t 
Ts ( x ) − To ( x )
ht
=
−0.5 =
−0.5 Bi
Ts ( x ) − T∞ ( x )
k
−
(3)
where Bi = ht/k, the Biot number, represents the ratio of the conduction to the convection thermal
resistances,
=
Bi
R ′′cd t / k
=
R ′′cv 1/ h
(4)
(c) The transverse temperature difference (Ts – To) will be negligible compared to the temperature
difference between the surface and the environment (Ts - T∞) when Bi << 1, say, 0.1, an order of
magnitude smaller. This is the criterion to validate the one-dimensional assumption used to model
extended surfaces.
COMMENTS: The coefficient 0.5 in Eq. (3) is a consequence of the parabolic distribution
assumption. This distribution represents the simplest polynomial expression that could approximate
the real distribution.
PROBLEM 3.106
KNOWN: Length, diameter, base temperature and environmental conditions associated with a brass rod.
FIND: Temperature at specified distances along the rod.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4)
Negligible radiation, (5) Uniform convection coefficient h.
)
(
 C : k 133 W/m ⋅ K.
PROPERTIES: Table A-1, Brass
=
T 110=
ANALYSIS: Evaluate first the fin parameter
1/ 2
1/ 2
1/ 2
 hP 
 hπ D 
 4h 
=
=
m  =



=
kD 
 kπ D 2 / 4 
 kAc 
1/ 2
 4 × 30 W/m 2 ⋅ K 


133 W/m ⋅ K × 0.005m 
m = 13.43 m-1.
Hence, m L = (13.43)×0.1 = 1.34 and from the results of Example 3.9, it is advisable not to make the
infinite rod approximation. Thus from Table 3.4, the temperature distribution has the form
θ=
cosh m ( L − x ) + ( h/mk ) sinh m ( L − x )
cosh mL + ( h/mk ) sinh mL
θb
Evaluating the hyperbolic functions, cosh mL = 2.04 and sinh mL = 1.78, and the parameter
h
30 W/m 2 ⋅ K
= = 0.0168,
mk 13.43m-1 (133 W/m ⋅ K )
with θb = 180°C the temperature distribution has the form
θ=
cosh m ( L − x ) + 0.0168 sinh m ( L − x )
(180 C).
2.07
The temperatures at the prescribed locations are tabulated below.
x(m)
cosh m(L-x)
sinh m(L-x)
θ
T(°C)
x1 = 0.025
1.55
1.19
136.5
156.5
x2 = 0.05
1.24
0.725
108.9
128.9
L = 0.10
1.00
0.00
87.0
107.0
<
<
<
COMMENTS: If the rod were approximated as infinitely long: T(x1) = 148.7°C, T(x2) =
112.0°C, and T(L) = 67.0°C. The assumption would therefore result in significant
underestimates of the rod temperature.
PROBLEM 3.107
KNOWN: Thickness, length, thermal conductivity, and base temperature of a rectangular fin. Fluid
temperature and convection coefficient.
FIND: (a) Heat rate per unit width, efficiency, effectiveness, thermal resistance, and tip temperature
for different tip conditions, (b) Effect of convection coefficient and thermal conductivity on the heat
rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along fin, (3) Constant
properties, (4) Negligible radiation, (5) Uniform convection coefficient, (6) Fin width is much longer
than thickness (w >> t).
ANALYSIS: (a) The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.77), (3.81) and
2 1/2
2
1/2
(3.85), where M ≈ (2 hw tk) (Tb - T∞) = (2 × 100 W/m ⋅K × 0.001m × 180 W/m⋅K) (75°C) w =
1/2
2
1/2
-1
-1
450 w W, m≈ (2h/kt) = (200 W/m ⋅K/180 W/m⋅K ×0.001m) = 33.3m , mL ≈ 33.3m × 0.010m
2
-1
= 0.333, and (h/mk) ≈ (100 W/m ⋅K/33.3m × 180 W/m⋅K) = 0.0167. From Table B-1, it follows that
sinh mL ≈ 0.340, cosh mL ≈ 1.057, and tanh mL ≈ 0.321. From knowledge of qf, Eqs. (3.91), (3.86)
and (3.88) yield
hf ≈
q ′f
, εf ≈
q ′f
q
b
, R ′t,f =
′f
h ( 2L + t )qq
ht
q
b
b
Case A: From Eq. (3.77), (3.91), (3.86), (3.88) and (3.75),
q ′f
=
M sinh mL + ( h / mk ) cosh mL
0.340 + 0.0167 × 1.057
W/m
151 W / m
= 450=
w cosh mL + ( h / mk ) sinh mL
1.057 + 0.0167 × 0.340
=
ηf
151 W / m
= 0.96
2
100 W / m ⋅ K ( 0.021m ) 75°C
=
εf
<
75°C
R ′t,f
= 20.2,=
= 0.50 m ⋅ K / W
2
151 W / m
100 W / m ⋅ K ( 0.001m ) 75°C
151 W / m
T ( L ) = T∞ +
θb
cosh mL + ( h / mk ) sinh mL
= 25°C +
<
75°C
1.057 + ( 0.0167 ) 0.340
= 95.6°C
<
<
Case B: From Eqs. (3.81), (3.91), (3.86), (3.88) and (3.80)
=
q ′f
M
=
tanh mL 450 W /=
m ( 0.321) 144 W / m
w
=
ηε
0.92,
=
R ′t,f 0.52 m ⋅ K / W
f
f 19.3,=
T ( L ) = T∞ +
θb
cosh mL
= 25°C +
75°C
1.057
= 96.0°C
<
<
<
Continued …
PROBLEM 3.107 (Cont.)
Case D (L → ∞): From Eqs. (3.85), (3.91), (3.86), (3.88) and (3.84)
′f
q=
M
= 450 W / m
w
<
ηε
0, f =
60.0, R ′t,f =
0.167 m ⋅ K / W, T ( L ) =
T∞ =
25°C
f =
<
(b) The effect of h on the heat rate is shown below for the aluminum and stainless steel fins.
Variation of qf' with h (k=180 W/m.K)
Heat rate, qf'(W/m)
1500
1000
500
0
0
200
400
600
800
1000
Convection coefficient, h(W/m^2.K)
qfA'
qfB'
qfD'
Variation of qf' with h (k=15W/m.K)
Heat rate, qf'(W/m)
400
300
200
100
0
0
200
400
600
800
1000
Convection coefficient, h(W/m^2.K)
qfA'
qfB'
qfD'
For both materials, there is little difference between the Case A and B results over the entire range of
h. The difference (percentage) increases with decreasing h and increasing k, but even for the worst
2
case condition (h = 10 W/m ⋅K, k = 180 W/m⋅K), the heat rate for Case A (15.7 W/m) is only slightly
larger than that for Case B (14.9 W/m). For aluminum, the heat rate is significantly over-predicted by
the infinite fin approximation over the entire range of h. For stainless steel, it is over-predicted for
2
small values of h, but results for all three cases are within 1% for h > 500 W/m ⋅K.
COMMENTS: From the results of Part (a), we see there is a slight reduction in performance (smaller
values of q ′f , ηε
f and f , as well as a larger value of R ′t,f ) associated with insulating the tip.
Although hf = 0 for the infinite fin, q ′f and ef are substantially larger than results for L = 10 mm,
indicating that performance may be significantly improved by increasing L.
PROBLEM 3.108
PROBLEM 3.109
PROBLEM 3.109 (Cont.)
PROBLEM 3.110
PROBLEM 3.111
KNOWN: Dimensions of a nanospring, dependence of pitch upon temperature.
FIND: Actuation distance of the spring in response to heating of its end, accuracy to which the
actuation length can be controlled.
SCHEMATIC:
L
2r = 60 nm
x
D = 15 nm
Tb = 50°C
Lc = 425 nm
s = 25 nm
T∞ = 25°C
h = 106 W/m2∙K
ASSUMPTIONS: (1) Constant properties, (2) Steady-state conditions, (3) One-dimensional heat
transfer, (4) Adiabatic tip, (5) Negligible radiation heat transfer, (6) Negligible impact of nanoscale
heat transfer effects.
PROPERTIES: Table A.2, silicon carbide (300 K): k = 490 W/m⋅K.
ANALYSIS: When the nanospring is at Ti = 25°C, the spring length is
Li =
s
Lc
2
2π r + ( S
2π
)2
=
25 × 10-9 m
×
2π
425 × 10-9 m
-9


(30 × 10 m) +  25 × 10 m) 
2π


-9
2
2
= 55.9 × 10-9 m = 55.9 nm
Since the average spring pitch varies linearly with the average temperature, the average pitch of the
heated spring is
s = si +
ds
(T - Ti )
dT
(1)
The average excess temperature is
θ = T - T∞ =
1
Lc
Lc
∫ θ(x)dx
where, from Eq. 3.80,
x=0
Continued…
PROBLEM 3.111 (Cont.)
θb
Lc
θ =
θ =
Lc
∫
x=0
θb
cosh m (L - x)
L
dx = sinh m (Lc - x) 0 c
cosh ml
mLc (cosh mLc )
θb
θ
× (0 - sinh mLc ) = b tanh (mLc )
mLc (cosh mLc )
mLc
For a particular spring,
1/2
 hP 
mLc = 

 kA c 
1/2
 4h 
Lc = 

 kD 
1/2


4 × 106 W/m 2 ⋅ K
Lc = 
 490 W/m ⋅ K × 15 × 10-9 m 


× 425 × 10-9 m = 0.314
(50 - 25)°C
tanh (0.314) = 24.2°C
0.314
T = θ + T∞ = 24.2°C + 25°C = 49.2°C
Therefore θ =
and
From Eq. (1),
s = 25 × 10-9 m + 0.1 × 10-9 m/K × (49.2 - 25)°C = 27.4 × 10-9 m
Therefore,
27.4 × 10-9 m
425 × 10-9 m
L2 =
×
= 61.1 × 10-9 m = 61.1 nm
-9
2π
(30 × 10-9 m) 2 + ( 27.4 × 10 m ) 2
2π
and the actuation length is
ΔL = L 2 - Li = 61.1 nm - 55.9 nm = 5.2 nm
<
If the base temperature can be controlled to within 1 degree Celsius, the resolution of the actuation
1 degree C
= 0.2 nm
length is: R = ΔL ×
25 degree C
<
COMMENTS: (1) The actuation distance and its resolution are extremely small. (2) Application of
other tip conditions will lead to different predictions of the actuation distance.
PROBLEM 3.112
PROBLEM 3.113
PROBLEM 3.114
KNOWN: Array of aluminum fins with specified dimensions. Base and environment temperatures.
Dependence of heat transfer coefficient on number of fins.
FIND: Total rate of heat transfer for 0, 3, 6, and 9 fins.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature nearly uniform across fin cross
section, (3) Constant properties, (4) Negligible radiation, (5) Uniform h.
PROPERTIES: Table A.1, Aluminum (pure), T ≅ 330 K: k ≅ 238 W/m⋅K.
ANALYSIS: Equations 3.103 and 3.107 can be used to calculate the heat transfer rate from a fin
array. The calculations are performed here for N = 3 fins. The heat transfer coefficient is:

N 
3

2
2
=
h h max 1 − =
 50 W/m ⋅ K × 1 −=
 35 W/m ⋅ K
 10 
 N max 
Then
P= 2(w + t)= 2(0.02 m+0.002 m)= 0.044 m
Ac =
wt =
0.02 m × 0.002 m =
4 ×10-5 m 2
1/2
 hP 
=
m =

 kA c 
Lc =
L+t /2=
20 mm + 2 mm / 2 =
21 mm,
M = ( hPkA c )
1/2
=
(
1/2
 35 W/m 2 ⋅ K × 0.044 m 


=
12.7 m-1
 238 W/m ⋅ K × 4 ×10-5 m 2 


mLc =
12.7 m-1 × 0.021 m =
0.267
θb
35 W/m 2 ⋅ K × 0.044 m × 238 W/m ⋅ K × 4 ×10−5 m 2
)
1/2
(95 −=
20)°C 9.08 W
Continued …
PROBLEM 3.114 (Cont.)
The heat transfer rate for a single fin is
qf =
M tanh(mLc ) =
9.08 W × tanh(0.267) =
2.37 W
With A b = (H − Nt)w = (0.02 m − 3 × 0.002 m) × 0.02 m= 0.00028 m 2 ,
q tot Nq fin + hA bq b
=
= 3 × 2.37 W + 35 W/m 2 ⋅ K × 0.00028 m 2 × (95 − 20)°C = 7.84 W
Repeating the calculations for different numbers of fins, the results are given in the table below.
N
0
3
6
9
<
qtot (W)
1.50
7.84
8.44
3.12
COMMENTS: There is a trade-off between increasing the number of fins and decreasing the heat
transfer coefficient because of blockage. In the case considered, the maximum heat transfer rate is for
N = 5 fins and is qtot = 8.89 W.
PROBLEM 3.115
KNOWN: Arrangement of fins between parallel plates. Temperature and convection coefficient of
air flow in finned passages. Maximum allowable plate temperatures.
FIND: (a) Expressions relating fin heat transfer rates to end temperatures, (b) Maximum power
dissipation for each plate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fins, (3) Constant
properties, (4) Negligible radiation, (5) Uniform h, (6) Negligible variation in T∞, (7) Negligible
contact resistance.
PROPERTIES: Table A.1, Aluminum (pure), 375 K: k = 240 W/m⋅K.
ANALYSIS: (a) The general solution for the temperature distribution in a fin is
= C1emx + C2e-mx
θ ( x ) ≡ T ( x ) - T∞
θ (0) =
θo =
To − T∞ ,
Boundary conditions:
Hence
θo =
C1 + C2
θ (L) =
θL =
TL − T∞ .
θL =
C1emL + C2e-mL
mL + θ - C e-mL
θ=
( o 1)
L C1e
θ L --θ o e-mL
θ L θ o e-mL θ o emL θ L
C1 =
C2 =
θo = .
emL --e-mL
emL e-mL emL e-mL
Hence
θ (x) =
θ (x) =
m x-L
m L-x
θ Lemx - θ o e ( ) + θ o e ( ) - θ Le-mx
emL - e-mL
(
m L-x
-m L-x
θ o e ( ) - e ( )  + θ L emx - e-mx


)
emL - e-mL
θ sinh m ( L-x ) + θ Lsinh mx
θ (x) = o
.
sinh mL
The fin heat transfer rate is then
−kA c
qf =
dT
q m
 q m

=
−kDt  − o
cosh m ( L − x ) + L
cosh mx  .
dx
sinh mL
 sinh mL

q m 
 qom
− L

 tanh mL sinh mL 
<
qLm 
 qom
−
.
 sinh mL tanh mL 
<
Hence =
q f,o kDt 
=
q f,L kDt 
Continued …
PROBLEM 3.115 (Cont.)
1/ 2
(b)
 hP 
m= 

 kA c 
1/ 2
 150 W/m 2 ⋅ K ( 2 × 0.1 m+2 × 0.001 m ) 
=

35.5 m-1


240 W/m ⋅ K × 0.1 m × 0.001 m


mL = 35.5 m-1 × 0.012 m = 0.43
=
sinh mL 0.439 =
tanh mL 0.401
=
θ o 100
=
K
θ L 50 K
 100 K × 35.5 m-1 50 K × 35.5 m-1 

=
q f,o 240 W/m ⋅ K × 0.1 m × 0.001 m 


0.401
0.439


q f,o = 115.4 W
(from the top plate)
 100 K × 35.5 m-1 50 K × 35.5 m-1 

=
q f,L 240 W/m ⋅ K × 0.1 m × 0.001 m 


0.439
0.401


q f,L = 87.8 W.
(into the bottom plate)
Maximum power dissipations are therefore
q o,max
= N f q f,o + ( W − N f t ) Dhq o
q o,max = 50 ×115.4 W+ ( 0.200 − 50 × 0.001) m × 0.1 m ×150 W/m 2 ⋅ K ×100 K
<
=
q o,max 5770
=
W+225 W 5995 W
q L,max =
− N f q f,L + ( W − N f t ) Dhq o
q L,max =
−50 × 87.8W + ( 0.200 − 50 × 0.001) m × 0.1 m ×150 W/m 2 ⋅ K × 50 K
<
q L,max =
−4390 W+112W =
−4278 W.
COMMENTS: (1) It is of interest to determine the air velocity needed to prevent excessive heating of the air as
it passes between the plates. If the air temperature change is restricted to
 air
m
=
∆T∞ = 5 K, its flowrate must be
q tot
1717 W
=
= 0.34 kg/s.
cp ∆T∞ 1007 J/kg ⋅ K × 5 K
Its mean velocity is then
 air
m
0.34 kg/s
=
Vair =
= 163 m/s.
3
rair Ac 1.16 kg/m × 0.012 m ( 0.2 − 50 × 0.001) m
Such a velocity would be impossible to maintain. To reduce it to a reasonable value, e.g. 10
m/s, Ac would have to be increased substantially by increasing W (and hence the space
between fins) and by increasing L. The present configuration is impractical from the
standpoint that 1717 W could not be transferred to air in such a small volume.
(2) A negative value of qL,max implies that the bottom plate must be cooled externally to
maintain the plate at 350 K.
PROBLEM 3.116
PROBLEM 3.117
(Continued)
PROBLEM 3.117 (Cont.)
(Continued)
PROBLEM 3.117 (Cont.)
PROBLEM 3.118
(Continued)
PROBLEM 3.118 (Cont.)
(Continued)
PROBLEM 3.118 (Cont.)
(Continued)
PROBLEM 3.118 (Cont.)
PROBLEM 3.119
PROBLEM 3.119 (Cont.)
PROBLEM 3.120
PROBLEM 3.120 (Cont.)
Problem 3.121
KNOWN: Two finned heat sinks, Designs A and B, prescribed by the number of fins in the array, N,
fin dimensions of square cross-section, w, and length, L, with different convection coefficients, h.
FIND: Determine which fin arrangement is superior. Calculate the heat rate, qf, efficiency, hf, and
effectiveness, ef, of a single fin, as well as, the total heat rate, qt, and overall efficiency, ho, of the
array. Also, compare the total heat rates per unit volume.
SCHEMATIC:
Fin dimensions
Cross section
Length
w x w (mm)
L (mm)
3x3
30
1x1
7
Design
A
B
Number of
fins
6x9
14 x 17
Convection
coefficient
(W/m2⋅K)
125
375
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fins, (3)
Convection coefficient is uniform over fin and prime surfaces, (4) Fin tips experience convection, and
(5) Constant properties.
ANALYSIS: Following the treatment of Section 3.6.5, the overall efficiency of the array, Eq.
(3.103), is
ho
=
qt
qt
=
q max hA tq b
(1)
where At is the total surface area, the sum of the exposed portion of the base (prime area) plus the fin
surfaces, Eq. 3.104,
A t =N ⋅ A f + A b
(2)
where the surface area of a single fin and the prime area are
Af = 4 ( L × w ) + w 2
(3)
A b = b1× b2 − N ⋅ A c
(4)
Combining Eqs. (1) and (2), the total heat rate for the array is
=
q t Nhf hA f qq
b + hA b b
(5)
where hf is the efficiency of a single fin. From Table 3.4, Case A, for the tip condition with
convection, the single fin efficiency based upon Eq. 3.91,
hf =
qf
hA f q b
(6)
Continued...
PROBLEM 3.121 (Cont.)
where
qf = M
=
M
sinh(mL) + ( h mk ) cosh(mL)
(7)
cosh(mL) + ( h mk ) sinh(mL)
θb
m ( hP
( hPkAc )1/ 2=
kA c )=
P 4w
=
Ac w 2
1/ 2
(8,9,10)
The single fin effectiveness, from Eq. 3.86,
εf =
qf
hA cq b
(11)
Additionally, we want to compare the performance of the designs with respect to the array volume,
′′′f q t ∀
= qt
q=
( b1⋅ b2 ⋅ L )
(12)
The above analysis was organized for easy treatment with equation-solving software. Solving Eqs. (1)
through (11) simultaneously with appropriate numerical values, the results are tabulated below.
Design
ho
hf
ef
q′′′f
qt
qf
(W)
(W)
A
113
1.80
0.804
0.779
31.9
1.25×106
B
165
0.475
0.909
0.873
25.3
7.81×106
(W/m3)
COMMENTS: (1) Both designs have good efficiencies and effectiveness. Clearly, Design B is
superior because the heat rate is nearly 50% larger than Design A for the same board footprint. Further,
the space requirement for Design B is four times less (∀ = 2.12×10-5 vs. 9.06×10-5 m3) and the heat rate
per unit volume is 6 times greater.
(2) Design A features 54 fins compared to 238 fins for Design B. Also very significant to the
performance comparison is the magnitude of the convection coefficient which is 3 times larger for
Design B. Estimating convection coefficients for fin arrays (and tube banks) is discussed in Chapter
7.6. Of concern is how the upstream fins alter the flow past the downstream fins and whether the
convection coefficient is uniform over the array.
(3) The IHT Extended Surfaces Model, for a Rectangular Pin Fin Array could have been used to solve
this problem.
PROBLEM 3.122
KNOWN: Dimensions of a fin array and dust layer. Aluminum and dust thermal conductivities. Base
temperature. Air temperature and heat transfer coefficient.
FIND: Allowable heat rate for dust layer thickness in the range of 0 ≤ Ld ≤ 5 mm.
SCHEMATIC:
T∞ = 25ºC
h = 375 W/m2·K
L = 7 mm
0 ≤ Ld ≤ 5 mm
Fin
kf = 175 W/m·K
Dust
kd = 0.032 W/m·K
1 mm x 1 mm
cross section
Tb = 75ºC
ASSUMPTIONS: (1) Steady-state, (2) Negligible temperature variation across fin thickness, (3)
Constant properties, (4) Uniform heat transfer coefficient, including over fin tips.
ANALYSIS: There are two heat transfer paths, one through the dust and into the air, and the other
through the fin. The thermal circuit is
Rconv
Rd,cond
T∞
Tb
T∞
Rfins
Rf,cond
The thermal resistances are given by
Ld
Ld
R d,cond =
=
k d Ad
k d (A p - NA c )
where Ap = 53 × 10-3 m × 57 × 10-3 m = 3.02 ×10-3 m2 , N = 14 × 17 = 238 and Ac = w2 = 10-6 m2.
R t,f
Ld
1
R conv =
, R f,cond =
and R fins =
k f NA c
hA d
N
where from Equation 3.88
θ
R t,f = b
qf
where qf is given by Equation 3.77,
cosh(mLf ) + (h / mk f ) sinh(mLf )
R t,f =
4hw 3 k (sinh(mLf ) + (h / mk f ) cosh(mLf ))
Continued…
PROBLEM 3.122 (Cont.)
m = (4h / kfw)1/2 = (4 × 375 W/m2∙K / 175 W/m∙K × 10-3 m)1/2 = 92.6 m-1
Here,
and
Lf = L - Ld.
Finally,
q = qdust + qfin
Tb - T∞
Tb - T∞
=
+
R d,cond + R conv
R f,cond + R fins
Performing the calculation for a dust layer thickness of Ld = 5 mm yields
Rd,cond = 56.1 K/W
Rconv = 0.958 K/W
Rf,cond = 0.120 K/W
Rt,f = 301 K/W, Rfins = 1.26 K/W
q=
75°C - 25°C
75°C - 25°C
+
= 0.876 W + 36.1 W = 37.0 W
(56.1 + 0.958) K/W
0.120 + 1.26
<
The figure shows the variation of the allowable heat rate as the dust layer thickness varies.
180
160
140
Heat rate, q (W)
120
100
80
60
40
20
0
0
0.001
0.002
0.003
Dust layer thickness, Ld (m)
0.004
0.005
COMMENTS: The figure below shows the two contributions to the heat rate, qdust and qfin. The heat
transfer through the dust layer decreases rapidly as the dust layer thickness increases and insulates the
surface. The fin heat transfer also decreases with increasing dust layer thickness as more of the fin
surface is insulated by the dust.
Continued…
PROBLEM 3.122 (Cont.)
180
q
qdust
qfin
160
140
Heat rate (W)
120
100
80
60
40
20
0
0
0.001
0.002
0.003
Dust thickness, Ld (m)
0.004
0.005
PROBLEM 3.123
PROBLEM 3.123 (Cont.)
PROBLEM 3.124
PROBLEM 3.125
PROBLEM 3.125 (Cont.)
PROBLEM 3.125 (Cont.)
\
PROBLEM 3.125 (Cont.)
PROBLEM 3.126
KNOWN: Dimensions, base temperature and environmental conditions associated with a triangular,
aluminum fin.
FIND: (a) Fin efficiency and effectiveness, (b) Heat dissipation per unit width.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties, (4) Negligible radiation and base contact resistance, (5) Uniform convection coefficient.
PROPERTIES: Table A-1, Aluminum, pure (T ≈ 400 K): k = 240 W/m⋅K.
ANALYSIS: (a) With Lc = L = 0.006 m, find
A p = Lt 2=
(
( 0.006 m )( 0.002 m )
)
2= 6 × 10−6 m 2 ,
1/ 2
1/ 2
L3c/ 2 h kA p


40 W m 2 ⋅ K
0.006 m )
0.077


(=
 240 W m ⋅ K × 6 × 10−6 m 2 


3/ 2
and from Fig. 3.19, the fin efficiency is
<
ηf ≈ 0.99 .
From Eq. 3.91 and Table 3.5, the fin heat rate is
1/ 2
2
 2
=
q f h=
f q max hf hA f=
(tri)q b 2hf hw  L + ( t 2 ) 
qb .
From Eq. 3.86, the fin effectiveness is
1/ 2
1/ 2
2
2
 2
2hf hw  L2 + ( t 2 ) 
qf

 q b 2hf  L + ( t 2 ) 


εf
=
=
=
hA c,bqq
h (w ⋅ t) b
t
b
2 1/ 2
2 × 0.99 ( 0.006 ) + ( 0.002 2 ) 

 m
=
6.02
0.002 m
2
εf
<
(b) The heat dissipation per unit width is
=
q′f
qf w )
(=
1/ 2
2
2hf h  L2 + ( t 2 ) 


qb
1/ 2
2
2
q′f =
2 × 0.99 × 40 W m 2 ⋅ K ( 0.006 ) + ( 0.002 2 ) 



m × ( 250 − 20 ) C =
110.8 W m .
COMMENTS: The parabolic profile is known to provide the maximum heat dissipation per unit fin
mass.
<
PROBLEM 3.127
PROBLEM 3.128
KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Heat flux and
ambient air conditions. Contact resistance.
FIND: Surface and interface temperatures (a) without and (b) with an interface contact
resistance.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3)
Uniform h over surfaces, (4) Negligible radiation.
ANALYSIS: The analysis may be performed per unit length of cylinder or for a 4 mm long
section. The following calculations are based on a unit length. The inner surface temperature
may be obtained from
Ti − T∞
q′
ri ) 105 W/m 2 × 2π × 0.06=
m 37, 700 W/m
=
= q′′i ( 2π=
R ′tot
where R ′tot =R ′c + R ′t,c + R ′w + R ′equiv ;
−1
R ′equiv =(1/ R ′f + 1/ R ′b ) .
R ′c , Conduction resistance of cylinder wall:
ln ( r / r )
ln ( 66/60 )
R ′c = 1 i =
=
3.034 ×10−4 m ⋅ K/W
2π k
2π ( 50 W/m ⋅ K )
R ′t,c , Contact resistance:
R ′t,c =
R ′′t,c / 2π r1 =
10−4 m 2 ⋅ K/W/2π × 0.066 m =
2.411×10−4 m ⋅ K/W
R ′w , Conduction resistance of aluminum base:
ln ( r / r )
ln ( 70/66 )
R ′w = b 1 =
=
3.902 ×10−5 m ⋅ K/W
2π k
2π × 240 W/m ⋅ K
R ′b , Resistance of prime or unfinned surface:
1
1
R ′b =
=
=
454.7 ×10−4 m ⋅ K/W
2
hA′b 100 W/m ⋅ K × 0.5 × 2π ( 0.07 m )
R ′f , Resistance of fins: The fin resistance may be determined from
=
R ′f
Tb − T∞
1
=
q′f
hf hA′f
The fin efficiency may be obtained from Fig. 3.20,
r2c =ro + t/2 =0.096 m
Lc =L+t/2 =0.026 m
Continued …
PROBLEM 3.128 (Cont.)
(
Ap =
Lc t =
5.2 ×10−5 m 2 r2c / r1 =
1.45 L3/2
c h/kA p
)
1/ 2
=
0.375
Fig. 3.20 → hf ≈ 0.88.
The total fin surface area per meter length
 m 2 6.78 m.
A′f 250 π ro2 - rb2=
=
× 2  250 m-1  2π 0.0962 - 0.07 2=




(
)
)
(
−1
Hence
R ′f = 0.88 ×100 W/m 2 ⋅ K × 6.78 m  = 16.8 ×10−4 m ⋅ K/W


=
1/ R ′equiv
⋅K
(1/16.8 ×10−4 + 1/ 454.7 ×10−4 ) W/m=
617.2 W/m ⋅ K
R ′equiv =
16.2 ×10−4 m ⋅ K/W.
Neglecting the contact resistance,
R ′tot =( 3.034 + 0.390 + 16.2 )10−4 m ⋅ K/W =19.6 ×10−4 m ⋅ K/W
-4
= 393.9 K
T=
i q′R ′tot + T∞= 37, 700 W/m × 19.6 × 10 m ⋅ K/W+320 K
T1 =
Ti - q′R ′w =
393.9 K - 37, 700 W/m × 3.034 ×10-4 m ⋅ K/W =
382.5 K
Tb =
T1 - q′R ′b =
382.5 K - 37, 700 W/m × 3.902 ×10-5 m ⋅ K/W =
381.0 K.
Including the contact resistance,
(
<
<
<
)
R ′tot = 19.6 ×10−4 + 2.411×10−4 m ⋅ K/W = 22.0 ×10−4 m ⋅ K/W
=
Ti 37, 700 W/m × 22.0 ×10-4 m ⋅ K/W+320
=
K 402.9 K
T1,i= 402.9 K - 37, 700 W/m × 3.034 ×10-4 m ⋅ K/W
= 391.5 K
<
<
T1,o= 391.5 K - 37, 700 W/m × 2.411×10-4 m ⋅ K/W= 382.4 K
<
Tb= 382.4 K - 37, 700 W/m × 3.902 ×10-5 m ⋅ K/W
= 380.9 K.
<
COMMENTS: (1) The effect of the contact resistance is small.
(2) The effect of including the aluminum fins may be determined by computing Ti without the
fins. In this case R ′tot
= R ′c + R ′conv , where
1
1
=
=
R ′conv =
241.1×10−4 m ⋅ K/W.
2
h2π r1 100 W/m ⋅ K 2π ( 0.066 m )
Hence, R tot =
244.1×10−4 m ⋅ K/W, and
-4
= 1240 K.
T=
i q′R ′tot + T∞= 37, 700 W/m × 244.1× 10 m ⋅ K/W+320 K
Hence, the fins have a significant effect on reducing the cylinder temperature.
(3) The overall surface efficiency is
ηη
1 − ( A′f / A′t )(1 − f ) =
1 − 6.78 m/7.00 m (1 − 0.88 ) =
0.884.
o=
It follows that q′=ho h o A′tq b = 37, 700 W/m, which agrees with the prescribed value.
PROBLEM 3.129
PROBLEM 3.129 (Cont.)
PROBLEM 3.130
KNOWN: Dimensions and thermal conductivities of a muscle layer and a skin/fat layer. Skin
emissivity and surface area. Metabolic heat generation rate and perfusion rate within the muscle layer.
Core body and arterial temperatures. Blood density and specific heat. Ambient conditions.
FIND: Perspiration rate to maintain same skin temperature as in Example 3.11.
SCHEMATIC:
Muscle
Tc = 37ºC
q m = 5600 W/m3
q p
km =
0.5 W/m·K
Skin/fat
Ti = 34.8ºC
e = 0.95
Tsur = 24ºC
Ts
ksf = 0.3 W/m·K
ω = 0.0005 s-1
T∞ = 297 K
h = 2 W/m2⋅K
Lm = 30 mm
Lsf = 3 mm
Air
x
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer through the muscle
and skin/fat layers, (3) Metabolic heat generation rate, perfusion rate, arterial temperature, blood
properties, and thermal conductivities are all uniform, (4) Radiation heat transfer coefficient is known
from Example 1.7, (5) Solar radiation is negligible, (6) Conditions are the same everywhere on the
torso, limbs, etc., (7) Perspiration on skin has a negligible effect on heat transfer from the skin to the
environment, that is, it adds a negligible thermal resistance and doesn’t change the emissivity.
ANALYSIS: First we need to find the skin temperature, Ts, for the conditions of Example 3.11, in the
air environment. Both q and Ti, the interface temperature between the muscle and the skin/fat layer, are
known. The rate of heat transfer across the skin/fat layer is given by
k A ( Ti − Ts )
(1)
q = sf
Lsf
Thus, the skin temperature is
qLsf
142 W × 0.003 m
= 34.8°C = 34.0°C
k sf A
0.3 W/m ⋅ K × 1.8 m 2
Now the heat transfer rate will change because of the increased metabolic heat generation rate. Heat
transfer in the muscle layer is governed by Equation 3.114. In Example 3.11, this equation was solved
subject to specified temperature boundary conditions, and the rate at which heat leaves the muscle and
enters the skin/fat layer was found to be
Ts = Ti -
 c
q x = L = - k m Amθ
m
 m -1
(θi /θc )coshmL
 m
sinhmL
(2)
Continued…
PROBLEM 3.130 (Cont.)
This must equal the rate at which heat is transferred across the skin/fat layer, given by Equation (1).
Equating Equations 1 and 2 and solving for Ti, recalling that Ti also appears in θi, yields


L 
q m 
 m + kmm
 sf θc +  Ta +
 m
Ts sinhmL
 coshmL
k sf 
ωρ b c b 


Ti =
Lsf
 m + kmm

 m
sinhmL
coshmL
k sf
where
1/2
 = ωρ b c b /k m = 0.0005 s-1 × 1000 kg/m3 × 3600 J/kg ⋅ K 0.5 W/m ⋅ K 
m


= 60 m-1
 m ) = cosh(60 m-1 × 0.03 m) = 3.11
 m ) = sinh(60 m-1 × 0.03 m) = 2.94 ; cosh(mL
sinh(mL
θc = Tc - Ta -
q m
q m
5600 W/m3
=== - 3.11 K
ωρ b c b
ωρ b c b
0.0005 s-1 × 1000 kg/m3 × 3600 J/kg ⋅ K
The excess temperature can be expressed in kelvins or degrees Celsius, since it is a temperature
difference. Thus
0.003 m
-3.11°C + ( 37°C + 3.11°C ) × 3.11
0.3 W/m ⋅ K 
0.003 m
2.94 + 0.5 W/m ⋅ K × 60 m-1 ×
× 3.11
0.3 W/m ⋅ K
34.0°C × 2.94 + 0.5 W/m ⋅ K × 60 m-1 ×
Ti =
Ti = 35.2°C
and again from Equation (1)
k A ( Ti - Ts ) 0.3 W/m ⋅ K × 1.8 m 2 (35.2°C - 34.0°C)
q = sf
=
= 222 W
Lsf
0.003 m
Since the skin temperature is unchanged from Example 3.11, the rate of heat transfer to the environment
by convection and radiation will remain the same, and is therefore still 142 W. The difference of 80 W
must be removed from the skin by perspiration, therefore
 per h fg = 80 W
q per = m
Assuming the properties of perspiration are the same as that of water, evaluated at the skin temperature
of 307 K, then from Table A.6 hfg = 2421 kJ/kg and r = 994 kg/m3. Thus the volume rate of
perspiration is
∀
m
q
80 W
∀ = per = per =
∀
= 3.3 × 10-8 m3 /s = 3.3 × 10-5 /s
3
3
ρ
ρh fg
994 kg/m × 2421 × 10 J/kg
<
COMMENTS: (1) This is a moderate rate of perspiration. In one hour, it would account for around 0.1
 . (2) In reality, our bodies adjust in many ways to maintain core and skin temperatures. Exercise will
likely cause an increase in perfusion rate near the skin surface, to locally elevate the temperature and
increase the rate of heat transfer to the environment.
PROBLEM 3.131
KNOWN: Dimensions and thermal conductivities of a muscle layer and a skin/fat layer. Skin
emissivity and surface area. Skin temperature. Perfusion rate within the muscle layer. Core body and
arterial temperatures. Blood density and specific heat. Ambient conditions.
FIND: Metabolic heat generation rate to maintain skin temperature at 33ºC.
SCHEMATIC:
e = 0.95
Muscle
Tc = 37ºC
Skin/fat
Ti
km =
0.5 W/m·K
Tsur = 15ºC
Ts = 33ºC
ksf = 0.3 W/m·K
ω = 0.0005 s-1
T∞ = 288 K
h = 2 W/m2⋅K
Lm = 30 mm
Lsf = 3 mm
Air
x
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer through the muscle
and skin/fat layers, (3) Metabolic heat generation rate, perfusion rate, arterial temperature, blood
properties, and thermal conductivities are all uniform, (4) Solar radiation is negligible, (5) Conditions
are the same everywhere on the torso, limbs, etc.
ANALYSIS: Since we know the skin temperature and environment temperature, we can find the heat
loss rate from the skin surface to the environment:
4
q = hA (Ts - T∞ ) + εσA(Ts4 - Tsur
)
(
)
= 2 W/m 2 ⋅ K × 1.8 m 2 (33 - 15)°C + 0.95 × 5.67 × 10-8 W/m 2 ⋅ K 4 × 1.8 m 2 3064 - 2884 K 4
= 248 W
We can then find Ti, the interface temperature between the skin/fat layer and the muscle layer, by
analyzing heat transfer through the skin/fat layer:
Ti = Ts +
qLsf
248 W × 0.003 m
= 33°C +
= 34.4°C
k sf A
0.3 W/m ⋅ K × 1.8 m 2
Heat transfer in the muscle layer is governed by Equation 3.114. In Example 3.11, this equation was
solved subject to specified surface temperature boundary conditions, and the rate at which heat leaves
the muscle and enters the skin/fat layer was found to be
Continued…
PROBLEM 3.131 (Cont.)

q = - k m Am
 m - θc
θi coshmL
 m
sinhmL
This must equal the rate at which heat is transferred across the skin/fat layer, as calculated above.
Inserting the definitions of θi and θc, we can solve for the metabolic heat generation rate:
q
 m + (Ti - Ta )coshmL
 m + (Tc - Ta )
sinhmL

k m Am
(1)
q m = ωρ b c b
 m +1
coshmL
where
1/2
 = ωρ b c b /k m = 0.0005 s-1 × 1000 kg/m3 × 3600 J/kg ⋅ K 0.5 W/m ⋅ K 
m


= 60 m-1
 m ) = sinh(60 m-1 × 0.03 m) = 2.94 ; cosh(mL
 m ) = cosh(60 m-1 × 0.03 m) = 3.11
sinh(mL
With Tc = Ta, Equation (1) yields
q m = 0.0005 s-1 × 1000 kg/m3 × 3600 J/kg ⋅ K
248 W


 0.5 W/m ⋅ K × 1.8 m 2 × 60 m-1 × 2.94 + (34.4 - 37)°C × 3.11 
× 

3.11 + 1




= 2341 W/m3
<
COMMENT: (1) Shivering can increase the metabolic heat generation rate by up to five to six times
the resting metabolic rate. The value found here is approximately three times the metabolic heat
generation rate given in Example 3.11, so it is well within what can be produced by shivering. (2) In the
water environment, even with the original 24ºC water temperature, shivering would be insufficient to
maintain a comfortable skin temperature.
PROBLEM 3.132
KNOWN: Dimensions and thermal conductivities of a muscle layer and a skin/fat layer. Metabolic
heat generation rate and perfusion rate within the muscle layer. Arterial temperature. Blood density
and specific heat. Ambient conditions.
FIND: Heat loss rate from body and temperature at inner surface of the skin/fat layer.
SCHEMATIC:
Muscle
Skin/fat
e = 0.95
Ti
Tsur = 24ºC
3
km = 0.5
W/m·K
= 700 W/m
ksf = 0.3 W/m·K
ω = 0.0005 s-1
rm = 5 cm
δsf = 3 mm
T∞ = 297 K
h = 2 W/m2⋅K
Air
r
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer through the muscle
and skin/fat layers, (3) Metabolic heat generation rate, perfusion rate, arterial temperature, blood
properties, and thermal conductivities are all uniform, (4) Radiation heat transfer coefficient is known
from Example 1.5.
ANALYSIS:
(a) Conduction with heat generation is expressed in radial coordinates by Equation 3.54. With
metabolic heat generation and perfusion, this becomes
1 d  dT  q m + ωρ b c b (Ta - T)
=0
r
+
r dr  dr 
k
<
The boundary conditions of symmetry at the centerline and specified temperature at the outer surface of
the muscle are expressed as
dT
= 0, T(r1 ) = Ti
dr r = 0
<
Defining an excess temperature, θ ≡ T - Ta - q m /ωρ b c b , the differential equation becomes
1 d  dθ 
 2θ = 0
r  - m
r dr  dr 
Continued…
PROBLEM 3.132 (Cont.)
 2 = ωρ b c b /k . The general solution to the differential equation is given in Section 3.6.4 as
where m
 + c2 K 0 (mr)

θ = c1I0 (mr)
Applying the boundary condition at r = 0 yields
dT
dθ
 1 (0) - c2 mK
 1 (0) = 0
=
= c1mI
dr r = 0 dr r = 0
Since K1(0) is infinite, we must have c2 = 0. Applying the specified temperature boundary condition at
r = r1 yields
q m
 1)
T(r1 ) = Ti , θ(r1 ) = Ti - Ta ≡ θi = c1I0 (mr
ωρ b c b
Solving for c1 we now have the complete solution for θ:

I (mr)
θ = θi 0

I0 (mr1 )
(1)
<
(b) The heat flux at the outer surface of the muscle is given by
q′′i = - k m
 )
I (mr
dT
dθ
 1 1
= - km
= - k m θi m
 1)
dr r = r
dr r = r
I0 (mr
1
(2)
1
This must be equated to the heat flux through the skin/fat layer and into the environment. In terms of
the heat transfer rate per unit length of forearm, q′i , and the total resistance for a unit length, R ′tot ,
q′′i =
q′i
1 Ti - T∞
=
2πr1
2πr1 R ′tot
(3)
As in Example 3.1 and for exposure of the skin to the air, R ′tot accounts for conduction through the
skin/fat layer in series with heat transfer by convection and radiation, which act in parallel with each
other. Here the conduction resistance is for a radial geometry. Thus, it is
-1
ln(r2 /r1 )
ln(r2 /r1 )
1  1
1 
1  1 
R ′tot =
+
+
+

 =


2πk sf
2πr2  1/h 1/h r 
2πk sf
2πr2  h + h r 
Using the values from Example 1.6 for air,
R ′tot =


ln(0.053 m/0.05 m)
1
1
+
= 0.41 m ⋅ K/W


2 × π × 0.3 W/m ⋅ K
2 × π × 0.053 m  (2 + 5.9) W/m 2 ⋅ K 
Combining Equations 2 and 3 yields
Continued…
PROBLEM 3.132 (Cont.)

- k m θi m
 1)
I1 (mr
1 Ti - T∞
=
 1)
I0 (mr
2πr1 R ′tot
This expression can be solved for Ti, recalling that Ti also appears in θi.

q m 
 1 ) + k m 2πr1mR
 ′tot  Ta +
 1)
T∞ I0 (mr
 I1 (mr
ωρ b c b 

Ti =
 1 ) + k m 2πr1mR
 ′tot I1 (mr
 1)
I0 (mr
where
1/2
 = ωρ b c b /k m = 0.0005 s-1 × 1000 kg/m3 × 3600 J/kg ⋅ K 0.5 W/m ⋅ K 
m


3
q m
700 W/m
=
= 0.389 K
-1
ωρ b c b
0.0005 s × 1000 kg/m3 × 3600 J/kg ⋅ K
= 60 m-1
and from Table B.5
 1 ) = I0 (60 m-1 × 0.05 m) = I0 (3) = 4.88 ;
I0 (mr
 1 ) = I1 (60 m-1 × 0.05 m) = I1 (3) = 3.95
I1 (mr
Thus,
  24°C × 4.88 + 0.5 W/m ⋅ K



-1
Ti =  × 2 × π × 0.05 m × 60 m

 × 0.41 m ⋅ K/W 37 + 0.389 °C × 3.95
(
)
 

 4.88 + 0.5 W/m ⋅ K

-1
× 2 × π × 0.05 m × 60 m
× 0.41 m ⋅ K/W × 3.95



  = 34.2°C

 
<
(c) The maximum temperature occurs at the centerline of the forearm, r = 0, thus from Equation 1, with
I0(0) = 1,
T = Ta +

q m
q m
+  Ti - Ta ωρ b c b
ωρ b c b

 1

 1)
 I0 (mr
<
1
= 37°C + 0.389°C + ( 34.2°C - 37°C - 0.389°C ) ×
= 36.7°C
4.88
COMMENTS: (1) The maximum temperature is very close to the core body temperature of 37ºC, as
would be expected. (2) Pennes [17] conducted an experimental investigation of the temperature
distribution in human forearms, by inserting thermocouples into living subjects.
PROBLEM 3.133
KNOWN: Thermoelectric module properties and performance, as given in Example 3.12.
FIND: (a) The thermodynamic efficiency, htherm ≡ PM=1/q1, (b) the figure of merit ZT for one module,
and the thermoelectric efficiency, hTE. (c) the Carnot efficiency, hCarnot = 1 - T2/T1, (d) the value of hTE
based upon the inappropriate use of T∞,1 and T∞,2 (e) the thermoelectric efficiency based upon the
correct usage of T1 and T2 in Equation 3.128, and the Carnot efficiency for the case where h1 = h2 → ∞.
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Negligible contact resistances, (3)
Negligible radiation exchange and gas phase conduction inside the module, (4) Negligible conduction
resistance due to metallic contacts and ceramic insulators, (5) The properties of the two semiconductors
are identical and Sp = - Sn.
ANALYSIS: (a) From Example 3.12 the electrical power per module is PM = 1 = Ptot/M = 46.9 W/48 =
0.9773 W. The heat input to one module may be evaluated from Equation 3 of the solution to the
example problem as
=
q1 h1W 2 (T∞ ,1=
− T1 ) 40 W/m 2 ⋅ K × (0.054m) 2 × ( 550 + 273)K − (173 + 273)K
=
) 43.92 W
<
Therefore, the thermodynamic efficiency is htherm = PM = 1/q1 = 0.9773 W/43.92 W = 0.022
(b) From Equations 3.121 and 3.125 (or 3.122 and 3.126), we note that
=
S p −n S=
0.1435 volts/K
=
/100 0.001435 volts/K
p − n ,eff / N
and
Re ,eff As
8 Ω ×1.2 ×10−5 m 2
=
= 9.6 ×10−5 Ω ⋅ m
ρe=
,s
−3
4 NL
4 × 100 × 2.5 ×10 m
From Section 3.7.2, S = Sp = - Sn and for Sp = - Sn, S = Sp = Sp-n/2 = 0.0007175 volts/K,
=
Z
S2
(0.0007175 volts/K) 2
=
= 0.004469 K -1
ρ e,s k 9.6 × 10-5 W ⋅ m × 1.2 W/m ⋅ K
For T1 = 173°C + 273 K = 446 K and T2 = 134°C + 273 K = 407 K, as determined in the example
(T1 + T2 ) / 2 = (446 K + 407 K) / 2 = 426.5 K .
problem, the average module temperature is T =
Therefore, the figure of merit is
=
ZT 0.004469 K −1 × 426.5=
K 1.908.
<
The thermoelectric efficiency is therefore,
Continued...
PROBLEM 3.133 (Cont.)

T 
1 + ZT − 1
1 + 1.908 − 1
 407 K 
<
=
=
0.024
1 −

T1  1 + ZT + T / T  446 K  1 + 1.908 + 407 K / 446 K
2
1
2
ηTE =
1 − 

We note that the thermodynamic efficiency is less than the thermoelectric efficiency based on the figure of
merit and the surface temperatures of the module. The thermoelectric efficiency is the maximum possible
efficiency for the case when the load resistance is optimized.
<
(c) The Carnot efficiency is hCarnot = 1 - 407K /446 K = 0.087
(d) The value of hTE based upon T∞,1 = 550°C + 273 K = 823 K, T∞,2 = 105°C + 273 K = 378 K, and
=
ZT 0.004469 K −1 × 600.5=
K 2.684 is

T 
1 + ZT − 1
1 + 2.684 − 1
 378 K 
<
0.21
=
=
1 −

T∞ ,1  1 + ZT + T∞ ,2 / T∞ ,1  823 K  1 + 2.684 + 378 K / 823 K
∞ ,2
ηTE =
 1 −


<
The Carnot efficiency is hCarnot = 1 - 378K /823 K = 0.54
(e) For h1 = h2 → ∞, the surface temperatures of the module are T1 = T∞,1 = 823 K and T2 = T∞,2 = 378
K, respectively. Therefore, from part (d) hTE = 0.21 and hCarnot = 0.54.
<
COMMENTS: (1) The conversion efficiency for the thermoelectric modules of Example 3.12 is quite
small, approximately 2%. (2) The conversion efficiency can be increased by an order of magnitude (to
21%) by utilizing thermal management approaches that will increase the temperature difference across
the module. (c) The incorrect usage of T∞,1 and T∞,2 in the expression for the thermoelectric efficiency
as in part (d) provides an efficiency (21%) that far exceeds the Carnot efficiency of 8.7% found in part
(c). Hence, use of the incorrect temperatures in the thermoelectric efficiency expression can lead to
grossly exaggerated levels of thermodynamic performance that violate the second law of
thermodynamics. Reporting the efficiency of a thermoelectric module or thermoelectric material based
upon fluid (or surrounding) temperatures is meaningless.
PROBLEM 3.134
KNOWN: Dimensions of thermoelectric module and heat sinks. Convection conditions, heat sink
thermal conductivity, thermoelectric module performance parameters, load electrical resistance. Contact
resistance between thermoelectric module and heat sinks.
FIND: (a) Sketch of the equivalent thermal circuit and electrical power generated without the heat
sinks. (b) Sketch of the equivalent thermal circuit and electrical power generated with the heat sinks.
SCHEMATIC:
Cover plate
Air
T∞,1, = 450°C
h1 = 80W/m2K
Heat
sink 1
Re,load = 4 W
k = 180 W/m2K
Re,load = 4 W
Air
T∞,1, = 450°C
h1 = 80W/m2K
Thermoelectric module
Lb = 4 mm
Thermoelectric module
2L=5mm
W = 54 mm
2L=5mm
Lf = 20 mm
Air
T∞,2 = 20°C,
h2 = 80 W/m2K
(a)
Heat sink 2,
k = 180 W/m∙K
t = 2.182 mm
Cover plate
(b)
Air
T∞,2 = 20°C,
h2 = 80 W/m2K
S = 5.182 mm
Solder,
Rt",c =
2.5 × 10−6 m 2 ⋅ K/W
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3)
Negligible radiation, (4) Adiabatic fin tips for part (b), (5) Convection coefficients same in parts (a) and
(b) and the same on the sides of the fin arrays.
ANALYSIS: (a) Without the heat sinks, the equivalent thermal circuit is shown in Figure 3.24b as
replicated below.
<
Continued…
PROBLEM 3.134 (Cont.)
The analysis can proceed as in Example 3.12. The conduction resistance of one module is the same as
in the example, namely
Rt=
,cond,mod
L
2.5 × 10-3 m
=
=
NAc , s k s 100 × 1.2 × 10-5 m 2 × 1.2 W/m ⋅ K
1.736 K/W
From Equations 3.125 and 3.126,
=
q1
=
q2
1
Rt ,cond,mod
1
Rt ,cond,mod
I 2 Re,eff
(T1 - T2 )
(T1 - T2 ) + IS p -n ,eff T1 -=
+ I × 0.1435 V/K × T1 - I 2 × 4 W
2
1.736 K/W
(1)
I 2 Re,eff
(T1 - T2 )
(T1 - T2 ) + IS p -n ,eff T2 +=
+ I × 0.1435 V/K × T2 + I 2 × 4 W
2
1.736 K/W
(2)
Newton’s law of cooling may be written at each surface as
2
=
- T1 ) 80 W/m 2 ⋅ K × (0.054 m) 2 × ( 450 + 273) K - T1 
q1 hW
(T∞ ,1 =
1
(3)
q2 h2W 2 (T2 − T=
80 W/m 2 ⋅ K × (0.054 m) 2 × T2 − ( 20 + 273) K 
=
∞ ,2 )
(4)
The electric power produced by a single module, PN, is equal to the electric power dissipated in the load
resistance. Equating the expression for PN from Equation 3.127 to the electric power dissipated in the load
gives
=
PN IS p -n ,eff (T1 -T2 ) I 2=
Re,eff I 2 Re,load
I × 0.1435 V/K × (T1 − T2 ) − I 2 × 8 Ω= I 2 × 4 Ω
(5)
Equations 1 through 5 may be solved simultaneously, for example using IHT, to yield I = 0.38 A, and
PN = I2Re,load = (0.38 A)2 × 4 W = 0.59 W
<
(b) The thermal circuit associated with the thermoelectric module is unchanged, but each convection
resistance must be replaced with the total thermal resistance, Rtot, associated with the contact resistance,
fin array base, and overall resistance of the fin array, as shown below. Also, qconv,1 and qconv,2 have been
replaced with the more general terms q1 and q2.
Continued…
PROBLEM 3.134 (Cont.)
The total thermal resistance is given by
Rtot =
Rt′′,c
W
2
+ Rbase + Rt ,o =
Rt′′,c
W
2
+
Lb
1
+
2
ho hAt
kW
where Equation 3.108 has been used to express the overall fin resistance. From Equations 3.104, 3.107,
and 3.95,
tanh mL f
ho At = Ab + NAf h f = ( N − 1)W ( S − t ) + 2 NWL f
mL f
where
m=
hP / kAc =
2h / kt =
2 × 80 W/m 2 ⋅ K / (180 W/m ⋅ K × 0.002182 m) = 20.2 m −1
Thus,
ho At =
10 × 0.054 m × 0.003 m + 2 × 11 × 0.054 m × 0.020 m
tanh(20.2 m −1 × 0.02 m)
=
0.0242 m 2
20.2 m −1 × 0.02 m
The total thermal resistance is then
2.5 × 10−6 K/W
0.004 m
1
+
+
=
Rtot =
0.526 K/W
2
2
2
(0.054 m)
180 W/m ⋅ K(0.054 m)
0.0242 m × 80 W/m 2 ⋅ K
Continued…
PROBLEM 3.134 (Cont.)
The system of equations from part (a) applies here, except that Equations 3 and 4 are replaced by the
revised versions
=
q1
T∞ ,1 − T1
=
Rtot
( 450 + 273) K − T1
q2
=
T2 − T∞ ,2
=
Rtot
T2 − ( 20 + 273) K
0.526 K/W
0.526 K/W
(3r)
(4r)
Equations 1, 2, 3r, 4r, and 5 may be solved simultaneously to yield I = 2.04 A, and
PN = I2Re,load = (2.04 A)2 × 4 W = 16.7 W
This is 28 times larger than the result of part (a).
<
<
COMMENTS: By adding the two heat sinks to the thermoelectric module, the power produced by the
module increases by a factor of 28. Not only does the power depend on the semiconductor properties of
the thermoelectric material, but also strongly on the thermal management of the module through, as in
this problem, addition of heat sinks.
PROBLEM 3.135
KNOWN: Dimensions of thermoelectric module. Convection conditions, thermoelectric module
performance parameters, load electrical resistance, contact resistance between thermoelectric module
and stove surface, emissivity of the exposed surface of the thermoelectric module, temperature of
surroundings.
FIND: Sketch of the equivalent thermal circuit and electrical power generated by the module.
SCHEMATIC:
2L = 5 mm
Wood
stove
surface
Ts = 375°C
Tsur = 25°C
e = 0.9
T∞ = 25°C
h = 15 W/m2∙K
W = 54 mm
Re,load = 3W
Rt",c =
5 × 10−6 m 2 ⋅ K/W
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) Large
surroundings.
ANALYSIS: The portion of the equivalent thermal circuit that describes the thermoelectric module is
the same as shown in Figure 3.24b. However, the external thermal resistances are different. The high
temperature side of the TEM is exposed to the stove surface through a contact resistance, Rt,c. The low
temperature side exchanges heat with the surroundings through radiation and convection. Since T∞ =
Tsur, the radiation and convection thermal resistances can be combined into a single resistance,
Rrad,conv,as shown below. Also, qconv,1 and qconv,2 have been replaced with the more general terms q1 and
q2.
<
Continued…
PROBLEM 3.135 (Cont.)
The two external resistances can be calculated as follows:
Rt ,c =
Rt′′,c / W 2 =
5 × 10−6 m 2 ⋅ K/W / (0.054 m) 2 =
1.71 × 10−3 K/W
=
Rrad,conv 1 / [(hconv + hr )W 2 ]
hr =εs (T2 + Tsur )(T22 + Tsur2 )
(1)
The radiation heat transfer coefficient, hr, depends on the unknown TEM surface temperature, T2. This
can be left as an unknown in solving the simultaneous equations.
The analysis proceeds as in Example 3.12. The conduction resistance of one module is the same as in
the example, namely
Rt=
,cond,mod
L
2.5 × 10-3 m
=
=
NAc , s k s 100 × 1.2 × 10-5 m 2 × 1.2 W/m ⋅ K
1.736 K/W
From Equations 3.125 and 3.126,
=
q1
q2
=
1
Rt ,cond,mod
1
Rt ,cond,mod
2
Re ,eff / 2
(T1 - T2 ) + IS p -n ,eff T1 - I =
2
(T1 - T2 ) + IS p -n ,eff T2 + I =
Re,eff / 2
(T1 - T2 )
+ I × 0.1435 V/K × T1 - I 2 × 4 W
1.736 K/W
(2)
(T1 - T2 )
+ I × 0.1435 V/K × T2 + I 2 × 4 W
1.736 K/W
(3)
Additional relationships can be written by considering heat transfer through the external resistances.
( 375 + 273) K - T1  / 1.71 × 10-3 K/W
q1 =
(Ts - T1 ) / Rt ,c =
q2 =
(T2 − T∞ ) / Rrad,conv =
(T2 − T∞ )(h + hr )W 2
= T2 − ( 25 + 273) K  × (15 W/m 2 ⋅ K + hr ) × (0.054 m) 2
(4)
(5)
The electric power produced by the single module, PN, is equal to the electric power dissipated in the load
resistance. Equating the expression for PN from Equation 3.127 to the electric power dissipated in the load
gives
=
PN IS p -n ,eff (T1 -T2 ) I 2=
Re,eff I 2 Re,load
I × 0.1435 V/K × (T1 − T2 ) − I 2 × 8 Ω = I 2 × 3 Ω
(6)
Continued…
PROBLEM 3.135 (Cont.)
Equations 1 through 6 may be solved simultaneously, for example using IHT, to yield I = 0.27 A, and
PN = I2Re,load = (0.27 A)2 × 3 W = 0.22 W
<
COMMENTS: (1) Radiation is significant. If radiation heat transfer were neglected, the electrical
power output would be decreased to 0.036 W. (2) The electrical power output is quite low. The power
output could be raised by increasing the temperature difference across the module. For example, the
electrical power could be used to rotate a small fan to increase the value of the heat transfer coefficient.
If h were to increase to 30 W/m2∙K, for example, the electrical power would increase to 0.40 W. A
tradeoff exists between the extra power provided by the fan and the power needed to operate the fan.
PROBLEM 3.136
KNOWN: Thermal energy generation rate. Dimensions of thermoelectric modules and total number of
modules. Thermoelectric module performance parameters, load electrical resistance, emissivity of the
exposed surface of the thermoelectric modules, deep space temperature.
FIND: Electrical power generated by the device. Surface temperatures of the modules.
SCHEMATIC:
Tsur = 4 K
W = 54 mm
2L = 5 mm
Heat source,
E g = 1, 10, 100 kW
Insulation
Thermoelectric
module, e = 0.93
Re, load = 250 W
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) Large
surroundings.
ANALYSIS: The portion of the equivalent thermal circuit that describes the thermoelectric module is
the same as shown in Figure 3.24b. The energy generated in the uranium is known, and under steadystate conditions q1 = E g / M . As a consequence, knowledge of the thermal resistance between the
uranium and the inner surface of the TEMs is not needed. The low temperature side of the TEMs
exchanges heat with the surroundings through radiation. Thus, the equivalent thermal circuit is as
shown below. Note that qconv,1 and qconv,2 have been replaced with the more general terms q1 and q2.
<
Continued…
PROBLEM 3.136 (Cont.)
The analysis proceeds as in Example 3.12. The conduction resistance of one module is the same as in
the example, namely
Rt=
,cond,mod
L
2.5 × 10-3 m
=
=
NAc , s k s 100 × 1.2 × 10-5 m 2 × 1.2 W/m ⋅ K
1.736 K/W
From Equations 3.125 and 3.126,
=
q1
=
q2
1
Rt ,cond,mod
1
Rt ,cond,mod
I 2 Re,eff
(T1 - T2 )
(T1 - T2 ) + IS p -n ,eff T1 -=
+ I × 0.1435 V/K × T1 - I 2 × 4 W
2
1.736 K/W
(1)
I 2 Re,eff
(T1 - T2 )
(T1 - T2 ) + IS p -n ,eff T2 +=
+ I × 0.1435 V/K × T2 + I 2 × 4 W
2
1.736 K/W
(2)
An additional relationship can be written by considering heat transfer by radiation to deep space.
q2 =hrW 2 (T2 − Tsur ) =hr × (0.054 m) 2 × (T2 − 4 K)
(3)
where
hr = εs (T2 + Tsur )(T22 + Tsur2 ) = 0.93 × 5.67 × 10−8 W/m 2 ⋅ K 4 × (T2 + 4 K) × (T22 + (4 K) 2 )
(4)
The radiation heat transfer coefficient, hr, depends on the unknown TEM surface temperature, T2. This
can be left as an unknown in solving the simultaneous equations.
The electric power produced by all 80 modules, Ptot, is equal to the electric power dissipated in the load
resistance. Making use of Equation 3.127 and equating the total electrical power generated in the M
modules to the electric power dissipated in the load gives
=
Ptot
=
MP
I 2 Re ,load
N
M  IS p -n ,eff (T1 -T2 ) I 2 Re ,eff  =
I 2 Re ,load
80  I × 0.1435 V/K × (T1 − T2 ) − I 2 × 8 Ω  = I 2 × 250 Ω
(5)
With q1 known from q1 = E g / M , Equations 1 through 5 can be solved for the five unknowns, T1, T2, I,
q2, and hr. Solving the equations numerically using IHT yields the following results for the three
different values of E g :
Continued…
PROBLEM 3.136 (Cont.)
E g (kW)
1
10
100
I (A)
0.10
0.67
3.99
Ptot (W)
2.63
114
3990
T2 (K)
534
947
1671
h = PN/q1
0.0026
0.011
0.040
<
COMMENTS: (1) The temperature for the highest thermal energy generation rate is unacceptably
high. (2) The electrical energy generated by the device is relatively high, but the efficiency is quite low.
The efficiency increases as a function of the thermal generation rate because of larger temperature
differences across the module, which are ∆T = 8, 52, and 310 K for the low, medium and high energy
generation rates. (3) Numerical solution of the equations requires a good initial guess. One can be
obtained by assuming that the current is zero, resulting in q1 = q2 and enabling direct calculation of the
temperatures due to conduction across the TEM and radiation to the surroundings. (4) In this
application, thermal generation can occur continuously for many years, providing reliable electrical
power to the satellite over its lifetime. (5) What steps could be taken to increase the electrical power
generated for each thermal energy generation rate?
PROBLEM 3.137
PROBLEM 3.137 (Cont.)
PROBLEM 3.138
KNOWN: Size and temperatures of parallel aluminum plates. Spacing between the plates. Air between
the plates.
FIND: The conduction heat transfer rate through the air.
SCHEMATIC:
ASSUMPTIONS: (1) Ideal gas behavior.
PROPERTIES: Table A.4 (T = 300 K): Air; cp = 1007 J/kg⋅K, k = 0.0263 W/m∙K. Figure 2.8: Air; M =
28.97 kg/kmol, d = 0.372 × 10-9 m.
ANALYSIS: For air, the ideal gas constant, specific heat at constant volume, and ratio of specific heats
are:
=
R
R 8.315 kJ/kmol ⋅ K
kJ
=
= 0.287
;
M
28.97 kg/kmol
kg ⋅ K
cv = c p − R = 1.007
c
kJ
kJ
kJ
1.007
− 0.287
= 0.720
= 1.399
; g = p =
cv 0.720
kg ⋅ K
kg ⋅ K
kg ⋅ K
From Equation 2.11 the mean free path of air is
k BT
1.381 × 10-23 J/K × 300 K
66.5 × 10-9 m =
66.5 nm
=
=
2
2
-9
5
2
2p d p
2p ( 0.372 × 10 m ) (1.0133 × 10 N/m )
λmfp =
For L = 1 mm,
Rt ,m=
−m
=
Rt ,m − s
L
1 × 10−3 m
=
= 380.2 K/W
kA 0.0263 W/m ⋅ K × 10 × 10−3 m × 10 × 10−3 m
λmfp  2 − α t   9γ − 5 
66.5 × 10−9 m
 2 − 0.92   9 × 1.399 − 5 
=


−6
2 


kA  α t   γ + 1  0.0263 W/m ⋅ K × 100 × 10 m  0.92   1.399 + 1 
= 0.09392 K/W
Hence, the conduction heat rate is
Continued...
PROBLEM 3.138 (Cont.)
=
q
Ts ,1 − Ts ,2
=
( Rt ,m−m + Rt ,m−s )
305K − 295K
= 0.0263 W
( 380.2 K/W + 0.09392 K/W )
<
For L = 1 µm,
Rt ,m=
−m
L
1 × 10−6 m
=
= 0.3802 K/W
kA 0.0263 W/m ⋅ K × 10 × 10−3 m × 10 × 10−3 m
Rt ,m − s = 0.09392 K/W
Hence, the conduction heat rate is
=
q
Ts ,1 − Ts ,2
=
( Rt ,m−m + Rt ,m−s )
305K − 295K
= 21.09 W
( 0.3802 K/W + 0.09392 K/W )
<
For L = 10 nm,
Rt ,m=
−m
L
10 × 10−9 m
=
= 0.0038 K/W
kA 0.0263 W/m ⋅ K × 10 × 10−3 m × 10 × 10−3 m
Rt ,m − s = 0.09392 K/W
Hence, the conduction heat rate is
=
q
Ts ,1 − Ts ,2
=
( Rt ,m−m + Rt ,m−s )
305K − 295K
= 102.3 W
( 0.0038 K/W + 0.09392 K/W )
<
COMMENT: If the molecule-surface collision resistance were to be neglected, the heat rates would
be q = 0.0263 W, 26.3 W, and 2632 W for the L = 1 mm, 1 mm and 10 nm plate spacings, respectively.
Hence, molecule-surface collisions are negligible for large plate spacings, and dominant at small plate
spacings.
PROBLEM 3.139
KNOWN: Air or helium between steel and aluminum parallel plates, respectively. Gas temperature
and pressure. Thermal accommodation coefficient values.
FIND: The separation distance, L, above which Rt,m-s/Rt,m-m is less than 0.01 for (a) air and (b) helium.
SCHEMATIC:
ASSUMPTIONS: (1) Ideal gas behavior, (2) One-dimensional conduction.
PROPERTIES: Table A.4 (T = 300 K): Air; cp = 1007 J/kg⋅K, k = 0.0263 W/m∙K. He; cp = 5193
J/kg⋅K, k = 0.170 W/m∙K Figure 2.8: Air; M = 28.97 kg/kmol, d = 0.372 × 10-9 m. He; M = 4.003
kg/kmol, d = 0.219 × 10-9 m.
ANALYSIS:
(i) For air, the ideal gas constant, specific heat at constant volume, and ratio of specific heats are:
=
R
R 8.315 kJ/kmol ⋅ K
kJ
=
= 0.287
;
M
28.97 kg/kmol
kg ⋅ K
cv = c p − R = 1.007
c
kJ
kJ
kJ
1.007
− 0.287
= 0.720
= 1.399
; g = p =
cv 0.720
kg ⋅ K
kg ⋅ K
kg ⋅ K
From Equation 2.11 the mean free path of air is
k BT
1.381 × 10-23 J/K × 300 K
66.5 × 10-9 m =
66.5 nm
=
=
2
-9
5
2
2p d 2 p
2p ( 0.372 × 10 m ) (1.0133 × 10 N/m )
λmfp =
From Section 3.7.3 the plate separation, L, is
L = λmfp
Rt ,m −m  2 − α t   9γ − 5 


Rt ,m − s  α t   γ + 1 
= 66.5 × 10
−9
 2 − 0.92   9 × 1.399 − 5 
m × 100 × 
= 2.47 × 10−5 m = 24.7 m m



0.92
1.399
1
+



<
Continued…
PROBLEM 3.139 (Cont.)
(ii) For He, the ideal gas constant, specific heat at constant volume, and ratio of specific heats are:
=
R
R 8.315 kJ/kmol ⋅ K
kJ
=
= 2.077
;
M
4.003 kg/kmol
kg ⋅ K
cv = c p − R = 5.193
c
kJ
kJ
kJ
5.193
− 2.077
= 3.116
; g = p =
= 1.667
kg ⋅ K
kg ⋅ K
kg ⋅ K
cv 3.116
The mean free path is
λmfp =
k BT
1.381 × 10-23 J/K × 300 K
=
=1.919 × 10-7 m =192 nm
2
2
-9
5
2
2p d p
2p ( 0.219 × 10 m ) (1.0133 × 10 N/m )
The plate separation, L, is
L = λmfp
Rt ,m −m  2 − α t   9γ − 5 


Rt ,m − s  α t   γ + 1 
= 1.919 × 10−7
 2 − 0.02   9 × 1.667 − 5 
= 0.0071 m= 7.1 mm
m × 100 × 
 0.02   1.667 + 1 
<
COMMENTS: The critical plate separation associated with helium is 7.1 × 10-3 m/ 24.7 × 10-6 m =
290 times greater than that for air. The thermal resistance associated with molecule-surface interactions
can become significant for gases of small molecular diameter and for gas-surface material combinations
that have a small thermal accommodation coefficient, even at relatively large plate separation distances
(7.1 mm).
PROBLEM 3.140
KNOWN: Thickness of of parallel aluminum plates and air layers. Wall surface temperatures.
FIND: The conduction heat flux through (a) aluminum wall, (b) air layer, (c) air layer contained
between two aluminum sheets and (d) composite wall consisting of 8 aluminum sheets and 7 air layers.
SCHEMATIC:
Ts,1= 301 K
Ts,1= 301 K
Aluminum
Air
Ts,2= 299 K
Ts,2= 299 K
x
Case A
Ts,1= 301 K
Ltot= 600 µm
or nm
x
Case B
Ltot= 600 µm
or nm
Ts,1= 301 K
δ = 40 µm
or nm
Air
Aluminum
δ = 40 µm
or nm
Ts,2= 299 K
Ts,2= 299 K
Air
x
Case C
Ltot= 600 µm
or nm
x
Aluminum
Case D
Ltot= 600 µm
or nm
ASSUMPTIONS: (1) Ideal gas behavior. (2) Nanoscale effects within the solid are not important.
PROPERTIES: Table A.4 (T = 300 K): Air; cp = 1007 J/kg⋅K, kAir = 0.0263 W/m∙K. Figure 2.8: Air; M
= 28.97 kg/kmol, d = 0.372 × 10-9 m. Table A.1 (T = 300 K): Pure Aluminum, kAl = 237 W/m∙K.
ANALYSIS:
(a) Case A: Aluminum Wall
For Ltot = 600 µm, the heat flux is
"
q=
x
k Al ( Ts ,1 − Ts ,2 ) 237 W/m ⋅ K ( 301 K − 299 K )
=
= 7.9 × 105 W/m 2
−6
600 × 10 m
Ltot
Similarly, for Ltot = 600 nm, the heat flux is q"x = 7.9 × 108 W/m2
<
<
Continued...
PROBLEM 3.140 (Cont.)
(b) Case B: Air Layer
For Ltot = 600 µm, the heat flux is
=
q"x
k Air ( Ts ,1 − Ts ,2 ) 0.0263 W/m ⋅ K ( 301 K − 299 K )
=
= 87.7 W/m 2
Ltot
600 × 10−6 m
<
<
Similarly, for Ltot = 600 nm, the heat flux is q"x = 87.7 × 103 W/m2
(c) Case C: Air Layer between two Aluminum Sheets
This case involves a resistance due to molecule-molecule interactions, as well as molecule-surface
collisions. For air, the ideal gas constant, specific heat at constant volume, and ratio of specific heats
are:
=
R
R 8.315 kJ/kmol ⋅ K
kJ
=
= 0.287
;
M
28.97 kg/kmol
kg ⋅ K
cv = c p − R = 1.007
c
kJ
kJ
kJ
1.007
− 0.287
= 0.720
= 1.399
; g = p =
cv 0.720
kg ⋅ K
kg ⋅ K
kg ⋅ K
From Equation 2.11 the mean free path of air is
k BT
1.381 × 10-23 J/K × 300 K
66.5 × 10-9 m =
66.5 nm
=
=
2
2
-9
5
2
2p d p
2p ( 0.372 × 10 m ) (1.0133 × 10 N/m )
λmfp =
For Ltot = 600 µm, the air layer is L = Ltot - 2δ = 600 µm - 2 × 40 µm = 520 µm thick.
Rt",m=
−m
L
520 × 10−6 m
=
= 0.0198 K ⋅ m 2 /W
k Air 0.0263 W/m ⋅ K
λmfp  2 -α t   9γ 5 
66.5 × 10-9 m  2 - 0.92   9 × 1.399 - 5 
=
=
R
k Air  α t   γ + 1  0.0263 W/m ⋅ K  0.92   1.399 + 1 
"
t ,m - s
= 9.39 × 10-6 K ⋅ m 2 /W
In addition, the aluminum sheets pose a cumulative thermal resistance of
"
t ,cond
R
2d 2 × 40 × 10−6 m
= =
=
3.38 × 10−7 K ⋅ m 2 /W
237 W/m ⋅ K
k Al
Hence, the conduction heat flux is
Continued...
PROBLEM 3.140 (Cont.)
q"x =
(R
"
t ,m − m
Ts ,1 − Ts ,2
+ Rt",m − s + Rt",cond )
301 K − 299 K
( 0.0198 K ⋅ m /W + 9.39 × 10−6 K ⋅ m2 /W + 3.38 × 10−7 K ⋅ m2 /W )
=
2
<
= 101 W/m 2
For Ltot = 600 nm, the air layer is L = Ltot - 2δ = 600 nm - 2 × 40 nm = 520 nm thick.
L
520 × 10−9 m
Rt",m −m = =
=
19.8 × 10−6 K ⋅ m 2 /W
k Air 0.0263 W/m ⋅ K
The aluminum sheets pose a cumulative thermal resistance of
2d 2 × 40 × 10−9 m
3.38 × 10−10 K ⋅ m 2 /W
Rt",cond = =
=
237 W/m ⋅ K
k Al
Hence, the conduction heat flux is
q"x =
=
(R
"
t ,m − m
Ts ,1 − Ts ,2
+ Rt",m − s + Rt",cond )
(19.8 × 10
−6
301 K − 299 K
K ⋅ m /W + 9.39 × 10−6 K ⋅ m 2 /W + 3.38 × 10−10 K ⋅ m 2 /W )
2
<
= 68.6 × 103 W/m 2
(d) Case D: Seven Air Layers between Eight Aluminum Sheets
This case involves multiple resistances due to molecule-molecule interactions, as well as moleculesurface collisions at multiple surfaces.
For Ltot = 600 µm, each air layer is L = Ltot × (1/15)= 600 µm × (1/15) = 40 µm thick. Hence, for each
air layer
L
40 × 10−6 m
Rt",m −m = =
=
1.52 × 10−3 K ⋅ m 2 /W
k Air 0.0263 W/m ⋅ K
In addition, the aluminum sheets pose a cumulative thermal resistance of
8d 8 × 40 × 10−6 m
1.35 × 10−6 K ⋅ m 2 /W
Rt",cond = =
=
237 W/m ⋅ K
k Al
Continued...
PROBLEM 3.140 (Cont.)
Hence, the conduction heat flux is
q"x =
Ts ,1 − Ts ,2
(7R
"
t ,m − m
+ 7 Rt",m − s + Rt",cond )
301 K − 299 K
( 7 × 1.52 × 10 K ⋅ m /W + 7 × 9.39 × 10−6 K ⋅ m2 /W + 1.35 × 10−6 K ⋅ m2 /W )
=
−3
2
<
= 186.7 W/m 2
For Ltot = 600 nm, each air layer is L = Ltot × (1/15)= 600 nm × (1/15) = 40 nm thick. Hence, for each air
layer
L
40 × 10−9 m
Rt",m −m = =
=
1.52 × 10−6 K ⋅ m 2 /W
k Air 0.0263 W/m ⋅ K
In addition, the aluminum sheets pose a cumulative thermal resistance of
"
t ,cond
R
8d 8 × 40 × 10−9 m
1.35 × 10−9 K ⋅ m 2 /W
= =
=
237 W/m ⋅ K
k Al
Hence, the conduction heat flux is
q"x =
=
(7R
"
t ,m − m
Ts ,1 − Ts ,2
+ 7 Rt",m − s + Rt",cond )
( 7 × 1.52 × 10
−6
301 K − 299 K
K ⋅ m /W + 7 × 9.39 × 10−6 K ⋅ m 2 /W + 1.35 × 10−9 K ⋅ m 2 /W )
2
<
= 26.2 × 103 W/m 2
The predicted heat fluxes are summarized below.
Case
Ltot = 600 µm
Ltot = 600 nm
A
"
q=
7.9 ×105 W/m 2
x
"
q=
7.9 ×108 W/m 2
x
B
q"x = 87.7 W/m 2
q"x 87.7 ×103 W/m 2
=
C
q"x = 101 W/m 2
=
q"x 68.6 ×103 W/m 2
D
q"x = 186.7 W/m 2
=
q"x 26.2 ×103 W/m 2
Continued...
PROBLEM 3.140 (Cont.)
COMMENTS: (1) For the Ltot = 600 µm cases, it is readily evident that the highest heat flux
corresponds to Case A in which conduction occurs exclusively through the high thermal conductivity
aluminum. The lowest heat flux is associated with conduction through the pure air layer (Case B). For
the case involving two aluminum sheets (Case C) the heat flux is increased relative to Case B primarily
in response to replacing some low thermal conductivity air with high thermal conductivity metal.
However, as more aluminum sheets are added, the thermal resistance across the entire layer is reduced,
leading to increases in the heat flux for Case D. In each case, the resistance posed by molecule-surface
interactions is not significant. Specifically, heat transfer rates for Cases C and D, calculated without
accounting for the molecule-surface collisions, are 101 W/m2 and 187.8 W/m2, respectively.
(2) For the Ltot = 600 nm cases, we again observe that the largest heat flux is associated with conduction
exclusively within the aluminum (Case A). However, consideration of the other three cases reveals
nanoscale behavior that would be unexpected from the macroscale point-of-view. Specifically, Case B
involving conduction through pure air is no longer characterized by the lowest heat flux. Rather, we
observe that as more sheets of high thermal conductivity metal are added to the composite layer, the
heat flux is reduced, with the minimum heat flux associated with the most aluminum sheets, Case D.
The reduction in the conduction resistance due to the replacement of low thermal conductivity air with
high thermal conductivity metal is more than offset with the increase in the total thermal resistance that
is associated with molecule-surface interactions at the interfaces between the aluminum sheets and the
air. The molecule-surface interactions can have a profound effect on nanoscale heat transfer.
(3) Nanoscale effects could become important in the solid as the thickness of the solid approaches the
mean free path. See Table 2.1.
PROBLEM 3.141
KNOWN: Knudsen number, specific heat ratio and thermal accommodation coefficient for an ideal
gas and solid surface.
FIND: Expression for the the ratio of the thermal resistance due to molecule-surface collisions to the
thermal resistance associated with molecule-molecule collisions, Rt,m-s/Rt,m-m.
ASSUMPTIONS: (1) Ideal gas behavior.
ANALYSIS: The expressions for Rt,m-m and Rt,m-s are
=
Rt ,m −m
λmfp  2 − aγ

L
t  9 − 5
=
and Rt ,m − s

  +1 
kA
kA  aγ

t

therefore,
Rt ,m − s λmfp  2 − α t   9γ − 5 
 2 − α t   9γ − 5 
=
=
Kn


 α   γ +1 
Rt ,m −m
L  α t   γ + 1 

 t 
<
Associating the critical Knudsen number, Kncrit, with Rt,m-s/Rt,m-m = 0.01, we may plot the value of the
critical Knudsen number for g = 1.4 and 1.67 over the range 0.01 ≤ αt ≤ 1 as shown below.
g = 1.4
g = 1.67
Thermal accommodation coefficient, alphaT
COMMENTS: (1) Relatively large Knudsen numbers are associated with more significant surfacemolecule collisions. (2) The critical Knudsen number is relatively insensitive to the specific heat ratio,
g.
PROBLEM 3.142
KNOWN: Thickness of alternating tungsten and aluminum oxide layers, interface thermal resistance,
thermal conductivities of tungsten and aluminum oxide thin films.
FIND: (a) Effective thermal conductivity of the nanolaminate. Comparison with bulk thermal
conductivities of aluminum oxide and tungsten, (b) Effective thermal conductivity of the nanolaminate
using bulk values of the thermal conductivity of aluminum oxide and tungsten.
SCHEMATIC:
Aluminum oxide
2δ
x
Tungsten
Nanolaminate
δ = 0.5 nm
Interfaces
ASSUMPTIONS: (1) Steady-state, one-dimensional conditions, (2) Constant properties.
PROPERTIES: Table A.1, tungsten (300 K): kT = 174 W/m⋅K. Table A.2, aluminum oxide (300 K): kA
= 36 W/m⋅K.
ANALYSIS: (a) Consider a unit cell consisting of one layer of aluminum oxide, one layer of tungsten,
and two interfaces of unit cell thickness 2δ = 1.0 nm as shown in the schematic. The sum of the thermal
resistances is
∑
2( Rt",i ) +
Rt" =
δ
kA
+
δ
kT
2 × 3.85 × 10−9
=
m 2 ⋅ K 0.5 × 10−9 m 0.5 × 10−9 m
m2 ⋅ K
8.08 × 10−9
+
+
=
W
1.65W/m ⋅ K 6.10W/m ⋅ K
W
The effective thermal conductivity is
=
keff
L
=
Rt"
∑
2δ
=
Rt"
∑
2 × 0.5 × 10−9 m
W
0.123
=
2
m⋅K
m ⋅K
8.08 × 10−9
W
<
The value of the effective thermal conductivity is 0.123/174 × 100 = 0.07% that of bulk tungsten and
0.123/36 × 100 = 0.34% that of bulk aluminum oxide.
W
(b) Repeating part (a) using kT = 174 W/m⋅K and kA = 36 W/m⋅K yields keff = 0.129
<
m⋅K
COMMENTS: (1) The effective thermal conductivity is dominated by the interface resistances and is
relatively insensitive to the thermal conductivity of the two materials. Although the interface resistance
is very small compared to typical contact resistance values (see Table 3.2), by using extremely thin
layer thicknesses, many such interfaces may be packed into the laminated structure, resulting in very
small values of the effective or bulk thermal conductivity. The material service temperature would be
limited to values less than approximately 1000°C, due to the tendency of the material to lose thermal
stability at high temperatures and, in turn, lose its nanolaminated structure. (2) From Table 2.1, λmfp =
5.08 nm for aluminum oxide. Since δ/λmfp = 0.5 nm/5.08 nm ≅ 0.1, Equation 2.9a cannot be used to
calculate the thermal conductivity of the aluminum oxide layers. (3) See Costescu, Cahill, Fabreguette,
Sechrist, and George, “Ultra-Low Thermal Conductivity in W/AL2O3 Nanolaminates,” Science, Vol.
303, pp. 989 – 990, 2004, for additional information.
PROBLEM 3.143
KNOWN: Dimensions of an aluminum tube. Thickness and number of alternating aluminum and
aluminum oxide layers used as insulating coating. Aluminum mean free path.
FIND: Thermal resistance for unit tube length for wall of aluminum tube and nanocomposite coating.
Comment on effectiveness of coating as thermal insulator.
SCHEMATIC:
N = 500 alternating layers
of aluminum and aluminumoxide, each of thickness
δ = 60 nm.
Di = 1 mm
Biological fluid
Tube wall
Biological
t = 0.1 mm
fluid
ASSUMPTIONS: (1) Steady-state, one-dimensional conditions, (2) Constant properties, (3) The
coating can be treated as a plane wall since it is thin in relation to the tube diameter.
PROPERTIES: Table 3.6, Aluminum/aluminum oxide: Rt′′,b = 95 × 10-10 m2⋅K/W. Table A.1, pure
aluminum (300 K): kA,bulk = 237 W/m⋅K; Table A.2, sapphire (300 K): kS,bulk = 46 W/m⋅K. Given, λmfp,A
= 35 nm.
ANALYSIS: The thermal resistance of the tube wall (for a unit length) is:
ln(r / r ) ln(0.6 mm / 0.5 mm)
1.22 × 10−4 m ⋅ K/W
=
Rw′ = o i =
2π kA,bulk
2π × 237 W/m ⋅ K
Referring to Table 2.1, for aluminum oxide the layer thickness of 60 nm is greater than the critical value
Lcrit,x, therefore no correction is needed to its bulk thermal conductivity. For aluminum, however,
considering the discussion beneath Equations 2.9, with δ/λmfp,A = 60/35 = 1.7 < 7, the correction is
significant. The thermal conductivity of the thin layers of aluminum, in the direction normal to the
layers, can be calculated from Equation 2.9a:
 lmfp,A 
35 nm 

kA,r kA,bulk 1 − =
=
=
 237 W/m ⋅ K × 1 −
 191 W/m ⋅ K
3δ 
 3 × 60 nm 

The total resistance of the nanocomposite layer can now be calculated, accounting for the thermal
resistance of 250 layers of aluminum and aluminum oxide and 500 interfacial resistances:
Continued …
PROBLEM 3.143 (Cont.)
Rt′′,i
N
N δ
δ
Rt′ =
N
+
+
π Do 2 kA,r π Do 2 kSπ Do
=

 60 × 10−9 m 60 × 10−9 m  
1
−10
m 2 ⋅ K/W + 250 × 
+
500 × 95 × 10
 191 W/m ⋅ K 46 W/m ⋅ K  
π × 0.0012 m 


=
1.37 × 10−3 m ⋅ K/W
<
The nanocomposite coating has a thermal resistance about an order of magnitude greater than the 0.1
mm thick aluminum tube wall even though its total thickness is only 30 µm.
COMMENTS: The thermal resistance of the nanocomposite is dominated by the interface resistances
and is relatively insensitive to the thermal conductivity of the two materials. Although the interface
resistance is very small compared to typical contact resistance values (see Table 3.2), by using
extremely thin layer thicknesses, many such interfaces may be packed into the nanocomposite structure,
resulting in a large total thermal resistance.
PROBLEM 4.1
KNOWN: Method of separation of variables for two-dimensional, steady-state conduction.
FIND: Show that negative or zero values of λ2, the separation constant, result in solutions which cannot
satisfy the boundary conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: From Section 4.2, identification of the separation constant λ2 leads to the two ordinary
differential equations, 4.6 and 4.7, having the forms
d2X
d2Y
2
λ
=
+
X
0
=
− λ 2Y 0
(1,2)
2
2
dx
dy
and the temperature distribution is
θ (=
x,y ) X ( x ) ⋅ Y ( y ) .
(3)
Consider now the situation when λ2 = 0. From Eqs. (1), (2), and (3), find that
X=
C1 + C2 x, Y =
C3 + C4 y and θ ( x,y ) =
( C1 + C2 x ) ( C3 + C4 y ) .
(4)
Evaluate the constants - C1, C2, C3 and C4 - by substitution of the boundary conditions:
x = 0:
θ ( 0,y ) = ( C1 + C2 ⋅ 0 ) ( C3 + C4 ⋅ y ) = 0
C1 = 0
y =0 :
θ ( x,0 ) =( 0 + C2 ⋅ x ) ( C3 + C4 ⋅ 0 ) =0
C3 =0
θ ( L,0 ) =( 0 + C2 ⋅ L )( 0 + C4 ⋅ y ) =0
x =L:
C2 =0
y = W:
θ ( x,W ) = ( 0 + 0 ⋅ x )( 0 + C4 ⋅ W ) = 1
0 ≠1
The last boundary condition leads to an impossibility (0 ≠ 1). We therefore conclude that a λ2 value of
zero will not result in a form of the temperature distribution which will satisfy the boundary conditions.
Consider now the situation when λ2 < 0. The solutions to Eqs. (1) and (2) will be
(5,6)
X=
C5 e - λ x + C 6 e + λ x ,
Y=
C7 cos λ y + C8sin λ y
and
C5e-λ x + C6 e + λ x  [ C7 cos λ y + C8sin λ y ].
θ ( x,y ) =


Evaluate the constants for the boundary conditions.
y = 0 : θ ( x,0 ) = C5e-λ x + C6 e-λ x  [ C7 cos 0 + C8sin 0] = 0


0
0

x=
0 : θ ( 0,y ) =
C e + C6 e [ 0 + C8sin λ y ] =
0
 5

(7)
C7 = 0
C8 =
0
If C8 = 0, a trivial solution results or C5 = -C6.
θ ( L,y ) =
x=
L:
C5 e-xL - e +xL  C8sin λ y =
0.


From the last boundary condition, we require C5 or C8 is zero; either case leads to a trivial solution with
either no x or y dependence.
PROBLEM 4.2
PROBLEM 4.3
KNOWN: Temperature distribution in the two-dimensional rectangular plate of Problem 4.2.
FIND: Expression for the heat rate per unit thickness from the lower surface (0 ≤ x ≤ 2, 0) and result
based on first five non-zero terms of the infinite series.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: The heat rate per unit thickness from the plate along the lower surface is
=
x 2=
x 2
=
x 2
∂T
∂q
− ∫ dq′y ( x, 0 ) =
− ∫ −k
q′out =
dx =
k ( T2 − T1 ) ∫
dx
∂ y y =0
∂y
=
x 0=
x 0
=
x 0
y =0
(1)
where from the solution to Problem 4.2,
n +1
+ 1  nπ x  sinh ( nπ y L )
T − T1 2 ∞ ( −1)
θ≡
=
sin 
.
∑

T2 − T1 π
n
L  sinh ( nπ W L )

n =1
(2)
Evaluate the gradient of θ from Eq. (2) and substitute into Eq. (1) to obtain
′out k ( T2 − T1 )
q=
x =2
∫
x =0
n +1
+ 1  nπ x  ( nπ L ) cosh ( nπ y L )
2 ∞ ( −1)
sin 
∑

π
n
sinh ( nπ W L )
 L 
n =1
dx
y =0
n +1
2 

+1
2 ∞ ( −1)
1
 − cos  nπ x 

q′out =
k ( T2 − T1 ) ∑
π n =1
n
sinh ( nπ W L ) 
 L  x =0 


n +1
+1
2 ∞ ( −1)
1
1 − cos ( nπ ) 
q′out =
k ( T2 − T1 ) ∑
π
n
sinh ( nπ L ) 
n =1
<
To evaluate the first five, non-zero terms, recognize that since cos(nπ) = 1 for n = 2, 4, 6 ..., only the nodd terms will be non-zero. Hence,
Continued …
PROBLEM 4.3 (Cont.)

q′out = 50 W m ⋅ K (150 − 50 ) C
+
( −1)6 + 1
5
⋅
1
sinh ( 5π 2 )
( −1) + 1 ⋅
2  ( −1) + 1
1
1
⋅
⋅ ( 2)
( 2) +

π
1
sinh (π 2 )
3
sinh ( 3π 2 )

( 2) +
2
( −1)8 + 1
7
4
⋅
1
sinh ( 7π 2 )
( 2) +
( −1)10 + 1
9
⋅
1
sinh ( 9π 2 )
=
q′out 3.183kW m [1.738 + 0.024 + 0.00062
=
+ (...) ] 5.611kW m
x =2
∫
dq′y ( x, W ) , it would follow that
x =0
n +1
+1
2 ∞ ( −1)
q′in =
k ( T2 − T1 ) ∑
coth ( nπ 2 ) 1 − cos ( nπ ) 
π
n
n =1
However, with coth(nπ/2) ≥ 1, irrespective of the value of n, and with
∞
∑ ( −1)n +1 + 1
n =1

<
COMMENTS: If the foregoing procedure were used to evaluate the heat rate into the upper surface,
q′in = −

( 2 )
n being a
divergent series, the complete series does not converge and q′in → ∞ . This physically untenable
condition results from the temperature discontinuities imposed at the upper left and right corners.
PROBLEM 4.4
KNOWN: Rectangular plate subjected to prescribed boundary conditions.
FIND: Steady-state temperature distribution and heat flux distribution along top edge.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.
ANALYSIS: The solution follows the method of Section 4.2. The product solution is
(
T ( x,y ) = X ( x ) ⋅ Y ( y ) = ( C1cosλ x + C2sinλ x ) C3e-λ y + C4e+λ y
)
and the boundary conditions are: T(0,y) = 0, T(a,y) = 0, T(x,0) = 0, T(x.b) = Ax. Applying
BC#1, T(0,y) = 0, find C1 = 0. Applying BC#2, T(a,y) = 0, find that λ = nπ/a with n = 1,2,….
Applying BC#3, T(x,0) = 0, find that C3 = -C4. Hence, the product solution is
 nπ  +λ y -λ y
-e
T ( x,y ) = X ( x ) ⋅ Y ( y ) = C2C4 sin 
x e
.
 a 
Combining constants and using superposition, find
)
(
∞
 nπ x 
 nπ y 
Cn sin 
sinh 
.

a 
a 


n =1
To evaluate Cn and satisfy BC#4, use orthogonal functions with Equation 4.16 to find
a
 nπ x 
 nπ b  a
2  nπ x 
Cn =
∫0 Ax ⋅ sin  a  ⋅ dx/sinh  a  ∫0 sin  a  dx,
noting that y = b. The numerator, denominator and Cn, respectively, are:
T ( x,y ) =
∑
 a
a
nπ x
⋅ dx= A  
A x ⋅ sin
0
a
  nπ
∫

a
2
Aa 2
Aa 2

 nπ x  ax
 nπ x  
−
=
−
=
sin
cos
cos
n
π
( −1)n+1 ,
[ ( )]

 a  nπ
 a  
nπ
nπ

0
a
a
a
 nπ b  a 2 n π x
 nπ b   1
 2nπ x  
 nπ b 
sin
⋅
dx
=
sinh
x
−
sin
=
⋅
sinh
∫


 a   2
 a  
 a  ,
a
4nπ
2
 a  0
0
sinh 
Aa 2
a
nπ b
 nπ b 
n+1
Cn =
2Aa ( −1)
/ nπ sinh 
( −1)n+1 / sinh   =
.
nπ
2
 a 
 a 
Hence, the temperature distribution is
 nπ y 
sinh 
∞ ( −1)n+1
2 Aa
 nπ x 
 a  .
T ( x,y )
=
⋅ sin 
∑

n
π n =1
 a  sinh  nπ b 
 a 
<
Continued...
PROBLEM 4.4 (Cont.)
The heat flux normal to the top edge can be found from Fourier’s Law:
nπ
 nπ b 
cosh 
∞ ( −1)n+1
∂T
2Aa
 nπ x  a
 a 
−k
=
⋅
q′′y ( x, y =
b) =
sin
∑ n
 a 
∂y y = b
π n =1

 sinh  nπ b 
 a 
∞
( −1)n+1
 nπ x 
⋅ sin 
.
<
nπ b 
a 


n =1 tanh
 a 
COMMENTS: The heat transfer rate out of the top edge can be found by integrating the heat
flux expression above over the x-direction. If this is done similarly for all four surfaces, and the
results summed, the net heat transfer rate leaving the rectangle must come out to zero since it is
at steady-state with no heat generation.
q′′y ( x, =
y b=
) 2A ∑
PROBLEM 4.5
KNOWN: Boundary conditions on four sides of a rectangular plate.
FIND: Temperature distribution.
SCHEMATIC:
y
q′′s
W
T1
T1
0
x
0
L
T1
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: This problem differs from the one solved in Section 4.2 only in the boundary condition at
the top surface. Defining θ = T – T∞, the differential equation and boundary conditions are
∂ 2θ
∂ 2θ
+
=0
∂x 2
∂y 2
∂θ
=
θ(0, y) 0=
θ(L, y) 0=
θ(x,0) 0 =
k
q′′s
∂y y = W
The solution is identical to that in Section 4.2 through Equation (4.11),
∞
nπx
nπy
θ = ∑ Cn sin
sinh
L
L
n=1
(1a,b,c,d)
(2)
To determine Cn, we now apply the top surface boundary condition, Equation (1d). Differentiating
Equation (2) yields
Continued…
PROBLEM 4.5 (Cont.)
∂θ
∂y
∞
=
∑ Cn
n=1
y=W
nπ
nπx
nπW
sin
cosh
L
L
L
(3)
Substituting this into Equation (1d) results in
q′′s
=
k
∞
∑ A n sin
n=1
nπx
L
(4)
where An = Cn(nπ/L)cosh(nπW/L). The principles expressed in Equations (4.13) through (4.16) still
apply, but now with reference to Equation (4) and Equation (4.14), we should choose f(x) = q′′s /k ,
nπx
g n (x) = sin
. Equation (4.16) then becomes
L
q′′s
nπx
sin
dx
∫
k 0
L
L
An =
L
∫ sin
0
2
nπx
dx
L
=
q′′s 2 (-1) n+1 + 1
n
k π
Thus
Cn = 2
q′′s L
(-1) n+1 + 1
k n 2 π 2 cosh(nπW/L)
The solution is given by Equation (2) with Cn defined by Equation (5).
(5)
PROBLEM 4.6
PROBLEM 4.7
KNOWN: Boundary conditions on four sides of a square plate.
FIND: Expressions for shape factors associated with the maximum and average top surface temperatures.
Values of these shape factors. The maximum and average temperatures for specified conditions.
SCHEMATIC:
y
q′′s
W
T1
T1
0
x
0
T1
W
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: We must first find the temperature distribution as in Problem 4.5. Problem 4.5 differs from
the problem solved in Section 4.2 only in the boundary condition at the top surface. Defining θ = T – T∞,
the differential equation and boundary conditions are
∂ 2θ
∂ 2θ
+
=0
∂x 2
∂y 2
∂θ
=
θ(0, y) 0=
θ(L, y) 0=
θ(x,0) 0 =
k
q′′s
∂y y = W
The solution is identical to that in Section 4.2 through Equation (4.11),
∞
nπx
nπy
θ = ∑ Cn sin
sinh
L
L
n=1
(1a,b,c,d)
(2)
To determine Cn, we now apply the top surface boundary condition, Equation (1d). Differentiating
Equation (2) yields
∂θ
∂y
∞
=
y=W
∑ Cn
n=1
nπ
nπx
nπW
sin
cosh
L
L
L
(3)
Continued …
PROBLEM 4.7 (Cont.)
Substituting this into Equation (1d) results in
∞
q′′s
nπx
= ∑ A n sin
k
L
n=1
(4)
where An = Cn(nπ/L)cosh(nπW/L). The principles expressed in Equations (4.13) through (4.16) still
apply, but now with reference to Equation (4) and Equation (4.14), we should choose f(x) = q′′s /k ,
nπx
g n (x) = sin
. Equation (4.16) then becomes
L
q′′s
nπx
sin
dx
∫
k 0
L
L
An =
L
∫ sin
0
2
=
nπx
dx
L
q′′s 2 (-1) n+1 + 1
k π
n
Thus
Cn = 2
q′′s L
(-1) n+1 + 1
k n 2 π 2 cosh(nπW/L)
(5)
The solution is given by Equation (2) with Cn defined by Equation (5). We now proceed to evaluate the
shape factors.
(a) The maximum top surface temperature occurs at the midpoint of that surface, x = W/2, y = W. From
Equation (2) with L = W,
θ(W/2,W) = T2,max - T1 =
∞
∑ Cn sin
n=1
nπ
sinh nπ =
2
∑ Cn (-1)(n-1)/2sinh nπ
n odd
where
Cn = 2
q′′s W (-1) n+1 + 1
k n 2 π 2 cosh nπ
Thus
-1
Smax
2

4

q′′s Wd
(-1)n+1 + 1
(-1)(n-1)/2
(n-1)/2
=
=  ∑
(-1)
tanh
nπ
=
tanh nπ 


∑
2
2
2
2
k(T2,max - T1 )  d n odd n π

 d n odd n π

-1
where d is the depth of the rectangle into the page.
Continued…
<
PROBLEM 4.7 (Cont.)
(b) The average top surface temperature is given by
θ(y = W) = T2 - T1 =
∞
∑ Cn
n=1
1
W
W
∫ sin
0
nπx
dx sinh nπ =
W
∞
∑ Cn
n=1
1 - (-1) n
sinh nπ
nπ
Thus
 2 ∞ [(-1)n+1 + 1][1- (-1)n ]

q′′s Wd
S=
=  ∑
tanh
nπ

k(T2 - T1 )  d n=1
n 3 π3

-1
8

1
=  ∑ 3 3 tanh nπ 
 d n odd n π

-1
<
(c) Evaluating the expressions for the shape factors yields
-1


Smax
(-1)(n-1)/2
= 4 ∑
tanh
nπ
2.70
 =
2 2
d
 n odd n π


S 
1
= 8 ∑ 3 3 tanh nπ 
d  n odd n π

<
-1
= 3.70
<
The temperatures can then be found from
T2,max = T1 +
T2 = T1 +
q
Smax k
= T1 +
q′′s Wd
1000 W/m 2 × 0.01 m
= 0°C +
= 0.19°C
Smax k
2.70 × 20 W/m ⋅ K
q′′Wd
q
1000 W/m 2 × 0.01 m
= T1 + s
= 0°C +
= 0.14°C
Sk
Sk
3.70 × 20 W/m ⋅ K
<
<
PROBLEM 4.8
PROBLEM 4.9
KNOWN: Shape of objects at surface of semi-infinite medium.
FIND: Shape factors between object at temperature T1 and semi-infinite medium at temperature T2.
SCHEMATIC:
D
D
T1
D
T1
T1
T2
T2
(a)
(b)
T2
(c)
ASSUMPTIONS: (1) Steady-state, (2) Medium is semi-infinite, (3) Constant properties, (4) Surface of
semi-infinite medium is adiabatic.
ANALYSIS: Cases 12 -15 of Table 4.1 all pertain to objects buried in an infinite medium. Since they all
possess symmetry about a horizontal plane that bisects the object, they are equivalent to the cases given in
this problem for which the horizontal plane is adiabatic. In particular, the heat flux is the same for the
cases of this problem as for the cases of Table 4.1. Note, that when we use Table 4.1 to determine the
dimensionless conduction heat rate, q*ss , we must be consistent and use the surface area of the “entire”
object of Table 4.1, not the “half” object of this problem. Then
q* k(T - T )
q
q′′ =
= ss 1 2
As
Lc
where Lc = ( As 4π)1/2 and As is the area given in Table 4.1
When we calculate the shape factors we must account for the fact that the surface areas and heat transfer
rates for the objects of this problem are half as much as for the objects of Table 4.1.
q′′ As 2
q* A
q* (4πAs )1/2
q
S=
=
= ss s = ss
k(T1 - T2 )
k(T1 - T2 )
2Lc
2
where As is still the area in table 4.1 and the 2 in the denominator accounts for the area being halved.
Thus, finally,
S = q*ss (πAs )1/2
(a)
<
S = 1 ⋅ (π ⋅ πD 2 )1/2 = πD
2 1/2
2 2  πD
π⋅
 = 2D
π 
2 
This agrees with Table 4.1a, Case 10.
<
(b)
S=
(c)
S = 0.932(π ⋅ 2D 2 )1/2 = 2π (0.932)D = 2.34D
<
(d) The height of the “whole object” is d = 2D. Thus
(
)
1/2
S = 0.961  π 2D 2 + 4D ⋅ 2D 


= 10π (0.961)D = 5.39D
<
PROBLEM 4.10
PROBLEM 4.11
KNOWN: Dimensions of a two-dimensional object, applied boundary conditions and thermal
conductivity.
FIND: (a) Shape factor for the object if the dimensions are a = 10 mm, b = 12 mm. (b) Shape factor for a
= 10 mm, b = 15 mm. (c) Shape factor for cases (a) and (b) using the alternative conduction analysis (d)
For T1 = 100°C and T2 = 60°C, determine the heat transfer rate per unit depth for k = 15 W/m⋅K for cases
(a) and (b).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties.
PROPERTIES: Given: k = 15 W/m⋅K.
ANALYSIS: (a) The geometry and applied boundary conditions correspond to Case 11 of Table 4.1(a).
Noting that the diagonals of the square channel of Case 11 are adiabats, the shape factor for b/a = W/w =
12/10 = 1.2 is one-fourth the shape factor given in Table 4.1(a),
2π L
2.00 L
2.00 × 0.10m
S=
0.25 ×
1.097
== =
0.785ln(W / w) ln(W / w)
ln(1.2)
<
(b) For b/a = W/w = 15/10 = 1.5,
2π L
1.69 L
1.69 × 0.10m
==
=
S=
0.25 ×
0.480
<
0.930ln(W / w) − 0.050 ln(W / w) − 0.0534 ln(1.5) − 0.0534
(c) From the one-dimensional alternative conduction analysis with the top surface described by y = x and
A = 2yL,
dT
dT
qx =
− kA
=
−2kLx
dx
dx
Separating variables and integrating yields
b
T2
dx
2kL
= −
dT or ln(
=
b / a ) 2kL (T1 − T2 )
qx
x
qx
x =a
T1
∫
∫
Hence, S1-D = 2L/[ln(b/a)].
Continued…
PROBLEM 4.11 (Cont.)
For b/a = 1.2, S1-D = 2 × 0.1m/[ln(1.2)] = 1.097. For b/a = 1.5, S1-D = 2 × 0.1m/[ln(1.5)] = 0.493.
<
As b/a becomes small, the influence of the lateral edges is diminished and one-dimensional conditions are
approached. Hence, the shape factor estimated using the one-dimensional alternative conduction solution
is nearly the same as for the two-dimensional shape factor for b/a = 1.2. As b/a is increased, the lateral
edge effects become more important, and the shape factors obtained by the two methods begin to diverge
in value. As two-dimensional conduction in the object becomes more pronounced, the heat transfer rate is
decreased relative to that associated with the assumed one-dimensional conditions.
(d) For b/a = 1.2, the heat transfer rate is
=
q Sk (T1 − T=
C 658 W
2 ) 1.097 × 15 W / m ⋅ K × (100 − 60)°=
<
for b/a = 1.5, the heat transfer rate is
=
q Sk (T1 − T=
C 288 W
2 ) 0.480 × 15 W / m ⋅ K × (100 − 60)°=
<
COMMENTS: The heat transfer rate is independent of the individual values of a or b. As either b or a is
increased while maintaining a fixed b/a ratio, the cross-sectional area for heat transfer increases, but the
increase is offset by increased thickness through which the conduction occurs. The offsetting effects
balance one another, and the net result is no change in the heat transfer rate.
PROBLEM 4.12
PROBLEM 4.13
PROBLEM 4.14
PROBLEM 4.15
PROBLEM 4.16
KNOWN: Power, size and shape of laser beam. Material properties.
FIND: Maximum surface temperature for a Gaussian beam, maximum temperature for a flat beam, and
average temperature for a flat beam.
Flat
Gaussian
q′′′
SCHEMATIC:
q′′′
rb = 0.1 mm
rb
P = 1W
α =0.45
Tmax
Tmax
T2 = 25°C
k = 27 W/m•K
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Semi-infinite medium, (4)
Negligible heat loss from the top surface.
ANALYSIS: The shape factor is defined in Eq. 4.20 and is q = Sk∆T1-2
(1)
From the problem statement and Section 4.3, the shape factors for the three cases are:
Beam Shape
Gaussian
Shape Factor
2 πrb
T1,avg or T1,max
T1,max
Flat
πrb
T1,max
Flat
3π2 rb / 8
T1,avg
For the Gaussian beam, S1 = 2 π × 0.1 × 10-3 m = 354 × 10-6 m
For the flat beam (max. temperature), S2 = π × 0.1 × 10-3 m = 314 × 10-6 m
For the flat beam (avg. temperature), S3 = (3/8) × π2 × 0.1 × 10-3 m = 370 × 10-6 m
The temperature at the heated surface for the three cases is evaluated from Eq. (1) as
T1 = T2 + q/Sk = T2 + Pα/Sk
<
), T1,max = 25°C + 1 W × 0.45 / (314 × 10-6 m × 27 W/m ⋅ K) = 78.1°C <
), T1,avg = 25°C + 1 W × 0.45 / (370 × 10-6 m × 27 W/m ⋅ K) = 70.0°C <
For the Gaussian beam, T1,max = 25°C + 1 W × 0.45 / (354 × 10-6 m × 27 W/m ⋅ K) = 72.1°C
For the flat beam (Tmax
For the flat beam (Tavg
COMMENTS: (1) The maximum temperature occurs at r = 0 for all cases. For the flat beam, the
maximum temperature exceeds the average temperature by 78.1 – 70.0 = 8.1 degrees Celsius.
PROBLEM 4.17
PROBLEM 4.18
PROBLEM 4.19
PROBLEM 4.20
PROBLEM 4.21
PROBLEM 4.22
KNOWN: Dimensions and temperature of thermocouple bead and wires. Manipulator temperature,
distance between bead and surface.
FIND: Surface temperature.
SCHEMATIC:
Manipulator, Tm = 23°C
d = 25µm
Air
∼
∼
∼∼
L = 300µm
qcond,Ch
qcond,Al
D = 120 µm Thermocouple
Ttc = 29°C bead
z = 100µm
qcond,a
Ts
ASSUMPTIONS: (1) Constant properties, (2) Negligible radiation and convection, (3) Isothermal
thermocouple bead, (4) Air behaves as a semi-infinite medium, (5) Steady state conditions.
PROPERTIES: Table A.4, Air (310 K): ka = 0.027 W/m⋅K.
ANALYSIS: An energy balance on the thermocouple bead yields
q=
qcond,Al + qcond,Ch
cond,air
(1)
where the conduction heat transfer rates through the alumel and chromel wires are denoted as qcond,Al and
qcond.Ch, respectively. Conduction from the surface through the air to the thermocouple bead, qcond,air, may
be determined by use of the shape factor S = (2πD)/(1 - D/4z)of Case 1 of Table 4.1. Therefore, Equation
(1) may be written as
Ska ( Ts=
− Ttc ) k Al
πd2
4L
(Ttc − Tm ) + kCh
πd2
4L
(Ttc − Tm )
(2)
which may be rearranged to yield
1 − D / 4z d 2
 kAl (Ttc − Tm ) + kCh (Ttc − Tm )  + Ttc
⋅
2 Dka 4 L 
=
Ts
=
1 − 120 / ( 4 × 100 )
( 25 ×10
⋅
−6
m
)
2
 29 W/m ⋅ K ( 29 − 23) °C + 19W/m ⋅ K ( 29 − 23) °C  + 29°C
2 × 120 × 10−6 m × 0.027 W/m ⋅ K 4 × 300 × 10−6 m 
= 45.2°C
<
Continued…
PROBLEM 4.22 (Cont.)
COMMENTS: The required surface temperature to induce the specified thermocouple temperature and
its dependence on the separation distance, z, is shown below. As expected, the required surface
temperature becomes greater as the separation distance increases.
50
48
46
Ts (°C)
44
42
40
70
110
150
zmicron
z (µm)
190
230
PROBLEM 4.23
KNOWN: Cylinder extending between two walls. Highly polished surfaces. Walls are at specified
temperatures far from the cylinder.
FIND: (a) Thermal resistance network. (b) Expression for shape factor corresponding to conduction
between Tb,1 and Tb,2. (3) Value of shape factor for specified cylinder diameter, length, and thermal
conductivity.
SCHEMATIC:
L
D
Tb,1
D = 0.01 m
L = 0.5 m
k = 23 W/m∙K
Tb,2
ASSUMPTIONS: (1) Steady-state conditions. (2) Constant properties. (3) Walls are semi-infinite. (4)
Walls and cylinder have same thermal conductivity.
ANALYSIS: (a) Heat conducting along the cylinder will alter the wall temperature in the vicinity of the
cylinder. This can be accounted for by using the shape factor for a disk on a semi-infinite medium, Case
10 in Table 4.1. It introduces a thermal resistance at each of the two cylinder-wall interfaces. Thus, the
resistance network is as shown below.
Tb,1
Tb,2
<
L
1
1
2
2Dk
2Dk
kπD /4
(b) According to the resistance network, when kw = kc = k, the heat transfer rate between the two walls is
given by
(Tb ,1 − Tb ,2 )
q=
1
1
L
+
+
2
2 Dk kπ D / 4 2 Dk
Then from Equation 4.20, the shape factor is
=
S
q
=
k (Tb ,1 − Tb ,2 )
1
D2
=
1
1
L
D + 4L / π
+
+
2
2D π D / 4 2D
<
(c) The value of the shape factor is
=
S
(0.01 m) 2
= 1.55 × 10−4 m
0.01 m + 4(0.5 m) / π
<
COMMENTS: (1) As long as the materials all have the same thermal conductivity, the value of the shape
factor is independent of the thermal conductivity. (2) The effect of the cylinder on the wall temperature
becomes negligible as the thermal resistance of the cylinder increases, for example by increasing L.
PROBLEM 4.24
PROBLEM 4.24 (Cont.)
PROBLEM 4.25
PROBLEM 4.26
KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient air.
Inlet temperature of water flow through the duct.
FIND: (a) Heat loss per duct length near inlet, (b) Minimum allowable flow rate corresponding to
maximum allowable temperature rise of water.
SCHEMATIC:
Air
Too = 0oC
h = 25 W/m2-K
Concrete
k = 1.4 W/m-K
T1, D = 0.15 m
T1
w = 0.30 m
m, Ti = 90oC
7 J/kg-K
cp = 4200
L = 100 m
Rcond(2D)
Too
Rconv
q
Water
ASSUMPTIONS: (1) Steady state, (2) Negligible water-side convection resistance, pipe wall conduction
resistance, and pipe/concrete contact resistance (temperature at inner surface of concrete corresponds to
that of water), (3) Constant properties, (4) Negligible flow work and kinetic and potential energy changes.
ANALYSIS: (a) From the thermal circuit, the heat loss per unit length near the entrance is
=
q′
Ti − T∞
Ti − T∞
=
R ′cond ( 2D ) + R ′conv ln (1.08 w / D )
1
+
2π k
h ( 4w )
where R ′cond ( 2D ) is obtained by using the shape factor of Case 6 from Table 4.1 with Eq. (4.21).
Hence,
q′
90°C
( 90 − 0 ) °C
= =
ln (1.08 × 0.3m / 0.15m )
1
( 0.0876 + 0.0333 ) K ⋅ m / W
+
2
2π (1.4 W / m ⋅ K )
25 W / m ⋅ K (1.2m )
745 W / m
<
5°C,
(b) From Eq. (1.12d), with q = q′L and ( Ti − To ) =

=
m
745 W / m (100m )
q′L
q′L
=
=
= 3.54 kg / s
u i − u o c ( Ti − To ) 4207 J / kg ⋅ K ( 5°C )
<
COMMENTS: The small reduction in the temperature of the water as it flows from inlet to outlet
induces a slight departure from two-dimensional conditions and a small reduction in the heat rate per unit
 is therefore obtained in part (b).
length. A slightly conservative value (upper estimate) of m
PROBLEM 4.27
KNOWN: Long constantan wire butt-welded to a large copper block forming a thermocouple junction
on the surface of the block.
FIND: (a) The measurement error (Tj - To) for the thermocouple for prescribed conditions, and (b)
Compute and plot (Tj - To) for h = 5, 10 and 25 W/m2⋅K for block thermal conductivity 15 ≤ k ≤ 400
W/m⋅K. When is it advantageous to use smaller diameter wire?
SCHEMATIC:
T∞ = 25°C
h = 25 W/m2∙K
ASSUMPTIONS: (1) Steady-state conditions, (2) Thermocouple wire behaves as a fin with constant
heat transfer coefficient, (3) Copper block has uniform temperature, except in the vicinity of the junction.
PROPERTIES: Table A-1, Copper (pure, 400 K), kb = 393 W/m⋅K; Constantan (350 K), kt ≈ 25 W/m⋅K.
ANALYSIS: The thermocouple wire behaves as a long fin permitting heat to flow from the surface
thereby depressing the sensing junction temperature below that of the block To. In the block, heat flows
into the circular region of the wire-block interface; the thermal resistance to heat flow within the block is
approximated as a disk of diameter D on a semi-infinite medium (kb, To). The thermocouple-block
combination can be represented by a thermal circuit as shown above. The thermal resistance of the fin
follows from the heat rate expression for an infinite fin, Rfin = (hPktAc)-1/2.
From Table 4.1, the shape factor for the disk-on-a-semi-infinite medium is given as S = 2D and hence
Rblock = 1/kbS = 1/2kbD. From the thermal circuit,
R block
1.27

To −=
Tj
C 0.0016 (125 − 25 ) =
C 0.16 C
( To − T=
(125 − 25 )=
∞)
R fin + R block
805 + 1.27
with P = πD and Ac = πD2/4 and the thermal resistances as
)
(
)
(
<
−1/ 2
3

=
R fin  25 W m 2 ⋅ K π 2 4 25 W m ⋅ K × 1 × 10−3 m =
805 K W



R block= 1 / (2 × 393 W m ⋅ K × 10−3 m)= 1.27 K W .
(b) We keyed the above equations into the IHT workspace, performed a sweep on kb for selected values
of h and created the plot shown. When the block thermal conductivity is low, the error (To - Tj) is larger,
increasing with increasing convection coefficient. A smaller diameter wire will be advantageous for low
values of kb and higher values of h.
5
Error, To-Tj (C)
4
3
2
1
0
0
100
200
300
Block thermal conductivity, kb (W/m.K)
h
h
h
h
=
=
=
=
25 W/m^2.K; D = 1 mm
10 W/m^2.K; D = 1mm
5 W/m^2.K; D = 1mm
25 W/m^2.K; D = 5 mm
400
PROBLEM 4.28
PROBLEM 4.29
PROBLEM 4.29 (Cont.)
PROBLEM 4.30
KNOWN: Long fin of aluminum alloy with prescribed convection coefficient attached to different base
materials (aluminum alloy or stainless steel) with and without thermal contact resistance R ′′t, j at the
junction.
FIND: (a) Heat rate qf and junction temperature Tj for base materials of aluminum and stainless steel, (b)
Repeat calculations considering thermal contact resistance, R ′′t, j , and (c) Plot as a function of h for the
range 10 ≤ h ≤ 1000 W/m2⋅K for each base material.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Infinite fin.
PROPERTIES: (Given) Aluminum alloy, k = 240 W/m⋅K, Stainless steel, k = 15 W/m⋅K.
ANALYSIS: (a,b) From the thermal circuits, the heat rate and junction temperature are
=
qf
Tb − T∞
Tb − T∞
=
R tot
R b + R t, j + R f
(1)
T
=
j T∞ + q f R f
(2)
and, with P = πD and Ac = πD2/4, from Tables 4.1 and 3.4 find
Rb =
1 Sk b =
1 ( 2Dk b ) =
( 2 × 0.005 m × k b )
−1
2
R t, j =
R ′′t, j A c =
3 ×10−5 m 2 ⋅ K / W / π ( 0.005m ) / 4  =
1.528K / W


R f = ( hPkA c )
Base
Al alloy
St. steel
−1/ 2
3
= 50 W m 2 ⋅ K π 2 ( 0.005 m ) 240 W m ⋅ K 4 


Rb (K/W)
0.417
6.667
Without R ′′t, j
qf (W)
Tj (°C)
4.46
98.2
3.26
78.4
−1/ 2
= 16.4 K W
With R ′′t, j
qf (W)
4.09
3.05
Tj (°C)
92.1
75.1
(c) We used the IHT Model for Extended Surfaces, Performance Calculations, Rectangular Pin Fin to
calculate qf for 10 ≤ h ≤ 100 W/m2⋅K by replacing R ′′tc (thermal resistance at fin base) by the sum of the
contact and spreading resistances, R ′′t, j + R ′′b .
Continued...
PROBLEM 4.30 (Cont.)
6
Fin heat rate, qf (W)
5
4
3
2
1
0
20
40
60
80
100
Convection coefficient, h (W/m^2.K)
Base material - aluminum alloy
Base material - stainless steel
COMMENTS: (1) From part (a), the aluminum alloy base material has negligible effect on the fin heat
rate and depresses the base temperature by only 2°C. The effect of the stainless steel base material is
substantial, reducing the heat rate by 27% and depressing the junction temperature by 25°C.
(2) The contact resistance reduces the heat rate and increases the temperature depression relatively more
with the aluminum alloy base.
(3) From the plot of qf vs. h, note that at low values of h, the heat rates are nearly the same for both
materials since the fin is the dominant resistance. As h increases, the effect of R ′′b becomes more
important.
PROBLEM 4.31
KNOWN: Cylinder of specified diameter within square bakelite coating of dimension w on a side.
Cylinder temperature, environment temperature, and heat transfer coefficient.
FIND: Heat transfer rate per unit length of cylinder. Plot heat transfer rate for 10 mm < w < 100 mm.
Explain dependence of heat transfer rate on bakelite thickness. Find critical value of w that maximizes
heat transfer from cylinder.
SCHEMATIC:
Bakelite
D
w
+
Cylinder, T1
T∞, h
A-A
Section A-A
D = 10 mm
w = 11 mm
or
10 mm ≤ w ≤ 100 mm
T1 = 100°C
T∞ = 25°C
h = 100 W/m2∙K
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties.
PROPERTIES: Table A-3, Bakelite, (T = 300 K): k = 1.4 W/m⋅K.
ANALYSIS: The resistance network between the cylinder and the environment is shown below, where
Rt,cond(2D) = 1/Sk (Equation 4.21), and the shape factor S for heat transfer through the bakelite is given in
Table 4.1, Case 6.
T1
T∞
Rt,cond(2D)
Rt,cond(2D) = 1/Sk = ln(1.08w/D)/2πkL
Rt,conv = 1/hAs = 1/4hwL
Rt,conv
Thus the heat transfer rate per unit cylinder length is
q=′
(T1 − T2 )
(100 − 25)°C
q
=
=
= 304 W/m
ln(1.08
/
)
1
ln(1.08
11
mm
/
10
mm)
1
w
D
×
L
+
+
2π k
2π × 1.4 W/m ⋅ K
100 W/m 2 ⋅ K × 4 × 11 m
h ⋅ 4w
<
The heat transfer rate per unit length is plotted below as a function of the bakelite dimension, w. It is
apparent that there is a maximum value of the heat transfer rate associated with a critical value of the
bakelite dimension wc, and that the heat transfer rate decreases as w increases beyond wc.
<
The existence of a maximum heat transfer rate is because of the competing effects that as w
increases, the thermal resistance of the bakelite increases, but the thermal resistance for
convection decreases because of the increasing surface area.
Continued…
Heat transfer rate per unit length (W/m)
PROBLEM 4.31 (Cont.)
Heat Transfer Rate per Unit Length
400
375
350
325
300
275
250
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
Bakelite dimension w (m)
0.1
<
Heat transfer is maximized when thermal resistance is minimized. Therefore the critical bakelite
dimension can be determined by differentiating the thermal resistance with respect to w and
setting the result to zero:
′
dRtot
1  1/ w
1
d  ln(1.08w / D)
0
= 
+
=
= −
2π k
dw dw 
h ⋅ 4 w  2π k h ⋅ 4 w2
=
w
c
πk
= 0.022 m
2h
<
COMMENTS: The critical bakelite thickness is very similar to the concept of a critical thickness of
insulation on a cylinder.
PROBLEM 4.32
PROBLEM 4.32 (Cont.)
PROBLEM 4.33
PROBLEM 4.34
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. 4.41.
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 – the shaded area of unit thickness normal to the page – 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:
E in −=
E out 0
k ( ∆y ⋅1)
q1 + q 2 + q3 + q 4 + q5=
+ q6 0
Tm-1,n -Tm,n
 ∆x  Tm,n-1 Tm,n
 ∆y 
+ k  ⋅1
+ h  ⋅1 T∞ - Tm,n
∆x
∆y
 2 
 2 
(
)
Tm,n+1 − Tm,n
 ∆y  Tm+1,n − Tm,n
+ 0 + k  ⋅1
+ k ( ∆x ⋅1)
=0.
∆x
∆y
 2 
Letting ∆x = ∆y, and regrouping, find
(
) (
)
2 Tm-1,n + Tm,n+1 + Tm+1,n + Tm,n-1 +
h∆x
h∆x 

T∞ - 6 +
Tm,n =
0.
k
k 

<
(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,
(
) (
)
2 Tm-1,n + Tm,n+1 + Tm+1,n + Tm,n-1 - 6 Tm,n =
0.
Note that this expression compares exactly with Equation 4.41 when h = 0, which corresponds to
insulated boundaries.
<
PROBLEM 4.35
PROBLEM 4.36
KNOWN: Boundary conditions that change from specified heat flux to convection.
FIND: The finite difference equation for the node at the point where the boundary condition changes.
SCHEMATIC:
h, T∞
q′′s
m -1,n
m,n
∆y/2
∆y
q1
m +1,n
q2
q3
q4
q5
∆x
∆x
m, n-1
ASSUMPTIONS: (1) Two dimensional, steady-state conduction with no generation, (2) Constant
properties.
ANALYSIS: Performing an energy balance on the control volume ∆x • ∆y/2,
q1 + q2 + q3 + q4 + q5 = 0
E in - E out = 0
Expressing q1 in terms of the specified heat flux, q2 in terms of the known heat transfer coefficient and
environment temperature, and the remaining heat rates using the conduction rate equation,
q1 = q′′s
Δx
⋅1
2
q 2 = h(T∞ - Tm,n )
q3 =
q4 =
q5 =
Δx
⋅1
2
k(Tm - 1,n - Tm,n ) Δy
Δx
2
⋅1
k(Tm + 1,n - Tm,n ) Δy
⋅1
Δx
2
k(Tm ,n - 1 - Tm,n )
Δx ⋅ 1
Δy
Letting ∆x = ∆y, substituting these expressions into the energy balance, and rearranging yields
q′′Δx
hΔx 
hΔx

Tm - 1,n + Tm + 1,n + 2Tm,n - 1 -  4 +
Tm,n +
T∞ + s
=0

k 
k
k

<
PROBLEM 4.37
KNOWN: Control volume and nodal configuration in the vicinity of the interface between two
materials.
FIND: Expressions for control surface heat rates. Finite difference equation at node m,n.
∆x
SCHEMATIC:
∆y
•
(m,n+1)
q(m-1,n)→(m,n)
•
(m-1,n)
Material A
kA
q(m,n+1)→(m,n)
•
(m,n)
(m+1,n)
•
q(m+1,n)→(m,n)
q(m,n-1)→(m,n)
•
(m,n-1)
Material B
kB
ASSUMPTIONS: Steady-state, two-dimensional heat transfer, no heat generation, negligible contact
resistance.
ANALYSIS: Conduction from Node (m,n+1) to Node (m,n) occurs exclusively in Material A.
Therefore,
q( m,n +1)→( m,n ) kA L
=
∆x
Tm,n +1 − Tm,n 
∆y 
Likewise for conduction from Node (m,n-1) to Node (m,n),
∆x
Tm,n −1 − Tm,n 
q( m,n −1)→( m,n ) k B L
=
∆y 
<
<
Conduction from Node (m-1,n) to Node (m,n) occurs in both Material A and Material B. In Material A,
∆y / 2
Tm −1,n − Tm,n 
=
qA( m −1,n )→( m,n ) kA L
∆x 
Likewise for conduction in Material B,
∆y / 2
Tm −1,n − Tm,n 
=
qB( m −1,n )→( m,n ) k B L
∆x 
For both materials,
q( m −1,n )→( m,n ) qA( m −1,n )→( m,n ) + qB( m −1,n )→( m,n )
=
∆y / 2
∆y / 2
Tm −1,n − Tm,n  + k B L
Tm −1,n − Tm,n 
∆x 
∆x 
∆y / 2
Tm −1,n − Tm,n 
= ( kA + k B ) L
∆x 
= kA L
<
Similarly for conduction from Node (m+1,n) to (m,n),
q( m +1,n )→( m,n ) =
( kA + k B ) L
∆y / 2
Tm +1,n − Tm,n 
∆x 
<
Continued...
PROBLEM 4.37 (Cont.)
An energy balance on node m,n yields
q( m −1,n )→( m,n ) + q( m +1,n )→( m,n ) + q( m,n −1)→( m,n ) + q( m,n +1)→( m,n ) =
0
or
kA
∆x
∆x
∆y / 2
Tm,n +1 − Tm,n  + k B
Tm,n −1 − Tm,n  + ( k A + k B )
Tm −1,n + Tm +1,n − 2Tm,n  =
0
∆y
∆y
∆x 
COMMENTS: How would you modify the analysis if the contact resistance is significant?
<
PROBLEM 4.38
KNOWN: Conduction in a one-dimensional (radial) cylindrical coordinate system with volumetric
generation.
FIND: Finite-difference equation for (a) Interior node, m, and (b) Surface node, n, with convection.
SCHEMATIC:
(a) Interior node, m
(b) Surface node with convection, n
ASSUMPTIONS: (1) Steady-state, one-dimensional (radial) conduction in cylindrical coordinates, (2)
Constant properties.
ANALYSIS: (a) The network has nodes spaced at equal ∆r increments with m = 0 at the center; hence, r
V 2π r ⋅ ∆r=
⋅  2π ( m∆r ) ∆r ⋅ . The energy balance is
= m∆r (or n∆r). The control volume is =
 =0
E in + E g = q a + q b + qV
  ∆r   T
  ∆r   T
-T
-T
k  2π  r -    m-1 m + k  2π  r+    m+1 m + q  2π ( m∆r ) ∆r  =
0.
2 
∆r
∆r
 
  2 
Recognizing that r = m∆r, canceling like terms, and regrouping find
 ∆r 2
1
1
qm


+
+
=
m
T
m+
T
2mT
0.
m

 m-1 
 m+1

2

2
k
V 2π r ⋅ ( ∆r/2 ) ⋅ . The energy balance is
(b) The control volume for the surface node is =
<

Use Fourier’s law to express qd and Newton’s law of cooling for qconv
E in + E g = q d + q conv + qV=0.
to obtain
  ∆r   T - T
∆r 

k  2π  r -    n-1 n + h [ 2π r ] ( T∞ - Tn ) + q  2π ( n∆r )   =
0.
2 
2 
∆r

 
Let r = n∆r, cancel like terms and regroup to find
 ∆r 2 hn∆r

1
1  hn∆r 
qn

+
+
Tn-1   n
Tn +
T∞ =
0.
 n --2 
2 
k 
2k
k


<
COMMENTS: (1) Note that when m or n becomes very large compared to ½, the finite-difference
equation becomes independent of m or n. Then the cylindrical system approximates a rectangular one.
(2) The finite-difference equation for the center node (m = 0) needs to be treated as a special case. The
control volume is
V
= π ( ∆r / 2 )  and the energy balance is
2
 ∆r

∆r  T − T

 = k 2π    1 0 + q π     = 0.
E in + E g = q a + qV
  2   ∆r
 


  2  
2
Regrouping, the finite-difference equation is − To + T1 +
2
q ∆r
4k
=
0.
PROBLEM 4.39
KNOWN: Two-dimensional cylindrical configuration with prescribed radial (∆r) and angular (∆φ)
spacings of nodes.
FIND: Finite-difference equations for nodes 2, 3 and 1.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in cylindrical
coordinates (r,φ), (3) Constant properties.
ANALYSIS: The method of solution is to define the appropriate control volume for each node, to
identify relevant processes and then to perform an energy balance.
(a) Node 2. This is an interior node with control volume as shown above. The energy balance is
E in = q′a + q′b + q′c + q′d = 0. Using Fourier’s law for each process, find
( T3 − T2 ) +

3   (T − T )
k   ri + ∆r  ∆φ  5 2 + k ( ∆r )
2  
∆r
( ri + ∆r ) ∆φ

( T − T ) 0.

1   (T − T )
+ k   ri + ∆r  ∆φ  i 2 + k ( ∆r ) 1 2 =
2  
∆r
( ri + ∆r ) ∆φ

Canceling terms and regrouping yields,
2
2


∆r )
∆r )
1
3 
1
(
(

 T2 +  ri + ∆r  T5 +
−2 ( ri + ∆r ) +
0.
( T3 + T1 ) +  ri + ∆r  Ti =
2
2
2 
2 



( ri + ∆r )( ∆φ )
( ∆φ ) ( ri + ∆r ) 
(b) Node 3. The adiabatic surface behaves as a symmetry surface. We can utilize the result of Part (a) to
write the finite-difference equation by inspection as

−2 ( ri + ∆r ) +

2
2 ( ∆r )
3 
1 
( ∆r )2 1 


⋅ T2 +  ri + ∆r  Ti =
T3 +  ri + ∆r  T6 +
0.
2 
2 


( ri + ∆r )( ∆φ )2
( ∆φ )2 ( ri + ∆r ) 
0. Substituting,
(c) Node 1. The energy balance is q′a + q′b + q′c + q′d =
( T2 − T1 ) +

3  ∆φ  ( T4 − T1 )
k   ri + ∆r 
+ k ( ∆r )

2  2 
∆r
( ri + ∆r ) ∆φ


1  ∆φ  ( Ti − T1 )
+ k   ri + ∆r 
+ h ( ∆r )( T∞ − T1 ) =
0
2  2  ∆r

This expression could now be rearranged.
<
PROBLEM 4.40
KNOWN: Heat generation and thermal boundary conditions of bus bar. Finite-difference grid.
FIND: Finite-difference equations for selected nodes.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties.
ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)(∆y/2)⋅1, find the FDE for
node 1,
k ( ∆y/2 ⋅1)
To − T1
 ∆x 
+ hu 
⋅1 ( T∞ − T1 ) +
( T2 − T1 )
R ′′t,c / ( ∆y/2 )1
∆x
 2 
k ( ∆x/2 ⋅1)
+
0
( T6 − T1 ) + q ( ∆x/2 )( ∆y/2 )1 =
∆y
∆x/kR ′′t,c To + ( h u ∆x/k ) T∞ + T2 + T6
(
)
(
)
2
+ q ( ∆x ) / 2k −  ∆x/kR ′′t,c + ( h u ∆x/k ) + 2  T1 =
0.


<
(b) Performing an energy balance on the control volume, (∆x)(∆y/2)⋅1, find the FDE for node 13,
h l ( ∆x ⋅1)( T∞ − T13 ) + ( k/∆x )( ∆y/2 ⋅1)( T12 − T13 )
+ ( k/∆y )( ∆x ⋅1)( T8 − T13 ) + ( k/∆x )( ∆y/2 ⋅1)( T14 − T13 ) + q ( ∆x ⋅ ∆y/2 ⋅1) =0
0.
( h l∆x/k ) T∞ + 1/ 2 ( T12 + 2T8 + T14 ) + q ( ∆x )2 / 2k − ( h l∆x/k + 2 ) T13 =
<
COMMENTS: For fixed To and T∞, the relative amounts of heat transfer to the air and heat sink are
determined by the values of h and R ′′t,c .
PROBLEM 4.41
KNOWN: Nodal point configurations corresponding to a diagonal surface boundary subjected to a
convection process and to the tip of a machine tool subjected to constant heat flux and convection
cooling.
FIND: Finite-difference equations for the node m,n in the two situations shown.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.
ANALYSIS: (a) The control volume about node m,n has triangular shape with sides ∆x and ∆y while the
diagonal (surface) length is 2 ∆x. The heat rates associated with the control volume are due to
conduction, q1 and q2, and to convection, qc. Performing an energy balance, find
E in -=
E out 0
q1 + q 2=
+ qc 0
Tm,n-1 -Tm,n
Tm+1,n Tm,n
k ( ∆x ⋅1)
+ k ( ∆y ⋅1)
+h
∆y
∆x
(
)(
)
2 ∆x ⋅1 T∞ - Tm,n =0.
Note that we have considered the solid to have unit depth normal to the page. Recognizing that ∆x = ∆y,
dividing each term by k and regrouping, find
Tm,n-1 + Tm+1,n + 2 ⋅
h∆x
h∆x 

T∞ -  2 + 2 ⋅
Tm,n =
0.
k
k 

<
(b) The control volume about node m,n has triangular shape with sides ∆x/2 and ∆y/2 while the lower
diagonal surface length is
2 ( ∆x/2 ) . The heat rates associated with the control volume are due to the
constant heat flux, qa, to conduction, qb, and to the convection process, qc. Perform an energy balance,
+ qc 0
E in −=
E out 0
q a + q b=
T
∆x 
 ∆x 
 ∆y  m+1,n − Tm,n

+ h ⋅  2 ⋅  T∞ − Tm,n =
q′′o ⋅  ⋅1 + k ⋅  ⋅1
0.
∆x
2 
 2 
 2 

(
)
Recognizing that ∆x = ∆y, dividing each term by k/2 and regrouping, find
Tm+1,n + 2 ⋅
h∆x
∆x 
h∆x 
⋅ T∞ + q′′o ⋅
− 1 + 2 ⋅
0.
 Tm,n =
k
k 
k 
COMMENTS: Note the appearance of the term h∆x/k in both results, which is a dimensionless
parameter (the Biot number) characterizing the relative effects of convection and conduction.
<
PROBLEM 4.42
KNOWN: Cutting tool with tip angle of 60°. Upper surface exposed to constant heat flux, diagonal
surface exposed to convection with known environment temperature and heat transfer coefficient.
FIND: Derive nodal finite difference equation for node (m,n).
SCHEMATIC:
qo“
qa
m,n
m+1,n
qb
qc
Δy
60°
T∞, h
Δx
m+1,n-1
ASSUMPTIONS: (1) Steady-state conditions. (2) Uniform properties.
ANALYSIS: It is convenient to choose ∆y/∆x = tan60° = 3, so that the control volume about node m,n
has a triangular shape as shown in the schematic. The heat rates associated with the control volume are
due to the constant heat flux on the upper surface, qa, to conduction from the right, qb, and to the
convection process, qc. Performing an energy balance,
= 0, qa + qb +=
Ein − Eout
qc 0
(Tm +1, n − Tm, n )
∆x
∆y
∆x
qo′′ ⋅ ⋅ 1 + k ⋅
⋅1 ⋅
+ h⋅
⋅ 1 ⋅ (T∞ − Tm, n ) =0
2
2
cos(60°)
∆x
which can be rewritten as
Tm +1, n +
4 h∆x
∆x 
4 h∆x 
− 1 +
T∞ + qo′′
0
 Tm, n =
3 k
3k 
3 k 
<
COMMENTS: Note the appearance of the term h∆x/k, which is a dimensionless parameter (the Biot
number) characterizing the relative effects of convection and conduction.
PROBLEM 4.43
KNOWN: Two-dimensional grid for a system with no internal volumetric generation.
FIND: Expression for heat rate per unit length normal to page crossing the isothermal boundary.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) Constant
properties.
ANALYSIS: Identify the surface nodes (Ts) and draw control volumes about these nodes.
Since there is no heat transfer in the direction parallel to the isothermal surfaces, the heat rate out
of the constant temperature surface boundary is
q′ = q′a + q′b + q′c + q′d + q′e + q′f
For each q′i , use Fourier’s law and pay particular attention to the manner in which the crosssectional area and gradients are specified.
T1 − Ts
T − Ts
T −T
+ k ( ∆y ) 2
+ k ( ∆y ) 3 s
∆x
∆x
∆x
T5 − Ts
T6 − Ts
T7 − Ts
+ k ( ∆x )
+ k ( ∆x )
+ k ( ∆x/2 )
∆y
∆y
∆y
q′ = k ( ∆y/2 )
Regrouping with ∆x = ∆y, find
q′ k [ 0.5T1 + T2 + T3 + T5 + T6 + 0.5T7 − 5Ts ].
=
<
COMMENTS: Looking at the corner node, it is important to recognize the areas associated
with q′c and q′d (∆y and ∆x, respectively).
PROBLEM 4.44
KNOWN: One-dimensional fin of uniform cross section insulated at one end with prescribed base
temperature, convection process on surface, and thermal conductivity.
FIND: Finite-difference equation for these nodes: (a) Interior node, m and (b) Node at end of fin, n,
where x = L.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction.
ANALYSIS: (a) The control volume about node m is shown in the schematic; the node spacing and
control volume length in the x direction are both ∆x. The uniform cross-sectional area and fin perimeter
are Ac and P, respectively. The heat transfer process on the control surfaces, q1 and q2, represent
conduction while qc is the convection heat transfer rate between the fin and ambient fluid. Performing an
energy balance, find
E in -=
E out 0
q1 + q 2=
+ qc 0
Tm-1 -Tm
Tm+1 Tm
+ kA c
+ hP∆x ( T∞ - Tm ) =
kA c
0.
∆x
∆x
Multiply the expression by ∆x/kAc and regroup to obtain
Tm-1 + Tm+1 +


hP
hP
⋅ ∆x 2T∞ -  2 +
∆x 2  Tm = 0
kA c
kA c


1<m<n
<
Considering now the special node m = 1, then the m-1 node is Tb, the base temperature. The finitedifference equation would be
Tb + T2 +


hP
hP
0
∆x 2T∞ −  2 +
∆x 2  T1 =
kA c
kA c


m=1
<
(b) The control volume of length ∆x/2 about node n is shown in the schematic. Performing an energy
balance,
E in -=
E out 0
q3 + q 4=
+ qc 0
∆x
Tn-1 - Tn
+ 0 + hP
kA c
( T∞ - Tn ) =0.
∆x
2
Note that q4 = 0 since the end (x = L) is insulated. Multiplying by ∆x/kAc and regrouping,
 hP ∆x 2 
hP ∆x 2
Tn-1 +
T∞ - 
0.
⋅
⋅
+ 1 Tn =
kAc 2
 kA c 2

<
COMMENTS: The value of ∆x will be determined by the selection of n; that is, ∆x = L/n. Note that the
grouping, hP/kAc, appears in the finite-difference and differential forms of the energy balance.
PROBLEM 4.45
PROBLEM 4.46
PROBLEM 4.46 (Cont.)
PROBLEM 4.46 (Cont.)
PROBLEM 4.46 (Cont.)
PROBLEM 4.47
KNOWN: Square channels of known dimension, evenly spaced along centerline of plate of known
thickness and thermal conductivity. Hot and cold fluids with known temperatures and heat transfer
coefficients flowing through alternate channels. N = 50 channels. Use ∆x = ∆y = 5 mm.
FIND: Maximum and minimum temperatures within plate. Heat transfer rate per unit plate length from
hot to cold fluid.
SCHEMATIC:
•19
•20
•21
•22
•23
•24
Δy
•13
•14
•15
•16
•17
•18
•7
•8
•9
•10
•11
• 12
•3
•4
•5
•6
T∞,h, h
1•
•2
y, n
T∞,c, h
x, m
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties, (3) No internal heat generation,
(4) Two-dimensional conduction.
ANALYSIS: The discretized solution domain is shown in the schematic. The finite difference equations
are:

T4 − T1
∆y T2 − T1
+k
+ h 
∆y
2 ∆x


T −T
∆y T1 − T2
Node 2 : k ∆x 5 2 + k
+ h 
2 ∆x
∆y

T −T
T −T
Node 3 : k ∆x 8 3 + k ∆y 4 3 + h
∆y
∆x
T −T
T − T6
Node 6 : k ∆x 11 6 + k ∆y 5
+h
∆y
∆x
Node 1: k ∆x
(
(
2∆x 
0
 (T∞ , h − T1 ) =
2 
2∆x 
0
 (T∞ ,c − T2 ) =
2 
)
2∆x (T∞ , h − T3 ) =
0
)
2∆x (T∞ ,c − T6 ) =
0
 2∆x 
T −T
∆x T13 − T7
0
+ k ∆y 8 7 + h 
 (T∞ , h − T7 ) =
2 ∆y
∆x
 2 
 2∆x 
T −T
∆x T18 − T12
Node 12 : k
0
+ k ∆y 11 12 + h 
 (T∞ ,c − T12 ) =
2
∆y
∆x
 2 
Node 7 : k
Continued…
PROBLEM 4.47 (Cont.)
T −T
∆x T19 − T13
∆x T7 − T13
+ k ∆y 14 13 + k
=
0
2
∆y
∆x
2 ∆y
T −T
∆x T24 − T18
∆x T12 − T18
+k
+ k ∆y 17 18 =
Node 18 : k
0
∆y
∆y
∆x
2
2
∆y T20 − T19
∆x T13 − T19
Node 19 : k
+k
=
0
2
∆x
2
∆y
∆x T18 − T24
∆y T23 − T24
Node 24 : k
+k
=
0
2
∆y
2
∆x
Node 13 : k
Tm , n +1 ---Tm , n
T
T
T 1 Tm , n
T
T
+ k ∆y m +1, n m , n + k ∆x m , n -+ k ∆y m 1, n m , n =
0
∆y
∆x
∆y
∆x
Tm , n
T 1 Tm , n
∆y Tm +1, n --∆y Tm 1, n Tm , n
Nodes 20-23: k
0
+ k ∆x m , n -+k
=
2
2
∆x
∆y
∆x
Nodes 4, 5, 8-11, 14-17 : k ∆x
Solving the preceding equations for k = 14 W/m⋅K, h = 40 W/m2⋅K, T∞,h = 120°C, T∞,c = 20°C, ∆x = ∆y =
5 mm yields the nodal temperatures shown below.
T19 = 70.48°C
T13 = 70.58°C
T7 = 70.96°C
T20 = 70.38°C
T14 = 70.44°C
T8 = 70.65°C
T3 = 70.99°C
T21 = 70.14°C
T15 = 70.16°C
T9 = 70.23°C
T4 = 70.33°C
T1 = 70.41°C
T22 = 69.86°C
T16 = 69.84°C
T10 = 69.77°C
T5 = 69.67°C
T2 = 69.59°C
T23 = 69.62°C
T17 = 69.56°C
T11 = 69.35°C
T6 = 69.01°C
T24 = 69.52°C
T18 = 69.42°C
T12 = 69.04°C
Hence the maximum and minimum temperatures are:
Tmax = T3 = 70.99°C
Tmin = T6 = 69.01°C
<
<
We elect to calculate the heat rate per unit length by first evaluating the convection rate from a hot
channel to the solid along the edge with nodes 1, 3, and 7.
 2∆x 
′
=
qedge
h 
 (T∞ , h − T1 ) + h
 2 
(
 2∆x 
2∆x (T∞ , h − T3 ) + h 
(T∞ , h − T7 ) 27.8 W/m
=
 2 
)
Under steady-state conditions, with the plate insulated, all of the heat that is transferred from the hot fluid
to the solid must then enter the cold fluid. For N = 50 channels, the total heat transfer rate from hot to cold
fluid is
′ =
′ N /2=
qtot
4qedge
4 × 27.8 W/m × 50 / 2 =
2.78 kW/m
<
where the factor of 4 accounts for the fact that each channel has four edges.
′ may be determined by
COMMENTS: The heat rate can be evaluated in several ways. For example, qedge
′ may be evaluated by calculating the
calculating the convection loss to the cold fluid. Alternatively, qedge
conduction rate perpendicular to the vertical centerline of the domain.
PROBLEM 4.48
KNOWN: Volumetric heat generation in a rectangular rod of uniform surface temperature.
FIND: (a) Temperature distribution in the rod, and (b) With boundary conditions unchanged, heat
generation rate causing the midpoint temperature to reach 600 K.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniform
volumetric heat generation.
ANALYSIS: (a) From symmetry it follows that six unknown temperatures must be determined. Since
all nodes are interior ones, the finite-difference equations may be obtained from Eq. 4.35 written in the
form
=
Ti 1 4 ∑ Tneighbors + 1 4 ( q ( ∆x∆y ) k ) .
With q ( ∆x∆y ) 4k = 62.5 K, the system of finite-difference equations is
=
T1 0.25 ( Ts + T2 + T4 + Ts ) + 15.625
(1)
=
T2 0.25 ( Ts + T3 + T5 + T1 ) + 15.625
(2)
=
T3 0.25 ( Ts + T2 + T6 + T2 ) + 15.625
(3)
=
T4 0.25 ( T1 + T5 + T1 + Ts ) + 15.625
(4)
=
T5 0.25 ( T2 + T6 + T2 + T4 ) + 15.625
(5)
=
T6 0.25 ( T3 + T5 + T3 + T5 ) + 15.625
(6)
With Ts = 300 K, the set of equations was written directly into the IHT workspace and solved for the
nodal temperatures,
T1
T2
T3
T4
T5
T6 (K)
348.6
368.9
374.6
362.4
390.2
398.0
<
(b) With the boundary conditions unchanged, the q required for T6 = 600 K can be found using the same
set of equations in the IHT workspace, but with these changes: (1) replace the last term on the RHS
(15.625) of Eqs. (1-6) by q (∆x∆y)/4k = (0.005 m)2 q /4×20 W/m⋅K = 3.125 × 10-7 q and (2) set T6 =
600 K. The set of equations has 6 unknown, five nodal temperatures plus q . Solving find
=
q 1.53 ×108 W m3
<
PROBLEM 4.49
PROBLEM 4.49 (Cont.)
PROBLEM 4.50
KNOWN: Steady-state temperatures (K) at three nodes of a long rectangular bar.
FIND: (a) Temperatures at remaining nodes and (b) heat transfer per unit length from the bar using nodal

temperatures; compare with result calculated using knowledge of q.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.
ANALYSIS: (a) The finite-difference equations for the nodes (1,2,3,A,B,C) can be written by inspection
using Eq. 4.35 and recognizing that the adiabatic boundary can be represented by a symmetry plane.
2
7
3
q ∆x 2 5 ×10 W/m ( 0.005m )
Tneighbors − =
4Ti + q ∆x 2 / k 0 =
and
= 62.5K.
∑
k
20 W/m ⋅ K
Node A (to find T2):
2T2 + 2TB − 4TA + q ∆x 2 / k =0
<
Node 3 (to find T3):
1
T2 =( −2 × 374.6 + 4 × 398.0 − 62.5 ) K =
390.2K
2
Tc + T2 + TB + 300K − 4T3 + q ∆x 2 / k =0
<
Node 1 (to find T1):
1
( 348.5 + 390.2 + 374.6 + 300 + 62.5) K= 369.0K
4
300 + 2TC + T2 − 4T1 + q ∆x 2 / k =0
T3=
=
T1
1
=
( 300 + 2 × 348.5 + 390.2 + 62.5
) 362.4K
4
<
(b) The heat rate out of the bar is determined by calculating the heat rate out of each control volume
around the 300 K nodes. Consider the node in the upper left-hand corner; from an energy balance
E in − E out + E g = 0
Hence, for the entire bar

q′a = q′a,in + E g where E g = qV.
q′bar = q′a + q′b + q′c + q′d + q′e + q′f , or
or
T − 300
 ∆y T1 − 300   ∆x ∆y   
 ∆x
    ∆x ∆y  
=
q′
+q
⋅
+  k∆y C
+ q 
⋅ ∆y   +  q 
⋅
+
k 2


bar
∆x
∆x
 2 2  a 
 2
  b   2 2   c

TC − 300
T3 − 300
∆y  
∆y  


 ∆x TB − 300


 ∆x ∆y  
+ q 
⋅
 .
 k∆x ∆y + q  ∆x ⋅ 2   +  k∆x ∆y + q  ∆x ⋅ 2   +  k 2
∆y

 d 

 e 
 2 2   f

Substituting numerical values, find q′bar = 7,502.5 W/m. From an overall energy balance on the bar,
  = q ( 3∆x ⋅ 2∆y ) = 5 ×107 W/m3 × 6 ( 0.005m )2 = 7,500 W/m.
q′bar = E ′g = qV/
As expected, the results of the two methods agree. Why must that be?
<
PROBLEM 4.51
KNOWN: Dimensions and thermal conductivity distribution within a two-dimensional solid. Applied
boundary conditions.
FIND: (a) Spatially-averaged thermal conductivity and heat rate per unit length based upon this value,
(b) Heat rate per unit length for case 1 boundary conditions and comparison to estimated heat rate per unit
length based upon the spatially-averaged thermal conductivity, (c) Heat rate per unit length for case 2
boundary conditions and comparison to estimated heat rate per unit length based upon the spatiallyaveraged thermal conductivity.
ASSUMPTIONS: Steady-state, one-dimensional heat transfer.
ANALYSIS: (a) The thermal conductivity varies only in the x-direction. Hence,
L
L
∫(
)
1
1
2
k =
k ( x )dx =a + bx 3/ 2 dx =
a + bL3/ 2
L
L
5
∫
=x 0 =x 0
<
W
2
W
W
3/ 2
28
= 20
+ × 7070 5/2
× (0.02m)
=
m⋅K 5
m⋅K
m ⋅K
Using this value, the heat rate per unit length is
q ' = kL( ∆T ) / L = 28
W
× 50°C = 1400W/m
m⋅K
<
(b) The nodal network is shown below. Note that the heat transfer is one-dimensional.
∆x
q´(m-1)-m
q´1-2
•
1
•
2
•
•
•
• •
m
m-1
•
•
•
•
11
m+1
L
For any control surface, Eq. 4.46 may be combined with Fourier’s law and written as
=
q'( m −1)−m
Tm −1 − Tm
Tm −1 − Tm
=
'
 ∆x / 2 ∆x / 2 
R tot
+


 Lkm −1 Lkm 
(1)
where the thermal conductivities, km-1 and km are evaluated at the left (m – 1) and right (m) nodes,
respectively. At steady state, the heat rate per unit length is constant. Hence, we may write Eqn. (1) for
each pair of nodal points from m = 1 to m = 11 using ∆x = 2 mm. The resulting heat rate per unit length is
found by solving the 11 simultaneous equations for q′ and the temperatures Tm for 2 ≤ m ≤ 10 yielding
q′ = 1339 W/m
<
Continued…
PROBLEM 4.51 (Cont.)
The predicted heat rate pure unit length is smaller than that of part (a).
(c) When the applied boundary conditions are changed to those of case 2, we may simply evaluate the
heat transfer from the hot surface to the cool surface by evaluating the heat transfer in 11 different lanes
and summing the results. For the interior lanes the width is ∆x resulting in
( ∆T )km
∆x for 2 ≤ m ≤ 10
L
=
qm'
where ∆T is the overall temperature difference across the domain. The thermal conductivities are
evaluated at the locations of the nodes. For the lanes adjacent to the adiabatic boundaries,
q1' = k1∆T ( ∆x / 2) / L
and
'
q11
k11∆T ( ∆x / 2) / L
and we evaluate the thermal conductivities k1 and k11 at the nodal points, x = 0 and 20 mm, respectively.
The heat rate per unit length of the object is
q' =
11
∑ qm'
<
or q′ = 1401 W/m
m =1
The heat rate per unit length is nearly identical to that of part (a).
COMMENTS: (1) The agreement between the results of parts (a) and (c) is expected since
L
q' =
∫
L
q "( x )dx =
∫ ( k ( x)∆T / L ) dx = k ( L / L ) ∆T . The minor difference is due to the evaluation of the
=
x 0=
x 0
thermal conductivity for each lane at the nodal point. The answers would become exactly the same as the
spatial resolution of the numerical solution is increased. (2) In part (b) heat transfer is in the x-direction,
the same direction in which thermal conductivity varies. This reduces heat transfer rates relative to the
value calculated in parts (a) and (c). This is because the resistance expressed in Eq. (1) is composed of
two values in series. The total resistance will be dominated by the higher of the two individual
resistances. (3) Temperatures calculated for case 1 and heat rates in each lane for case 2 are shown in the
table below.
Node or Lane
Temperature, °C (case 1)
Heat rate per unit length, W/m (case 2)
1
2
3
4
5
6
7
8
9
10
11
100.00
93.41
87.09
81.14
75.60
70.45
65.69
61.30
57.24
53.48
50.00
50
103.2
108.9
116.4
125.3
135.3
146.5
158.6
171.5
185.4
99.99
1401
PROBLEM 4.52
KNOWN: Steady-state temperatures at selected nodal points of the symmetrical section of a flow
channel with uniform internal volumetric generation of heat. Inner and outer surfaces of channel
experience convection.
FIND: (a) Temperatures at nodes 1, 4, 7, and 9, (b) Heat rate per unit length (W/m) from the outer
surface A to the adjacent fluid, (c) Heat rate per unit length (W/m) from the inner fluid to surface B, and
(d) Verify that results are consistent with an overall energy balance.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) The nodal finite-difference equations are obtained from energy balances on control
volumes about the nodes shown in the schematics below.
Node 1
Node 4
Continued …
PROBLEM 4.52 (Cont.)
<
Node 7
<
Node 9
<
(b) The heat rate per unit length from the outer surface A to the adjacent fluid,
convection heat rates from the outer surfaces of nodes 7, 8, 9 and 10.
is the sum of the
Continued …
PROBLEM 4.52 (Cont.)
<
q′A = 1117 W / m
(c) The heat rate per unit length from the inner fluid to the surface B, q ′B , is the sum of the convection
heat rates from the inner surfaces of nodes 2, 4, 5 and 6.
(
)
(
)
(
)
(
)
=
q ′B h i ( ∆y / 2 ) T∞ ,i − T2 + ( ∆y / 2 + ∆x / 2 ) T∞ ,i − T4 + ∆x T∞ ,i − T5 + ( ∆x / 2 ) T∞ ,i − T6 
=
q′B 500 W / m 2 ⋅ K ( 25 / 2 )( 50 − 95.47 ) + ( 25 / 2 + 25 / 2 )( 50 − 94.50 )
+25 ( 50 − 79.79 ) + ( 25 / 2 )( 50 − 77.29 )  ×10−3 m ⋅ K
q′B = −1383 W / m
<
(d) From an overall energy balance on the section, we see that our results are consistent since the
conservation of energy requirement is satisfied.
−q′A + q′B + E ′gen =
E ′in − E ′out + E ′gen =
(−1117 − 1383 + 2500)W / m =
0
∀ ′ =q∀ ∀′ =106 W / m3 [ 25 × 50 + 25 × 50] ×10−6 m 2 =2500 W / m
where E
gen
COMMENTS: The nodal finite-difference equations for the four nodes can be obtained by using IHT
Tool Finite-Difference Equations | Two-Dimensional | Steady-state. Options are provided to build the
FDEs for interior, corner and surface nodal arrangements including convection and internal generation.
The IHT code lines for the FDEs are shown below.
/* Node 1: interior node; e, w, n, s labeled 2, 2, 3, 3. */
0.0 = fd_2d_int(T1,T2,T2,T3,T3,k,qdot,deltax,deltay)
/* Node 4: internal corner node, e-n orientation; e, w, n, s labeled 5, 3, 2, 8. */
0.0 = fd_2d_ic_en(T4,T5,T3,T2,T8,k,qdot,deltax,deltay,Tinfi,hi,q″a4
q″a4 = 0
// Applied heat flux, W/m^2; zero flux shown
/* Node 7: plane surface node, s-orientation; e, w, n labeled 8, 8, 3. */
0.0 = fd_2d_psur_s(T7,T8,T8,T3,k,qdot,deltax,deltay,Tinfo,ho,q″a7
q″a7=0
// Applied heat flux, W/m^2; zero flux shown
/* Node 9: plane surface node, s-orientation; e, w, n labeled 10, 8, 5. */
0.0 = fd_2d_psur_s(T9, T10, T8, T5,k,qdot,deltax,deltay,Tinfo,ho,q″a9
q″a9 = 0
// Applied heat flux, W/m^2; zero flux shown
PROBLEM 4.53
PROBLEM 4.53 (Cont.)
PROBLEM 4.54
KNOWN: Dimensions of a two-dimensional object with isothermal and adiabatic boundaries.
FIND: Conduction heat transfer rate per unit depth from the hot surface to the cold surface.
SCHEMATIC:
Tc = 0°C
ro = 50 mm
y
ri = 30 mm
H = 30 mm
W = 20 mm
x
Th = 20°C
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) No internal generation, (4)
Two-dimensional conduction.
ANALYSIS: We may combine heat fluxes determined from Fourier’s law with expressions for the size
of the control surfaces of the various control volumes to determine the heat rate per unit depth into each
control volume in the discretized domain, which is shown on the next page. Application of conservation
of energy for each control volume yields the expression E in = 0 . Note that energy balances for nodes 10,
11, and 12 are included in both the rectangular and cylindrical sub-domains. These energy balances
couple the solutions together.
Rectangular Sub-Domain. For the rectangular sub-domain, application of Fourier’s law in Cartesian
coordinates along with conservation of energy yields the following finite difference equations.
Nodes 1, 2 and 3: T1 = T2 = T3 = Th = 20°C.
Node 4:
k
(T1 − T4 ) ∆x
(T − T )
(T − T ) ∆x
+ k 5 4 ∆y + k 7 4
=0
∆y
2
∆x
∆y
2
Node 5:
k
(T2 − T5 )
(T − T )
(T − T )
(T − T )
∆x + k 6 5 ∆y + k 8 5 ∆x + k 4 5 ∆y =0
∆y
∆x
∆y
∆x
Node 6:
k
(T3 − T6 ) ∆x
(T − T )
(T − T ) ∆x
+ k 5 6 ∆y + k 9 6
=0
∆y
∆x
∆y
2
2
Node 7:
k
(T4 − T7 ) ∆x
(T − T )
(T − T ) ∆x
+ k 8 7 ∆y + k 10 7
=0
∆y
2
∆x
∆y
2
Continued…
PROBLEM 4.54 (Cont.)
Node 8:
k
(T5 − T8 ) ∆x
(T − T )
(T − T )
(T − T )
+ k 9 8 ∆y + k 11 8 ∆x + k 7 8 ∆y =
0
∆y
2
∆x
∆y
∆x
Node 9:
k
(T6 − T9 ) ∆x
(T − T )
(T − T ) ∆x
+ k 8 9 ∆y + k 12 9
=0
∆y
2
∆x
∆y
2
Node 10:
k
(T7 − T10 ) ∆x
(T − T ) ∆y
+ k 11 10
− q1' =
0
∆y
2
∆x
2
Node 11:
k
(T8 − T11 )
(T − T ) ∆y
(T − T ) ∆y
∆x + k 10 11
+ k 12 11
− q2' =0
∆y
∆x
2
∆x
2
Node 12:
k
(T9 − T12 ) ∆x
(T − T ) ∆y
+ k 11 12
− q3' =
0
∆y
2
∆x
2
• 22
•19
• 20
•16
•21
•17
• 18
•
r = 35 mm
•
13
•15
14
• 23
• 24
Tc = 0°C
r = 30 mm
r = 45 mm
r = 50 mm
10 •
y
•
•
•
•
•
•
•
•
•
•
•
7
4
1
q1´ q2´ 11 q3´ 12
8
5
2
9
∆x = ∆y = ∆r = 10 mm
6
x
3
Th = 20°C
Continued…
PROBLEM 4.54 (Cont.)
Cylindrical Sub-Domain. We begin by recalling that Fourier’s law for the cylindrical coordinate system
yields
∂T
∆T
k ∂T
k ∆T
qr" = − k
≈ −k
; qφ" = −
≈−
∂r
∆r
r ∂φ
r ∆φ
and the areas through which conduction occurs in the radial direction increase as the radius increases.
Nodes 10, 13, 16, 19, 22, 23 and 24: T10 = T13 = T16 = T19 = T22 = T23 = T24 = Tc = 0°C
Node 11: q2' + k
(T12 − T11 )
∆φ
(T − T )
(T − T )
∆φ
⋅ 3.5∆r ⋅
+ k 14 11 ⋅ ∆r + k 10 11 ⋅ 4.5∆r ⋅
=0
∆r
2
4 ∆r∆φ
∆r
2
Node 12: q3' + k
(T11 − T12 )
∆φ
(T − T ) ∆r
⋅ 3.5∆r ⋅
+ k 15 12 ⋅
=0
∆r
2
3∆r∆φ
2
Node 14: k
(T11 − T14 )
(T − T )
(T − T )
(T − T )
⋅ ∆r + k 15 14 ⋅ 3.5∆r∆φ + k 17 14 ∆r + k 13 14 ⋅ 4.5∆r∆φ =0
4 ∆r∆φ
∆r
4 ∆r∆φ
∆r
Node 15: k
(T12 − T15 ) ∆r
(T − T )
(T − T ) ∆r
+ k 14 15 ⋅ 3.5∆r∆φ + k 18 15
=0
∆r
3∆r∆φ 2
3∆r∆φ 2
Node 17: k
(T14 − T17 )
(T − T )
(T − T )
(T − T )
⋅ ∆r + k 18 17 ⋅ 3.5∆r∆φ + k 20 17 ∆r + k 16 17 ⋅ 4.5∆r∆φ =0
4 ∆r∆φ
∆r
4 ∆r∆φ
∆r
Node 18: k
(T15 − T18 ) ∆r
(T − T )
(T − T ) ∆r
+ k 17 18 ⋅ 3.5∆r∆φ + k 21 18
=0
3∆r∆φ 2
3∆r∆φ 2
∆r
Node 20: k
(T17 − T20 )
(T − T )
(T − T )
∆r + k 21 20 ⋅ 3.5∆r∆φ + k 23 20 ∆r =0
4 ∆r∆φ
∆r
4 ∆r∆φ
Node 21: k
(T18 − T21 ) ∆r
(T − T )
(T − T ) ∆r
+ k 20 21 ⋅ 3.5∆r∆φ + k 24 21
=0
∆r
3∆r∆φ 2
3∆r∆φ 2
Note that energy balances for nodes 10, 11 and 12 are included in both the rectangular and cylindrical
sub-domains. These energy balances couple the solutions for the two sub-domains together.
The preceding finite difference equations may be solved simultaneously with the IHT code provided in
the Comment yielding the following temperatures and q ' = 114.5 W/m .
The nodal temperatures are:
T1 = 20.00°C
T7 = 7.86°C
T13 = 0.00°C
T19 = 0.00°C
T2 = 20.00°C
T8 = 8.66°C
T14 = 1.04°C
T20 = 0.10°C
T3 = 20.00°C
T9 = 9.03°C
T15 = 1.59°C
T21 = 0.16°C
T4 = 14.11°C
T10 = 0.00°C
T16 = 0.00°C
T22 = 0.00°C
T5 = 14.29°C
T11 = 3.46°C
T17 = 0.33°C
T23 = 0.00°C
<
T6 = 14.41°C
T12 = 4.40°C
T18 = 0.54°C
T24 = 0.00°C
Continued…
PROBLEM 4.54 (Cont.)
COMMENTS: (1) The IHT code is listed below. For each control volume, we note that E in = 0 and ∆y =
∆x, yielding the following energy balances for all but the isothermal nodes.
// Input Parameters
Th = 20
Tc = 0
k = 10
deltaphi = pi/8
// Node Equations for Rectangle
//Node 1
T1 = Th
//Node 2
T2 = Th
//Node 3
T3 = Th
//Node 4
(T1 - T4)/2 + (T5 - T4) + (T7 - T4)/2 = 0
//Node 5
(T2 - T5) + (T6 - T5) + (T8 - T5) + (T4 - T5) = 0
//Node 6
(T3 - T6)/2 + (T5 - T6) + (T9 - T6)/2 = 0
//Node 7
(T4 - T7)/2 + (T8 - T7) + (T10 - T7)/2 = 0
//Node 8
(T5 - T8) + (T9 - T8) + (T11 - T8) + (T7 - T8) = 0
//Node 9
(T6 - T9)/2 + (T8 - T9) + (T12 - T9)/2 = 0
//Nodes Common to Both Sub-Domains (Rectangle)
//Node 10
(T7 - T10)/2 + (T11 - T10)/2 -qprime1/k = 0
//Node 11
(T8 - T11)+(T10 - T11)/2 + (T12 - T11)/2 - qprime2/k = 0
//Node 12
(T9 - T12)/2 + (T11 - T12)/2 - qprime3/k = 0
//Nodes Common to Both Sub-Domains (Cylindrical)
//Node 10
T10 = Tc
//Node 11
qprime2/k + (T12 - T11)*3.5*deltaphi/2 + (T14 - T11)/4/deltaphi + (T10 - T11)*4.5*deltaphi/2 = 0
//Node 12
qprime3/k + (T11 - T12)*3.5*deltaphi/2 + (T15 - T12)/3/deltaphi/2 = 0
// Node Equations for Cylindrical Region
//Node 13
T13 = Tc
//Node 14
(T11 - T14)/4/deltaphi + (T15 - T14)*3.5*deltaphi + (T17 - T14)/4/deltaphi + (T13 - T14)*4.5*deltaphi = 0
//Node 15
(T12 - T15)/3/deltaphi/2 + (T14 - T15)*3.5*deltaphi + (T18 - T15)/3/deltaphi/2 = 0
//Node 16
T16 = Tc
//Node 17
(T14 - T17)/4/deltaphi + (T18 - T17)*3.5*deltaphi + (T20 - T17)/4/deltaphi + (T16 - T17)*4.5*deltaphi = 0
//Node 18
(T15 - T18)/3/deltaphi/2 + (T17 - T18)*3.5*deltaphi + (T21 - T18)/3/deltaphi/2 = 0
Continued…
PROBLEM 4.54 (Cont.)
//Node 19
T19 = Tc
//Node 20
(T17 - T20)/4/deltaphi + (T21 - T20)*3.5*deltaphi + (T23 - T20)/4/deltaphi + (T19 - T20)*4.5*deltaphi = 0
//Node 21
(T18 - T21)/3/deltaphi/2 + (T20 - T21)*3.5*deltaphi + (T24 - T21)/3/deltaphi/2 = 0
//Node 22
T22 = Tc
//Node 23
T23 = Tc
//Node 24
T24 = Tc
qprime = qprime1 + qprime2 + qprime3
=
(2) The shape factor for the geometry
is S
q'
114.5 W/m
=
= 0.573
k (Th − Tc ) 10W/m ⋅ K × 20K
PROBLEM 4.55
KNOWN: Dimensions of a tube of non-circular cross section that can be broken into rectangular and
cylindrical sub-domains. Fluid temperature and heat transfer coefficient, external surface temperature and
tube wall thermal conductivity.
FIND: Heat transfer rate per unit length of tube.
SCHEMATIC:
• 25
•22
•19
Adiabatic surfaces
• 20
k = 15 W/m·K
•16
ro = 40 mm
q´1
•6
x
•
1
•21
17
•
q´2
• 11 12•
y
•
•
7
•
2
23
•24 • 27
ri = 20 mm
•18
• 14•
13
q´3
•8
• 26
•
9
•
15
•
10
•
•
•
3
4
5
T∞,i = 20°C
hi = 240 W/m2·K
r
∆x = ∆y = ∆r = 10 mm
∆θ = π/8
r1 = 25 mm
r2 = 30 mm
r3 = 35 mm
Ts = 100°C
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) No internal generation, (4)
Two-dimensional conduction.
ANALYSIS: We may combine heat fluxes determined from Fourier’s law with expressions for the size
of the control surfaces of the various control volumes to determine the heat rate per unit depth into each
control volume in the discretized domain. Application of conservation of energy for each control volume
yields the expression E in = 0 .
Rectangular Sub-Domain. For the rectangular sub-domain, application of Fourier’s law in Cartesian
coordinates along with conservation of energy yields the finite difference equations that are listed in the
IHT code included in COMMENT (1).
Cylindrical Sub-Domain. For the cylindrical sub-domain, application of Fourier’s law in cylindrical
coordinates along with conservation of energy yields the finite difference equations that are listed in the
IHT code included in COMMENT (1).
Coupling of Domains. Note that energy balances for Nodes 11, 12 and 13 are included in both the
rectangular and cylindrical sub-domains. These energy balances couple the two solutions together.
Continued…
PROBLEM 4.55 (Cont.)
The equations are solved simultaneously to yield the following nodal temperatures in degrees Celsius.
59.50
60.82
58.89
60.20
64.80
55.79
64.14
57.05
60.78
71.62
70.90
67.38
81.60
80.54
77.90
79.53
79.88
90.45
90.10
89.42
89.69
89.82
100
100
100
100
100
The heat transfer rate per unit depth may be expressed as
 k ∆x 
=
q' 2
 [(T1 − T6 ) / 2 + (T2 − T7 ) + (T3 − T8 ) + (T4 − T9 ) + (T5 − T10 ) / 2]
 ∆y 
(100 − 90.45)°C / 2 + (100 − 90.10)°C + 

= 2 × 15W/m ⋅ K × (100 − 89.42)°C + (100 − 89.69)°C


 + (100 − 89.82)°C / 2

<
= 1220 W/m
COMMENTS: (1) The IHT code is listed below. For each control volume, we note that E in = 0 , yielding
the energy balances for the rectangular and cylindrical sub-domains.
ri = 20/1000
ro = 40/1000
r1 = 25/1000
r2 = 30/1000
r3 = 35/1000
dx = 10/1000
dy = 10/1000
dr = 10/1000
dtheta = pi/8
Continued…
PROBLEM 4.55 (Cont.)
k = 15
h = 240
Tinf = 20
T1 = 100
T2 = 100
T3 = 100
T4 = 100
T5 = 100
//Rectangular Domain
//Node 6
k*(T1 - T6)*(dx/2)/dy + k*(T7 - T6)*dy/dx + k*(T11 - T6)*(dx/2)/dy = 0
//Node 7
k*(T6 - T7)*dy/dx + k*(T12 - T7)*dx/dy + k*(T8 - T7)*dy/dx + k*(T2 - T7)*dx/dy = 0
//Node 8
k*(T7 - T8)*dy/dx + k*(T13 - T8)*dx/dy + k*(T9 - T8)*dy/dx + k*(T3 - T8)*dx/dy = 0
//Node 9
k*(T8 - T9)*dy/dx + k*(T14 - T9)*dx/dy + k*(T10 - T9)*dy/dx + k*(T4 - T9)*dx/dy = 0
//Node 10
k*(T9 - T10)*dy/dx + k*(T15 - T10)*(dx/2)/dy + k*(T5 - T10)*(dx/2)/dy = 0
//Node 11
k*(T6 - T11)*(dx/2)/dy + k*(T12 - T11)*(dy/2)/dx = qprime1
//Node 12
k*(T11 - T12)*(dy/2)/dx + k*(T7 - T12)*dx/dy + k*(T13 - T12)*(dy/2)/dx = qprime2
//Node 13
k*(T12 - T13)*(dy/2)/dx + k*(T14 - T13)*(dy/2)/dx + k*(T8 - T13)*dx/dy + h*(dx/2)*(Tinf - T13) = qprime3
//Node 14
k*(T13 - T14)*(dy/2)/dx + k*(T9 - T14)*dx/dy + k*(T15 - T14)*(dy/2)/dx + h*dx*(Tinf - T14)=0
//Node 15
k*(T14 - T15)*(dy/2)/dx + k*(T10 - T15)*(dx/2)/dy + h*(dx/2)*(Tinf - T15) = 0
//Cylindrical Domain
//Node 11
k*(T16 - T11)*(dr/2)/(ro*dtheta) + k*(T12 - T11)*(r3*dtheta/2)/dr + qprime1 = 0
//Node 12
k*(T11 - T12)*(r3*dtheta/2)/dr + k*(T17 - T12)*dr/(r2*dtheta) + k*(T13 - T12)*(r1*dtheta/2)/dr + qprime2 = 0
//Node 13
k*(T12 - T13)*(r1*dtheta/2)/dr + k*(T18 - T13)*(dr/2)/(ri*dtheta) + qprime3 + h*(ri*dtheta/2)*(Tinf - T13) = 0
//Node 16
k*(T11 - T16)*(dr/2)/(ro*dtheta) + k*(T17 - T16)*(r3*dtheta)/dr + k*(T19 - T16)*(dr/2)/(ro*dtheta) = 0
//Node 17
k*(T16 - T17)*(r3*dtheta)/dr + k*(T20 - T17)*dr/(r2*dtheta) + k*(T18 - T17)*(r1*dtheta)/dr + k*(T12 T17)*dr/(r2*dtheta) = 0
//Node 18
k*(T13 - T18)*(dr/2)/(ri*dtheta) + k*(T17 - T18)*(r1*dtheta)/dr + k*(T21 - T18)*(dr/2)/(ri*dtheta) +
h*(ri*dtheta)*(Tinf - T18) = 0
//Node 19
k*(T16 - T19)*(dr/2)/(ro*dtheta) + k*(T20 - T19)*(r3*dtheta)/dr + k*(T22 - T19)*(dr/2)/(ro*dtheta) = 0
//Node 20
k*(T19 - T20)*(r3*dtheta)/dr + k*(T23 - T20)*dr/(r2*dtheta) + k*(T21 - T20)*(r1*dtheta)/dr + k*(T17 T20)*dr/(r2*dtheta) = 0
//Node 21
k*(T18 - T21)*(dr/2)/(ri*dtheta) + k*(T20 - T21)*(r1*dtheta)/dr + k*(T24 - T21)*(dr/2)/(ri*dtheta) +
h*(ri*dtheta)*(Tinf - T21) = 0
//Node 22
k*(T19 - T22)*(dr/2)/(ro*dtheta) + k*(T23 - T22)*(r3*dtheta)/dr + k*(T25 - T22)*(dr/2)/(ro*dtheta) = 0
//Node 23
k*(T22 - T23)*(r3*dtheta)/dr + k*(T26 - T23)*dr/(r2*dtheta) + k*(T24 - T23)*(r1*dtheta)/dr + k*(T20 T23)*dr/(r2*dtheta) = 0
//Node 24
k*(T21 - T24)*(dr/2)/(ri*dtheta) + k*(T23 - T24)*(r1*dtheta)/dr + k*(T27 - T24)*(dr/2)/(ri*dtheta) +
h*(ri*dtheta)*(Tinf - T24) = 0
Continued…
PROBLEM 4.55 (Cont.)
//Node 25
k*(T22 - T25)*(dr/2)/(ro*dtheta) + k*(T26 - T25)*(r3*dtheta/2)/dr = 0
//Node 26
k*(T25 - T26)*(r3*dtheta/2)/dr + k*(T23 - T26)*dr/(r2*dtheta) + k*(T27 - T26)*(r1*dtheta/2)/dr = 0
//Node 27
k*(T24 - T27)*(dr/2)/(ri*dtheta) + k*(T26 - T27)*(r1*dtheta/2)/dr + h*(ri*dtheta/2)*(Tinf - T24) = 0
//Heat Transfer Rate Per Unit Depth (W/m)
qprime = 2*((k/dy)*((T1 - T6)*dx/2 + (T2 - T7)*dx + (T3 - T8)*dx + (T4 - T9)*dx + (T5 - T10)*dx/2))
(2) For an isothermal tube at Ts = 100°C, the heat transfer per unit length is
q '= h(π ri + 2 ri ) × (Ts −=
T ) 240W / m ⋅ K × (π × 20 × 10−3 m + 2 × 20 × 10−3 m) × (100 − 20)°C
∞
= 1975 W/m. The conduction resistance posed by the tube wall is responsible for decreasing
the actual heat transfer rate below this value.
PROBLEM 4.56
PROBLEM 4.57
PROBLEM 4.57 (Cont.)
PROBLEM 4.58
KNOWN: Long bar of square cross section, three sides of which are maintained at a constant
temperature while the fourth side is subjected to a convection process.
FIND: (a) The mid-point temperature and heat transfer rate between the bar and fluid; a numerical
technique with grid spacing of 0.2 m is suggested, and (b) Reducing the grid spacing by a factor of 2, find
the midpoint temperature and the heat transfer rate. Also, plot temperature distribution across the surface
exposed to the fluid.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) Considering symmetry, the nodal network is shown above. The matrix inversion
method of solution will be employed. The finite-difference equations are:
Nodes 1, 3, 5 Nodes 2, 4, 6 Nodes 7, 8 -
Interior nodes, Eq. 4.29; written by inspection.
Also can be treated as interior points, considering symmetry.
On a plane with convection, Eq. 4.42; noting that h∆x/k =
10 W/m2⋅K × 0.2 m/2W/m⋅K = 1, find
Node 7: (2T5 + 300 + T8) + 2×1⋅100 - 2(1+2)T7 = 0
Node 8: (2T6 + T7 + T7) + 2×1⋅100 - 2(1+2)T8 = 0
The solution matrix [T] can be found using a stock matrix program using the [A] and [C] matrices shown
below to obtain the solution matrix [T] (Eq. 4.48). Alternatively, the set of equations could be entered
into the IHT workspace and solved for the nodal temperatures.
 −4 1 1 0 0 0 0 0 
 2 −4 0 1 0 0 0 0 
 1 0 −4 1 1 0 0 0 

2 −4 0 1 0 =
0
=
A 0 1
C
0 0 1 0 −4 1 1 0 
0 0 0 1
2 −4 0 1 
 0 0 0 0 2 0 −6 1 
 0 0 0 0 0 2 2 −6 
 −600 
 −300 
 −300 
 0 
T
=
−300 
 0 
 −500 
 −200 
 292.2 
 289.2 
 279.7 
 272.2 
 254.5 
 240.1
 198.1 
179.4 
From the solution matrix, [T], find the mid-point temperature as
T4 = 272.2°C
<
Continued...
PROBLEM 4.58 (Cont.)
The heat rate by convection between the bar and fluid is given as,
q′conv= 2 ( q′a + q′b + q′c )
q′conv= 2  h ( ∆x 2 ) ( T8 − T∞ ) + h ( ∆x ) ( T7 − T∞ ) + h ( ∆x 2 )( 300 − T∞ ) 
=
q′conv 2 10 W m 2 ⋅ K × ( 0.2 m 2 ) (179.4 − 100 ) + 2 (198.1 − 100 ) + ( 300 − 100 )  K 


<
q′conv = 952 W m .
(b) Reducing the grid spacing by a factor of 2, the nodal arrangement will appear as shown. The finitedifference equation for the interior and centerline nodes were written by inspection and entered into the
IHT workspace. The IHT Finite-Difference Equations Tool for 2-D, SS conditions, was used to obtain
the FDE for the nodes on the exposed surface.
The midpoint temperature T13 and heat rate for the finer mesh are
T13 = 271.0°C
q′ = 834 W/m
<
COMMENTS: The midpoint temperatures for the coarse and finer meshes agree closely, T4 = 272°C vs.
T13 = 271.0°C, respectively. However, the estimate for the heat rate is substantially influenced by the
mesh size; q′ = 952 vs. 834 W/m for the coarse and finer meshes, respectively.
PROBLEM 4.59
PROBLEM 4.59 (Cont.)
PROBLEM 4.59 (Cont.)
PROBLEM 4.59 (Cont.)
PROBLEM 4.60
KNOWN: Rectangular air ducts having surfaces at 80°C in a concrete slab with an insulated bottom and
upper surface maintained at 30°C.
FIND: Heat rate from each duct per unit length of duct, q′.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) No internal
volumetric generation, (4) Constant properties.
PROPERTIES: Concrete (given): k = 1.4 W/m⋅K.
ANALYSIS: Taking advantage of symmetry, the
nodal network, using the suggested grid spacing
∆x = 2∆y = 37.50 mm
∆y = 0.125L = 18.75 mm
where L = 150 mm, is shown in the sketch. To
evaluate the heat rate, we need the temperatures T1,
T2, T3, T4, and T5. All the nodes may be treated as
interior nodes ( considering symmetry for those nodes on
insulated boundaries), Eq. 4.29. Use matrix notation, Eq. 4.48,
[A][T] = [C], and perform the inversion.
The heat rate per unit length from the prescribed section of the
duct follows from an energy balance on the nodes at the top
isothermal surface.
q′ = q1′ + q′2 + q′3 + q′4 + q′5
T −T
T − Ts
T −T
T − Ts
T − Ts
q=′ k ( ∆x/2 ) 1 s + k ⋅ ∆x 2
+ k ⋅ ∆x 3 s + k ⋅ ∆x 4
+ k ( ∆x/2 ) 5
∆y
∆y
∆y
∆y
∆y
q=′ k ( T1 − Ts ) + 2 ( T2 − Ts ) + 2 ( T3 − Ts ) + 2 ( T4 − Ts ) + ( T5 − Ts ) 
q′ 1.4 W/m ⋅ K ( 41.70 − 30 ) + 2 ( 44.26 − 30 ) + 2 ( 53.92 − 30 ) + 2 ( 54.89 − 30 ) + ( 54.98 − 30 ) 
=
q′ = 228 W/m.
Since the section analyzed represents one-half of the region about an air duct, the heat loss per unit length
for each duct is,
q′duct
= 2xq
=′ 456 W/m.
<
Continued …
PROBLEM 4.60 (Cont.)
Coefficient matrix [A]
PROBLEM 4.61
KNOWN: Dimensions of long cylinder, thickness of metal sheathing, volumetric generation rate within
the sheathing, thermal conductivity of sheathing and convection heat transfer coefficient dependence
upon angle θ.
FIND: (a) Temperature distribution within the thin sheathing neglecting θ-direction conduction heat
transfer, (b) temperature distribution in the sheathing accounting for θ-direction conduction heat transfer
in the metal.
SCHEMATIC:
t = 50 µm
Air
T∞ = 25°C
D = 25 mm
θ
Metal sheathing
.
q = 5×106 W/m3
k = 25 W/m∙K
Insulation
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform internal generation,
(4) Metal sheathing is very thin relative to cylinder diameter, (5) One-dimensional conduction, (6)
Negligible radiation.
ANALYSIS: (a) Neglecting conduction in the θ-direction in the sheathing, an energy balance at any θ
location yields
 h(q )(Ts − T∞ ) or T=
 / h(q ) and
=
qt
T∞ + qt
s
h(θ) = 26 + 0.637θ - 8.92θ 2 for 0 ≤ θ < π/2; h(θ) = 5 for π/2 ≤ θ ≤ π
These equations may be solved to yield the temperature distribution that is plotted on the next page where
x = θD/2.
(b) Since the sheathing is thin relative to the cylinder diameter, we may evaluate one-dimensional
conduction in the x-direction using the Cartesian coordinate system. The finite difference equations are
derived by combining expressions for heat fluxes based upon Fourier’s law and Newton’s law of cooling,
along with conservation of energy for each control volume within the discretized domain. Application of
0 . The discretized
conservation of energy for each control volume yields the expression E in + E g =
domain is shown below.
L = πD/2
x
∆x
•
•
•
•
•
•
•
•
•
•
•
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15 16 17
•
•
•
•
•
t
•
•
•
•
•
18 19 20 21
dqconv
m -1 •
qcond,l
•
m
• m+1
qcond,r
Continued…
PROBLEM 4.61 (Cont.)
The finite difference equations are as follows.
(T2 − T1 )
t + h( ∆x / 2)(T∞ − T1 ) + q ( ∆x / 2)t =
0
∆x
Node 1:
k
Nodes 2 - 20: k
(Tm −1 − Tm )
(T − T )
t + k m +1 m t + h( ∆x )(T∞ − Tm ) + q ( ∆x )t =
0
∆x
∆x
Node 21:
k
(T20 − T21 )
t + h( ∆x / 2)(T∞ − T21 ) + q ( ∆x / 2)t =
0
∆x
The temperature distribution is plotted below.
80
70
T (C)
60
50
Without conduction
40
With conduction
T (C)
30
T (C)
20
0
10
20
30
40
50
x (mm)
COMMENTS: (1) Conduction in the θ-direction within the sheathing smears the temperature
distribution, increasing the low temperatures on the upstream half of the cylinder and lowering the
temperatures on the downstream half of the cylinder. (2) The IHT code is listed below.
qdot = 5*10^6
t = 50*10^(-6)
Tinf = 25 + 273
k = 25
D = 25/1000
L = pi*D/2
dx = L/20
//W/m^3
//m
//K
//W/m⋅K
//m
//m
//m
h1 = 26
h2 = 25.88
h3 = 25.32
h4 = 24.32
h5 = 22.88
h6 = 21.00
h7 = 18.68
h8 = 15.92
h9 = 12.71
h10 = 9.07
h11 = 5
h12 = 5
h13 = 5
h14 = 5
h15 = 5
h16 = 5
h17 = 5
h18 = 5
//W/m^2⋅K
Continued…
PROBLEM 4.61 (Cont.)
h19 = 5
h20 = 5
h21 = 5
//Node 1
k*(T2 - T1)*t/dx + h1*(dx/2)*(Tinf - T1) + qdot*dx*t/2 = 0
//Node 2
k*(T1 - T2)*t/dx + k*(T3 - T2)*t/dx + h2*dx*(Tinf - T2) + qdot*dx*t = 0
//Node 3
k*(T2 - T3)*t/dx + k*(T4 - T3)*t/dx + h3*dx*(Tinf - T3) + qdot*dx*t = 0
//Node 4
k*(T3 - T4)*t/dx + k*(T5 - T4)*t/dx + h4*dx*(Tinf - T4) + qdot*dx*t = 0
//Node 5
k*(T4 - T5)*t/dx + k*(T6 - T5)*t/dx + h5*dx*(Tinf - T5) + qdot*dx*t = 0
//Node 6
k*(T5 - T6)*t/dx + k*(T7 - T6)*t/dx + h6*dx*(Tinf - T6) + qdot*dx*t = 0
//Node 7
k*(T6 - T7)*t/dx + k*(T8 - T7)*t/dx + h7*dx*(Tinf - T7) + qdot*dx*t = 0
//Node 8
k*(T7 - T8)*t/dx + k*(T9 - T8)*t/dx + h8*dx*(Tinf - T8) + qdot*dx*t = 0
//Node 9
k*(T8 - T9)*t/dx + k*(T10 - T9)*t/dx + h9*dx*(Tinf - T9) + qdot*dx*t = 0
//Node 10
k*(T9 - T10)*t/dx + k*(T11 - T10)*t/dx + h10*dx*(Tinf - T10) + qdot*dx*t = 0
//Node 11
k*(T10 - T11)*t/dx + k*(T12 - T11)*t/dx + h11*dx*(Tinf - T11) + qdot*dx*t = 0
//Node 12
k*(T11 - T12)*t/dx + k*(T13 - T12)*t/dx + h12*dx*(Tinf - T12) + qdot*dx*t = 0
//Node 13
k*(T12 - T13)*t/dx + k*(T14 - T13)*t/dx + h13*dx*(Tinf - T13) + qdot*dx*t = 0
//Node 14
k*(T13 - T14)*t/dx + k*(T15 - T14)*t/dx + h14*dx*(Tinf - T14) + qdot*dx*t = 0
//Node 15
k*(T14 - T15)*t/dx + k*(T16 - T15)*t/dx + h15*dx*(Tinf - T15) + qdot*dx*t = 0
//Node 16
k*(T15 - T16)*t/dx + k*(T17 - T16)*t/dx + h16*dx*(Tinf - T16) + qdot*dx*t = 0
//Node 17
k*(T16 - T17)*t/dx + k*(T18 - T17)*t/dx + h17*dx*(Tinf - T17) + qdot*dx *t= 0
//Node 18
k*(T17 - T18)*t/dx + k*(T19 - T18)*t/dx + h18*dx*(Tinf - T18) + qdot*dx*t = 0
//Node 19
k*(T18 - T19)*t/dx + k*(T20 - T19)*t/dx + h19*dx*(Tinf - T19) + qdot*dx*t = 0
//Node 20
k*(T19 - T20)*t/dx + k*(T21 - T20)*t/dx + h20*dx*(Tinf - T20) + qdot*dx*t = 0
//Node 21
k*(T20 - T21)*t/dx +h21*(dx/2)*(Tinf - T21) + + qdot*dx*t/2 = 0
PROBLEM 4.62
KNOWN: Bar of known thermal conductivity and trapezoidal cross section. Temperatures of left and
right faces.
FIND: Heat transfer rate per unit bar length using ∆x = ∆y = 10 mm. Compare to heat rate for
rectangular cross section.
SCHEMATIC:
10
Th
6
x
1
12
11
∆x
7
∆y
Tc
8
9
y
2
3
4
5
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties, (3) No internal heat generation,
(4) Two-dimensional conduction.
ANALYSIS: The discretized solution domain is shown in the schematic. The finite difference equations
are:
Node 1: T1 = Th
T − T2
∆y T3 − T2
∆y T1 − T2
Node 2 : k ∆x 7
0
+k
+k
=
2 ∆x
2 ∆x
∆y
T −T
∆y T2 − T3
∆y T4 − T3
Node 3 : k
+ k ∆x 8 3 + k
=
0
2 ∆x
∆y
2 ∆x
T − T4
∆y T5 − T4
∆y T3 − T4
+k
+k
=
Node 4 : k ∆x 9
0
∆y
2 ∆x
2 ∆x
Node 5 : T5 = Tc
Node 6 : T6 = Th
T −T
T −T
T − T7
T − T7
Node 7 : k ∆x 11 7 + k ∆y 8 7 + k ∆x 2
+ k ∆y 6
=
0
∆y
∆x
∆y
∆x
T −T
T −T
T −T
T −T
Node 8 : k ∆x 12 8 + k ∆y 9 8 + k ∆x 3 8 + k ∆y 7 8 =
0
∆y
∆x
∆y
∆x
Node 9 : T9 = Tc
Node 10 : T10 = Th
T −T
∆y T12 − T11
∆y T10 − T11
Node 11: k
0
+ k ∆x 7 11 + k
=
2
2
∆x
∆y
∆x
Node 12 : T12 = Tc
Continued…
PROBLEM 4.62 (Cont.)
Solving the preceding equations yields
T10 = 100°C
T6 = 100°C
T1 = 100°C
T11 = 54.84°C
T7 = 59.68°C
T2 = 61.90°C
T12 = 0°C
T8 = 21.98°C
T3 = 28.23°C
T9 = 0°C
T4 = 7.06°C
T5 = 0°C
The heat rate per unit length may be calculated as
=
q′ k
T −T
∆y T10 − T11
∆y T1 − T2
+ k ∆y 6 7 + k
2
∆x
∆x
2 ∆x
=×
1.06 104 W/m =
10.6 kW/m
<
The heat rate for the 20 mm × 30 mm rectangular cross section bar is
′ = k ⋅ 2∆y
qrect
Th − Tc
= 9.33 × 103 W/m = 9.33 kW/m
3∆x
<
COMMENTS: (1) The trapezoidal and rectangular bars are of the same mass. Hence, the trapezoidal bar
has a greater heat transfer rate per unit mass. (2) The heat rate per unit length can be calculated in various
ways. For example, it could be determined by calculating the heat rate in the horizontal direction per unit
length between nodes 11 and 12, 7 and 8, plus that between nodes 2 and 3.
PROBLEM 4.63
PROBLEM 4.63 (Cont.)
PROBLEM 4.64
PROBLEM 4.64 (Cont.)
PROBLEM 4.65
KNOWN: Upper surface and grooves of a plate are maintained at a uniform temperature T1, while the
lower surface is maintained at T2 or is exposed to a fluid at T∞.
FIND: (a) Heat rate per width of groove spacing (w) for isothermal top and bottom surfaces using a
finite-difference method with ∆x = 40 mm, (b) Effect of grid spacing and convection at bottom surface.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) Using a space increment of ∆x = 40 mm, the symmetrical section shown in schematic
(b) corresponds to one-half the groove spacing. There exist only two interior nodes for which finitedifference equations must be written.
Node a:
Node b:
4Ta − ( T1 + Tb + T2 + T1 ) =
0
4Ta − ( 200 + Tb + 20 + 200 ) =
0
or
4Ta − Tb =
420
(1)
4Tb − ( T1 + Ta + T2 + Ta ) =
0
4Tb − ( 200 + 2Ta + 20 ) =
0
or
−2Ta + 4Tb =
220
(2)
Multiply Eq. (2) by 2 and add to Eq. (1) to obtain
7Tb = 860
or
Tb = 122.9°C
From Eq. (1),
4Ta - 122.9 = 420
or
Ta = (420 + 122.9)/4 = 135.7°C.
The heat transfer through the symmetrical section is equal to the sum of heat flows through control
volumes adjacent to the lower surface. From the schematic,
T − T2
 ∆x  T1 − T2
 ∆x  Tb − T2
q′ = q1′ + q′2 + q′3 = k 
+ k ( ∆x ) a
+ k
.


∆y
 2  ∆y
 2  ∆y
Continued...
PROBLEM 4.65 (Cont.)
Noting that ∆x = ∆y, regrouping and substituting numerical values, find
1
1

=
q′ k  ( T1 − T2 ) + ( Ta − T2 ) + ( Tb − T2 ) 
2
2

1
1
=
q′ 15 W m ⋅ K  ( 200 − 20 ) + (135.7 − 20 ) + (122.9 − 20=
) 3.86 kW m .
2
2

<
For the full groove spacing, q′total = 2 × 3.86 kW/m = 7.72 kW/m.
(b) Using the Finite-Difference Equations option from the Tools portion of the IHT menu, the following
two-dimensional temperature field was computed for the grid shown in schematic (b), where x and y are
in mm and the nodal temperatures are in °C. Nodes 2-54 are interior nodes, with those along the
symmetry adiabats characterized by Tm-1,n = Tm+1,n, while nodes 55-63 lie on a plane surface.
y\x
0
10
20
30
40
50
60
70
80
0
200
141.4
97.09
57.69
20
10
200
175.4
134.3
94.62
56.83
20
20
30
200
182.4
160.3
125.7
90.27
55.01
20
200
186.7
169.5
148.9
118.0
85.73
52.95
20
40
200
191
177.2
160.1
140.1
111.6
81.73
51.04
20
50
200
186.6
171.2
153.4
133.5
106.7
78.51
49.46
20
60
200
184.3
167.5
149.0
128.7
103.1
76.17
48.31
20
70
200
183.1
165.5
146.4
125.7
100.9
74.73
47.60
20
80
200
182.8
164.8
145.5
124.4
100.1
74.24
47.36
20
The foregoing results were computed for h = 107 W/m2⋅K (h → ∞) and T∞ = 20°C, which is tantamount to
prescribing an isothermal bottom surface at 20°C. Agreement between corresponding results for the
coarse and fine grids is surprisingly good (Ta = 135.7°C ↔ T23 = 140.1°C; Tb = 122.9°C ↔ T27 =
124.4°C). The heat rate is
q′ =
2 × k ( T46 − T55 ) 2 + ( T47 − T56 ) + ( T48 − T57 ) + ( T49 − T58 ) + ( T50 − T59 )
+ ( T51 − T60 ) + ( T52 − T61 ) + ( T53 − T62 ) + ( T54 − T63 ) 2
q′ =
2 ×15 W m ⋅ K [18.84 + 36.82 + 35.00 + 32.95 + 31.04 + 29.46
<

+28.31 + 27.6 + 13.68] C =
7.61kW m
The agreement with q¢ = 7.72 kW/m from the coarse grid of part (a) is excellent and a fortuitous
consequence of compensating errors. With reductions in the convection coefficient from h → ∞ to h =
1000, 200 and 5 W/m2⋅K, the corresponding increase in the thermal resistance reduces the heat rate to
values of 6.03, 3.28 and 0.14 kW/m, respectively. With decreasing h, there is an overall increase in nodal
temperatures, as, for example, from 191°C to 199.8°C for T2 and from 20°C to 196.9°C for T55.
NOTE TO INSTRUCTOR: To reduce computational effort, while achieving the same educational
objectives, the problem statement has been changed to allow for convection at the bottom, rather than the
top, surface.
PROBLEM 4.66
PROBLEM 4.67
KNOWN: Edge of adjoining walls (k = 1 W/m⋅K) represented by symmetrical element bounded by the
diagonal symmetry adiabat and a section of the wall thickness over which the temperature distribution is
assumed to be linear.
FIND: (a) Temperature distribution, heat rate and shape factor for the edge using the nodal network with
= ∆x = ∆y = 10 mm; compare shape factor result with that from Table 4.1; (b) Assess the validity of
assuming linear temperature distributions across sections at various distances from the edge.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties, and (3) Linear
temperature distribution at specified locations across the section.
ANALYSIS: (a) Taking advantage of symmetry along the adiabat diagonal, all the nodes may be treated
as interior nodes. Across the left-hand boundary, the temperature distribution is specified as linear. The
finite-difference equations required to determine the temperature distribution, and hence the heat rate, can
be written by inspection.
=
T3 0.25 ( T2 + T4 + T6 + Tc )
=
T4 0.25 ( T2 + T5 + T7 + T3 )
=
T5 0.25 ( T2 + T2 + T4 + T4 )
=
T6 0.25 ( T3 + T7 + T8 + Tb )
=
T7 0.25 ( T4 + T4 + T6 + T6 )
=
T8 0.25 ( T6 + T6 + Ta + Ta )
The heat rate for both surfaces of the edge is
q′tot= 2 [ q′a + q′b + q′c + q′d ]
q′tot = 2 [ k ( ∆x 2 )( Tc − T2 ) ∆y + k∆x ( T3 − T2 ) ∆y + k∆x ( T4 − T2 ) ∆y + k∆x ( T5 − T2 ) ∆x ]
The shape factor for the full edge is defined as
=
q′tot kS′ ( T1 − T2 )
Solving the above equation set in IHT, the temperature (°C) distribution is
Continued...
PROBLEM 4.67 (Cont.)
0
25
0
0
0
0
18.75 12.5 6.25
50
75
37.5 25.0
56.25
<
100
and the heat rate and shape factor are
<
=
q′tot 100
=
W m
S 1
From Table 4.1, the edge shape factor is 0.54, considerably below our estimate from this coarse grid
analysis.
(b) The effect of the linear temperature distribution on the shape factor estimate can be explored using a
more extensive grid as shown below. The FDE analysis was performed with the linear distribution
imposed as the different sections a, b, c, d, e. Following the same approach as above, find
Location of linear distribution
Shape factor, S
(a)
0.797
(b)
0.799
(c)
0.809
(d)
0.857
(e)
1.00
The shape factor estimate decreases as the imposed linear temperature distribution section is located
further from the edge. We conclude that assuming the temperature distribution across the section directly
at the edge is a poor-one.
COMMENTS: The grid spacing for this analysis is quite coarse making the estimates in poor agreement
with the Table 4.1 result. However, the analysis does show the effect of positioning the linear
temperature distribution condition.
PROBLEM 4.68
PROBLEM 4.68 (Cont.)
PROBLEM 4.69
PROBLEM 4.69 (Cont.)
PROBLEM 4.69 (Cont.)
PROBLEM 4.70
PROBLEM 4.70 (Cont.)
PROBLEM 4.71
PROBLEM 4.71 (Cont.)
PROBLEM 4.72
PROBLEM 4.72 (Cont.)
PROBLEM 4.73
PROBLEM 4.73 (Cont.)
PROBLEM 4.73 (Cont.)
PROBLEM 4.73 (Cont.)
PROBLEM 4S.1
KNOWN: Long furnace of refractory brick with prescribed surface temperatures and material thermal
conductivity.
FIND: Shape factor and heat transfer rate per unit length using the flux plot method
SCHEMATIC:
ASSUMPTIONS: (1) Furnace length normal to page, , >> cross-sectional dimensions, (2) Twodimensional, steady-state conduction, (3) Constant properties.
ANALYSIS: Considering the cross-section, the cross-hatched area represents a symmetrical element.
Hence, the heat rate for the entire furnace per unit length is
q=′
q
S
= 4 k ( T1 − T2 )


(1)
where S is the shape factor for the symmetrical section. Selecting three temperature increments ( N = 3),
construct the flux plot shown below.
From Equation 4S.7,
and from Equation (1),
=
S
M
N
or
S M 8.5
= = = 2.83
 N
3
q′ =
4 × 2.83 ×1.2
W
7.34 kW/m.
( 600 − 60 ) C =
m⋅K
<
<
COMMENTS: The shape factor can also be estimated from the relations of Table 4.1. The symmetrical
section consists of two plane walls (horizontal and vertical) with an adjoining edge. Using the
appropriate relations, the numerical values are, in the same order,
S=
0.75m
0.5m
 + 0.54 +
= 3.04
0.5m
0.5m
Note that this result compares favorably with the flux plot result of 2.83.
PROBLEM 4S.2
KNOWN: Hot pipe embedded eccentrically in a circular system having a prescribed thermal
conductivity.
FIND: The shape factor and heat transfer per unit length for the prescribed surface
temperatures.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Length 
>> diametrical dimensions.
ANALYSIS: Considering the cross-sectional view of the pipe system, the symmetrical section
shown above is readily identified. Selecting four temperature increments (N = 4), construct the
flux plot shown below.
For the pipe system, the heat rate per unit length is
q
W

q′ = =kS ( T1 − T2 ) =0.5
× 4.26 (150 − 35 ) C =245 W/m.
m⋅K

<
COMMENTS: Note that in the lower, right-hand quadrant of the flux plot, the curvilinear
squares are irregular. Further work is required to obtain an improved plot and, hence, obtain a
more accurate estimate of the shape factor.
PROBLEM 4S.3
KNOWN: Structural member with known thermal conductivity subjected to a temperature difference.
FIND: (a) Temperature at a prescribed point P, (b) Heat transfer per unit length of the strut, (c) Sketch
the 25, 50 and 75°C isotherms, and (d) Same analysis on the shape but with adiabatic-isothermal
boundary conditions reversed.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant properties.
ANALYSIS: (a) When constructing the flux plot, note that the line of symmetry which passes through
the point P is an isotherm as shown above. It follows that
T (P) =
50 C .
( T1 + T2 ) 2 =
(100 + 0 ) C 2 =
<
(b) The flux plot on the symmetrical section is now constructed to obtain the shape factor from which the
heat rate is determined. That is, from Equation 4S.6 and 4S.7,
q =−
kS ( T1 T2 ) and S =
M N .
(1,2)
From the plot of the symmetrical section,
=
So 4.2
=
 4 1.05 .
For the full section of the strut,
M M
=
=
o 4.2
but N = 2No = 8. Hence,
=
S S=
o 2 0.53
and with q′ = q  , giving
=
q′  75 W m ⋅ K × 0.53 (100 − 0=
) C 3975 W m .
<
(c) The isotherms for T = 50, 75 and 100°C are shown on the flux plot. The T = 25°C isotherm is
symmetric with the T = 75°C isotherm.
(d) By reversing the adiabatic and isothermal boundary conditions, the two-dimensional shape appears as
shown in the sketch below. The symmetrical element to be flux plotted is the same as for the strut, except
the symmetry line is now an adiabat.
Continued...
PROBLEM 4S.3 (Cont.)
From the flux plot, find Mo = 3.4 and No = 4, and from Equation (2)
=
So M o  N
=
4 0.85
o 3.4=
=
S 2S
=
o 1.70
and the heat rate per unit length from Equation (1) is

=
q′ 75 W m ⋅ K ×1.70 (100 − 0=
) C 12, 750 W m
<
From the flux plot, estimate that
T(P) ≈ 40°C.
<
COMMENTS: (1) By inspection of the shapes for parts (a) and (b), it is obvious that the heat rate for the
latter will be greater. The calculations show the heat rate is greater by more than a factor of three.
(2) By comparing the flux plots for the two configurations, and corresponding roles of the adiabats and
isotherms, would you expect the shape factor for parts (a) to be the reciprocal of part (b)?
PROBLEM 4S.4
KNOWN: Relative dimensions and surface thermal conditions of a V-grooved channel.
FIND: Flux plot and shape factor.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant
properties.
ANALYSIS: With symmetry about the midplane, only one-half of the object need be
considered as shown below.
Choosing 6 temperature increments (N = 6), it follows from the plot that M ≈ 7. Hence from
Equation 4S.7, the shape factor for the half section is
M
7
=
S
=
 =
 1.17.
N
6
For the complete system, the shape factor is then
S = 2.34.
<
PROBLEM 4S.5
KNOWN: Long conduit of inner circular cross section and outer surfaces of square cross section.
FIND: Shape factor and heat rate for the two applications when outer surfaces are insulated or
maintained at a uniform temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties and (3)
Conduit is very long.
ANALYSIS: The adiabatic symmetry lines for each of the applications is shown above. Using the flux
plot methodology and selecting four temperature increments (N = 4), the flux plots are as shown below.
For the symmetrical sections, S = 2So, where So = M  /N and the heat rate for each application is q =
2(So/  )k(T1 - T2).
Application
A
B
M
10.3
6.2
N
4
4
So/ 
2.58
1.55
q′ (W/m)
11,588
6,975
COMMENTS: (1) For application A, most of the heat lanes leave the inner surface (T1)
on the upper portion. (2) For application B, most of the heat flow lanes leave the inner
surface on the upper portion (that is, lanes 1-4). Because the lower, right-hand corner is
insulated, the entire section experiences small heat flows (lane 6 + 0.2). Note the shapes
of the isotherms near the right-hand, insulated boundary and that they intersect the
boundary normally.
<
<
PROBLEM 4S.6
KNOWN: Shape and surface conditions of a support column.
FIND: (a) Heat transfer rate per unit length. (b) Height of a rectangular bar of equivalent
thermal resistance.
SCHEMATIC:
ASSUMPTIONS: (1)Steady-state conditions, (2) Negligible three-dimensional conduction
effects, (3) Constant properties, (4) Adiabatic sides.
PROPERTIES: Table A-1, Steel, AISI 1010 (323K): k = 62.7 W/m⋅K.
ANALYSIS: (a) From the flux plot for the
half section, M ≈ 5 and N ≈ 8. Hence for the
full section
M
≈ 1.25
N
=
q Sk ( T1 − T2 )
W
q′ ≈ 1.25 × 62.7
(100 − 0 ) C
m⋅K
=
S 2
q′ ≈ 7.8 kW/m.
<
(b) The rectangular bar provides for one-dimensional heat transfer. Hence,
( T − T ) k 0.3 ( T1 − T2 )
q k=
A 1 2
=
( )
H
H
(
)

0.3k ( T1 − T2 ) 0.3m ( 62.7 W/m ⋅ K ) 100 C
Hence,
=
= 0.24m.
H =
q′
7800 W/m
<
COMMENTS: The fact that H < 0.3m is consistent with the requirement that the thermal
resistance of the trapezoidal column must be less than that of a rectangular bar of the same height
and top width (because the width of the trapezoidal column increases with increasing distance, x,
from the top). Hence, if the rectangular bar is to be of equivalent resistance, it must be of smaller
height.
PROBLEM 4S.7
KNOWN: Hollow prismatic bars fabricated from plain carbon steel, 1m in length with
prescribed temperature difference.
FIND: Shape factors and heat rate per unit length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties.
PROPERTIES: Table A-1, Steel, Plain Carbon (400K), k = 57 W/m⋅K.
ANALYSIS: Construct a flux plot on the symmetrical sections (shaded-regions) of each of the
bars.
The shape factors for the symmetrical sections are,
So,A
=
M 4
= =
 1
N 4
So,B
=
M 3.5
=
=
 0.88.
N
4
Since each of these sections is ¼ of the bar cross-section, it follows that
SA =4 ×1 =4
SB =4 × 0.88 =3.5.
<
q′ q/=
 k ( S/ )( T1 − T2 ) ,
The heat rate per unit length is =
q′A= 57
W
× 4 ( 500 − 300 ) K= 45.6 kW/m
m⋅K
<
q′B = 57
W
× 3.5 ( 500 − 300 ) K = 39.9 kW/m.
m⋅K
<
PROBLEM 4S.8
KNOWN: Two-dimensional, square shapes, 1 m to a side, maintained at uniform temperatures as
prescribed, perfectly insulated elsewhere.
FIND: Using the flux plot method, estimate the heat rate per unit length normal to the page if the thermal
conductivity is 50 W/m⋅K
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: Use the methodology of Section 4S.1 to construct the flux plots to obtain the shape factors
from which the heat rates can be calculated. With Figure (a), begin at the lower-left side making the
isotherms almost equally spaced, since the heat flow will only slightly spread toward the right. Start
sketching the adiabats in the vicinity of the T2 surface. The dashed line represents the adiabat which
separates the shape into two segments. Having recognized this feature, it was convenient to identify
partial heat lanes. Figure (b) is less difficult to analyze since the isotherm intervals are nearly regular in
the lower left-hand corner.
The shape factors are calculated from Equation 4S.7 and the heat rate from Equation 4S.6.
=
S′
M 0.5 + 3 + 0.5 + 0.5 + 0.2
=
N
6
=
S′
M 4.5
= = 0.90
N
5
S′ = 0.70
q′ kS′ ( T1 − T2 )
=
q′ kS′ ( T1 − T2 )
=
=
q′ 50 W m ⋅ K × 0.70 (100 − =
0 ) K 3500 W m =
q′ 50 W m ⋅ K × 0.90 (100 − =
0 ) K 4500 W m
<
COMMENTS: Using a finite-element package with a fine mesh, we determined heat rates of 4780 and
4575 W/m, respectively, for Figures (a) and (b). The estimate for the less difficult Figure (b) is within
2% of the numerical method result. For Figure (a), our flux plot result was 27% low.
PROBLEM 5.1
KNOWN: Electrical heater attached to backside of plate while front surface is exposed to
convection process (T∞,h); initially plate is at a uniform temperature of the ambient air and
suddenly heater power is switched on providing a constant q′′o .
FIND: (a) Sketch temperature distribution, T(x,t), (b) Sketch the heat flux at the outer surface,
q′′x ( L,t ) as a function of time.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible
heat loss from heater through insulation.
ANALYSIS: (a) The temperature distributions for four time conditions including the initial
distribution, T(x,0), and the steady-state distribution, T(x,∞), are as shown above.
Note that the temperature gradient at x = 0, -dT/dx)x=0, for t > 0 will be a constant since the
flux, q′′x ( 0 ) , is a constant. Noting that To = T(0,∞), the steady-state temperature distribution
will be linear such that
T − T ( L,∞ )
q′′o k o
=
= h T ( L,∞ ) − T∞  .
L
(b) The heat flux at the front surface, x = L, is given by q′′x ( L,t ) = −k ( dT/dx ) x=L . From the
temperature distribution, we can construct the heat flux-time plot.
COMMENTS: At early times, the temperature and heat flux at x = L will not change from
their initial values. Hence, we show a zero slope for q′′x ( L,t ) at early times. Eventually, the
value of q′′x ( L,t ) will reach the steady-state value which is q′′o .
PROBLEM 5.2
KNOWN: Plane wall whose inner surface is insulated and outer surface is exposed to an
airstream at T∞. Initially, the wall is at a uniform temperature equal to that of the airstream.
Suddenly, a radiant source is switched on applying a uniform flux, q′′o , to the outer surface.
FIND: (a) Sketch temperature distribution on T-x coordinates for initial, steady-state, and two
intermediate times, (b) Sketch heat flux at the outer surface, q′′x ( L,t ) , as a function of time.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal
generation, E g = 0, (4) Surface at x = 0 is perfectly insulated, (5) All incident radiant power is
absorbed and negligible radiation exchange with surroundings.
ANALYSIS: (a) The temperature distributions are shown on the T-x coordinates and labeled
accordingly. Note these special features: (1) Gradient at x = 0 is always zero, (2) gradient is
more steep at early times and (3) for steady-state conditions, the radiant flux is equal to the
convective heat flux (this follows from an energy balance on the CS at x = L),
q ′′o q ′′conv
=
= h [ T ( L,∞ ) − T∞ ] .
(b) The heat flux at the outer surface, q′′x ( L,t ) , as a function of time appears as shown above.
COMMENTS: The sketches must reflect the initial and boundary conditions:
T(x,0) = T∞
∂ T
0
−k
x=0 =
∂ x
∂ T
−k
= h T ( L,t ) − T∞  − q′′o
x=L
∂ x
uniform initial temperature.
insulated at x = 0.
surface energy balance at x = L.
PROBLEM 5.3
KNOWN: Microwave and radiant heating conditions for a slab of beef.
FIND: Sketch temperature distributions at specific times during heating and cooling.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Uniform internal heat generation
for microwave, (3) Uniform surface heating for radiant oven, (4) Heat loss from surface of
meat to surroundings is negligible during the heating process, (5) Symmetry about midplane.
ANALYSIS:
COMMENTS: (1) With uniform generation and negligible surface heat loss, the temperature
distribution remains nearly uniform during microwave heating. During the subsequent surface
cooling, the maximum temperature is at the midplane.
(2) The interior of the meat is heated by conduction from the hotter surfaces during radiant
heating, and the lowest temperature is at the midplane. The situation is reversed shortly after
cooling begins, and the maximum temperature is at the midplane.
PROBLEM 5.4
KNOWN: Stainless steel disk of known thickness, radius, emissivity, and initial temperature.
Environment and surroundings temperatures and heat transfer coefficients.
FIND: Derive differential equation governing transient temperature distribution, T(r,t). List initial and
boundary conditions.
SCHEMATIC:
Tsur
ε
Initial temperature, Ti
ht, T∞
hb, T∞
Side view
Top view
r
ro
ASSUMPTIONS: (1) Uniform properties. (2) Temperature is nearly uniform across the disk thickness.
(3) The surroundings are large relative to the disk. (4) Convection and radiation from the rim are
negligible because it is so thin.
ANALYSIS: Consider a differential
control volume that is a ring of width dr
and thickness b, as shown. Assuming the
temperature variation through the
thickness is negligible, and applying
conservation of energy to the differential
control volume:
ht, T∞
hb, T∞
dqrad
dqconv, t
Side view
qr+dr
qr
•
dEst
,
E in − E out =
dt
dqconv, b
Est
r cV
qr − qr + dr − dqconv,b − dqconv,t − dqrad =
∂T
∂t
where
V=
2π rdrb,
−k 2π rb
qr =
dqconv,b =hb 2π rdr (T − T∞ ),
∂T
,
∂r
dqconv,t
∂qr
dr ,
∂r
∂
∂T 
qr − qr + dr =
 k 2π rb
 dr
∂r 
∂r 
=ht 2π rdr (T − T∞ ), dqrad =εs 4π rdr (T 4 − Tsur4 )
qr + dr =
qr +
Combining these terms,
∂
∂T 
∂T
4
4
r c 2π rdrb
 k 2π rb
 dr −  hb (T − T∞ ) + ht (T − T∞ ) + 2εs (T − Tsur )  2π rdr =
∂r 
∂r 
∂t
Dividing by 2π rdrb yields the differential equation for T(r,t):
Continued…
PROBLEM 5.4 (Cont.)
∂T 1 ∂  ∂T
rc
=
 kr
∂t r ∂r  ∂r
 1
4
4
 −  hb (T − T∞ ) + ht (T − T∞ ) + 2εs (T − Tsur ) 
 b
<
The initial condition expresses that the disk is initially at temperature Ti, while the boundary conditions
express symmetry about r = 0 and an adiabatic outer edge.
∂T
∂T
= 0,
= 0
∂r r 0=
∂r r ro
=
T (r ,=
t 0)= Ti ,
<
COMMENTS: (1) The assumption of negligible temperature variation through the thickness makes
this an extended surface problem, with transient behavior. (2) The boundary condition at the outer edge
could take convection and radiation into account. Assuming a heat transfer coefficient he for the edge,
∂T
the boundary condition would become −k = he (T (ro , t ) − T∞ ) + es (T 4 (ro , t ) − Tsur4 ) .
∂r r =ro
PROBLEM 5.5
PROBLEM 5.5 (Cont.)
PROBLEM 5.5 (Cont.)
PROBLEM 5.5 (Cont.)
PROBLEM 5.5 (Cont.)
PROBLEM 5.6
PROBLEM 5.7
PROBLEM 5.7 (Cont.)
PROBLEM 5.8
PROBLEM 5.9
PROBLEM 5.10
KNOWN: Thickness and properties of flaked food product. Conveyor length. Initial flake temperature.
Ambient temperature and convection heat transfer coefficient. Final product temperature.
FIND: Required conveyor velocities for thick and thin flakes.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties. (2) Lumped capacitance behavior. (3) Negligible radiation
heat transfer. (4) Negligible moisture evaporation from product. (5) Negligible conduction between
flake and conveyor belt.
PROPERTIES: Flake: r = 700 kg/m3, cp = 2400 J/kg⋅K, and k = 0.34 W/m⋅K.
ANALYSIS: The Biot number is
hL 55 W/m 2 ⋅ K × 0.7 × 10−3 m
Bi =
=
= 0.113
0.34 W/m ⋅ K
k
Hence the lumped capacitance assumption is valid. The required heating time is
t
=
Vc θi
Lc θi 700 kg/m3 × 0.7 × 10−3 m × 2400 J/kg ⋅ K ( 20 − 300 )
ρρ
ln
ln
ln
=
=
= 27 s
hAs θ
h
θ
( 220 − 300 )
55 W/m 2 ⋅ K
<
Therefore the required conveyor velocity is V = Lo/t = 3m/23s = 0.11 m/s.
If the flake thickness is reduced to 2L = 1 mm, the lumped capacitance approximation remains valid and
<
the heating time is 19 s. The associated conveyor velocity is 0.16 m/s.
COMMENTS: (1) Assuming large surroundings, a representative value of the radiation heat transfer
(
)
(
)
coefficient is hr = σ (Ti + To ) Ti2 + To2 = 5.67 × 10−8 W/m 2 ⋅ K 4 ( 293 + 573) 2932 + 5732 K 4 = 20.3
W/m ⋅K. Radiation heat transfer would be significant and would serve to increase the product heating
rate, increasing the allowable conveyor belt speed. (2) The food product is likely to enter the oven in a
moist state. Additional thermal energy would be required to remove the moisture during heating,
reducing the rate at which the product temperature increases. (3) The effects noted in Comments 1 and 2
would tend to offset each other. A detailed analysis would be required to assess the impact of radiation
and evaporation on the required conveyor velocity.
2
PROBLEM 5.11
KNOWN: Thickness, surface area, and properties of iron base plate. Heat flux at inner surface.
Temperature of surroundings. Temperature and convection coefficient of air at outer surface.
FIND: Time required for plate to reach a temperature of 135°C. Operating efficiency of iron.
SCHEMATIC:
Tsur =
Iron
r = 2800 kg/m3
c = 900 J/kg-K
25oC
k = 180 W/m-K
ε = 0.80
qh = 1.25x10
1.5 × 1044W/m
W/m2 2
As = 0.04 m2
Air
L = 0.007 m
Too = 25oC
h = 10 W/m2-K
ASSUMPTIONS: (1) Radiation exchange is between a small surface and large surroundings, (2)
Convection coefficient is independent of time, (3) Constant properties, (4) Iron is initially at room
temperature (Ti = T∞).
ANALYSIS: Biot numbers may be based on convection heat transfer and/or the maximum heat
transfer by radiation, which would occur when the plate reaches the desired temperature (T = 135°C).
(
2
2
)
From Eq. (1.9) the corresponding radiation transfer coefficient is hr = εs(T +Tsur) T + Tsur = 0.8 ×
-8
2
4
2
2
2
2
5.67 × 10 W/m ⋅K (408 + 298) K (408 + 298 ) K = 8.2 W/m ⋅K. Hence,
2
hL 10 W / m ⋅ K ( 0.007m )
Bi
= =
= 3.9 × 10−4
k
180 W / m ⋅ K
Bi=
r
2
h r L 8.2 W / m ⋅ K ( 0.007m )
=
= 3.2 ×10−4
k
180 W / m ⋅ K
With convection and radiation considered independently or collectively, Bi, Bir, Bi + Bir << 1 and the
lumped capacitance analysis may be used.
The energy balance, Eq. (5.15), associated with Figure 5.5 may be applied to this problem. With
∀
=
Eg
0 and
=
∀ As L, the equation is
(
)
4 =
q′′ − h ( T − T∞ ) − εs T 4 − Tsur
r cL
h
dT
dt
Integrating numerically using IHT, we obtain, for T = 135°C,
t = 138 s
COMMENTS: Note that, if heat transfer is by natural convection, h, like hr, will vary during the
process from a value of 0 at t = 0 to a maximum at t = 138 s.
<
PROBLEM 5.12
PROBLEM 5.13
PROBLEM 5.13 (Cont.)
PROBLEM 5.14
PROBLEM 5.15
PROBLEM 5.16
PROBLEM 5.17
PROBLEM 5.18
PROBLEM 5.18 (Cont.)
PROBLEM 5.19
KNOWN: Initial length, density and specific heat of self-assembled molecular chains. Time constant
of the molecules’ vibrational response.
FIND: Value of the contact resistance at the metal-molecule interface.
SCHEMATIC:
ASSUMPTIONS: (1) Molecules lose no thermal energy to surroundings. (2) Lumped capacitance
behavior, (3) Constant properties, (4) Vibrational intensity represents temperature at the molecular
scale, (5) Cylindrical molecule geometry.
=
t t R=
Rt",cCt / Ac where
ANALYSIS: From Equation 5.7,
=
Ct ρρ
=
Vc p
Ac Lc p is the lumped
t ,c Ct
thermal capacitance and Rt",c is the thermal contact resistance. Combining the preceding two equations
yields
=
Rt",c
tt
5 × 10-12s
=
= 4.6 × 10-9 m2⋅K/W
ρ Lc p 180 kg/m3 × 2 × 10-9 m × 3000 J/kg ⋅ K
<
COMMENTS: (1) The contact resistance is very small, compared to values typical of larger systems.
Nonetheless, the contact resistance may be larger than the conduction resistance within the moledule or
thin gold film. (2) The time response is very fast, as expected at these length scales. This suggests that
computational speed using such devices will be correspondingly fast. (3) See Z. Wang, J.A. Carter, A.
Lagutchev, Y.K. Koh, N.-H. Seong, D.G. Cahill, and D.D. Dlott, “Ultrafast Flash Thermal Conductance
of Molecular Chains,” Science, Vol. 317, pp. 787-790, 2007, for details.
PROBLEM 5.20
PROBLEM 5.21
KNOWN: Diameter and thermophysical properties of alumina particles. Convection conditions
associated with a two-step heating process.
FIND: (a) Time-in-flight (ti-f) required for complete melting, (b) Validity of assuming negligible
radiation.
SCHEMATIC:
ASSUMPTIONS: (1) Particle behaves as a lumped capacitance, (2) Negligible radiation, (3) Constant
properties.
ANALYSIS: (a) The two-step process involves (i) the time t1 to heat the particle to its melting point
and (ii) the time t2 required to achieve complete melting. Hence, ti-f = t1 + t2, where from Eq. (5.5),
t1
=
t1=
ρρ
T −T
p Vcp θ i
p Dpcp
ln
ln i ∞
=
hAs
θ
6h
Tmp − T∞
)
(
3970 kg m3 50 ×10−6 m 1560 J kg ⋅ K
(
6 30, 000 W m 2 ⋅ K
)
ln
( 300 − 10, 000 ) =
( 2318 − 10, 000 )
4 ×10−4 s
Performing an energy balance for the second step, we obtain
t1 + t 2
∫t1
q conv dt = ∆Est
where qconv = hAs(T∞ - Tmp) and ∆Est = rpVhsf. Hence,
(
)
3970 kg m3 50 ×10−6 m
ρp Dp
h sf
3.577 ×106 J kg
×
=
t2 = =
5 ×10−4 s
2
−
6h T∞ − Tmp
10,
000
2318
K
(
)
6 30, 000 W m ⋅ K
(
)
(
)
Hence t i − f =
9 ×10−4 s ≈ 1ms
<
8
2
(b) Contrasting the smallest value of the convection heat flux, q′′conv ,min =h ( T∞ − Tmp ) =2.3 × 10 W m
(
)
4
4
to the largest radiation flux, q′′rad,max
= 6.7 × 105 W/m2, with ε =0.41 from Table
= εs Tmp
− Tsur
A.11 for aluminum oxide at 1500 K, and Tsur = 300 K we conclude that radiation is, in fact, negligible.
COMMENTS: (1) Since Bi = (hrp/3)/k ≈ 0.02, the lumped capacitance assumption is good. (2) In an
actual application, the droplet should impact the substrate in a superheated condition (T > Tmp), which
would require a slightly larger ti-f.
PROBLEM 5.22
KNOWN: Dimensions and operating conditions of an integrated circuit.
FIND: Steady-state temperature and time to come within 1°C of steady-state.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties, (2) Negligible heat transfer from chip to substrate.
3
PROPERTIES: Chip material (given): r = 2000 kg/m , c = 700 J/kg⋅K.
ANALYSIS: At steady-state, conservation of energy yields
−E out + E g =
0
( )
( )
−h L2 ( Tf − T∞ ) + q L2 ⋅ t =
0

qt
T=
f T∞ +
h
9 ×106 W/m3 × 0.001 m
Tf =
20 C +
=
80 C.
2
150 W/m ⋅ K
<
From the general lumped capacitance analysis, Equation 5.15 reduces to
dT
ρ L2 ⋅ t c = q L2 ⋅ t − h ( T − T∞ ) L2 .
dt
With
( )
( )
h
=
a≡
ρ tc
q
=
b≡
ρc
(
(
150 W/m 2 ⋅ K
= 0.107 s-1
2000 kg/m3 ( 0.001 m )( 700 J/kg ⋅ K )
)
9 ×106 W/m3
= 6.429 K/s.
2000 kg/m3 ( 700 J/kg ⋅ K )
)
From Equation 5.24,
exp
=
( −at )
T − T∞ − b/a
=
Ti − T∞ − b/a
− 60 ) K
( 79 − 20=
( 20 − 20 − 60 ) K
ln ( 0.01667 )
t=
=
38.3 s.
0.107 s-1
0.01667
<
COMMENTS: Due to additional heat transfer from the chip to the substrate, the actual values
of Tf and t are less than those which have been computed.
PROBLEM 5.23
PROBLEM 5.23 (Cont.)
PROBLEM 5.24
PROBLEM 5.25
PROBLEM 5.25 (Cont.)
PROBLEM 5.26
KNOWN: Mass and initial temperature of frozen ground beef. Temperature and convection coefficient
of air. Rate of microwave power absorbed in beef.
FIND: (a) Time for beef to reach 0°C, (b) Time for beef to be heated from liquid at 0°C to 80°C, and
(c) Explain nonuniform heating in microwave and reason for low power setting for thawing.
SCHEMATIC:
Beef, 1 kg
Ti = -20°C
q
Air
T∞ = 30°C
h = 15 W/m2∙K
ASSUMPTIONS: (1) Beef is nearly isothermal, (2) Beef has properties of water (ice or liquid), (3)
Radiation is negligible, (4) Constant properties (different for ice and liquid water).
PROPERTIES: Table A.3, Ice (≈ 273 K): ρ = 920 kg/m3, c = 2040 J/kg∙K, Table A.6, Water (≈ 315
K): c = 4179 J/kg∙K.
ANALYSIS: (a) We apply conservation of energy to the beef
E in + E g = E st
dT
dt
The initial condition is T(0) = Ti. This differential equation can be solved by defining
q
θ = T - T∞ hAs
hAs
dθ
Then Eq.(1) becomes
=θ
dt
mc
Separating variables and integrating,
θ(t) dθ
hAs t
=dt
θ(0) θ
mc 0
 θ(t) 
hAs t
ln 
 =mc
 θ(0) 
hAs (T∞ - T) + q = mc
∫
(1)
∫

 T - T∞ - q/hA
hAs t
s 
ln 
 =
mc
 Ti - T∞ - q/hAs 
(2)
The heat generation rate is given by q = 0.03P = 0.03(1000 W) = 30 W. The radius of the sphere can be
found from knowledge of the mass and density:
Continued…
PROBLEM 5.26 (Cont.)
4
m = ρV = ρ π ro3
3
1/3
 3 m
ro = 

 4π ρ 
1/3
 3
1 kg 
= 
3

 4π 920 kg/m 
= 0.0638 m
Thus As = 4πro2 = 4π(0.0638 m) 2 = 0.0511 m 2
Substituting numerical values into Eq.(2), we can find the time at which the temperature reaches 0°C:
 0°C - 30°C - 30 W/(15 W/m 2 ⋅ K × 0.0511 m 2 ) 
15 W/m 2 ⋅ K × 0.0511 m 2
ln 
=
t
2
2 
1 kg × 2040 J/kg ⋅ K
 - 20°C - 3 0°C - 30 W/(15 W/m ⋅ K × 0.0511 m ) 
Thus t = 676 s = 11.3 min
<
(b) After all the ice is converted to liquid, the absorbed power is q = 0.95P = 950 W. The time for the
beef to reach 80°C can again be found from Eq.(2):
 80°C - 30°C - 950 W/(15 W/m 2 ⋅ K × 0.0511 m 2 ) 
15 W/m 2 ⋅ K × 0.0511 m 2
ln 
=
t
2
2 
1 kg × 4179 J/kg ⋅ K
 0°C - 3 0°C - 950 W/(15 W/m ⋅ K × 0.0511 m ) 
Thus t = 355 s = 5.9 min
<
(c) Microwave power is more efficiently absorbed in regions of liquid water. Therefore, if food or the
microwave irradiation is not homogeneous or uniform, the power will be absorbed nonuniformly,
resulting in a nonuniform temperature rise. Thawed regions will absorb more energy per unit volume
than frozen regions. If food is of low thermal conductivity, there will be insufficient time for heat
conduction to make the temperature more uniform. Use of low power allows more time for conduction
to occur.
COMMENTS: (1) The time needed to turn the ice at 0°C into liquid water at 0°C was not calculated.
The required energy is Q = mhfg = 1 kg × 2502 kJ/kg = 2502 kJ. The required time depends on how the
fraction of microwave power absorbed changes during the thawing process. The minimum possible
time would be tmin = 2502 kJ/950 W = 2600 s = 44 min. Therefore, the time to thaw is significant.
(2) Radiation may not be negligible. It depends on the temperature of the oven walls and the emissivity
of the beef. Radiation would contribute to heating the beef.
PROBLEM 5.27
PROBLEM 5.27 (Cont.)
PROBLEM 5.28
PROBLEM 5.29
KNOWN: Diameters, initial temperature and thermophysical properties of WC and Co in composite
particle. Convection coefficient and freestream temperature of plasma gas. Melting point and latent
heat of fusion of Co.
FIND: Times required to reach melting and to achieve complete melting of Co.
SCHEMATIC:
ASSUMPTIONS: (1) Particle is isothermal at any instant, (2) Radiation exchange with surroundings is
negligible, (3) Negligible contact resistance at interface between WC and Co, (4) Constant properties.
ANALYSIS: From Eq. (5.5), the time required to reach the melting point is
t1 =
( ρ Vc )tot
h p Do2
ln
Ti − T∞
Tmp − T∞
where the total heat capacity of the composite particle is

Vc ) tot = ( Vc )c + ( Vc )s = 16, 000 kg / m3 π
( ρρρ

(1.6 ×10−5 m ) / 6 300 J / kg ⋅ K
3
) (
(
3


+8900 kg / m3 π / 6  2.0 ×10−5 m − 1.6 ×10−5 m


)
(
)   750 J / kg ⋅ K
3 

= 1.03 ×10−8 + 1.36 ×10−8 J / K = 2.39 ×10−8 J / K
=
t1
(
2.39 × 10−8 J / K
)(
20, 000 W / m 2 ⋅ K π 2.0 × 10−5 m
)
2
ln
K
( 300 − 10, 000 )=
(1770 − 10, 000 ) K
1.56 × 10−4 s
<
The time required to melt the Co may be obtained by applying the first law, Eq. (1.12b) to a
control surface about the particle. It follows that
(
)
(
)
3
3
DEst =
Ein =
hpρp
Do2 T∞ − Tmp t 2 =
s ( / 6 ) Do − Di h sf
( 2 × 10 m ) − (1.6 × 10 m )  2.59 × 10
( 20, 000 W / m ⋅ K ) π ( 2 × 10 m ) (10, 000 − 1770 ) K
8900 kg / m
t2
=
3

−5
(π / 6 ) 
2
3
−5
−5
2
3
5
J / kg
= 2.28 × 10
−5
s
<
COMMENTS: (1) The largest value of the radiation coefficient corresponds to hr = εs (Tmp + Tsur)
(T
)
2
2
2
mp + Tsur . For the maximum possible value of ε = 1 and Tsur = 300K, hr = 378 W/m ⋅K << h =
2
20,000 W/m ⋅K. Hence, the assumption of negligible radiation exchange is excellent. (2) Despite the
large value of h, the small values of Do and Di and the large thermal conductivities (~ 40 W/m⋅K and 70
W/m⋅K for WC and Co, respectively) render the lumped capacitance approximation a good one. (3) A
detailed treatment of plasma heating of a composite powder particle is provided by Demetriou, Lavine
and Ghoniem (Proc. 5th ASME/JSME Joint Thermal Engineering Conf., March, 1999).
PROBLEM 5.30
KNOWN: Spherical ceramic particle of known properties, diameter, and initial temperature.
Environment temperature and heat transfer coefficient.
FIND: Particle temperature at t = 0.1 s for Tsur = 750 K, 850 K, and 950 K. Plot of particle temperature
vs. time.
SCHEMATIC:
Tsur
T ∞, h
+
D
Ti
D=300 μm
Ti = 1100K
T∞ = 1200K
Tsur = 750, 850, or 950 K
h = 500 W/m2•K
ρ= 2500 kg/m3
c = 750 J/kg•K
k = 1.4 W/m•K
ε= 0.94
ASSUMPTIONS: (1) Radiation can be approximated as a small object in large surroundings, (2)
Uniform properties, (3) The lumped capacitance approximation is valid.
PROPERTIES: Given, r = 2500 kg/m3, cp =750 J/kg⋅K, k = 1.4 W/m⋅K, ε = 0.94.
ANALYSIS: The Biot number is
Bi
=
hLc h(V / As )
=
=
k
k
1 hro 1 500 W/m 2 ⋅ K × 150 × 10−6 m
=
= 0.018
3 k
3
1.4 W/m 2 ⋅ K
Hence the lumped capacitance approximation is valid. The temperature of the sphere is governed by
Equation 5.15 with qs′′ and E g both zero. Thus,
rVc
dT
=
−  h(T − T∞ ) + εs (T 4 − Tsur4 )  As
dt
(1)
This differential equation can be solved numerically subject to the initial condition T(t = 0) = Ti = 1100
K. Using the Der function of IHT (see code in Comments section), the particle temperature at t = 0.1 s
for the three surroundings temperatures of 750, 850, and 950 K are:
Tsur (K)
750
850
950
T(t = 0.1 s) (K)
1092
1100
1111
<
The particle temperature as a function of time is shown below for the three surroundings temperatures.
Continued…
PROBLEM 5.30 (Cont.)
COMMENTS: (1) The IHT code is given below. (2) Depending on the surroundings temperature, the
particle temperature can decrease, stay the same, or increase. (3) The steady-state temperature can be
found by setting dT/dt = 0 in Eq. (1) and solving for Tss. The results for the three surroundings
temperatures are Tss = 1086, 1100, and 1119 K for Tsur = 750, 850, and 950 K, respectively.
D = 300e-6
ro = D/2
k = 1.4
cp = 750
rho = 2500
h = 500
Tinf = 1200
Ti = 1000
eps = 0.94
sigma=5.67e-8
Tsur1 = 750
Tsur2 = 850
Tsur3 = 950
rho*Vol*cp*Der(T1,t) = -Area*(h*(T1 - Tinf) + eps*sigma*(T1^4 - Tsur1^4))
rho*Vol*cp*Der(T2,t) = -Area*(h*(T2 - Tinf) + eps*sigma*(T2^4 - Tsur2^4))
rho*Vol*cp*Der(T3,t) = -Area*(h*(T3 - Tinf) + eps*sigma*(T3^4 - Tsur3^4))
Vol = (4/3)*pi*ro^3
Area = 4*pi*ro^2
PROBLEM 5.31
KNOWN: Diameter of highly polished aluminum rod. Temperature of rod initially and at two later
times. Room air temperature.
FIND: Values of constants C and n in Equation 5.26. Plot rod temperature vs. time for varying and
constant heat transfer coefficients.
SCHEMATIC:
t1 = 1250s t2 = 6700s
T1 = 65°C T2 = 30°C
D = 35 mm
Ti = 90°C
T∞ = 20°C
ASSUMPTIONS: (1) Constant properties, (2) Radiation negligible because rod is highly polished, (3)
Lumped capacitance approximation is valid.
PROPERTIES: Table A.1, Aluminum (T = 328 K): c = 916 J/kg⋅K, r = 2702 kg/m3, k = 238 W/m⋅K.
ANALYSIS: If the heat transfer coefficient is given by Equation 5.26, then the temperature as a
function of time is given by Equation 5.28:

θ  nCAs ,cθin
t + 1
= 
θi  ρVc

−1/ n
(1)
where θ= T − T∞ and As,c is the area exposed to convection, As,c = πDL. Since the rod temperature is
known at two different times, Equation (1) can be evaluated at these two times, making it possible to
solve for the two unknowns, C and n. The two equations are
−1/ n
n
n


θ 2  nCAs ,cθi
θ1  nCAs ,cθi
= 
t2 + 1
t1 + 1 = 
θi  ρVc
θi  ρVc


−1/ n
(2a,b)
These equations cannot be explicitly solved for C and n. They can be numerically solved in this form,
using IHT or some other software, or they can be further manipulated to solve for the times:
 θ − n  ρ cD
 θ − n  ρ cD
1
t1   − 1
=
t2  2  − 1
=
n
n
 θi 
 4nCθi
 θi 
 4nCθi
(3a,b)
where we have used V/As,c = D/4. Taking the ratio of Equations (3a) and (3b) yields
−n
 θ2 
  −1
t2  θ i 
=
t1  θ  − n
1
  −1
θ
 i
−n
 30°C − 20°C 
−1
6700 s  90°C − 20°C 
=
1250 s  65°C − 20°C  − n

 −1
 90°C − 20°C 
( 0.143) − 1
5.36 =
−n
( 0.643) − 1
−n
(4a,b,c)
Continued...
PROBLEM 5.31 (Cont.)
This can be iteratively or numerical solved for n, to find n = 0.25. Then C can be determined from
Equation (3a) or (3b):
 θ − n  ρ cD  45°C −0.25  2702 kg/m3 × 916 J/kg ⋅ K × 0.035 m
=
− 1
=2.8 W/m 2 ⋅ K1.25
C = 1  − 1

n
0.25
°
×
×
°
×
θ
θ
n
t
4
70
C
4
0.25
(70
C)
1250
s

 i 



i 1
<
=
C 2.8 W/m 2 ⋅ K1.25 , n = 0.25
Now that these constants are known, the validity of the lumped capacitance approximation can checked.
The maximum heat transfer coefficient occurs at the initial time,
h = C (T − T∞ ) n = 2.8 W/m 2 ⋅ K1.25 [ (90 − 20)K ]
0.25
= 8.1 W/m 2 ⋅ K
Thus, using the conservative definition, Bi = hD/2k = 6 × 10-4. The lumped capacitance approximation
is valid.
T
The heat transfer coefficient corresponding to a rod temperature of =
h = C (T − T∞ ) n = 2.8 W/m 2 ⋅ K1.25 [ (55 − 20)K ]
0.25
(Ti + T∞ ) / 2 = 55°C is
= 6.8 W/m 2 ⋅ K
The plot below shows the rod temperature as a function of time using Equation (1) above for
variable heat transfer coefficient, as well as the rod temperature assuming the constant value of
h = 6.8 W/m2⋅K, using text Equation 5.6.
1
0.9
0.8
constant h
0.7
variable h
0.6
θ/θ i 0.5
0.4
0.3
0.2
0.1
0
0
2000
4000
6000
8000
10000
t (s)
<
COMMENTS: (1) Since the heat transfer coefficient is temperature difference-dependent (variable h),
the initial cooling rates are larger when this dependence is accounted for. As the temperature difference
decreases, the variable h case cools slower relative to the constant h case. (2) The discrepancy between
the variable and constant heat transfer coefficient cases is not large under these conditions. The
difference would be greater if n were larger.
PROBLEM 5.32
PROBLEM 5.33
KNOWN: Ambient conditions, initial water droplet temperature and diameter.
FIND: Total time to completely freeze the water droplet for (a) droplet solidification at Tf = 0°C and (b)
rapid solidification of the droplet at Tf,sc.
SCHEMATIC:
T8 = - 40°C
h = 900 W/m2•K
Ti = 10°C
D = 50 µm
ASSUMPTIONS: (1) Isothermal particle, (2) Negligible radiation heat transfer, (3) Constant
properties.
PROPERTIES: Table A.6, liquid water (T = 0 °C): cp = 4217 J/kg⋅K, k = 0.569 W/m⋅K, ρ = 1000
kg/m3. Example 1.5: hsf = 334 kJ/kg.
ANALYSIS: We begin by evaluating the validity of the lumped capacitance method by determining the
value of the Biot number.
Bi =
hLc
hD/3
900 W/m 2 ⋅ K × 50 × 10-6 m/3
=
=
= 0.026 << 0.1
k
k
0.569 W/m ⋅ K
Hence, the lumped capacitance approach is valid.
Case A: Equilibrium solidification, Tf = 0°C.
The solidification process occurs in two steps. The first step involves cooling the drop to Tf = 0°C
while the drop is completely liquid. Hence, Equation 5.6 is used where
A = πD 2 = π × (50 × 10-6 m)2 = 7.85 × 10-9 m 2 and
V = 4π(D/2)3 3 = 4 × π × (50 × 10-6 m/2)3 /3 = 65.4 × 10-15 m3 . Equation 5.6 may be rearranged to
yield
(1)
ρVc  T - T∞ 
t1 = ln 

hA  Ti - T∞ 
=-
1000 kg/m3 × 65.4 × 10-15 m3 × 4217 J/kg ⋅ K
2
-9
900 W/m × 7.85 × 10 m
2
 0 - (- 40°C) 
× ln 

10°C - (- 40°C) 
t1 = 8.7 × 10-3 s = 8.7 ms
Continued…
PROBLEM 5.33 (Cont.)
The second step involves solidification of the ice, which occurs at Tf = 0°C. An energy balance on the
droplet yields
-E out = ΔEst or - hA(Tf - T∞ )t 2 = ρVh sf
which may be rearranged to provide
ρVh sf
t2 = hA(Tf - T∞ )
=
(2)
1000 kg/m3 × 65.4 × 10-15 m3 × 334,000 J/kg
2
900 W/m × 7.85 × 10
-9
2
= 77.3 × 10-3 s = 77.3 ms
m × (0°C - (- 40)°C)
The time needed to cool and solidify the particle is
<
t = t1 + t 2 = 8.7 ms + 77.3 ms = 86 ms
Case B: Rapid solidification at Tf,sc.
Using the expression given in the problem statement, the liquid droplet is supercooled to a temperature
of Tf,sc prior to freezing.
Tf,sc = - 28 + 1.87ln(50 × 10-6 m) = - 36.6°C
The solidification process occurs in multiple steps, the first of which is cooling the particle to Tf,sc = 36.6°C. Substituting T = Tf,sc into Equation 1 yields
t1 = 105 × 10-3 s = 105 ms
The second step involves rapid solidification of some or all of the supercooled liquid. An energy
balance on the particle yields
E st = 0 = ρVh sf f = ρVc(Tf - Tf,sc )
(3)
where f is the fraction of the mass in the droplet that is converted to ice. Solving the preceding equation
for f yields
c(Tf - Tf,sc )
4217 J/kg ⋅ K × (0°C - (- 36.6°C))
f=
=
= 0.462
h sf
334,000 J/kg
Hence, immediately after the rapid solidification, the water droplet is approximately 46% ice and 54%
liquid. The time required for the rapid solidification is t2 ≈ 0 s.
The third stage of Case B involves the time required to freeze the remaining liquid water, t3. Equation 2
is modified accordingly to yield
(1-f)ρVh sf
t3 = hA(Tf - T∞ )
=-
(1 - 0.462) × 1000 kg/m3 × 65.4 × 10-15 m3 × 334,000 J/kg
2
-9
2
900 W/m × 7.85 × 10 m × (0°C - (- 40)°C)
= 42 × 10-3 s = 42 ms
Continued…
PROBLEM 5.33 (Cont.)
The total time to solidify the particle is
<
t = t1 + t2 + t3 = 105 ms + 0 + 42 s = 147 ms
The temperature histories associated with Case A and Case B are shown in the sketch below.
20
f=1
T, °C
0
Case A
f = 0.462
-20
Case B
-40
-60
0
50
100
150
200
t, ms
COMMENTS: (1) Equation 3 may be derived by assuming a reference temperature of Tf = 0 °C and a
liquid reference state. The energy of the particle prior to the rapid solidification is E1 = ρVc(Tf,sc - Tf).
The energy of the particle after the rapid solidification is E2 = -fρVhsf + (1 - f) ρVc(Tf – Tf) = -fρVhsf.
Setting E1 = E2 yields Equation 3. (2) The average temperature of the supercooled particle is
significantly lower than the average temperature of the particle of Case A. Hence, the rate at which the
supercooled particle of Case B is cooled by the cold air is, on average, much less than the particle of
Case A. Since both particles ultimately reach the same state (all ice at T = 0 °C), it takes longer to
completely solidify the supercooled particle. (3) For Case A, the ice particle at T = 0 °C will be a solid
sphere, sometimes referred to as sleet. For Case B, the rapid solidification will result in a snowflake.
PROBLEM 5.34
KNOWN: Series solution, Eq. 5.42, for transient conduction in a plane wall with convection.
FIND: Midplane (x*=0) and surface (x*=1) temperatures θ* for Fo=0.1 and 1, using Bi=0.1, 1 and 10
with only the first four eigenvalues. Based upon these results, discuss the validity of the approximate
solutions, Eqs. 5.43 and 5.44.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties.
ANALYSIS: The series solution, Eq. 5.42a, is of the form,
θ* =
∞
∑ Cn
n =1
) (
(
exp -ζ n2 Fo cos ζ n x*
)
where the eigenvalues, ζ n , and the constants, Cn, are from Eqs. 5.39b and 5.39c.
ζ n tan ζ n Bi
Cn 4sin ζ n / ( 2ζ n + sin ( 2ζ n ) ) .
=
=
The eigenvalues are tabulated in Appendix B.3; note, however, that ζ1 and C1 are available from
Table 5.1. The values of ζ n and Cn used to evaluate θ* are as follows:
Bi
ζ1
0.1
1
10
0.3111
0.8603
1.4289
C1
ζ2
C2
ζ3
C3
1.0160
1.1191
1.2620
3.1731
3.4256
4.3058
-0.0197
-0.1517
-0.3934
6.2991
6.4373
7.2281
0.0050
0.0466
0.2104
ζ4
C4
9.4354
9.5293
10.2003
-0.0022
-0.0217
-0.1309
Using ζ n and Cn values, the terms of θ * , designated as θ1* , θ 2* , θ 3* and θ 4* , are as follows:
Fo=0.1
Bi=1.0
Bi=0.1
x*
0
1
0
Bi=10
1
0
1
θ1*
θ 2*
1.0062
0.9579
1.0393
0.6778
1.0289
0.1455
-0.0072
0.0072
-0.0469
0.0450
-0.0616
0.0244
θ 3*
0.0001
0.0001
0.0007
0.0007
0.0011
0.0006
θ 4*
-2.99×10-7
3.00×10-7
2.47×10-6
2.46×10-7
-3.96×10-6
2.83×10-6
θ*
0.9991
0.9652
0.9931
0.7235
0.9684
0.1705
Continued …
PROBLEM 5.34 (Cont.)
Fo=1
Bi=0.1
x*
θ1*
θ 2*
θ 3*
θ 4*
*
θ
0
Bi=1.0
1
0.9223
0
0.8780
-7
8.35×10
0.5339
-7
0
0.3482
-5
8.35×10
Bi=10
1
-1.22×10
1.17×10
1
0.1638
-6
0.0232
-9
3.49×10
1.38×10-9
7.04×10-20
-
4.70×10-20
-
4.30×10-24
-
4.77×10-42
-
7.93×10-42
-
8.52×10-47
-
0.9223
0.8780
0.5339
0.3482
0.1638
0.0232
)
(
The tabulated results for θ * = θ * x* , Bi, Fo demonstrate that for Fo=1, the first eigenvalue is
sufficient to accurately represent the series. However, for Fo=0.1, three eigenvalues are required
for accurate representation.
A more detailed analysis would show that a practical criterion for representation of the series
solution by one eigenvalue is Fo > 0.2. For these situations the approximate solutions, Eqs. 5.43
and 5.44, are appropriate. For the midplane, x*=0, the first two eigenvalues for Fo=0.2 are:
Fo=0.2
x*=0
Bi
0.1
1.0
10
θ1*
0.9965
0.9651
0.8389
θ 2*
-0.00226
-0.0145
-0.0096
θ*
0.9939
0.9506
0.8293
+0.26
+1.53
+1.16
Error,%
The percentage error shown in the last row of the above table is due to the effect of the second
term. For Bi = 0.1, neglecting the second term provides an error of 0.26%. For Bi = 1, the error
is 1.53%.
Hence we conclude that the approximate series solutions (with only one eigenvalue) provides
systematically high results, but by less than 1.5%, for the Biot number range from 0.1 to 10.
PROBLEM 5.35
PROBLEM 5.36
PROBLEM 5.37
PROBLEM 5.38
KNOWN: Thickness and initial temperature of acrylic sheet.
FIND: Time needed to bring the external surface of the acrylic to its softening temperature.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) One-term approximate
solution is valid.
PROPERTIES: Acrylic (given): r = 1990 kg/m3, c = 1470 J/kg⋅K and k = 0.21 W/m⋅K.
ANALYSIS: For the constant temperature boundary condition, the Biot number is Bi → ∞. Hence, from
Table 5.1 for the plane wall, ζ1 = 1.5708, C1 = 1.2733. The dimensionless external surface temperature at
the time of interest is
90°C − 300°C
=0.75 =C1 exp( −ζ 12 Fo) =
1.2733exp( −1.57082 Fo)
20°C − 300°C
θo* =
From which Fo = 0.214. Hence,
t = FoL2/α = FoL2rc/k = [0.214 × (0.004 m)2 × 1990 kg/m3 × 1470 J/kg⋅K]/0.21 W/m⋅K = 48 s
<
COMMENTS: (1) Since Fo = 0.214 is greater than 0.2, the one-term approximation is valid. (2) A
contact resistance would be present at the interface between the acrylic and the substrate. However, as the
acrylic softens and deforms locally to make better contact with the substrate, the thermal contact
resistance would decrease in value.
PROBLEM 5.39
PROBLEM 5.40
KNOWN: Initial temperature of concrete slabs. Air temperature and convection heat transfer
coefficient.
FIND: Slab thickness required so that Q/Qo = 0.90 for t = 8 h.
SCHEMATIC:
Q/Qo = 0.9
at t = 8 h
T∞ = 200°C, h = 35 W/m2∙K
T∞, h
2L
From
solar
collector
Concrete slabs
T∞, h
2L
T∞, h
Ti = 40°C
ASSUMPTIONS: (1) Properties at 300 K are satisfactory at higher temperature, (2) One-dimensional
conduction, (3) Constant convection heat transfer coefficient, (4) Negligible radiation because each
concrete slab is surrounded by others at the same temperature.
PROPERTIES: Table A.3, Concrete (stone mix) (T = 300 K): r = 2300 kg/m3, c = 880 J/kg⋅K, k = 1.4
W/m⋅K.
ANALYSIS: Without knowing the thickness, we cannot determine in advance whether the lumped
capacitance approximation is valid. Considering the small thermal conductivity of concrete, we might
anticipate that the Biot number will not be small. Considering the long time period, we may anticipate
that the Fourier number is large. Therefore, we will begin by using the first-term approximation of the
series solution and check its validity later. From Equation 5.49
sin ζ 1 *
Q
1−
0.9
θo =
=
ζ1
Qo
(1)
where from Equation 5.44
=
θ o* C1 exp ( −ζ 12 Fo )
(2)
In these equations, C1 and ζ1 are functions of Bi:
ζ 1 tan ζ 1 = Bi ,
C1 =
4sin ζ 1
2ζ 1 + sin(2ζ 1 )
(3,4)
where ζ1 is the smallest root of Equation (3). Both Bi and Fo are unknown because L is unknown. They
are given by
Bi =
hL
,
k
Fo =
αt
L2
(5,6)
Continued…
PROBLEM 5.40 (Cont.)
where α = k/rc = 6.92 × 10-7 W/m2⋅K. These six simultaneous equations can be solved
iteratively. One approach is to guess a value for ζ1, from which C1 can be calculated from
Equation (4). Equation (1) can be used to find the required value of θ o* . Then Equation (2) can
be used to determine Fo and a new value of L can be determined from Equation (6). Finally, Bi
can be calculated from Equation (5) and a new value of ζ1 can be found from Equation (3).
Beginning this approach with a guessed value of ζ1 = 1, the iterations proceed as follows:
Guess:
Equation (4):
Equation (1):
Equation (2):
Equation (6):
Equation (5):
Equation (3):
ζ1 = 1
C1 = 1.16
θ o* = 0.119
Fo = 2.28
L = 0.0936 m
Bi = 2.34
ζ1 = 1.1231
Repeating with the new value of ζ1 and iterating until L converges to two significant digits, we find
L = 0.11 m
<
with Bi = 2.7, Fo = 1.7. Thus the lumped capacitance approximation is not appropriate, and the first-term
approximation is valid.
COMMENTS: This hand-solution is time-consuming, especially since Equation (3) must itself be
solved iteratively. A much faster approach would be to solve these six equations simultaneously using
IHT or other software.
PROBLEM 5.41
KNOWN: Properties and thickness L of ceramic coating on rocket nozzle wall. Convection conditions.
Initial temperature and maximum allowable wall temperature.
FIND: (a) Maximum allowable engine operating time, tmax, for L = 10 mm, (b) Coating inner and outer
surface temperature histories for L = 10 and 40 mm.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3)
Negligible thermal capacitance of metal wall and heat loss through back surface, (4) Negligible contact
resistance at wall/ceramic interface, (5) Negligible radiation.
ANALYSIS: (a) Subject to assumptions (3) and (4), the maximum wall temperature corresponds to the
ceramic temperature at x = 0. Hence, for the ceramic, we wish to determine the time tmax at which T(0,t)
= To(t) = 1500 K. With Bi = hL/k = 5000 W/m2⋅K(0.01 m)/10 W/m⋅K = 5, the lumped capacitance
method cannot be used. Assuming Fo > 0.2, obtaining ζ1 = 1.3138 and C1 = 1.2402 from Table 5.1, and
*
evaluating θ o = ( To − T∞ ) ( Ti − T∞ ) = 0.4, Equation 5.44 yields
)
(
ln θ o* C1
ln ( 0.4 1.2402 )
Fo =
0.656
−
=
−
=
2
ζ12
(1.3138)
confirming the assumption of Fo > 0.2. Hence,
( )
2
Fo L2
0.656 ( 0.01m )
=
t max =
= 10.9s
a
6 ×10−6 m 2 s
<
(b) Using the IHT Lumped Capacitance Model for a Plane Wall, the inner and outer surface temperature
histories were computed and are as follows:
2300
Temperature, T(K)
1900
1500
1100
700
300
0
30
60
90
120
150
Time, t(s)
L
L
L
L
=
=
=
=
0.01,
0.01,
0.04,
0.04,
x
x
x
x
=
=
=
=
L
0
L
0
Continued...
PROBLEM 5.41 (Cont.)
The increase in the inner (x = 0) surface temperature lags that of the outer surface, but within t ≈ 45s both
temperatures are within a few degrees of the gas temperature for L = 0.01 m. For L = 0.04 m, the
increased thermal capacitance of the ceramic slows the approach to steady-state conditions. The thermal
response of the inner surface significantly lags that of the outer surface, and it is not until t ≈ 137s that the
inner surface reaches 1500 K. At this time there is still a significant temperature difference across the
ceramic, with T(L,tmax) = 2240 K.
COMMENTS: The allowable engine operating time increases with increasing thermal capacitance of
the ceramic and hence with increasing L.
PROBLEM 5.42
KNOWN: Plate of thickness 2L = 25 mm at a uniform temperature of 600°C is removed from a hot
pressing operation. Case 1, cooled on both sides; case 2, cooled on one side only.
FIND: (a) Calculate and plot on one graph the temperature histories for cases 1 and 2 for a 500-second
cooling period; use the IHT software; Compare times required for the maximum temperature in the plate
to reach 100°C; and (b) For both cases, calculate and plot on one graph, the variation with time of the
maximum temperature difference in the plate; Comment on the relative magnitudes of the temperature
gradients within the plate as a function of time.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the plate, (2) Constant properties, and (3) For
case 2, with cooling on one side only, the other side is adiabatic.
3
PROPERTIES: Plate (given): r = 3000 kg/m , c = 750 J/kg⋅K, k = 15 W/m⋅K.
ANALYSIS: (a) From IHT, call up Plane Wall, Transient Conduction from the Models menu. For case
1, the plate thickness is 25 mm; for case 2, the plate thickness is 50 mm. The plate center (x = 0)
temperature histories are shown in the graph below. The times required for the center temperatures to
reach 100°C are
t1 = 164 s
<
t2 = 367 s
(b) The plot of T(0, t) – T(1, t), which represents the maximum temperature difference in the plate during
the cooling process, is shown below.
Plate center temperature histories
Temperature difference history
150
600
T(0,t) - T(L,t) (C)
T(0,t) (C)
500
400
300
200
100
50
100
0
0
0
100
200
300
Time, t (s)
Cooling - both sides
Cooling - one side only
400
500
0
100
200
300
400
500
Time (s)
Cooling - both sides
Cooling - one side only
COMMENTS: (1) From the plate center-temperature history graph, note that it takes more than twice as
long for the maximum temperature to reach 100°C with cooling on only one side.
(2) From the maximum temperature-difference graph, as expected, cooling from one side creates a larger
maximum temperature difference during the cooling process. The effect could cause microstructure
differences, which could adversely affect the mechanical properties within the plate.
PROBLEM 5.43
KNOWN: Thickness and initial temperatures of two plates of the same material.
*
*
FIND: (a) Steady-state dimensionless temperatures of the two plates, Tss,1
and Tss,2
, as well as the
*
interface temperature, Tint
, (b) Expression for the effective dimensionless overall heat transfer coefficient
(
*
*
)
*
for the two-plate system, U eff
,2 ≡ q * / T 2 − T 1 for Fo > 0.2.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible thermal
contact resistance.
ANALYSIS: (a) Since the two plates are of the same thickness and have the same properties, both
plates will reach a steady-state temperature that is the average of T1 and T2. In terms of the dimensionless
*
*
= Tss,2
= 0.5.
temperature, T*(Fo) ≡ (T – T1)/(T2 – T1), this implies that Tss,1
<
*
Accounting for the symmetry about x = 0, the dimensionless interface temperature will be Tint
= 0.5 at any
<
time.
(b) Taking advantage of the geometrical symmetry about x = 0, we may simplify the problem by
analyzing just one of the plates, accounting for one adiabatic surface and a second surface being held at a
constant temperature. For the constant temperature boundary condition, Bi = hL/k → ∞ and from Table
5.1 ζ1 = π/2, C1 = 1.2733. Equations 5.44 and 5.49 may be combined to yield
Q
sin ζ 1
=
1−
C exp( −ζ 12 Fo)
Qo
ζ1 1
(1)
The spatially-averaged dimensionless temperature for one plate is
Continued...
PROBLEM 5.43 (Cont.)
θ *=
1 T ( x, t ) − Tint
dV
V
Ti − Tint
∫
where Tint is the interface temperature. From Eq. 5.46b, θ * = 1 – Q/Qo.
(2)
From Eq. (1),
d (Q / Qo )
= q=
* 2C1 ( sin ζ 1 ) exp( −ζ 12 Fo)
dFo
and from Eq. (2)
=
θ*
C1 sin ζ 1
ζ1
*
Therefore, for one plate, U eff
=
,1
exp( −ζ 12 Fo)
d (Q / Qo)
* 2=
θ=
ζ1 π
dFo
The dimensionless temperature difference for the two-plate system is T2* − T1* =
2θ * .
Hence, for the two-plane system, U*eff = U*eff, 2 /2 = U*eff,1/2 = π/2.
<
COMMENTS: (1) For this case, the heat transfer rate between the two plates is proportional to the
difference in the average temperatures of the plates. If Fo < 0.2, it may be shown that U*eff is initially
infinite and decreases with time. This behavior becomes evident if one considers the situation
immediately after the plates make contact when the heat transfer between the plates is very large, but the
average plate temperatures have not been affected significantly by the heat transfer in the vicinity of the
interface. (2) A proportionality between the dimensionless heat transfer rate and the difference in the
average dimensionless plate temperatures also exists at large Fo, even if the plates are not identical.
PROBLEM 5.44
KNOWN: Initial temperature, thickness and thermal diffusivity of glass plate. Prescribed
surface temperature.
FIND: (a) Time to achieve 75% reduction in midplane temperature, (b) Maximum temperature
gradient at that time.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.
ANALYSIS: Prescribed surface temperature is analogous to h → ∞ and T∞ = Ts. Hence, Bi =
∞. Assume validity of one-term approximation to series solution for T (x,t).
(a) At the midplane,
=
θo∗
(
To − Ts
= 0.75
= C1exp −ζ12 Fo
Ti − Ts
)
ζ1tanζ1 = Bi = ∞ → ζ1 = π / 2 = 1.5708
Hence
C=
1
4sinζ1
4
= = 1.273
2ζ1 + sin ( 2ζ1 ) π
)
(
ln θo∗ / C1
Fo =
0.214
−
=
ζ12
FoL2 0.214 ( 0.01 m )
=
t =
= 3.6 s.
α
6 ×10−7 m 2 / s
2
(
<
)
(b) With
=
θ ∗ C1exp −ζ12 Fo cosζ1x∗
(T − T )
∂ T ( Ti − Ts ) ∂θ ∗
=
=
− i s ζ1C1exp −ζ12 Fo sinζ1x∗
∂ x
L
L
∂ x∗
(
∂ T/∂ x
max
)
300a C π
=
∂ T/∂ x x∗ =1 =
−
0.75 =
−3.53 ×104 aC/m.
0.01 m 2
COMMENTS: Validity of one-term approximation is confirmed by Fo > 0.2.
<
PROBLEM 5.45
KNOWN: Thickness and properties of rubber tire. Convection heating conditions. Initial and final
midplane temperature.
FIND: (a) Time to reach final midplane temperature. (b) Effect of accelerated heating.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3)
Negligible radiation.
ANALYSIS: (a) With Bi = hL/k = 200 W/m2⋅K(0.01 m)/0.14 W/m⋅K = 14.3, the lumped capacitance
method is clearly inappropriate. Assuming Fo > 0.2, Eq. (5.44) may be used with C1 = 1.265 and ζ1 ≈
1.458 rad from Table 5.1 to obtain
θ o* =
)
(
To − T∞
= C1 exp −ζ12 Fo = 1.265exp ( −2.126 Fo )
Ti − T∞
− ln ( 0.286 1.265 ) 2.126 =
0.70 =
α tf L
With θ o =−
( To T∞ ) ( Ti − T∞ ) = (-50)/(-175) = 0.286, Fo =
2
*
0.7 ( 0.01m )
= 1100s
6.35 ×10−8 m 2 s
2
=
tf
<
(b) The desired temperature histories were generated using the IHT Transient Conduction Model for a
Plane Wall, with h = 5 × 104 W/m2⋅K used to approximate imposition of a surface temperature of 200°C.
Temperature, T(C)
200
150
100
50
0
0
400
200
600
800
1000
1200
Time, t(s)
x
x
x
x
=
=
=
=
0,
L,
0,
L,
h
h
h
h
=
=
=
=
200 W/m^2.K
200 W/m^2.K
5E4 W/m^2.K
5E4W/m^2.K
The fact that imposition of a constant surface temperature (h → ∞) does not significantly accelerate the
heating process should not be surprising. For h = 200 W/m2⋅K, the Biot number is already quite large (Bi
= 14.3), and limits to the heating rate are principally due to conduction in the rubber and not to convection
at the surface. Any increase in h only serves to reduce what is already a small component of the total
thermal resistance.
COMMENTS: The heating rate could be accelerated by increasing the steam temperature, but an upper
limit would be associated with avoiding thermal damage to the rubber.
PROBLEM 5.46
PROBLEM 5.47
KNOWN: Long rod with prescribed diameter and properties, initially at a uniform temperature, is heated
in a forced convection furnace maintained at 750 K with a convection coefficient of h = 1000 W/m2⋅K.
FIND: (a) The corresponding center temperature of the rod, T(0, to), when the surface temperature T(ro,
to) is measured as 550 K, (b) Effect of h on centerline temperature history.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial conduction in rod, (2) Constant properties, (3) Rod, when
initially placed in furnace, had a uniform (but unknown) temperature, (4) Fo ≥ 0.2.
ANALYSIS: (a) Since the rod was initially at a uniform temperature and Fo ≥ 0.2, the approximate
solution for the infinite cylinder is appropriate. From Eq. 5.52b,
(
( )
)
θ * r* , Fo = θ o* ( Fo ) J 0 ζ1r*
(1)
where, for r* = 1, the dimensionless temperatures are, from Eq. 5.34,
θ * (1, Fo ) =
T ( ro , t o ) − T∞
Ti − T∞
θ o* ( Fo ) =
T ( 0, t o ) − T∞
Ti − T∞
(2,3)
Combining Eqs. (2) and (3) with Eq. (1) and rearranging,
T ( ro , t o ) − T∞ T ( 0, t o ) − T∞
=
J 0 (ζ1 ⋅1)
Ti − T∞
Ti − T∞
1
T ( ro , t o ) − T∞ 
T ( 0, t o ) =
T∞ +
J 0 (ζ 1 ) 
(4)
The eigenvalue, ζ1 = 1.0185 rad, follows from Table 5.1 for the Biot number
2
hro 1000 W m ⋅ K ( 0.060 m 2 )
=
Bi =
= 0.60 .
k
50 W m ⋅ K
From Table B-4, with ζ1 = 1.0185 rad, J0(1.0185) = 0.7568. Hence, from Eq. (4)
T ( 0, t o ) =750 K +
1
[550 − 750] K =486 K
0.7568
<
(b) Using the IHT Transient Conduction Model for a Cylinder, the following temperature histories were
generated.
Continued...
PROBLEM 5.47 (Cont.)
Centerline temperature, To(K)
500
400
300
0
100
200
300
400
Time, t(s)
h = 100 W/m^2.K
h = 500 W/m^2.K
h = 1000 W/m^2.K
The times required to reach a centerline temperature of 500 K are 367, 85 and 51s, respectively, for h =
100, 500 and 1000 W/m2⋅K. The corresponding values of the Biot number are 0.06, 0.30 and 0.60.
Hence, even for h = 1000 W/m2⋅K, the convection resistance is not negligible relative to the conduction
resistance and significant reductions in the heating time could still be effected by increasing h to values
considerably in excess of 1000 W/m2⋅K.
COMMENTS: For Part (a), recognize why it is not necessary to know Ti or the time to. We require that
Fo ≥ 0.2, which for this sphere corresponds to t ≥ 14s. For this situation, the time dependence of the
surface and center are the same.
PROBLEM 5.48
KNOWN: A long cylinder, initially at a uniform temperature, is suddenly quenched in a large oil bath.
FIND: (a) Time required for the surface to reach 500 K, (b) Effect of convection coefficient on surface
temperature history.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Fo > 0.2.
ANALYSIS: (a) Check first whether lumped capacitance method is applicable. For h = 50 W/m2⋅K,
Bic
=
2
hLc h ( ro 2 ) 50 W m ⋅ K ( 0.015 m / 2 )
=
=
= 0.221 .
k
k
1.7 W m ⋅ K
Since Bic > 0.1, method is not suited. Using the approximate series solution for the infinite cylinder,
)
(
(
) ( )
θ * r* , Fo = C1 exp −ζ12 Fo × J 0 ζ1r*
(1)
Solving for Fo and setting r* = 1, find
 θ*

ln 

ζ12  C1J 0 (ζ1 ) 
T ( ro , t o ) − T∞
=
=
=
θ * (1,
Fo )
where
Ti − T∞
Fo = −
1
( 500 − 350=
)K
(1000 − 350 ) K
0.231 .
From Table 5.1, with Bi = 0.441, find ζ1 = 0.8882 rad and C1 = 1.1019. From Table B.4, find J0(ζ1) =
0.8121. Substituting numerical values into Eq. (2),
1
Fo =
−
ln [ 0.231 1.1019 × 0.8121] =
1.72 .
( 0.8882 )2
2
From the definition of the Fourier number, Fo = α t ro , and α = k/rc,
ro2
rc
t Fo = Fo ⋅ ro2
=
k
α
=
t 1.72 ( 0.015 m ) × 400 kg m3 ×1600 J kg ⋅ K 1.7
=
W m ⋅ K 145s .
2
<
(b) Using the IHT Transient Conduction Model for a Cylinder, the following surface temperature histories
were obtained.
Continued...
PROBLEM 5.48 (Cont.)
1000
Surface temperature, T(K)
900
800
700
600
500
400
300
0
50
100
150
200
250
300
Time, t(s)
h = 250 W/m^2.K
h = 50 W/m^2.K
Increasing the convection coefficient by a factor of 5 has a significant effect on the surface temperature,
greatly accelerating its approach to the oil temperature. However, even with h = 250 W/m2⋅K, Bi = 1.1
and the convection resistance remains significant. Hence, in the interest of accelerated cooling, additional
benefit could be achieved by further increasing the value of h.
COMMENTS: For Part (a), note that, since Fo = 1.72 > 0.2, the approximate series solution is
appropriate.
PROBLEM 5.49
PROBLEM 5.50
PROBLEM 5.50 (Cont.)
PROBLEM 5.51
PROBLEM 5.52
PROBLEM 5.53
PROBLEM 5.54
KNOWN: Initial temperature, density, specific heat and diameter of cylindrical rod. Convection
coefficient and temperature of air flow. Time for centerline to reach a prescribed temperature.
Dependence of convection coefficient on flow velocity.
FIND: (a) Thermal conductivity of material, (b) Effect of velocity and centerline temperature and
temperature histories for selected velocities.
SCHEMATIC:
ASSUMPTIONS: (1) Lumped capacitance analysis cannot be used but one-term approximation for an
infinite cylinder is appropriate, (2) One-dimensional conduction in r, (3) Constant properties, (4)
Negligible radiation, (5) Negligible effect of thermocouple hole on conduction.
∗
ANALYSIS: (a) With θ o =[To(0,1136s) - T∞]/(Ti - T∞) = (40 – 25)/(100 – 25) = 0.20, Eq. 5.52c yields
Fo =
αt
ro2
=
k t
r c p ro2
=
k (1136 s )
1200 kg / m × 1250 J / kg ⋅ K × ( 0.02 m )
3
2
= − ln ( 0.2 / C1 ) / ζ12
(1)
Because C1 and ζ1 depend on Bi = hro/k, a trial-and-error procedure must be used. For example, a value
of k may be assumed and used to calculate Bi, which may then be used to obtain C1 and ζ1 from Table
5.1. Substituting C1 and ζ1 into Eq. (1), k may be computed and compared with the assumed value.
Iteration continues until satisfactory convergence is obtained, with
<
k ≈ 0.30 W / m ⋅ K
and, hence, Bi = 3.67, C1 = 1.45, ζ1 = 1.87 and Fo = 0.568. For the above value of k,
− ln ( 0.2 / C1 ) / ζ12 =
0.567, which equals the Fourier number, as prescribed by Eq. (1).
2
0.618
0.618
2.618
(b) With h = 55 W/m ⋅K for V = 6.8 m/s, h = CV
yields a value of C = 16.8 W⋅s
/m
⋅K. The
desired variations of the centerline temperature with velocity (for t = 1136 s) and time (for V = 3, 10 and
20 m/s) are as follows:
Continued …..
PROBLEM 5.54 (Cont.)
Centerline temperature, To(C)
Centerline temperature, To(C)
50
45
40
35
30
0
5
10
15
100
75
50
25
1000
500
0
20
1500
Time, t(s)
Air velocity, V(m/s)
V=3 m/s
V=10 m/s
V=20 m/s
2
With increasing V from 3 to 20 m/s, h increases from 33 to 107 W/m ⋅K, and the enhanced cooling
reduces the centerline temperature at the prescribed time. The accelerated cooling associated with
increasing V is also revealed by the temperature histories, and the time required to achieve thermal
equilibrium between the air and the cylinder decreases with increasing V.
2
COMMENTS: (1) For the smallest value of h = 33 W/m ⋅K, Bi ≡ h (ro/2)/k = 1.1 >> 0.1, and use of the
lumped capacitance method is clearly inappropriate.
(2) The IHT Transient Conduction Model for a cylinder was used to perform the calculations of Part (b).
Because the model is based on the exact solution, Eq. 5.50a, it is accurate for values of Fo < 0.2, as well
as Fo > 0.2. Although in principle, the model may be used to calculate the thermal conductivity for the
conditions of Part (a), convergence is elusive and may only be achieved if the initial guesses are close to
the correct results.
PROBLEM 5.55
PROBLEM 5.56
PROBLEM 5.56 (Cont.)
PROBLEM 5.57
KNOWN: Properties, diameter, and initial temperature of sphere. Environment temperature and heat
transfer coefficient.
FIND: Time for sphere’s surface temperature to reach 26°C. Energy transferred from sphere up to that
time.
SCHEMATIC:
T ∞, h
+
D=100 mm
Ti = 50°C
T∞ = 20°C
h = 30 W/m2•K
D
Ti
ρ= 2500 kg/m3
c = 750 J/kg•K
k = 2.0 W/m•K
ASSUMPTIONS: (1) One-dimensional (radial) conduction in sphere. (2) Uniform properties. (3)
Negligible radiation.
PROPERTIES: Given, r = 2500 kg/m3, c = 750 J/kg⋅K, k = 2.0 W/m⋅K.
ANALYSIS: The Biot number is first calculated to determine if the lumped capacitance method is valid.
h(ro / 3) 30 W/m 2 ⋅ K(0.05 m/3)
=
= 0.25
k
2.0 W/m ⋅ K
=
Bi
The lumped capacitance method is not valid and the radial dependence of temperature must be accounted
for. We try the one-term approximation and will check the Fourier number later.
=
Bi
hro 30 W/m 2 ⋅ K(0.05 m)
=
= 0.75
k
2.0 W/m ⋅ K
From Table=
5.1, ζ 1 1.392,
=
C1 1.211. From the specified temperatures, θ * =
(T − T∞ ) / (Ti − T∞ ) = 6/30 =
0.2. Then from Equation 5.53a, with r* = 1,
1
θ* =
C1 exp(−ζ 12 Fo) sin(ζ 1 ),
ζ1
 0.2 × 1.392 
1  θ *ζ1 
1
− 2 ln 
−
Fo =
ln 
0.750
=
=
2
ζ 1  C1 sin(ζ 1 ) 
1.392
 1.211sin(1.392) 
Since Fo > 0.2, the one-term approximation is valid. Finally,
=
t
ro2 Fo
=
α
(0.05 m) 2 × 0.750
= 1760 s
( 2.0 W/m ⋅ K / (2500 kg/m3 × 750 J/kg ⋅ K) )
<
Continued…
PROBLEM 5.57 (Cont.)
The energy transferred can be found from Equation 5.55, with Qo = rcV(Ti – T∞) and θ o * from Equation
5.53c:
4

Qo ρp
cV (Ti −=
T∞ ) 2500 kg/m3 × 750 J/kg ⋅ K × 
=
=
°C 29, 450 J
(0.05 m)3  × 30
3

θ o * = C1 exp(−ζ 12 Fo) = 1.211exp(−(1.392) 2 × 0.750) = 0.283
 3θ *

Q= ρ cV (Ti − T∞ ) 1 − o3 ( sin ζ 1 − ζ 1 cos ζ 1 ) 
ζ1


 3 × 0.283
= 29, 450 × 1 −
sin(1.392) − 1.392cos(1.392)=
) 22,600 J
3 (
1.392


<
COMMENTS: If the Fourier number had been smaller than 0.2, more terms would have been required in
the infinite series of Equation 5.51a, and an iterative solution would have been required to find Fo.
PROBLEM 5.58
PROBLEM 5.59
KNOWN: Sphere quenching in a constant temperature bath.
FIND: (a) Plot T(0,t) and T(ro,t) as function of time, (b) Time required for surface to reach 415 K, t ′ , (c)
Heat flux when T(ro, t ′ ) = 415 K, (d) Energy lost by sphere in cooling to T(ro, t ′ ) = 415 K, (e) Steadystate temperature reached after sphere is insulated at t = t ′ , (f) Effect of h on center and surface
temperature histories.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Uniform initial
temperature.
ANALYSIS: (a) Calculate Biot number to determine if sphere behaves as spatially isothermal object,
=
Bi
2
hLc h ( ro 3) 75 W m ⋅ K ( 0.015 m 3)
=
=
= 0.22 .
k
k
1.7 W m ⋅ K
Hence, temperature gradients exist in the sphere and T(r,t) vs. t appears as shown above.
(b) The exact solution may be used to find t ′ when T(ro, t ′ ) = 415 K. We assume that the one-term
approximation is valid and check later. Hence, with
2
hro 75 W/m ⋅ K ( 0.015 m )
=
Bi =
= 0.662
1.7 W/m ⋅ K
k
from Table 5.1, ζ1 = 1.3188, C1 = 1.1877. Then
* (r* 1,=
Fo)
θ=
T(ro , t) − T∞ 415°C − 320°C
=
= 0.1979
Ti − T∞
800°C − 320°C
∗
and Equation 5.53b can be solved for θ o :
θ o* = θ *ζ1r * / sin(ζ1r*) = 0.1979 ×1.3188 ×1/ sin(1.3188) = 0.2695
Then Equation 5.53c can be solved for Fo:
1
1
Fo =
ln θ o∗ / C1 =
ln ( 0.2695 /1.1877 ) =
0.853
−
−
1.31882
ζ12
r cp 2
r2
400 kg m3 ×1600 J kg ⋅ K
2
⋅ ro = 0.853
× ( 0.015 m ) = 72 s
t ′ = Fo o = Fo ⋅
)
(
α
k
1.7 W m ⋅ K
<
Note that the one-term approximation is accurate, since Fo > 0.2.
Continued...
PROBLEM 5.59 (Cont.)
(c) The heat flux at the outer surface at time t ¢ is given by Newton’s law of cooling
2
q′′ h T ( ro , t ′ ) − T=
=
] K 7125 W / m2 .
∞  75 W m ⋅ K [ 415 − 320=
<
The manner in which q′′ is calculated indicates that energy is leaving the sphere.
(d) The energy lost by the sphere during the cooling process from t = 0 to t ′ can be determined from
Equation 5.55:
3θ o*
Q
3 × 0.2695
=
1−
1−
0.775
[sin(ζ1) − ζ1 cos(ζ1)] =
[sin(1.3188) − 1.3188cos(1.3188)] =
3
Qo
ζ1
1.31883
The energy loss by the sphere with V = (πD3)/6 is therefore, from Equation 5.47,
3
=
Q 0.775Q
=
o 0.775 ρp D 6 cp ( Ti − T∞ )
(
(
)
)
Q =0.775 × 400 kg m3 π [ 0.030 m ] 6 1600 J kg ⋅ K ( 800 − 320 ) K =3364 J
3
<
(e) If at time t ′ the surface of the sphere is perfectly insulated, eventually the temperature of the sphere
will be uniform at T(∞). Applying conservation of energy to the sphere over a time interval, Ein - Eout =
∆E ≡ Efinal - Einitial. Hence, -Q = rcV[T(∞) - T∞] - Qo, where Qo ≡ rcV[Ti - T∞]. Dividing by Qo and
regrouping, we obtain
T ( ∞ ) = T∞ + (1 − Q Qo ) ( Ti − T∞ ) = 320 K + (1 − 0.775 )( 800 − 320 ) K = 428 K
<
(f) Using the IHT Transient Conduction Model for a Sphere, the following graphical results were
generated.
800
90000
Heat flux, q''(ro,t) (W/m^2.K)
Temperature, T(K)
700
600
500
400
300
0
50
100
60000
30000
150
0
Time, t (s)
h = 75 W/m^2.K, r = ro
h = 75 W/m^2.K, r = 0
h = 200 W/m^2.K, r = ro
h = 200 W/m^2.K, r = 0
0
50
100
150
Time, t(s)
h = 75 W/m^2.K
h = 200 W/m^2.K
The quenching process is clearly accelerated by increasing h from 75 to 200 W/m2⋅K and is virtually
completed by t ≈ 100s for the larger value of h. Note that, for both values of h, the temperature difference
[T(0,t) - T(ro,t)] decreases with increasing t. Although the surface heat flux for h = 200 W/m2⋅K is
initially larger than that for h = 75 W/m2⋅K, the more rapid decline in T(ro,t) causes it to become smaller
at t ≈ 30s.
COMMENTS: Using the Transient Conduction/Sphere model in IHT based upon multiple-term series
solution, the following results were obtained: t′ = 72.1 s; Q/Qo = 0.7745, and T(∞) = 428 K.
PROBLEM 5.60
PROBLEM 5.61
PROBLEM 5.61 (Cont.)
PROBLEM 5.62
KNOWN: Initial temperature and properties of a solid sphere. Surface temperature after immersion in a
fluid of prescribed temperature and convection coefficient.
FIND: (a) Time to reach surface temperature, (b) Effect of thermal diffusivity and conductivity on
thermal response.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties.
ANALYSIS: (a) For k = 15 W/m⋅K, the Biot number is
h ( ro 3) 300 W m 2 ⋅ K ( 0.05 m 3)
=
Bi =
= 0.333 .
k
15 W m ⋅ K
Hence, the lumped capacitance method cannot be used. From Equation 5.53a,
)
) (
sin ζ1r*
T − T∞
2
.
= C1 exp −ζ1 Fo
Ti − T∞
ζ1r*
(
At the surface, r* = 1. From Table 5.1, for Bi = 1.0, ζ1 = 1.5708 rad and C1 = 1.2732. Hence,
60 − 75
sin 90
2
= 0.30
= 1.2732 exp −1.5708 Fo
25 − 75
1.5708
)
(
exp(-2.467Fo) = 0.370
Fo
=
αt
= 0.403
ro2
( 0.05 m )
ro2
=
t 0.403
=
0.403 = 100s
α
10−5 m 2 s
2
<
(b) Using the IHT Transient Conduction Model for a Sphere to perform the parametric calculations, the
effect of α is plotted for k = 15 W/m⋅K.
Continued...
75
75
65
65
Center temperature, T(C)
Surface temperature, T(C)
PROBLEM 5.62 (Cont.)
55
45
35
55
45
35
25
25
0
50
100
150
200
250
50
0
300
100
250
200
150
300
Time, t(s)
Time, t(s)
k = 15 W/m.K, alpha = 1E-4 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-6 m^2/s
k = 15 W/m.K, alpha = 1E-4 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-6m^2/s
For fixed k and increasing α, there is a reduction in the thermal capacity (rcp) of the material, and hence
the amount of thermal energy which must be added to increase the temperature. With increasing α, the
material therefore responds more quickly to a change in the thermal environment, with the response at the
center lagging that of the surface.
75
75
65
65
Center temperature, T(C)
Surface temperature, T(C)
The effect of k is plotted for α = 10-5 m2/s.
55
45
35
25
55
45
35
25
0
50
100
150
200
Time, t(s)
k = 1.5 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k = 150W/m.K, alpha = 1E-5 m^2/s
250
300
0
50
100
150
200
250
300
Time, t(s)
k = 1.5 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k =150 W/m.K, alpha = 1E-5m^2/s
With increasing k for fixed alpha, there is a corresponding increase in rcp, and the material therefore
responds more slowly to a thermal change in its surroundings. The thermal response of the center lags
that of the surface, with temperature differences, T(ro,t) - T(0,t), during early stages of solidification
increasing with decreasing k.
COMMENTS: Use of this technique to determine h from measurement of T(ro) at a prescribed t requires
an iterative solution of the governing equations.
PROBLEM 5.63
KNOWN: Temperature requirements for cooling the spherical material of Example 5.6 in air and in a
water bath.
FIND: (a) For step 1, the time required for the center temperature to reach T(0,t) = 335°C while cooling
2
in air at 20°C with h = 10 W/m ⋅K; find the Biot number; do you expect radial gradients to be
appreciable?; compare results with hand calculations in Example 5.6; (b) For step 2, time required for the
2
center temperature to reach T(0,t) = 50°C while cooling in water bath at 20°C with h = 6000 W/m ⋅K; and
(c) For step 2, calculate and plot the temperature history, T(x,t) vs. t, for the center and surface of the
sphere; explain features; when do you expect the temperature gradients in the sphere to the largest? Use
the IHT Models | Transient Conduction | Sphere model as your solution tool.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the radial direction, (2) Constant properties.
ANALYSIS: The IHT model represents the series solution for the sphere providing the temperatures
evaluated at (r,t). A selected portion of the IHT code used to obtain results is shown in the Comments.
(a) Using the IHT model with step 1 conditions, the time required for T(0,ta) = T_xt = 335°C
with r = 0 and the Biot number are:
<
=
t a 94.2
=
s
Bi 0.0025
Radial temperature gradients will not be appreciable since Bi = 0.0025 << 0.1. The sphere
behaves as space-wise isothermal object for the air-cooling process. The result is identical to the
lumped-capacitance analysis result of the Text example.
(b) Using the IHT model with step 2 conditions, the time required for T(0,tw) = T_xt = 50°C with r = 0
and Ti = 335°C is
<
t w = 3.0 s
Radial temperature gradients will be appreciable, since Bi = 1.5 >> 0.1. The sphere does not
behave as a space-wise isothermal object for the water-cooling process.
(c) For the step 2 cooling process, the temperature histories for the center and surface of the
sphere are calculated using the IHT model.
Continued …
PROBLEM 5.63 (Cont.)
Temperature-time history, Step 2
Temperature, T(r,t) (C)
400
300
200
100
0
0
1
2
3
4
5
6
Time, t (s)
Surface, r = ro
Center, r = 0
At early times, the difference between the center and surface temperature is appreciable. It is in
this time region that thermal stresses will be a maximum, and if large enough, can cause fracture.
Within 6 seconds, the sphere has a uniform temperature equal to that of the water bath.
COMMENTS: Selected portions of the IHT sphere model codes for steps 1 and 2 are shown below.
/* Results, for part (a), step 1, air cooling; clearly negligible gradient
Bi
Fo
t
T_xt
Ti
r
ro
0.0025
25.13
94.22
335
400
0
0.005 */
// Models | Transient Conduction | Sphere - Step 1, Air cooling
// The temperature distribution T(r,t) is
T_xt = T_xt_trans("Sphere",rstar,Fo,Bi,Ti,Tinf) //
T_xt = 335
// Surface temperature
/* Results, for part (b), step 2, water cooling; Ti = 335 C
Bi
Fo
t
T_xt
Ti
r
ro
1.5
0.7936 2.976
50
335
0
0.005 */
// Models | Transient Conduction | Sphere - Step 2, Water cooling
// The temperature distribution T(r,t) is
T_xt = T_xt_trans("Sphere",rstar,Fo,Bi,Ti,Tinf) //
//T_xt = 335
// Surface temperature from Step 1; initial temperature for Step 2
T_xt = 50
// Center temperature, end of Step 2
PROBLEM 5.64
PROBLEM 5.65
PROBLEM 5.66
PROBLEM 5.67
PROBLEM 5.67 (Cont.)
PROBLEM 5.68
KNOWN: Irreversible thermal injury (cell damage) occurs in living tissue maintained at T ≥ 48°C for a
duration ∆t ≥ 10s.
FIND: (a) Extent of damage for 10 seconds of contact with machinery in the temperature range 50 to
100°C, (b) Temperature histories at selected locations in tissue (x = 0.5, 1, 5 mm) for a machinery
temperature of 100°C.
SCHEMATIC:
ASSUMPTIONS: (1) Portion of worker’s body modeled as semi-infinite medium, initially at a uniform
temperature, 37°C, (2) Tissue properties are constant and equivalent to those of water at 37°C, (3)
Negligible contact resistance.
PROPERTIES: Table A-6, Water, liquid (T = 37°C = 310 K): r = 1/vf = 993.1 kg/m3, c = 4178 J/kg⋅K,
k = 0.628 W/m⋅K, α = k/rc = 1.513 × 10-7 m2/s.
ANALYSIS: (a) For a given surface temperature -- suddenly applied -- the analysis is directed toward
finding the skin depth xb for which the tissue will be at Tb ≥ 48°C for more than 10s? From Eq. 5.60,
T ( x b , t ) − Ts
 x 2 (α t )1/ 2  erf [η ] .
=
=
erf
 b

Ti − Ts
For the two values of Ts, the left-hand side of the equation is


48 − 50 ) C
48 − 100 ) C
(
(


=
Ts 50
=
C:
0.154
=
Ts 100
=
C:
0.825


( 37 − 100 )
( 37 − 50 )
C
C
The burn depth is
x b = [ w ] 2 (α t )
1/ 2
(
= [ w ] 2 1.513 ×10−7 m 2 s × t
)
1/ 2
= 7.779 ×10−4 [ w ] t1/ 2 .
Continued...
PROBLEM 5.68 (Cont.)
Using Table B.2 to evaluate the error function and letting t = 10s, find xb as
Ts = 100°C:
xb = 7.779 × 10-4 [0.96](10s)1/2 = 2.362 × 103 m = 2.36 mm
Ts = 50°C:
xb = 7.779 × 10-4 [0.137](10s)1/2 = 3.37 × 103 m = 0.34 mm
<
<
Recognize that tissue at this depth, xb, has not been damaged, but will become so if Ts is maintained for
the next 10s. We conclude that, for Ts = 50°C, only superficial damage will occur for a contact period of
20s.
(b) Temperature histories at the prescribed locations are as follows.
97
Temperature, T(C)
87
77
67
57
47
37
0
15
30
Time, t(s)
x = 0.5 mm
x = 1.0 mm
x = 2.0 mm
The critical temperature of 48°C is reached within approximately 1s at x = 0.5 mm and within 7s at x = 2
mm.
COMMENTS: Note that the burn depth xb increases as t1/2.
PROBLEM 5.69
PROBLEM 5.69 (Cont.)
PROBLEM 5.70
PROBLEM 5.70 (Cont.)
PROBLEM 5.71
KNOWN: Gel applied to ultrasound probe window. Gel and window properties. Initial probe
temperatures.
FIND: Required initial gel temperature to ensure gel remains below 12°C.
SCHEMATIC:
Gel
Tmax = 12°C
Window
Ti = 20°C
ρw = 1200 kg/m3
cp,w = 1200 J/kg•K
kw = 0.2 W/m•K
Gel
Window
x
ASSUMPTIONS: (1) Gel and window can be treated as semi-infinite solids. (2) Uniform properties.
PROPERTIES: Table A-6, Water (T = 285 K): rg =1000 kg/m3, cp= 4189 J/kg⋅K, k = 0.59 W/m⋅K.
Given, Window: rw =1200 kg/m3, cp= 1200 J/kg⋅K, k = 0.2 W/m⋅K.
ANALYSIS: The maximum gel temperature occurs at the gel-window interface. For two semi-infinite
materials brought into contact, the interface temperature is given by Equation 5.66:
Ts =
(k ρρ
c)1/g 2 Tg ,i + (k c)1/w 2 Tw,i
(k ρρ
c)1/g 2 + (k c)1/w 2
Solving for the initial gel temperature,
(k ρρρ
c)1/g 2 + (k c)1/w 2  Ts − (k c)1/w 2 Tw,i
Tg ,i = 
(k ρ c)1/g 2
The effusivities of the two materials are:
1/ 2
(k ρ=
c)1/g 2 (0.59 W/m ⋅ K × 1000 kg/m3 × 4189 J/kg ⋅ K)
=
1572 W ⋅ s1/ 2 /m 2 ⋅ K
1/ 2
(k ρ c=
)1/w 2 (0.2 W/m ⋅ K × 1200 kg/m3 × 1200 J/kg ⋅ K)=
537 W ⋅ s1/ 2 /m 2 ⋅ K
Thus, with the requirement Ts < 12°C, we find
Tg ,i <
[1572 + 537] × 12°C − 537 × 20°C =
9.3°C
1572
<
COMMENTS: The assumption that the two materials can be treated as semi-infinite is only valid for
short times, before the penetration depth, dp = 2.3(αt)1/2, exceeds the thickness of the material.
PROBLEM 5.72
KNOWN: Thick oak wall, initially at a uniform temperature of 25°C, is suddenly exposed to combustion
products at 800°C with a convection coefficient of 22 W/m2⋅K.
FIND: (a) Time of exposure required for the surface to reach an ignition temperature of 400°C, (b)
Temperature distribution at time t = 325s.
SCHEMATIC:
ASSUMPTIONS: (1) Oak wall can be treated as semi-infinite solid, (2) One-dimensional conduction,
(3) Constant properties, (4) Negligible radiation.
PROPERTIES: Table A-3, Oak, cross grain (300 K): r = 545 kg/m3, c = 2385 J/kg⋅K, k = 0.17 W/m⋅K,
α = k/rc = 0.17 W/m⋅K/545 kg/m3 × 2385 J/kg⋅K = 1.31 × 10-7 m2/s.
ANALYSIS: (a) This situation corresponds to Case 3 of Figure 5.7. The temperature distribution is
given by Eq. 5.63 or by Figure 5.8. Using the figure with
T ( 0, t ) − Ti 400 − 25
= = 0.48
T∞ − Ti
800 − 25
x
and
2 (α t )
1/ 2
we obtain h(αt)1/2/k ≈ 0.75, in which case t ≈ (0.75k/hα1/2)2. Hence,
(
t ≈ 0.75 × 0.17 W m ⋅ K 22 W m 2 ⋅ K (1.31× 10−7 m 2 s )
)
1/2 2
=0
=
257s
<
(b) Using the IHT Transient Conduction Model for a Semi-infinite Solid, the following temperature
distribution was generated for t = 325s.
400
Temperature, T(C)
325
250
175
100
25
0
0.005
0.01
0.015
0.02
0.025
0.03
Distance from the surface, x(m)
The temperature decay would become more pronounced with decreasing α (decreasing k, increasing rcp)
and in this case the penetration depth of the heating process corresponds to x ≈ 0.025 m at 325s.
COMMENTS: The result of part (a) indicates that, after approximately 5 minutes, the surface of the
wall will ignite and combustion will ensue. Once combustion has started, the present model is no longer
appropriate.
PROBLEM 5.73
KNOWN: Thickness, initial temperature and thermophysical properties of concrete firewall. Incident
radiant flux and duration of radiant heating. Maximum allowable surface temperatures at the end of
heating.
FIND: If maximum allowable temperatures are exceeded.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in wall, (2) Validity of semi-infinite medium
approximation, (3) Negligible convection and radiative exchange with the surroundings at the irradiated
surface, (4) Negligible heat transfer from the back surface, (5) Constant properties.
ANALYSIS: The thermal response of the wall is described by Eq. (5.62)
1/ 2
 − x 2  q′′ x
2 q′′o (α t / p )
 x 

 − o erfc 
=
+
T ( x, t ) Ti
exp



k
4α t
k
 2 αt 


−7 2
1/ 2
α k / ρ=
c p 6.92 × 10 m / s and=
min 1800s, 2q ′′o (α t / π )=
/ k 284.5 K. Hence, at x
=
for t 30
where,=
= 0,
T ( 0,30 min ) = 25°C + 284.5°C = 309.5°C < 325°C
(
<
)
1/ 2
0.25m, x / 4aa
t =
1, 786K, and x / 2 ( t )
3.54. Hence,
−12.54, q ′′o x / k =
=
At x =−
2
)
(
T ( 0.25m, 30 min ) = 25°C + 284.5°C 3.58 ×10−6 − 1786°C × ( ~ 0 ) ≈ 25°C
<
Both requirements are met.
COMMENTS: The foregoing analysis is conservative since heat transfer at the irradiated surface due to
convection and net radiation exchange with the environment have been neglected. If the emissivity of the
surface and the temperature of the surroundings are assumed to be ε = 1 and Tsur = 298K, radiation
(
)
4
exchange at Ts = 309.5°C would be q′′rad = εs Ts4 − Tsur
= 6, 080 W / m 2 ⋅ K, which is significant
(~ 60% of the prescribed radiation).
PROBLEM 5.74
KNOWN: Initial temperature of copper and glass plates. Initial temperature and properties of
finger.
FIND: Whether copper or glass feels cooler to touch.
SCHEMATIC:
ASSUMPTIONS: (1) The finger and the plate behave as semi-infinite solids, (2) Constant
properties, (3) Negligible contact resistance.
3
PROPERTIES: Skin (given): r = 1000 kg/m , c = 4180 J/kg⋅K, k = 0.625 W/m⋅K; Table A-1
3
(T = 300K), Copper: r = 8933 kg/m , c = 385 J/kg⋅K, k = 401 W/m⋅K; Table A-3 (T = 300K),
3
Glass: r = 2500 kg/m , c = 750 J/kg⋅K, k = 1.4 W/m⋅K.
ANALYSIS: Which material feels cooler depends upon the contact temperature Ts given by
Equation 5.66. For the three materials of interest,
2
1/ 2
=
×
×
= 1, 616 J/m 2 ⋅ K ⋅ s1/2
0.625
1000
4180
( kρ c )1/
(
)
skin
1/ 2
= 401× 8933 × 385 ) = 37,137 J/m 2 ⋅ K ⋅ s1/2
( kρ c )1/2
cu (
2
1/ 2
= (1.4 × 2500 × 750 ) = 1, 620 J/m 2 ⋅ K ⋅ s1/2 .
( kρ c )1/glass
Since ( kρρ
c )cu >> ( k c )glass , the copper will feel much cooler to the touch. From Equation
5.66,
1/2
1/2
1/2
1/2
c )A TA,i + ( k c )B TB,i
( kρρ
Ts =
1/2
1/2
c )A + ( k c )B
( kρρ
=
Ts( cu )
1, 616 ( 310 ) + 37,137 ( 300 )
= 300.4 K
1, 616 + 37,137
=
Ts( glass )
1, 616 ( 310 ) + 1, 620 ( 300 )
= 305.0 K.
1, 616 + 1, 620
<
<
COMMENTS: The extent to which a material’s temperature is affected by a change in its
1/2
thermal environment is inversely proportional to (krc) . Large k implies an ability to spread
the effect by conduction; large rc implies a large capacity for thermal energy storage.
PROBLEM 5.75
PROBLEM 5.76
KNOWN: Thickness and properties of liquid coating deposited on a metal substrate. Initial temperature
and properties of substrate.
FIND: (a) Expression for time required to completely solidify the liquid, (b) Time required to solidify an
alumina coating.
SCHEMATIC:
Coating:
r = 3970 kg/m3
hsf = 3577 kJ/kg
Tf = 2318 K
rhsfdS/dt
Liquid alumina
S(t)
Solid alumina
d = 2 mm
q″cond
Tungsten
x
Substrate: ks = 120 W/m∙K, αs = 4.0 × 10-5 m2/s
ASSUMPTIONS: (1) Substrate may be approximated as a semi-infinite medium in which there is onedimensional conduction, (2) Solid and liquid alumina layers remain at fusion temperature throughout
solidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contact
resistance at the coating/substrate interface, (4) Negligible solidification contraction, (5) Constant
properties.
ANALYSIS: (a) Performing an energy balance on the solid layer, whose thickness S increases with t, the
latent heat released at the solid/liquid interface must be balanced by the rate of heat conduction into the
solid. Hence, per unit surface area,
dS
1/ 2
= k ( Tf − Ti ) (πα t ) . It follows that
= q′′cond where, from Eq. 5.61, q′′cond
ρ h sf
dt
ρ h sf
dS k s ( Tf − Ti )
=
dt
(πα t )1/ 2
s
k s ( Tf − Ti )
d
∫o dS = ρ h
δ=
t=
t dt
1/ 2 ∫o t1/ 2
sf (παs )
 Tf − Ti  1/ 2

t
(παs )1/ 2  ρ hsf 
2k s
παs  δρ h sf 
2
<


4k s2  Tf − Ti 
(b) For the prescribed conditions,
π 4 ×10−5 m 2 s
t=
(
)  0.002 m × 3970 kg m3 × 3.577 ×106 J kg 2 = 0.43 s
2
4 (120 W m ⋅ K ) 
2018 K


COMMENTS: If solidification occurs over a short time resulting in a change of the solid’s
microstructure (relative to slow solidification), it is termed rapid solidification. See Problem 5.34.
<
PROBLEM 5.77
PROBLEM 5.78
PROBLEM 5.78 (Cont.)
PROBLEM 5.79
KNOWN: Closely-spaced buried tubing, annual temperature variation.
FIND: Depth associated with the soil behaving as an infinite medium.
Ambient temperature, T
SCHEMATIC:
t
Burial depth, d
Buried tubing
Soil surface
Tube temperature, T
1 year
t
1 year
ASSUMPTIONS: (1) Periodic conditions, (2) Constant properties, (3) One-dimensional heat transfer
(closely-spaced tubing), (4) Diurnal variation in ambient temperature is small relative to its annual
variation.
PROPERTIES: Table A.3: Soil (300 K): r = 2050 kg/m3, k = 0.52 W/m⋅K, cp = 1840 J/kg⋅K.
ANALYSIS: Since soil heating and cooling is associated with annual changes in both the ambient and
buried tubing temperatures, the burial depth, d, must be larger than twice the penetration depth associated
with periodic heating of the soil. Or,
d =2d p =2 × 4 α/ω =8 k /( ρ c pω )
The heating frequency is ω = 2π/tp = 2π/(1 yr × 365 days/yr × 24 h/day × 60 min/h × 60 s/min) = 200 ×
10-9 s-1. Therefore, the required burial depth is
(
)
=
d 8 0.52W/m 2 ⋅ K/ 2050kg/m3 × 1840 J/kg ⋅ K × 200 × 10-9s-1 = 6.6 m
<
COMMENTS: The installation cost increases as the burial depth increases. A tradeoff exists between
the installation cost and the attainment of constant soil temperature conditions.
PROBLEM 5.80
PROBLEM 5.80 (Cont.)
PROBLEM 5.81(a)
KNOWN: Cylinder with constant surface temperature.
FIND: Expression for Q/Qo as a function of Fo = αt/ro2.
SCHEMATIC:
Ti
r
2ro
Ts
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Validity of the
approximate solution of Table 5.2a.
ANALYSIS: From Table 5.2a for Fo < 0.2,
q*=
q"s ro
=
k (Ts − Ti )
1
αt
− 0.50 − 0.65Fo ; Fo= 2
π Fo
ro
Substituting the expression for Fo into the first equation yields
=
qs"
αt 
k (Ts − Ti )  ro −1/ 2
− 0.50 − 0.65 2 
t

ro
ro 
 πα
We desire an expression for Q/Qo. Hence,
t
2π ro
∫ qs dt
"
αt 
Q Q'
2α  ro −1/ 2
= ' = 2 t =0
= 2
− 0.50 − 0.65 2  dt
t

Qo Qo π ro r c(Ts − Ti ) ro
πα
ro 
t =0 
t
∫
or
αt 
Q 2α t 
ro
=
− 0.50 − 0.325 2 
2
2
Qo
ro  πα t
ro 
=
4
π
Fo − Fo − 0.65Fo 2
<
COMMENTS: (1) A plot of Q/Qo versus Fo is shown below. (2) The exact solution for Q/Qo involves
many terms that would need to be evaluated in the infinite series expression. (2) See Lavine and
Bergman, “Small and Large Time Solutions for Surface Temperature, Surface Heat Flux, and Energy
Input in Transient, One-Dimensional Conduction in Simple Geometries,” ASME Journal of Heat
Transfer, Vol 130, pp. 101302-1 to 101302-8, 2008 for details.
Continued…
PROBLEM 5.81(a) (Cont.)
Q/Qo vs. Fo (cylinder)
1
0.8
Q/Qo
0.6
0.4
0.2
0
0
0.05
0.1
Fo
0.15
0.2
PROBLEM 5.81(b)
KNOWN: Sphere with constant surface temperature.
FIND: Expression for Q/Qo as a function of Fo = αt/ro2.
SCHEMATIC:
Ti
r
2ro
Ts
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Validity of the
approximate solution of Table 5.2a.
ANALYSIS: From Table 5.2a for Fo < 0.2,
1
q" r
αt
q* = s o
=
− 1 ; Fo = 2
k (Ts − Ti )
π Fo
ro
Substituting the expression for Fo into the first equation yields
q"s
=
k (Ts − Ti )  ro −1/ 2 
− 1
 πα t
ro


We desire an expression for Q/Qo. Hence,
t
4π ro2
∫ qs dt
"
t
3α  ro −1/ 2 
Q
t =0
t
=
=
− 1 dt

3
Qo (4 / 3)π ro r c(Ts − Ti ) ro2
πα

t =0 
∫
or
Q 3α  2 r

= 2  o t1/ 2 − t 
Qo ro  πα

 2
= 3
 π
<

Fo − Fo 

COMMENTS: (1) A plot of Q/Qo versus Fo is shown below. (2) The exact solution for Q/Qo
involves many terms that would need to be evaluated in the infinite series expression.
Continued…
PROBLEM 5.81b (Cont.)
Q/Qo vs. Fo (sphere)
1
0.8
Q/Qo
0.6
0.4
0.2
0
0
0.05
0.1
Fo
0.15
0.2
PROBLEM 5.82
KNOWN: Plane wall, infinite cylinder and sphere each subjected to a constant surface heat flux.
FIND: The ratio of (i) the actual surface temperature minus the initial temperature, (Ts,act – Ti), to (ii) the
value of this temperature difference associated with lumped capacitance behavior, (Ts,lc – Ti), for each of
the three geometries. Criteria associated with (Ts,act – Ti)/(Ts,lc – Ti) ≤ 1.1.
SCHEMATIC:
Ti, k, r, c
Ti, k, r, c
Ts
q″ s
Ts
q″ s
q″ s
L
r
ro
L
x
Plane wall
Infinite cylinder or sphere
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Fo ≥ 0.2.
ANALYSIS: From Table 5.2b, the approximate solutions for all three geometries may be expressed as
 kt

q"s Lc
= a
+
b

k (Ts ,act − Ti )  ρ cL2c

−1
(1)
where Lc = L for the plane wall, and Lc = ro for the infinite cylinder and the sphere. The constants a and b
have the following values.
Geometry
Lc
a
b
Plane wall
Infinite cylinder
Sphere
L
ro
ro
1
2
3
1/3
1/4
1/5
For the lumped-capacitance analysis with constant heat flux, we note that in general,
=
q"s As t ρ cV (Ts ,lc − Ti ) . Substituting expressions for the surface area and volume of each of the three
geometries results in the general expression,
aq"s t / ρ cLc
(Ts,lc − Ti ) =
(2)
where the constant a is the same as in the table above. Solving Eq. (1) for (Ts,act – Ti) and dividing the
resulting expression by Eq. (2) results in
Continued…
PROBLEM 5.82 (Cont.)
 q" L  k t

+
b
 s c  a
  ρ cL
(Ts,act − Ti ) = k  ρ c L2c  c = 1 aFo + b =1 + b
]
[
aFo
aFo
aq"s t
(Ts,lc − Ti )
(3)
Therefore, for the plane wall,
(Ts,act − Ti ) =
1
b
1+
1+
=
3Fo
aFo
(Ts,lc − Ti )
<
(Ts,act − Ti ) =
1
b
1+
1+
=
8Fo
aFo
(Ts,lc − Ti )
<
(Ts,act − Ti ) =
b
1
1+
=
1+
aFo
15Fo
(Ts,lc − Ti )
<
for the infinite cylinder,
for the sphere,
The preceding equations may be solved for the Fourier number associated with (Ts,act – Ti)/(Ts,lc – Ti) = 1.1,
yielding critical Fourier numbers, Foc = 3.33, 1.25 and 0.667 for the plane wall, infinite cylinder, and
sphere, respectively. Lumped capacitance behavior is exhibited at dimensionless times greater than Foc.
COMMENTS: (1) The calculated values of the Fourier number are each greater than 0.2. Hence, use of
the approximate solutions for Fo ≥ 0.2 is justified. (2) The dimensionless time at which lumped
capacitance behavior is reached varies with the geometry.
PROBLEM 5.83
KNOWN: Desired minimum temperature response of a 3ω measurement.
FIND: Minimum sample thickness that can be measured.
ASSUMPTIONS: (1) Constant properties, (2) Two-dimensional conduction, (3) Semi-infinite medium,
(4) Negligible radiation and convection losses from the metal strip and the top surface of the sample.
PROPERTIES: (Example 5.10): k = 1.11 W/m∙K, a = 4.37 × 10-7 m2/s.
ANALYSIS: Equation 5.74 maybe rearranged to yield
 
ΔTLπk  
ω = 2exp  2  C2 
Δqs  
 
 
0.1°C × 3.5 × 10-3 m × π × 1.11 W/m ⋅ K  
ω = 2 × exp  2  5.35  

3.5 × 10-3 W
 
 
ω = 44.2 × 103 rad/s
α = 4.37 × 10-7 m 2 /s
Therefore
δ p = α/ω = 4.37 × 10-7 m 2 /s
44.2 × 103 rad/s = 3.1 × 10-6 m = 3.1 μm
The minimum sample thickness is therefore 3.1 µm.
<
COMMENTS: (1) To ensure the thickness of the sample is adequate, the actual minimum thickness
should be greater than the thermal penetration depth. (2) The sample thickness could be increased further
by increasing the amplitude of the heating rate, ∆qs. (3) It is commonly desired to measure very thin
samples to discern the effect of the top and bottom boundaries of a thin film on the conduction heat
transfer rate, as depicted in Figure 2.6. As the film becomes thinner, the experimental uncertainties
increase.
PROBLEM 5.84
KNOWN: Diameter and initial temperature of spheres with properties of Pyrex. Temperature of
environment with large heat transfer coefficient.
FIND: Rate of cooling of spheres at t = 0.1, 0.25, and 1 s.
SCHEMATIC:
T ∞, h
+
D=2 mm
Ti = 90°C
T∞=Ts = 5°C
h→∞
D
Ti
ASSUMPTIONS: (1) Uniform properties, (2) Heat transfer coefficient is infinite, therefore surface
temperature is equal to environment temperature, (3) Negligible radiation.
PROPERTIES: Table A-3, Pyrex, (T = 300 K): r = 2225 kg/m3, cp = 835 J/kg⋅K, k = 1.4 W/m⋅K.
ANALYSIS: The cooling rate can be found from an energy balance on a sphere, written in terms of the
average sphere temperature:
dT
mc p= qs′′As ,
dt
qs′′4p ro2
3qs′′
dT qs′′As
= =
=
3
dt mc p r c p (4p ro / 3) r c p ro
(1)
where qs′′ is the heat flux entering the sphere.
The thermal diffusivity is α = k/r cp = 7.54 × 10-7 m2/s. The Fourier numbers for the three times are:
Fo
=
at
=
ro2
7.54 × 10−7 m 2 /s
=
t 0.0754, 0.188, and 0.754
(0.001 m) 2
From Table 5.2a, the heat flux entering the sphere is related to q* according=
to qs′′ q * k (Ts − Ti ) / ro .
Combining this with Eq. (1) gives:
dT
3q *  k (Ts − Ti )  3α q * (Ts − Ti ) 3 × 7.54 × 10−7 m 2 /s(5 − 90)°C
=
=
− q * ×192°C/s
q* =

=
dt r c p ro 
ro
ro2
(0.001 m) 2

Then, for heat transfer in the interior of a sphere, the cases Fo = 0.0754 and Fo = 0.188 are in the
category Fo < 0.2, therefore
Continued…
(2)
PROBLEM 5.84 (Cont.)
For t = 0.1 s,
For t = 0.25 s,
=
q*
=
q*
1
=
−1
π Fo
1
=
− 1 1.055
π × 0.0754
1
=
−1
π Fo
1
=
− 1 0.300
π × 0.188
The case Fo = 0.754 is in the category Fo > 0.2, therefore
2
= 0.00118
q=
* 2exp(−pp
Fo=
) 2exp(− 2 × 0.0754)
From Eq. (2) above,
dT
=
− q * ×192°C/s =
−203, − 57.6, − 0.227°C/s
dt
Thus the cooling rates are 203, 57.6, and 0.227°C/s for t = 0.1, 0.25, and 1 s, respectively.
<
COMMENTS: (1) The cooling rate is highest at early times when the temperature difference between the
spheres and water is large, and decreases with time. (2) The middle case, Fo = 0.188, is close to the
border of Fo = 0.2 The result of the Fo > 0.2 formula yields a cooling rate of 60.0, within around 4% of
the more accurate result found earlier.
PROBLEM 5.85
KNOWN: Periodic soil surface temperature variation of known amplitude and frequency.
FIND: Thermal penetration depth and maximum soil surface heat flux.
SCHEMATIC:
ΔT=30°C
T(0,t)
δp
x
t
Soil
ASSUMPTIONS: (1) Uniform properties, (2) One-dimensional conduction.
PROPERTIES: Table A-3, Soil, (T = 300 K): r = 2050 kg/m3, cp = 1840 J/kg⋅K, k = 0.52 W/m⋅K, α =
k/rcp = 1.38 × 10-7 m2/s.
ANALYSIS: The period of temperature oscillations is one year, therefore ω = 2π/(1 year) = 1.99 × 10-7
s-1. The thermal penetration depth is given by
α
ω
=
δ p 4= 4
1.38 × 10−7 m 2 /s
= 3.33 m
1.99 × 10−7 s −1
<
From Equation 5.73, the maximum surface heat flux is given by
qs′′,max =k ∆T
ω
1.99 × 10−7 s −1
=0.52 W/m ⋅ K × 30°C
= 18.8 W/m 2
a
1.38 × 10−7 m 2 /s
<
COMMENTS: The assumption of one-dimensional heat transfer is probably poor since the soil surface
temperature will only experience the specified variation close to the heat pump-air conditioning system.
PROBLEM 5.86
p
KNOWN: Stability criterion for the explicit method requires that the coefficient of the Tm term
of the one-dimensional, finite-difference equation be zero or positive.
p+1
FIND: For Fo > 1/2, the finite-difference equation will predict values of Tm
Second law of thermodynamics. Consider the prescribed numerical values.
which violate the
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant properties, (3) No internal
heat generation.
ANALYSIS: The explicit form of the finite-difference equation, Eq. 5.81, for an interior node is
)
(
p+1
p
p
p
=
Tm
Fo Tm+1
+ Tm-1
+ (1 - 2 Fo ) Tm
.
p
The stability criterion requires that the coefficient of Tm be zero or greater. That is,
1
or
Fo ≤ .
(1 − 2 Fo ) ≥ 0
2
For the prescribed temperatures, consider situations for which Fo = 1, ½ and ¼ and calculate
p+1
Tm
.
=
Fo 1
=
Fo 1/2
=
Fo 1/4

p+1
Tm
= 1(100 + 100 ) C + (1 − 2 ×1) 50=
C 250C

p+1
 C 100C
T=
1/ 2 (100 + 100 ) C + (1 − 2 ×1/ 2 ) 50
=
m

p+1
=
Tm
1/ 4 (100 + 100 ) C + (1 − 2 ×1/ 4 ) =
50 C 75C.
p+1
Plotting these distributions above, note that when Fo = 1, Tm
p+1
is greater than 100°C, while for
Fo = ½ and ¼ , Tm ≤ 100°C. The distribution for Fo = 1 is thermodynamically impossible:
heat is flowing into the node during the time period ∆t, causing its temperature to rise; yet heat is
flowing in the direction of increasing temperature. This is a violation of the Second law. When
Fo = ½ or ¼, the node temperature increases during ∆t, but the temperature gradients for heat
flow are proper. This will be the case when Fo ≤ ½, verifying the stability criterion.
PROBLEM 5.87
KNOWN: Thin rod of diameter D, initially in equilibrium with its surroundings, Tsur, suddenly
passes a current I; rod is in vacuum enclosure and has prescribed electrical resistivity, re, and
other thermophysical properties.
FIND: Transient, finite-difference equation for node m.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are much
larger than rod, (3) Properties are constant and evaluated at an average temperature, (4) No
convection within vacuum enclosure.
ANALYSIS: The finite-difference equation is derived from
the energy conservation requirement on the control volume,
=
D2 / 4
and
P π D.
Ac∆x, where A c π=
The energy balance has the form
p+1
p
− Tm
Tm
2
q a + q b − q rad +=
I R e r cV
.
+ E g E st
E in − E out=
∆t
where E g = I2 R e and R =
e ρe ∆x/A c . Using Fourier’s law to express the conduction terms, qa
and qb, and Eq. 1.7 for the radiation exchange term, qrad, find
)
(
p
p
p
p
p
-Tm-1
Tm
Tm+1
Tm
x
T p+1 - Tm
4,p
4 + I 2 re ∆=
+ kA c
- e P∆xs Tm
- Tsur
kAc
r cAc ∆x m
.
∆x
∆x
∆t
Ac
p+1
Divide each term by rcAc ∆x/∆t, solve for Tm
p+1
Tm
=
and regroup to obtain

 p
k
∆t
p
p
Tm-1
+ Tm+1
- 2 ⋅
⋅
- 1 Tm
)
(
ρρ
c ∆x 2
c ∆x 2 

2
e Ps ∆t 4,p
4 + I r e ⋅ ∆t .
−
⋅
−
T
T
m
sur )
Ac r c (
A c2 r c
k
⋅
∆t
2
Recognizing that Fo = α ∆t/∆x , regroup to obtain
(
)
p+1
p
p
p
=
+ Tm+1
+ (1 -Tm
Fo Tm-1
2 Fo ) Tm
e Ps∆x 2
kA c
)
(
4,p
4 +
⋅ Fo Tm
- Tsur
I2 re ∆x 2
kA c2
⋅ Fo.
<
p
The stability criterion is based upon the coefficient of the Tm term written as
Fo ≤ ½.
COMMENTS: Note that we have used the forward-difference representation for the time derivative; see
Section 5.10.1. This permits convenient treatment of the non-linear radiation exchange term.
PROBLEM 5.88
PROBLEM 5.88 (Cont.)
PROBLEM 5.88 (Cont.)
PROBLEM 5.89
PROBLEM 5.90
KNOWN: Three-dimensional, transient conduction.
FIND: Explicit finite difference equation for an interior node, stability criterion.
SCHEMATIC:
m, n + 1, q
y
m, n, q - 1
∆y
m + 1, n, q
m - 1, n, q
x
z
m, n, q + 1
m, n - 1, q
Δx = Δy = Δz
∆z
∆x
ASSUMPTIONS: (1) Constant properties, (2) Equal grid spacing in all three directions, (3) No heat
generation.
ANALYSIS: We begin with the three-dimensional form of the transient heat equation, Equation 2.19
1 ∂T
∂ 2T
∂ 2T
∂ 2T
=
+
+
α ∂t
∂x 2
∂y 2
∂z 2
The finite-difference approximation to the time derivative is given by Equation 5.77:
p+1
p
Tm,n,q
- Tm,n,q
∂T
=
∂t m,n,q
Δt
The spatial derivatives for the x- and y- directions are given by Equations 4.27 and 4.28, with an extra
subscript q. By analogy, the z-direction derivative is approximated as
Tm,n,q+1 + Tm,n,q-1 - 2Tm,n,q
∂ 2T
≈
2
∂z m,n,q
(Δz) 2
Evaluating the spatial derivatives at time step p for the explicit method, assuming ∆x = ∆y = ∆z, yields
Continued…
PROBLEM 5.90 (Cont.)
p
p
p
p+1
p
Tm+1,n,q
+ Tm-1,n,q
- 2Tm,n,q
1 Tm,n,q - Tm,n,q
=
α
Δt
(Δx) 2
+
+
p
p
p
Tm,n+1,q
+ Tm,n-1,q
- 2Tm,n,q
(Δx) 2
p
p
p
Tm,n,q+1
+ Tm,n,q-1
- 2Tm,n,q
(Δx) 2
Solving for the nodal temperature at time step p+1 results in
p
p
p
p
p
p
p+1
Tm,n,q
= Fo(Tm+1,n,q
+ Tm-1,n,q
+ Tm,n+1,q
+ Tm,n-1,q
+ Tm,n,q+1
+ Tm,n,q-1
)
p
+ (1 - 6Fo)Tm,n,q
where Fo = αΔt/(Δx) 2 .
<
p
The stability criterion is determined by the requirement that the coefficient of Tm,n,q
≥ 0. Thus
Fo ≤ 1/6
<
COMMENTS: These results could also have been obtained using the energy balance method applied to
a control volume about the interior node.
PROBLEM 5.91
KNOWN: Nodal point located at boundary between two materials A and B.
FIND: Two-dimensional explicit, transient finite difference equation.
SCHEMATIC:
2
q2
∆y
q1A
Material A
kA, ρA, cA
q3A
0
1
3
q1B
q3B
∆x = ∆y
Material B
kB, ρB, cB
q4
4
∆x
ASSUMPTIONS: (1) Two-dimensional conduction, (2) No heat generation, (3) Constant properties
(different in each material).
ANALYSIS: We perform an energy balance on the control volume around node 0.
E in = E st
q + q + q + q + q + q = E
+ E
1A
1B
3A
3B
2
4
st,A
st,B
Using q1A as an example,
T - T Δy
q1A = k A 1 0
w = k A (T1 - T0 )w/2
Δx 2
where w is the depth into the page. The quantities q1B, q3A, and q3B can be found similarly. Then q2 is
given by
T2 - T0
Δx w = k A (T2 - T0 )w
Δy
and similarly for q4.
The storage term E st,A is given by
q2 = k A
p+1
p
Δy T0 - T0
E
st,A = ρ A c A Δx
2
Δt

E
and similarly for st,B .
Combining equations yields
Continued….
PROBLEM 5.91 (Cont.)
kA
T1 - T0
T -T
T -T
T -T
+ kB 1 0 + kA 3 0 + kB 3 0 +
2
2
2
2
k A (T2 - T0 ) + k B (T4 - T0 ) = (ρ A c A + ρ Bc B )
p+1
p
(Δx) 2 T0 - T0
2
Δt
Rearranging, we find
T0p+1 =
(Fo A + Fo B ) p
(T1 + T3p ) + Fo A T2p + Fo BT4p + [1 - 2(Fo A + Fo B ) ] T0p
2
<
where
Fo A =
2k A Δt
(ρ A c A + ρ Bc B )(Δx)
2
, Fo B =
2k BΔt
(ρ A c A + ρ Bc B )(Δx) 2
Note, that Fo A ≠ α A Δt/(Δx) 2 .
COMMENTS: Note that when the material properties are the same for materials A and B, the result
agrees with Equation 5.79.
PROBLEM 5.92
KNOWN: A 0.12 m thick wall, with thermal diffusivity 1.5 × 10-6 m2/s, initially at a uniform
temperature of 85°C, has one face suddenly lowered to 20°C while the other face is perfectly insulated.
FIND: (a) Using the explicit finite-difference method with space and time increments of ∆x = 30 mm
and ∆t = 300s, determine the temperature distribution within the wall 45 min after the change in surface
temperature; (b) Effect of ∆t on temperature histories of the surfaces and midplane.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties.
ANALYSIS: (a) The finite-difference equations for the interior points, nodes 0, 1, 2, and 3, can be
determined from Equation 5.81,
)
(
p +1
p
p
p
=
Tm
Fo Tm
−1 + Tm +1 + (1 − 2 Fo ) Tm
(1)
with
Fo = α∆t ∆x 2 = 1.5 ×10−6 m 2 s × 300s ( 0.03m ) = 1/ 2 .
2
(2)
Note that the stability criterion, Equation 5.82, Fo ≤ 1/2, is satisfied. Hence, combining Equations (1) and
p +1
p
p
(2),
=
Tm
1/ 2 Tm
−1 + Tm +1 for m = 0, 1, 2, 3. Since the adiabatic plane at x = 0 can be treated as a
)
(
symmetry plane, Tm-1 = Tm+1 for node 0 (m = 0). The finite-difference solution is generated in the table
below using t = p⋅∆t = 300 p (s) = 5 p (min).
p
0
1
2
3
4
5
6
7
8
9
t(min)
0
10
20
30
40
45
T0
85
85
85
85
76.9
76.9
68.8
68.8
61.7
61.7
T1
85
85
85
76.9
76.9
68.8
68.8
61.7
61.7
55.6
T2
85
85
68.8
68.8
60.7
60.7
54.6
54.6
49.5
49.5
T3
85
52.5
52.5
44.4
44.4
40.4
40.4
37.3
37.3
34.8
TL(°C)
20
20
20
20
20
20
20
20
20
20
<
The temperature distribution can also be determined from the one-term approximation of the exact
solution. The insulated surface is equivalent to the midplane of a wall of thickness 2L. Thus,
−6 2
α t 1.5 ×10 m s × ( 45 × 60 ) s
and
Bi → ∞.
Fo
= =
= 0.28
2
L2
( 0.12 m )
Continued...
PROBLEM 5.92 (Cont.)
From Table 5.1, ζ1 = 1.5707, C1 = 1.2733. Then from Equation 5.44,
θ o∗ = C1 exp(−ζ12 Fo)= 1.2733exp(−1.5707 2 × 0.28)= 0.64 or

To =T ( 0, t ) =T∞ + θ o∗ ( Ti − T∞ ) =20 C + 0.64 ( 85 − 20 ) C =61.5 C .
This value shows excellent agreement with 61.7°C for the finite-difference method.
(b) Using the IHT Finite-Difference Equation Tool Pad for One-Dimensional Transient Conduction,
temperature histories were computed and results are shown for the insulated surface (T0) and the
midplane, as well as for the chilled surface (TL).
85
Temperature, T(C)
75
65
55
45
35
25
15
0
2000
4000
6000
8000 10000 12000 14000 16000 18000
Time, t(s)
T0,
T2,
TL
T0,
T2,
deltat = 75 s
deltat = 75 s
deltat = 300 s
deltat = 300 s
Apart from small differences during early stages of the transient, there is excellent agreement between
results obtained for the two time steps. The temperature decay at the insulated surface must, of course,
lag that of the midplane.
PROBLEM 5.93
KNOWN: Initial temperature of thick slab of copper. Surface heat flux.
FIND: (a) Expression for implicit finite-difference equations for interior nodes using the Der function of
IHT. (b) Expression for implicit finite-difference equations for surface node using the Der function of
IHT. (c) Nodal temperatures at t = 120 s.
SCHEMATIC:
Δx = 75 mm
q“o
•0
•2
•1
•3
•4
•5
•6
•7
•8
•9
ASSUMPTIONS: (1) Uniform properties, (2) One-dimensional conduction.
PROPERTIES: Table A-1, Copper, (T = 300 K): k = 401 W/m⋅K, α = 117 × 10−6 m2/s.
ANALYSIS: (a) Performing an energy balance on an interior node m, as
shown, and evaluating the heat fluxes at time step p+1 for the implicit
method, gives:
∂T
mc = E in − E out
∂t
T p +1 − Tmp
T p +1 − Tmp +1
T p +1 − Tmp +1
ρ A∆xc m
= kA m −1
+ kA m +1
∆t
∆x
∆x
m-1
•
m
•
″
qcond
m+1
•
″
qcond
∆x
Since the Der function of IHT uses implicit time integration, the time derivative on the left side
can be replaced by the Der function. Dividing by A, we find the required form:
rDxcDer(Tmp +1 , t ) = k
Tmp−+11 − Tmp +1
T p +1 − Tmp +1
+ k m +1
DD
x
x
<
(b) For the surface node 0, an energy balance gives:
T p +1 − T0p +1
∆x T p +1 − T0p
ρA c 0
= qo′′A + kA 1
2
∆t
∆x
Replacing the derivative on the left with the Der function and dividing
by A gives:
r
0
•
m+1
•
″
qcond
qo″
∆x/2
T p +1 − T0p +1
Dx
) qo′′ + k 1
cDer(T0p +1 , t=
2
Dx
<
(c) The IHT code to solve for the temperatures at nodes 0 through 8 is given below, where the temperature
at node 9 is specified as 20°C.
Continued…
PROBLEM 5.93 (Cont.)
//Example 5.12 using Der function of IHT
//Properties, dimensions, heat flux
k = 401
alpha = 117e-6
rhoc = k/alpha
qflux = 3e5
delx = 0.075
//Node 0
rhoc*delx/2*Der(T0,t) = qflux + k*(T1 - T0)/delx
//Nodes 1 through 8
rhoc*delx*Der(T1,t) = k/delx*(T0 - T1) + k/delx*(T2 - T1)
rhoc*delx*Der(T2,t) = k/delx*(T1 - T2) + k/delx*(T3 - T2)
rhoc*delx*Der(T3,t) = k/delx*(T2 - T3) + k/delx*(T4 - T3)
rhoc*delx*Der(T4,t) = k/delx*(T3 - T4) + k/delx*(T5 - T4)
rhoc*delx*Der(T5,t) = k/delx*(T4 - T5) + k/delx*(T6 - T5)
rhoc*delx*Der(T6,t) = k/delx*(T5 - T6) + k/delx*(T7 - T6)
rhoc*delx*Der(T7,t) = k/delx*(T6 - T7) + k/delx*(T8 - T7)
rhoc*delx*Der(T8,t) = k/delx*(T7 - T8) + k/delx*(T9 - T8)
//Node 9
T9 = 20
With initial temperatures of 20°C and a time step of 24 s, the solution at t = 120 s is:
T0
114.7
T1
69.9
T2
T3
T4
44.16
30.91
24.66
T5
21.9
T6
T7
T8
T9
20.74
20.28
20.09
20.00
<
COMMENTS: The solution using the Der function of IHT agrees almost exactly with the solution of the
implicit finite-difference equations in Example 5.12 for Fo = ½ (∆t = 24 s), as it should.
PROBLEM 5.94
KNOWN: Thickness of aluminum alloy (2024-T6) absorber plate. Spacing of tubes attached to plate,
temperature of plate at tube location. Net radiation heat flux.
FIND: (a) Steady-state temperature distribution and rate of thermal energy delivered to a tube per unit
′′ = 800 W/m2. Maximum plate temperature. (b) Temperature
length, using finite difference model, for qrad
′′ is reduced to zero. Plot of maximum plate temperature vs.
distribution for the 500 s time period after qrad
time. Values of maximum plate temperature for t = 0, 10, and 100 s. (c) Plot of heat transfer rate to tube
per unit length for the 500 s time period. Time for heat transfer rate to tube to be reduced by 50%.
SCHEMATIC:
“ = 800 or 0 W/m2
qrad
qrad
“
Δx
T(0,t) = Tw •
0
•1
•
2
•3
•4
•5
•6
•7
•8
•
•
9
10
L/2
ASSUMPTIONS: (1) Uniform properties, (2) Negligible temperature variation through plate thickness,
(3) Uniform radiation absorption at plate surface, (4) Negligible losses by conduction through insulation,
(5) Negligible losses by convection at absorber plate surface, (6) Temperature of absorber plate at x = 0 is
approximately that of the water.
PROPERTIES: Table A-1, Aluminum alloy (2024-T6), (T = 325 K): r = 2770 kg/m3, cp = 873 J/kg⋅K, k
= 177 W/m⋅K, α = 7.33 × 10−5 m2/s.
∙
′
Est
ANALYSIS: The absorber plate acts as an extended surface. Finite
difference equations may be obtained by applying an energy balance
to one of the nodes, shown. For a unit length of tube, using the Der
function of IHT for the time derivative in the implicit method (and
using t for the plate thickness to distinguish from time t):
T p +1 − Tmp +1
T p +1 − Tmp +1
′′ Dx
r tDxcDer(
=
T , t ) kt m −1
+ kt m +1
+ qrad
DD
x
x
p +1
m
r t
p +1
10
T −T
DD
x
x
′′
=
cDer(T10p +1 , t ) kt 9
+ qrad
2
2
Dx
m
•
m+1 ~
•
t
′
qcond
∆x
(1)
For Node 10, shown, which is at the symmetry plane between tubes,
there is zero heat flux into the control volume through the symmetry
plane. Thus the finite difference equation is:
p +1
m-1
•
′
qcond
q′rad
(2)
∙′
Est
•9
′
qcond
′
qrad
~
t
• 10
∆x/2
Continued…
PROBLEM 5.94 (Cont.)
The heat transfer rate to a tube can be determined from an energy balance about Node 0, where the factor
of 2 accounts for heat coming from both directions (right and left):
 T p +1 − T0p +1
∆x 
′
′′
=
qtube
2  kt 1
+ qrad

∆x
2 

(3)
The IHT code that implements these equations is in the Comments section.
(a) For part (a), the radiation heat flux is set to 800 W/m2 and the solution is advanced until a steady-state
is reached. Alternatively, there is an analytical solution to the steady-state case, see Problem 3.89. The
steady-state solution from the IHT code is:
T0
T1
T2
T3
T4
T5
T6
T7
T8
T9
T10
60.00
60.71
61.35
61.92
62.41
62.82
63.16
63.42
63.61
63.72
63.76
<
<
The maximum temperature is at Node 10 and is 63.76°C .
From the IHT solution, using Eq. (3), the heat transfer rate to the tube (per unit tube length) is 160 W/m.
This can also be found from the fact that all of the radiation heat flux incident on the absorber plate is
′ =qrad
′′ L =800 W/m 2 × 0.2 m =160 W/m.
transferred to the tube under steady-state conditions, so qtube
<
(b) Starting from the steady-state temperature distribution, the IHT code is run with zero radiation heat
flux and a time step of 1 s. A plot of the maximum temperature (at Node 10) as a function of time is given
below.
Max. temperature (C)
64
63
62
61
60
59
0
100
200
300
400
500
Time (s)
The maximum temperature is Tmax = 63.76, 63.27, and 60.66°C at t = 0, 10 and 100 s.
<
Continued…
PROBLEM 5.94 (Cont.)
(c) The plot of the heat transfer rate to the tube per unit length, as a function of time, is shown below.
160
qprime (W/m)
140
120
100
80
60
40
20
0
0
100
200
300
400
500
Time (s)
Since the initial rate of heat transfer to the tube is 160 W/m, when it is reduced by 50% its value is 80
W/m. This occurs at t = 27 s.
<
COMMENTS: (1) The IHT code that implements Eqs. (1-3) is given below.
//Properties,dimensions
//Properties at 325 K
Tprop = 325
// Aluminum 2024 property functions : From Table A.1
// Units: Tprop(K)
rho = rho_300K("Aluminum 2024")
// Density, kg/m^3
alpha = k / (rho * cp)
// Thermal diffusivity, m^2/s
k = k_T("Aluminum 2024",Tprop)
// Thermal conductivity,W/m·K
cp = cp_T("Aluminum 2024",Tprop)
// Specific heat, J/kg·K
rhoc = rho*cp
dx = 0.01
//Radiative conditions
//qflux = 800
//Use for part (a) and run to steady-state.
qflux = 0
//Use for parts (b) and (c).
//Node 0
T0 = 60
//Nodes 1 through 9
Ac = 1*thick
thick = 0.006
rhoc*dx*Ac*Der(T1,t) = Ac*k/dx*(T0 - T1) + Ac*k/dx*(T2 - T1) + qflux*dx
rhoc*dx*Ac*Der(T2,t) = Ac*k/dx*(T1 - T2) + Ac*k/dx*(T3 - T2) + qflux*dx
rhoc*dx*Ac*Der(T3,t) = Ac*k/dx*(T2 - T3) + Ac*k/dx*(T4 - T3) + qflux*dx
rhoc*dx*Ac*Der(T4,t) = Ac*k/dx*(T3 - T4) + Ac*k/dx*(T5 - T4) + qflux*dx
rhoc*dx*Ac*Der(T5,t) = Ac*k/dx*(T4 - T5) + Ac*k/dx*(T6 - T5) + qflux*dx
rhoc*dx*Ac*Der(T6,t) = Ac*k/dx*(T5 - T6) + Ac*k/dx*(T7 - T6) + qflux*dx
Continued…
PROBLEM 5.94 (Cont.)
rhoc*dx*Ac*Der(T7,t) = Ac*k/dx*(T6 - T7) + Ac*k/dx*(T8 - T7) + qflux*dx
rhoc*dx*Ac*Der(T8,t) = Ac*k/dx*(T7 - T8) + Ac*k/dx*(T9 - T8) + qflux*dx
rhoc*dx*Ac*Der(T9,t) = Ac*k/dx*(T8 - T9) + Ac*k/dx*(T10 - T9) + qflux*dx
//Node 10
rhoc*(dx/2)*Ac*Der(T10,t) = Ac*k/dx*(T9 - T10) + qflux*(dx/2)
//Heat transfer per unit length to cold tube
qtubeprime= (k*Ac*(T1 - T0)/dx + qflux*(dx/2))*2
//Factor of 2 since tube is heated from both left and right
PROBLEM 5.95
KNOWN: Thickness, initial temperature and thermophysical properties of molded plastic part.
Convection conditions at one surface. Other surface insulated.
FIND: Surface temperatures after one hour of cooling.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in product, (2) Negligible radiation, at cooled
surface, (3) Negligible heat transfer at insulated surface, (4) Constant properties.
ANALYSIS: Adopting the implicit scheme, the finite-difference equation for the cooled surface node is
given by Eq. (5.96), from which it follows that
p +1
p +1
p
2 FoBi T∞ + T10
− 2Fo T=
(1 + 2 Fo + 2 FoBi ) T10
9
The general form of the finite-difference equation for any interior node (1 to 9) is given by Eq. (5.97),
p
Tm
(1 + 2 Fo ) Tmp +1 − Fo Tmp +−11 + Tmp ++11 =
)
(
The finite-difference equation for the insulated surface node may be obtained by applying the symmetry
p
p
requirement to Eq. (5.97); that is, Tm +1 = Tm −1. Hence,
Top
(1 + 2 Fo ) Top +1 − 2 Fo T1p +1 =
2
For the prescribed conditions, Bi = h∆x/k = 100 W/m ⋅K (0.006m)/0.30 W/m⋅K = 2. If the explicit
method were used, the most restrictive stability requirement would be given by Eq. (5.87). Hence, for Fo
2
-7 2
(1+Bi) ≤ 0.5, Fo ≤ 0.167. With Fo = α∆t/∆x and α = k/rc = 1.67 ×10 m /s, the corresponding
restriction on the time increment would be ∆t ≤ 36s. Although no such restriction applies for the implicit
method, a value of ∆t = 30s is chosen, and the set of 11 finite-difference equations is solved using the
Tools option designated as Finite-Difference Equations, One-Dimensional, and Transient from the IHT
Toolpad. At t = 3600s, the solution yields:
T10 ( 3600s ) =
24.1°C
T0 ( 3600s ) =
71.5°C
<
COMMENTS: (1) More accurate results may be obtained from the one-term approximation to the exact
solution for one-dimensional, transient conduction in a plane wall. With Bi = hL/k = 20, Table 5.1 yields
2
ζ 1 = 1.496 rad and C1 = 1.2699. With Fo = αt/L = 0.167, Eq. (5.41) then yields To = T∞ + (Ti - T∞) C1
(
)
72.4°C, and from Eq. (5.43b), Ts = T∞ + (Ti - T∞) cos (ζ1 ) = 24.5°C. Since the finiteexp −ζ 1 Fo =
2
difference results do not change with a reduction in the time step (∆t < 30s), the difference between the
numerical and analytical results is attributed to the use of a coarse grid. To improve the accuracy of the
numerical results, a smaller value of ∆x should be used.
Continued …
PROBLEM 5.95 (Cont.)
(2) Temperature histories for the front and back surface nodes are as shown.
80
Temperature (C)
70
60
50
40
30
20
0
600
1200
1800
2400
3000
3600
Time (s)
Insulated surface
Cooled surface
Although the surface temperatures rapidly approaches that of the coolant, there is a significant lag in the
thermal response of the back surface. The different responses are attributable to the small value of α and
the large value of Bi.
PROBLEM 5.96
KNOWN: Thickness and thermal diffusivity of a plane wall. Initial and boundary conditions.
FIND: (a) Time required for the left face temperature to reach 50% of its maximum possible
temperature reduction and (b) Time required for the left face temperature to recover to a 20% temperature
reduction when the right face temperature is returned to its initial value.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Constant properties.
PROPERTIES: Thermal diffusivity, α = 6 × 10-7 m2/s (given).
ANALYSIS: Using the IHT Finite-Difference Equation Tool Pad for One-Dimensional Transient
Conduction, the following temperatures may be computed using the IHT code in the Comments section.
Note that for this solution, the conditions at the right face have been incorporated by specifying a very
large convection coefficient at the right face so that T5 ≈ T∞.
(a) For cooling conditions, the initial temperature is arbitrarily set to unity, and T∞ = Ts,r = 0.5. If a steadystate solution were achieved, it would correspond to a uniform wall temperature of Tss = Ts,r = 0.5. Hence
we seek to determine the time when T0 = (Ti + Ts,r)/2 = 0.75. The following representative results are
obtained using a time step of ∆t = 2 s.
t(s)
T0
T1
T2
T3
T4
T5
20
40
60
0.952
0.852
0.765
0.937
0.835
0.752
0.888
0.787
0.714
0.798
0.710
0.656
0.665
0.611
0.582
0.5
0.5
0.5
64
0.750
0.738
0.702
0.647
0.577
0.5
<
(b) For the heating conditions, we set the initial conditions in IHT to the temperatures shown at t = 64 in
the table above. We set T∞ = 1 and monitor the transient response of the system. The following
representative results are obtained.
t(s)
T0
T1
T2
T3
T4
T5
20
40
60
80
0.735
0.789
0.840
0.880
0.741
0.798
0.848
0.886
0.763
0.826
0.870
0.903
0.812
0.872
0.906
0.930
0.893
0.932
0.950
0.963
1.0
1.0
1.0
1.0
94
0.902
0.907
0.921
0.943
0.970
1.0
<
Continued…
PROBLEM 5.96 (Cont.)
COMMENTS: (1) The IHT Code for cooling and heating is shown below. (2) Note the thermal
response of T0 after heating ensues at the right face. The value of T0 continues to decrease for a short time
before it begins to increase in value. Does this make sense to you? (3) Part (a) can be solved analytically
using the one-term approximation, yielding t = 63 s. The numerical and analytical solutions are in
agreement to within an uncertainty associated with the time step of ∆t = 2 s.
/* Node 0: surface node (w-orientation); transient conditions; e labeled 1. */
rho * cp * der(T0,t) = fd_1d_sur_w(T0,T1,k,qdot,deltax,Tinf,h1,qfla0)
/* Node 1: interior node; e and w labeled e and 0. */
rho*cp*der(T1,t) = fd_1d_int(T1,T2,T0,k,qdot,deltax)
/* Node 2: interior node; e and w labeled 3 and 1. */
rho*cp*der(T2,t) = fd_1d_int(T2,T3,T1,k,qdot,deltax)
/* Node 3: interior node; e and w labeled 4 and 2. */
rho*cp*der(T3,t) = fd_1d_int(T3,T4,T2,k,qdot,deltax)
/* Node 4: interior node; e and w labeled 4 and 3. */
rho*cp*der(T4,t) = fd_1d_int(T4,T5,T3,k,qdot,deltax)
/* Node 5: surface node (e-orientation); transient conditions; w labeled 4. */
rho * cp * der(T5,t) = fd_1d_sur_e(T5,T4,k,qdot,deltax,Tinf,h2,qfla5)
// Cooling
//Tinf = 0.5
//Heating
Tinf = 1.0
//Initial Conditions are T = 1.0 everywhere.
//Initial Conditions are the temperatures at t = 64 s.
h1 = 0
h2 = 1*10^10
qfla0 = 0
qfla5 = 0
deltax = 2/1000
qdot = 0
//Insulated left wall
//Right wall at Tinf since h2 is nearly infinite
//Zero applied heat flux at left wall
//Zero applied heat flux at right wall
//meters
//W/m^3
// Set k, rho and cp values so that alpha = k/(rho*cp) = 6*10^-7 m2/s
k = 6*10^-7
rho = 1
cp =1
//W/m-K
//kg
//J/kg-K
PROBLEM 5.97
PROBLEM 5.98
KNOWN: Dimensions and properties of acrylic and steel plates. Initial temperatures.
FIND: Time needed to bring external surface of the acrylic to its softening temperature. Plot of the
average acrylic and steel plate temperatures and acrylic surface temperature for 0 ≤ t ≤ 300 s.
SCHEMATIC:
2L = 10 mm
y
∆y/2
• •
1
• • • •
•••
2
Adiabatic
surface
∆y/2
∆y
9
10
11
• •
•••
12
19
20
Adiabatic
surface
Material B
Material A
Interface
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible contact
resistance.
PROPERTIES: Acrylic (given): rA = 1990 kg/m3, cA = 1470 J/kg⋅K and kA = 0.21 W/m⋅K. Steel
(given): rB = 7800 kg/m3, cB = 500 J/kg⋅K and kB = 45 W/m⋅K.
ANALYSIS: We begin by writing energy balances on each of the 20 control volumes using the implicit
method,
Node 1:
(
)
(
p +1
p
kA T2p +1 − T1p +1
∆y T1 − T1
ρ A cA
=
2
∆t
∆y
)
ρ A cA
(T
∆y
− Tnp
kA Tnp−+11 − Tnp +1
kA Tnp++11 − Tnp +1
=
+
∆t
∆y
∆y
Node 10:
ρ A cA
(T
∆y
− T10p
kA T9p +1 − T10p +1
T11p +1 − T10p +1
=
+
∆t
∆y
Rt"
Node 11:
ρ BcB∆y
Nodes 12 – 19:
(T
ρ c ∆y
Nodes 2 – 9:
Node 20:
p +1
n
p +1
10
(T
p +1
11
− T11p
∆t
p +1
n
B B
(
)
(
)
)
(
) (
) = (T
p +1
10
)
− T11p +1
Rt"
(
(
) + k (T
B
)
p +1
12
− T11p +1
)
)
∆y
(
− Tnp
k B Tnp−+11 − Tnp +1
k B Tnp++11 − Tnp +1
=
+
∆t
∆y
∆y
)
(
p +1
p
k B T19p +1 − T20p +1
∆y T20 − T20
ρ B cB
=
2
∆t
∆y
)
)
)
Continued…
PROBLEM 5.98 (Cont.)
=
Rt"
where
∆y / 2 ∆y / 2
+
kA
kB
Also, note that the average temperature of each material may be written as
1
T=
L
L
∫ T ( y )dy
y =0
or, in finite difference form,
=
TA
1
LA
 ∆y

T1 2 + T2 ∆y + T3∆y + T4 ∆y + T5∆y + T6 ∆y + T7 ∆y + T8 ∆y + T9 ∆y + T10 ∆y 
=
TB
1
LB
∆y 

T11∆y + T12 ∆y + T13∆y + T14 ∆y + T15∆y + T16 ∆y + T17 ∆y + T18 ∆y + T19 ∆y + T20 2 
and
The preceding equations were solved using IHT code that is listed in the Comments. The spatiallyaveraged temperatures of the two plates, as well as the external temperature of the acrylic, T1, are shown
in the plot below.
From the simulation, we also find that the surface of the acrylic reaches Tsoft = 90°C at t = 87 s.
<
<
Average Material and Acrylic Surface Temperatures
350
300
Steel average temperature
Temperature (C)
250
200
150
Acrylic average
temperature
100
50
Acrylic surface temperature
0
0
75
150
225
300
Time (s)
COMMENTS: (1) Ultimately, the temperatures of the two plates will reach the same steady-state value.
The steady-state temperature may be found by recognizing the energy gained by the acrylic is lost by the
steel, LAcA(Tss – Ti,A) = LBcB(Ti,B – Tss) yielding Tss = 180°C, as evident in the plot of the average
temperatures. (2) If the surface of the acrylic in contact with the metal is assumed to be of constant
temperature and equal to 300°C, the external surface of the acrylic reaches the softening temperature at t
= 74 s. (3) The IHT code is shown on the next page.
Continued…
PROBLEM 5.98 (Cont.)
//Geometry and Discretization
LA = 0.005
LB = 0.005
delyA = LA/9.5
delyB = LB/9.5
//m
//m
//m
//m
//Properties
kA = 0.21
kB = 45
cA = 1470
cB = 500
rhoA = 1990
rhoB = 7800
//W/mK
//W/mK
//J/kgK
//J/kgK
//kg/m^3
//kg/m^3
//Initial Temperatures
TiA = 20
TiB = 300
//C
//C
//Interface Resistance
Rt = delyA/kA/2 + delyB/kB/2
//Node 1
rhoA*cA*(delyA/2)*der(T1,t) = (kA/delyA)*(T2 - T1)
//Nodes 2 through 9
rhoA*cA*delyA*der(T2,t) = (kA/delyA)*(T1 - T2) + (kA/delyA)*(T3 - T2)
rhoA*cA*delyA*der(T3,t) = (kA/delyA)*(T2 - T3) + (kA/delyA)*(T4 - T3)
rhoA*cA*delyA*der(T4,t) = (kA/delyA)*(T3 - T4) + (kA/delyA)*(T5 - T4)
rhoA*cA*delyA*der(T5,t) = (kA/delyA)*(T4 - T5) + (kA/delyA)*(T6 - T5)
rhoA*cA*delyA*der(T6,t) = (kA/delyA)*(T5 - T6) + (kA/delyA)*(T7 - T6)
rhoA*cA*delyA*der(T7,t) = (kA/delyA)*(T6 - T7) + (kA/delyA)*(T8 - T7)
rhoA*cA*delyA*der(T8,t) = (kA/delyA)*(T7 - T8) + (kA/delyA)*(T9 - T8)
rhoA*cA*delyA*der(T9,t) = (kA/delyA)*(T8 - T9) + (kA/delyA)*(T10 - T9)
//Node 10
rhoA*cA*delyA*der(T10,t) = (kA/delyA)*(T9 - T10) + (1/Rt)*(T11 - T10)
//Node 11
rhoB*cB*delyB*der(T11,t) = (1/Rt)*(T10 - T11) + (kB/delyB)*(T12 - T11)
//Nodes 12 through 19
rhoB*cB*delyB*der(T12,t) = (kB/delyB)*(T11 - T12) + (kB/delyB)*(T13 - T12)
rhoB*cB*delyB*der(T13,t) = (kB/delyB)*(T12 - T13) + (kB/delyB)*(T14 - T13)
rhoB*cB*delyB*der(T14,t) = (kB/delyB)*(T13 - T14) + (kB/delyB)*(T15 - T14)
rhoB*cB*delyB*der(T15,t) = (kB/delyB)*(T14 - T15) + (kB/delyB)*(T16 - T15)
rhoB*cB*delyB*der(T16,t) = (kB/delyB)*(T15 - T16) + (kB/delyB)*(T17 - T16)
rhoB*cB*delyB*der(T17,t) = (kB/delyB)*(T16 - T17) + (kB/delyB)*(T18 - T17)
rhoB*cB*delyB*der(T18,t) = (kB/delyB)*(T17 - T18) + (kB/delyB)*(T19 - T18)
rhoB*cB*delyB*der(T19,t) = (kB/delyB)*(T18 - T19) + (kB/delyB)*(T20 - T19)
//Node 20
rhoB*cB*(delyB/2)*der(T20,t) = (kB/delyB)*(T19 - T20)
//Average Temperatures
TavgA*LA = T1*delyA/2 + T2*delyA + T3*delyA + T4*delyA + T5*delyA + T6*delyA + T7*delyA + T8*delyA
+ T9*delyA + T10*delyA
TavgB*LB = T11*delyB + T12*delyB + T13*delyB + T14*delyB + T15*delyB + T16*delyB + T17*delyB +
T18*delyB + T19*delyB + T20*delyB/2
//Steady State Temperature
LA*cA*rhoA*(TiA - Tss) = LB*cB*rhoB*(Tss - TiB)
PROBLEM 5.99
KNOWN: Conditions associated with heat generation in a rectangular fuel element with surface cooling.
See Example 5.11.
FIND: (a) The temperature distribution 1.5 s after the change in operating power; compare your results
with those tabulated in the example, (b) Calculate and plot temperature histories at the mid-plane (00) and
surface (05) nodes for 0≤ t ≤ 400 s; determine the new steady-state temperatures, and approximately how
long it will take to reach the new steady-state condition after the step change in operating power. Use the
IHT Tools | Finite-Difference Equations | One-Dimensional | Transient conduction model builder as your
solution tool.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Uniform generation,
and (3) Constant properties.
ANALYIS: The IHT model builder provides the transient finite-difference equations for the implicit
method of solution. Selected portions of the IHT code used to obtain the results tabulated below are
shown in the Comments.
(a) Using the IHT code, the temperature distribution (°C) as a function of time (s) up to 1.5 s
after the step power change is obtained from the summarized results copied into the workspace
1
2
3
4
5
6
t
0
0.3
0.6
0.9
1.2
1.5
T00
357.6
358.1
358.6
359.1
359.6
360.1
T01
356.9
357.4
357.9
358.4
358.9
359.4
T02
354.9
355.4
355.9
356.4
356.9
357.4
T03
351.6
352.1
352.6
353.1
353.6
354.1
T04
346.9
347.4
347.9
348.4
348.9
349.3
T05
340.9
341.4
341.9
342.3
342.8
343.2
(b) Using the code, the mid-plane (00) and surface (05) node temperatures are plotted as a
function of time.
Continued …
PROBLEM 5.99 (Cont.)
Note that at t ≈ 240 s, the wall has nearly reached the new steady-state condition for which the
nodal temperatures (°C) were found as:
T00
465
T01
463.7
T02
459.7
T03
453
T04
443.7
T05
431.7
COMMENTS: (1) Can you validate the new steady-state nodal temperatures from part (b) by
comparison against an analytical solution?
(2) Will using a smaller time increment improve the accuracy of the results? Use your code with
∆t = 0.15 s to justify your explanation.
(3) Selected portions of the IHT code to obtain the nodal temperature distribution using spatial and time
increments of ∆x = 2 mm and ∆t = 0.3 s, respectively, are shown below. For the solve-integration step,
the initial condition for each of the nodes corresponds to the steady-state temperature distribution with
q1.
// Tools | Finite-Difference Equations | One-Dimensional | Transient
/* Node 00: surface node (w-orientation); transient conditions; e labeled 01. */
rho * cp * der(T00,t) = fd_1d_sur_w(T00,T01,k,qdot,deltax,Tinf01,h01,q''a00)
q''a00 = 0
// Applied heat flux, W/m^2; zero flux shown
Tinf01 = 20
// Arbitrary value
h01 = 1e-8
// Causes boundary to behave as adiabatic
/* Node 01: interior node; e and w labeled 02 and 00. */
rho*cp*der(T01,t) = fd_1d_int(T01,T02,T00,k,qdot,deltax)
/* Node 02: interior node; e and w labeled 03 and 01. */
rho*cp*der(T02,t) = fd_1d_int(T02,T03,T01,k,qdot,deltax)
/* Node 03: interior node; e and w labeled 04 and 02. */
rho*cp*der(T03,t) = fd_1d_int(T03,T04,T02,k,qdot,deltax)
/* Node 04: interior node; e and w labeled 05 and 03. */
rho*cp*der(T04,t) = fd_1d_int(T04,T05,T03,k,qdot,deltax)
/* Node 05: surface node (e-orientation); transient conditions; w labeled 04. */
rho * cp * der(T05,t) = fd_1d_sur_e(T05,T04,k,qdot,deltax,Tinf05,h05,q''a05)
q''a05 = 0
// Applied heat flux, W/m^2; zero flux shown
Tinf05 = 250
// Coolant temperature, C
h05 = 1100
// Convection coefficient, W/m^2.K
// Input parameters
qdot = 2e7
// Volumetric rate, W/m^3, step change
deltax = 0.002
// Space increment
k = 30
// Thermophysical properties
alpha = 5e-6
rho = 1000
alpha = k / (rho * cp)
/* Steady-state conditions, with qdot1 = 1e7 W/m^3; initial conditions for step change
T_x = 16.67 * (1 - x^2/L^2) + 340.91
// See text
Seek T_x for x = 0, 2, 4, 6, 8, 10 mm; results used for Ti are
Node T_x
00
357.6
01
356.9
02
354.9
03
351.6
04
346.9
05
340.9
*/
PROBLEM 5.100
PROBLEM 5.100 (Cont.)
PROBLEM 5.101
KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a
convection cooling process (T∞,h).
FIND: Temperatures at the surface and a 45mm depth after 3 minutes using finite-difference
method with space and time increments of 15mm and 18s.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Plate approximates semiinfinite medium, (3) Constant properties.
ANALYSIS: The grid network representing the plate is shown above. The finite-difference
equation for node 0 is given by Eq. 5.90 for one-dimensional conditions or Eq. 5.85,
)
(
=
T0p+1 2 Fo T1p + Bi ⋅ T∞ + (1 − 2 Fo − 2 Bi ⋅ Fo ) T0p .
(1)
The numerical values of Fo and Bi are
5.6 ×10-6 m 2 / s ×18s
=
Fo =
= 0.448
∆x 2
( 0.015m )2
α∆t
(
)
2
-3
h∆x 100 W/m ⋅ K × 15 × 10 m
=
Bi =
= 0.075.
k
20 W/m ⋅ K
Recognizing that T∞ = 15°C, Eq. (1) has the form
T0p+1 = 0.0359 T0p + 0.897 T1p + 1.01.
(2)
It is important to satisfy the stability criterion, Fo (1+Bi) ≤ 1/2. Substituting values,
0.448 (1+0.075) = 0.482 ≤ 1/2, and the criterion is satisfied.
The finite-difference equation for the interior nodes, m = 1, 2…, follows from Eq. 5.78,
)
(
p+1
p
p
p
=
Tm
Fo Tm+1
+ Tm-1
+ (1 - 2Fo ) Tm
.
(3)
Recognizing that the stability criterion, Fo ≤ 1/2, is satisfied with Fo = 0.448,
(
)
p+1
p
p
p
T=
0.448 Tm+1
+ Tm-1
+ 0.104Tm
.
m
(4)
Continued …
PROBLEM 5.101 (Cont.)
The time scale is related to p, the number of steps in the calculation procedure, and ∆t, the time
increment,
t = p∆t.
(5)
The finite-difference calculations can now be performed using Eqs. (2) and (4). The results are
tabulated below.
p
0
1
2
3
4
5
6
7
8
9
10
t(s)
0
18
36
54
72
90
108
126
144
162
180
T0
325
304.2
303.2
294.7
293.0
287.6
285.6
281.8
279.8
276.7
274.8
T1
325
324.7
315.3
313.7
307.8
305.8
301.6
299.5
296.2
294.1
291.3
T2
325
325
324.5
320.3
318.9
315.2
313.5
310.5
308.6
306.0
304.1
T3
325
325
325
324.5
322.5
321.5
319.3
317.9
315.8
314.3
312.4
T4
325
325
325
325
324.5
323.5
322.7
321.4
320.4
319.0
T5
325
325
325
325
325
324.5
324.0
323.3
322.5
T6
T7(K)
325
325
325
325
325
325
325
325
325
325
325
325
324.5 325
324.2
Hence, find
10 275 C
T ( 0, 180s
=
) T=
0
10 312 C.
T ( 45mm, 180s
=
) T=
3
<
COMMENTS: (1) The above results can be readily checked against the analytical solution
represented in Fig. 5.8 (see also Eq. 5.63). For x = 0 and t = 180s, find
x
=0
1/ 2
2 (α t )
)
(
1/ 2
100 W/m 2 ⋅ K 5.60 ×10-6 m 2 / s ×180s
h (α t )
= 0.16
k
20 W/m ⋅ K
for which the figure gives
T − Ti
= 0.15
T∞ − Ti
so that,
1/ 2

T ( 0, 180s
=
Ti 0.15 (15 − 325 ) C + 325C
) 0.15 ( T∞ − Ti ) +=
T ( 0, 180s ) = 278 C.
For x = 45mm, the procedure yields T(45mm, 180s) = 316°C. The agreement with the numerical
solution is nearly within 1%.
PROBLEM 5.102
KNOWN: Thickness and thermophysical properties of wind turbine nacelle of Example 3.1. Rotational
speed and number of blades. Average value and amplitude of sinusoidal variation of heat transfer
coefficient versus time. Initial wall temperature distribution.
FIND: Transient wall temperature distribution for 40 ≤ t ≤ 50 s, using ∆x = 2 mm and ∆t = 0.05 s. Plot of
exterior nacelle surface temperature, exterior surface convection heat flux, and exterior surface radiation
heat flux for 40 ≤ t ≤ 50 s. Explain.
SCHEMATIC:
L = 20 mm
h(t) = h + Δh•sin(ωt)
T(x,t = 0) = Ax + B
Tsur
h = 35 W/m2•K
Δh = 15 W/m2•K
ω = 17 rpm
A = -3460°C/m
B = 212°C
h(t), T∞
ε
Thermoplastic
nacelle
L
dx
x
k = 1.5 W/m•K
ρ = 1250 kg/m3
c = 1500 J/kg•K
T(x = 0) = Ts
ASSUMPTIONS: (1) One-dimensional plane wall, (2) Uniform properties, (3) Large surroundings, α =
ε, (4) Ts = constant.
PROPERTIES: Given, r = 1250 kg/m3, c = 1500 J/kg⋅K, k = 1.5 W/m⋅K.
Tsur
ANALYSIS: The frequency of the oscillation of h(t) is:
α=ε
 1 min  blades 
=
(17
rev) 
 3
 2π rad 5.34 rad/s
 60 s  rev 
•
9
8
•
7
•
6
•
5
(1)
•
4
For the surface node 10, where heat is transferred by convection and
radiation, an energy balance gives the following finite difference
equation;
Δx
3
2
•
Tmp−1 − Tmp
T p −T p
+ kA m +1 m
DD
x
x
•
The nodal layout is shown in the schematic to the right. The finite difference
equation for internal nodes 1 through 9 is the same as in Example 5.11 (with
q = 0), except that we write the time derivative in terms of the Der function
of IHT:
r ADxcDer(Tmp , t ) = kA
•10
•
ω
h(t), T∞
•
1
T −T
Dx
cDer(=
T10p , t ) kA 9
− h p A(T10p − T∞ ) − es A ( (T10p ) 4 − Tsur4 )
2
Dx
p
10
(2)
To = Ts
0
•
rA
p
Continued…
L
PROBLEM 5.102 (Cont.)
Exterior surface temperature (deg. C)
Equations (1) and (2) are implemented in the IHT code shown in the Comments section. The
solution for the nacelle temperature, convection heat flux, and radiation heat flux at the exterior
surface are plotted below.
144
143
142
40
41
42
43
44
45 46
Time (s)
47
48
49
50
47
48
49
50
6,000
Convection flux (W/m^2)
5,500
5,000
4,500
4,000
3,500
3,000
2,500
2,000
40
41
42
43
44
45 46
Time (s)
1,063
Radiation flux (W/m^2)
1,062
1,061
1,060
1,059
1,058
1,057
40
41
42
43
44
45 46
Time (s)
47
48
49
50
Continued…
PROBLEM 5.102 (Cont.)
The initial predicted response (t ≅ 0) is not shown. The behavior for 40 ≤ t ≤ 50 s corresponds to quasisteady behavior. The time-varying convection heat transfer coefficient leads to time variation of the
exterior surface temperature, which in turn leads to time varying exterior heat fluxes and time variation of
temperatures internal to the wall.
COMMENTS: (1) The IHT code is shown below.
// T10 at exposed surface, T0 at inner surface
//Properties,dimensions
k = 1.5
cp = 1500
rho = 1250
rhoc = rho*cp
dx = 0.002
//Convective conditions outside nacelle
h = hbar + delh*sin(omega*t)
//17 rpm rotors, 3 blades
omega = 17*3*2*pi/60
hbar = 35
delh = 15
Tinf = 25
//Radiative conditions
eps = 0.83
sigma = 5.67e-8
Tsur = 20 + 273
//Node 0
T0 = 212
//Nodes 1 through 9
Ac = 1
rhoc*dx*Ac*Der(T1,t) = Ac*k/dx*(T0 - T1) + Ac*k/dx*(T2 - T1)
rhoc*dx*Ac*Der(T2,t) = Ac*k/dx*(T1 - T2) + Ac*k/dx*(T3 - T2)
rhoc*dx*Ac*Der(T3,t) = Ac*k/dx*(T2 - T3) + Ac*k/dx*(T4 - T3)
rhoc*dx*Ac*Der(T4,t) = Ac*k/dx*(T3 - T4) + Ac*k/dx*(T5 - T4)
rhoc*dx*Ac*Der(T5,t) = Ac*k/dx*(T4 - T5) + Ac*k/dx*(T6 - T5)
rhoc*dx*Ac*Der(T6,t) = Ac*k/dx*(T5 - T6) + Ac*k/dx*(T7 - T6)
rhoc*dx*Ac*Der(T7,t) = Ac*k/dx*(T6 - T7) + Ac*k/dx*(T8 - T7)
rhoc*dx*Ac*Der(T8,t) = Ac*k/dx*(T7 - T8) + Ac*k/dx*(T9 - T8)
rhoc*dx*Ac*Der(T9,t) = Ac*k/dx*(T8 - T9) + Ac*k/dx*(T10 - T9)
//Node 10
rhoc*dx/2*Ac*Der(T10,t) = Ac*k/dx*(T9 - T10) + h*Ac*(Tinf - T10) + eps*sigma*Ac*(Tsur^4 - (T10 +
273)^4)
qradflux =eps*sigma*Ac*((T10 + 273)^4 - Tsur^4)
qconflux = h*Ac*(T10 - Tinf)
Continued…
PROBLEM 5.102 (Cont.)
//Initial conditions. Enter into IHT initial condition window.
//T0i = 212
//T1i = T0i - 3460*dx = 205.1
//T2i = T1i - 3460*dx = 198.2
//T3i = T2i - 3460*dx = 191.2
//T4i = T3i - 3460*dx = 184.3
//T5i = T4i - 3460*dx = 177.4
//T6i = T5i - 3460*dx = 170.5
//T7i = T6i - 3460*dx = 163.6
//T8i = T7i - 3460*dx = 156.6
//T9i = T8i - 3460*dx = 149.7
//T10i = T9i - 3460*dx = 142.8
(2) The time-averaged exterior total heat flux is 5187 W/m2. (3) The magnitudes of the various fluctuating
quantities depend on the thermophysical properties of the nacelle. Try some parametric simulations!
PROBLEM 5.103
KNOWN: Sudden exposure of the surface of a thick slab, initially at a uniform temperature, to
convection and to surroundings at a high temperature.
FIND: (a) Explicit, finite-difference equation for the surface node in terms of Fo, Bi, Bir, (b)
Stability criterion; whether it is more restrictive than that for an interior node and does it change
with time, and (c) Temperature at the surface and at 30mm depth for prescribed conditions after
1 minute exposure.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Thick slab may be
approximated as semi-infinite medium, (3) Constant properties, (4) Radiation exchange is
between small surface and large surroundings.
ANALYSIS: (a) The explicit form of the FDE for
the surface node may be obtained by applying an
energy balance to a control volume about the node.
E ′′in − E ′′out = q′′conv + q′′rad + q′′cond = E ′′st
T1p − Top
p
p
h T∞ − To + h r Tsur − To + k ⋅1 ⋅
∆x
) (
(
)
p+1
p
 ∆x  To − To
(1)
= ρ c  ⋅1
∆t
 2 
where the radiation process has been linearized, Eq. 1.9.
2

2 .
hr =
h rp Top , Tsur =
εs Top + Tsur  T0p  + Tsur

  


Divide Eq. (1) by rc∆x/2∆t and regroup using these definitions to obtain the FDE:
) (
(
Fo ≡ ( k/r c ) ∆t/∆x 2
(
)
Bi ≡ h∆x/k
)
Bi r ≡ h r ∆x/k
(2)
(3,4,5)
Top+1
= 2Fo Bi ⋅ T∞ + Bi r ⋅ Tsur + T1p + (1 − 2 Bi ⋅ Fo − 2Bi r ⋅ Fo − 2Fo ) Top .
(6) <
p
(b) The stability criterion for Eq. (6) requires that the coefficient of To be positive.
1 − 2Fo ( Bi + Bi r + 1) ≥ 0
or
Fo ≤ 1/2 ( Bi + Bi r + 1) .
(7) <
The stability criterion for an interior node, Eq. 5.82, is Fo ≤ 1/2. Since Bi + Bir > 0, the stability
criterion of the surface node is more restrictive. Note that Bir is not constant but depends upon
p
hr which increases with increasing To (time). Hence, the restriction on Fo increases with
p
increasing To (time).
Continued …
PROBLEM 5.103 (Cont.)
(c) Consider the prescribed conditions with negligible convection (Bi = 0). The FDEs for the
thick slab are:
Surface (0)
(
)
Top+1
= 2Fo Bi ⋅ Fo + Bi r ⋅ Tsur + T1 + (1 − 2Bi ⋅ Fo − 2Bi r ⋅ Fo − 2Fo ) Top
p
)
(
p+1
p
p
p
=
Tm
Fo Tm+1
+ Tm-1
+ (1 - 2Fo ) Tm
Interior (m≥1)
(8)
(9,5,7,3)
The stability criterion from Eq. (7) with Bi = 0 is,
Fo ≤ 1/2 (1 + Bi r )
(10)
To proceed with the explicit, marching solution, we need to select a value of ∆t (Fo) that will
satisfy the stability criterion. A few trial calculations are helpful. A value of ∆t = 15s provides
2
Fo = 0.105, and using Eqs. (2) and (5), hr(300K, 1000K) = 72.3 W/m ⋅K and Bir = 0.482. From
p
the stability criterion, Eq. (10), find Fo ≤ 0.337. With increasing To , hr and Bir increase:
2
hr(800K, 1000K) = 150.6 W/m ⋅K, Bir = 1.004 and Fo ≤ 0.249. Hence, if
Top < 800K,
=
∆t 15s=
or Fo 0.105 satisfies the stability criterion.
Using ∆t = 15s or Fo = 0.105 with the FDEs, Eqs. (8) and (9), the results of the solution are
p
p
tabulated below. Note how h r and Bi r are evaluated at each time increment. Note that t = p⋅∆t,
where ∆t = 15s.
p
t(s)
To / h pr / Bi r
0
0
300
72.3
0.482
300
300
300
300
1
15
370.867
79.577
0.5305
300
300
300
300
2
30
426.079
85.984
0.5733
307.441
300
300
300
3
45
470.256
91.619
0.6108
319.117
300.781
300
300
4
60
502.289
333.061
302.624
300.082
300
After 60s(p = 4),
T1(K)
T2
T3
T4
To(0, 1 min) = 502.3K and T3(30mm, 1 min) = 300.1K.
….
<
COMMENTS: (1) The form of the FDE representing the surface node agrees with Eq. 5.90 if
this equation is reduced to one-dimension.
(2) We should recognize that the ∆t = 15s time increment represents a coarse step. To improve
the accuracy of the solution, a smaller ∆t should be chosen.
PROBLEM 5.104
PROBLEM 5.104 (Cont.)
PROBLEM 5.105
KNOWN: One-dimensional convective heating and cooling of a plane slab with Bi1 = 10 (for Heating
Phase 1, beginning at Fo = 0 and ending at Fo = Fo1 = 0.1), Bi2 = 1 (for Cooling Phase 2, beginning at Fo
= Fo1).
FIND: (a) Dimensionless form of the heat equation, initial and boundary conditions for Phase 1, (b)
Dimensionless form of the heat equation, initial and boundary conditions for Phase 2, (c) Finite difference
solution for θ 0* (x* = 1, Fo), θ 0* (x* = 0, Fo) and θ 0* (x* = 0.5, Fo) for Fo1 = 0.1 over the range 0 ≤ Fo ≤
0.5 using ∆x* = 0.1 and ∆Fo = 0.001, (d) Value of and time associated with the minimum dimensionless
temperature at the midplane of the wall.
SCHEMATIC:
-L
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) The dimensionless forms of the heat equation, initial and boundary conditions for
Phase 1 are given by Eqs. 5.37 through 5.40,
∂ 2θ * ∂θ *
=
∂x*2 ∂Fo
;
∂θ *
∂x* x*
=
θ * (x* ,0) = 1 ;
∂θ *
∂x* x*
0=
= 0
= − Bi1θ * (1, Fo)
;
<
1
where θ* = (T - T∞,1)/(Ti - T∞,1), x* = x/L, Fo = αt/L2 and Bi1 = h1L/k.
(b) For cooling (Phase 2), the dimensionless form of the heat equation and the boundary condition at x* =
0 are unchanged from part (a). However, the initial condition for Phase 2 is the temperature
distribution that exists at the conclusion of Phase 1, θ*(x*, Fo = Fo1) and the boundary condition
at x* = 1 is derived beginning with the energy balance at the x* = 1 control surface,
−k
<
<
∂T
= h2 T ( L, t ) − T∞=
,2 
 h2 [T ( L, t ) − Ti ]
∂x x = L
Substituting the expressions T = θ*(Ti - T∞,1) + T∞,1 and x = x*L into the preceding equation yields
∂ T∞,1 + θ * (Ti − T∞,1 ) 
= h T + θ * (T − T ) − T 
−k 
i
∞ ,1
i
2  ∞ ,1
∂ ( x* L)
which may be simplified to
Continued…
PROBLEM 5.105 (Cont.)
∂θ *
h2 L
(T∞,1 − Ti ) + θ * (Ti − T∞,1 ) 
=
−
*

k (Ti − T∞,1 ) 
∂x
<
hL
=
− 2 θ * − 1 =
Bi2 1 − θ * 



k 
(c) The dimensionless form of the energy equation, based upon the discretization shown below, was
solved using IHT software. The IHT code is included in the comments.
Adiabatic surface
•
•
1
•
2
Convectively heated/cooled surface
•
3
•
4
•
5
•
6
7
•
•
8
9
•
•
10 11
∆x* = 0.1
x*
0
1
The thermal response is shown in the graph below.
1
1
0.8 0.8
0.8
θ∗
Theta
1
Center
Quarter-Depth
Surface
0.6 0.6
0.4 0.4
0.6
0.4
θo* = 0.79
0.0
Bi1 = 10
Bi2 = 1
Fo1 = 0.1
at
Fo2 = 0.37
0.2
0.2 0.2
0
0
0
0
0.02
0.02
0.04
0.06
0.08
0.1
0.1
0.04Fo 0.06 0.08 0.1
Fo
0.2
0.3
0.4
0.5
0.2
0.3
0.4
0.5
Fo
Center
Middle
Surface
Note that the surface heats quickly in Phase 1 (θ* decreases rapidly) while the center midplane
temperature changes only slightly over the heating time. The quarter plane (or quarter depth) temperature
increases at an intermediate rate during heating. During cooling (Phase 2), the surface cools rapidly, but,
as noted during heating, it takes some time for inner regions of the wall to begin changing in temperature
in response to the new boundary conditions. Warm temperatures in the inner regions of the wall continue
to propagate to the midplane, increasing its temperature until a maximum temperature is reached. The
cooling response is slower than the heating response since Bi2 < Bi1. Ultimately (not shown), the entire
wall will return to its initial temperature of θ* = 1.
(d) The minimum dimensionless temperature at the midplane may be determined by inspecting the
predictions of the finite difference model. The minimum midplane temperature is θ* = 0.79, and occurs at
Fo = Fo2 = 0.37, as shown in the figure of part (c).
Continued…
PROBLEM 5.105 (Cont.)
COMMENTS: (1) The IHT code used in part (c) is listed below.
// Boundary Conditions.
Bi1 = 10
Bi2 = 1
//Initial Conditions.
//Initial Conditions for Phase 1 Heating
// Set all initial dimensionless temperatures to thN = 1 for N = 1 //through 11 in IHT solver.
//Initial Conditions for Phase 2 Cooling
//Set the initial conditions (at Fo = Fo1 = 0.1) equal to the predicted //temperatures from Phase //1
Heating simulation at Fo1 = 0.1.
//Control volume size.
dx = 0.1
//Node 1 is at the midplane.
//Node 11 is at the surface.
//Write energy balances for Nodes 1 through 10.
//Node 1
(th2 - th1)/dx = (dx/2)*der(th1,Fo)
//Node 2
(th1 - th2)/dx + (th3 - th2)/dx = (dx)*der(th2,Fo)
//Node 3
(th2 - th3)/dx + (th4 - th3)/dx = (dx)*der(th3,Fo)
//Node 4
(th3 - th4)/dx + (th5 - th4)/dx = (dx)*der(th4,Fo)
//Node 5
(th4 - th5)/dx + (th6 - th5)/dx = (dx)*der(th5,Fo)
//Node 6
(th5 - th6)/dx + (th7 - th6)/dx = (dx)*der(th6,Fo)
//Node 7
(th6 - th7)/dx + (th8 - th7)/dx = (dx)*der(th7,Fo)
//Node 8
(th7 - th8)/dx + (th9 - th8)/dx = (dx)*der(th8,Fo)
//Node 9
(th8 - th9)/dx + (th10 - th9)/dx = (dx)*der(th9,Fo)
//Node 10
(th9 - th10)/dx + (th11 - th10)/dx = (dx)*der(th10,Fo)
//Energy balance for Node 11 for Phase 1 Heating.
//Enable in Phase 1. Disable in Phase 2.
(th10 - th11)/dx - Bi1*th11 = (dx/2)*der(th11,Fo)
//Energy balance for Node 11 for Phase 2 Cooling.
//Disable for Phase 1. Enable for Phase 2.
//(th10 - th11)/dx - Bi2*(th11 - 1) = (dx/2)*der(th11,Fo)
Continued…
PROBLEM 5.105 (Cont.)
(2) If one is interested in determining the maximum midplane temperature in response to heating,
recognize that this maximum value does not occur at the moment heating is stopped. In this problem,
the increase in the midplane temperature from the onset of heating to the curtailment of heating is
∆θ*(x* = 0, Fo1 = 0.1) = 0.033 whereas the maximum change in the midplane temperature is ∆θ*(x*
= 0, Fo2 = 0.37) = 0.212 which is 6 times larger than the increase at Fo1. Furthermore, maximum
midplane temperatures are reached long after heating is stopped and cooling begins (Fo2/Fo1 = 3.7).
(3) The effects illustrated in this problem, specifically the continued heating of inner regions of the
wall and the long time delay before maximum wall temperatures are reached after heating has stopped
and cooling has begun, are important in many applications including materials processing, food
processing and thermal sterilization, and structural response (building collapse) to fire long after the
fire is extinguished. (4) See Bergman, “Extreme Midplane Wall Temperatures due to Sequential
Heating and Cooling, ASME Journal of Heat Transfer, vol. 130, pp. 094503-1 to 094503-4, 2008
for more information.
PROBLEM 5.106
PROBLEM 5.106 (Cont.)
PROBLEM 5.106 (Cont.)
PROBLEM 5.106 (Cont.)
PROBLEM 5.106 (Cont.)
PROBLEM 5.107
KNOWN: Insulated rod of prescribed length and diameter, with one end in a fixture at 200°C, reaches a
uniform temperature. Suddenly the insulating sleeve is removed and the rod is subjected to a convection
process.
FIND: (a) Time required for the mid-length of the rod to reach 100°C, (b) Temperature history T(x,t ≤
t1), where t1 is time at which the midlength reaches 50°C. Temperature distribution at 0, 200s, 400s and
t1.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction in rod, (2) Uniform h along rod and at end,
(3) Negligible radiation exchange between rod and surroundings, (4) Constant properties.
ANALYSIS: (a) Choosing ∆x = 0.016 m, the finite-difference equations for the interior and end nodes
are obtained.
p +1
p
Tm
− Tm
q a + q b + q c = ρ ⋅ A c ∆x ⋅ cp ⋅
Interior Point, m:
∆t
k ⋅ Ac
p
p
Tm
−1 − Tm
∆x
+ kA c
p
p
Tm
+1 − Tm
∆x
p
T p +1 − Tm
p
+ hP∆x T∞ − Tm
= ρ A c ∆xcp m
∆t
)
(
Regrouping,
p +1
p
Tm
= Tm
(1 − 2Fo − Bi ⋅ Fo ) + Fo Tmp −1 + Tmp +1 + Bi ⋅ FoT∞
(
where
Fo =
α∆t
∆x 2
(2)
)
Bi
= h  ∆x 2 ( A c P )  k .


(1)
(3)
From Eq. (1), recognize that the stability of the numerical solution will be assured when the first term on
the RHS is positive; that is
Continued...
PROBLEM 5.107 (Cont.)
(1 − 2Fo − Bi ⋅ Fo ) ≥ 0
Fo ≤ 1 ( 2 + Bi ) .
or
(4)
p
Nodal Point 1: Consider Eq. (1) for the special case that Tm −1 = To, which is independent of time.
Hence,
(5)
T1p +1= T1p (1 − 2Fo − Bi ⋅ Fo ) + Fo To + T2p + Bi ⋅ FoT∞ .
)
(
End Nodal Point 10:
p +1
p
T10
− T10
∆x
q a + q b + q c =ρ ⋅ Ac
⋅ cp
2
∆t
p
p +1
p
T p − T10
∆x
∆x T10 − T10
p
p
k ⋅ Ac 9
+ hA c T∞ − T10
+ hP
T∞ − T10
=
cp
ρ Ac
∆x
2
2
∆t
)
(
)
(
p +1
p
p
Regrouping, T10 = T10 (1 − 2Fo − 2N ⋅ Fo − Bi ⋅ Fo ) + 2FoT9 + T∞ ( 2N ⋅ Fo + Bi ⋅ Fo )
where N = h∆x/k.
(6)
The stability criterion is Fo ≤ 1/2(1 + N + Bi/2).
(8)
(7)
With the finite-difference equations established, we can now proceed with the numerical solution.
Having already specified ∆x = 0.016 m, Bi can now be evaluated. Noting that Ac = πD2/4 and P = πD,
giving Ac/P = D/4, Eq. (3) yields

2 0.010 m 
Bi 30 W m 2 ⋅ K ( 0.016 m )
14.8 W=
m ⋅ K 0.208
=
4 

(9)
From the stability criteria, Eqs. (4) and (8), for the finite-difference equations, it is recognized that Eq. (8)
requires the greater value of Fo. Hence
1
0.208 
Fo =1 + 0.0324 +
0.440
=
2
2 
(10)
30 W m 2 ⋅ K × 0.016 m
where from Eq. (7), N = 0.0324 .
=
14.8 W m ⋅ K
(11)
From the definition of Fo, Eq. (2), we obtain the time increment
2
Fo ( ∆x )
2
6 m 2 s 31.1s
=
∆t
= 0.440 ( 0.016 m ) 3.63 ×10−=
α
(12)
and the time relation is t = p∆t = 31.1t.
(13)
Using the numerical values for Fo, Bi and N, the finite-difference equations can now be written (°C).
Nodal Point m (2 ≤ m ≤ 9):
p +1
p
Tm
= Tm
(1 − 2 × 0.440 − 0.208 × 0.440 ) + 0.440 Tmp −1 + Tmp +1 + 0.208 × 0.440 × 25
)
(
(
)
p +1
p
p
p
Tm
= 0.029Tm
+ 0.440 Tm
−1 + Tm +1 + 2.3
Nodal Point 1:
p +1
T1=
0.029T1p + 0.440 200 + T2p +=
2.3 0.029T1p + 0.440T2p + 90.3
(
)
(14)
(15)
Nodal Point 10:
p +1
p
T10
= 0 × T10
+ 2 × 0.440T9p + 25 ( 2 × 0.0324 × 0.440 + 0.208 × 0.440 ) = 0.880T9p + 3.0 (16)
Continued...
PROBLEM 5.107 (Cont.)
Using finite-difference equations (14-16) with Eq. (13), the calculations may be performed to obtain
p
0
1
2
3
4
5
6
7
8
t(s)
0
31.1
62.2
93.3
124.4
155.5
186.6
217.7
248.8
T1
200
184.1
175.6
168.6
163.3
158.8
155.2
152.1
145.1
T2
200
181.8
166.3
154.8
145.0
137.1
130.2
124.5
119.5
T3
200
181.8
165.3
150.7
138.8
128.1
119.2
111.3
104.5
T4
200
181.8
165.3
150.7
137.0
125.3
114.8
105.7
97.6
T5
200
181.8
165.3
150.7
137.0
124.5
113.4
103.5
94.8
T6
200
181.8
165.3
150.7
137.0
124.3
113.0
102.9
T7
200
181.8
165.3
150.7
136.5
124.2
112.6
102.4
T8
200
181.8
165.3
149.7
136.3
123.4
112.3
T9
200
181.8
164.0
149.2
135.0
123.0
111.5
T10(°C)
200
179.0
163.0
147.3
134.3
121.8
111.2
<
Using linear interpolation between rows 7 and 8, we obtain T(L/2, 230s) = T5 ≈ 100°C.
(b) Using the option concerning Finite-Difference Equations for One-Dimensional Transient Conduction
in Extended Surfaces from the IHT Toolpad, the desired temperature histories were computed for 0 ≤ t ≤
t1 = 930s. A Lookup Table involving data for T(x) at t = 0, 200, 400 and 930s was created.
t(s)/x(mm)
0
200
400
930
0
200
200
200
200
16
200
157.8
146.2
138.1
32
200
136.7
114.9
99.23
48
200
127.0
97.32
74.98
64
200
122.7
87.7
59.94
80
200
121.0
82.57
50.67
96
200
120.2
79.8
44.99
112
200
119.6
78.14
41.53
128
200
118.6
76.87
39.44
144
200
117.1
75.6
38.2
160
200
114.7
74.13
37.55
and the LOOKUPVAL2 interpolating function was used with the Explore and Graph feature of IHT to
create the desired plot.
Temperature, T(C)
225
200
175
150
125
100
75
50
25
0
20
40
60
80
100
120
140
160
Fin location, x(mm)
t
t
t
t
= 0
= 200 s
= 400 s
= 930 s
Temperatures decrease with increasing x and t, and except for early times (t < 200s) and locations in
proximity to the fin tip, the magnitude of the temperature gradient, |dT/dx|, decreases with increasing x.
The slight increase in |dT/dx| observed for t = 200s and x → 160 mm is attributable to significant heat
loss from the fin tip.
COMMENTS: The steady-state condition may be obtained by extending the finite-difference
calculations in time to t ≈ 2650s or from Eq. 3.75.
PROBLEM 5.108
PROBLEM 5.108 (Cont.)
PROBLEM 5.109
KNOWN: Tantalum rod initially at a uniform temperature, 300K, is suddenly subjected to a
current flow of 80A; surroundings (vacuum enclosure) and electrodes maintained at 300K.
FIND: (a) Estimate time required for mid-length to reach 1000K, (b) Determine the steady-state
temperature distribution and estimate how long it will take to reach steady-state. Use a finitedifference method with a space increment of 10mm.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are much
larger than rod, (3) Properties are constant and evaluated at an average temperature.
=
PROPERTIES: Table A-1,
Tantalum ( T
300+1000 ) K/2
(=
650K ) : r = 16,600 kg/m , c =
3
3
147 J/kg⋅K, k = 58.8 W/m⋅K, and α = k/rc = 58.8 W/m⋅K/16,600 kg/m × 147 J/kg⋅K = 2.410 ×
-5 2
10 m /s.
ANALYSIS: The finite-difference equation is
)
(
(
)
2
2
2
p+1
p
p
p e Ps∆x
4,p
4 + I re ∆x ⋅ Fo
=
+ Tm+1
+ (1 --Tm
Fo Tm-1
2Fo ) Tm
Fo Tm
Tsur
kA c
kA c2
Fo = αDD
t/ x 2
where
Ac = π D2 / 4
P = π D.
(1)
(2,3,4)
From the stability criterion, let Fo = 1/2 and numerically evaluate terms of Eq. (1).
2
0.1× 5.67 ×10-8 W/m 2 ⋅ K 4 × ( 0.01m ) 4 1 4,p
4
p+1 1 p
p
Tm =
Tm-1 + Tm+1 ⋅ Tm - [300K ] +
2
58.8 W/m ⋅ K × ( 0.003m )
2
(
)
(
)
2
2
80A ) × 95 ×10-8 W ⋅ m ( 0.01m ) 1
(
+
⋅
(
58.8 W/m ⋅ K p [ 0.003m ] / 4
p+1
Tm
=
(
2
)
)
2
2
1 p
4,p
T
+ Tp
- 6.4285 ×10-12 Tm
+ 103.53.
2 m-1 m+1
(5)
Note that this form applies to nodes 0 through 5. For node 0, Tm-1 = Tm+1 = T1. Since Fo =
1/2, using Eq. (2), find that
∆t =∆x 2 Fo/α =( 0.01m ) ×1/ 2 / 2.410 ×10−5 m 2 / s =2.07s.
2
(6)
Hence, t = p∆t = 2.07p.
(7)
Continued …
PROBLEM 5.109 (Cont.)
(a) To estimate the time required for the mid-length to reach 1000K, that is To = 1000K, perform
the forward-marching solution beginning with Ti = 300K at p = 0. The solution, as tabulated
below, utilizes Eq. (5) for successive values of p. Elapsed time is determined by Eq. (7).
P
t(s)
T0
T1
T2
T3
T4
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
300
403.5
506.9
610.0
712.6
814.5
915.2
1010.9
1104.7
1190.9
1274.1
1348.2
1419.7
1479.8
1542.6
1605.3
300
403.5
506.9
610.0
712.6
814.5
911.9
1007.9
1096.8
1183.5
1261.6
1336.7
1402.4
1465.5
1538.2
1569.3
300
403.5
506.9
610.0
712.6
808.0
902.4
988.9
1073.8
1150.4
1224.9
1290.6
1353.9
1408.4
1460.9
1514.0
300
403.5
506.9
610.0
699.7
788.8
867.4
945.0
1014.0
1081.7
1141.5
1199.8
1250.5
1299.8
1341.2
1381.6
300
403.5
506.9
584.1
661.1
724.7
787.9
842.3
896.1
943.2
989.4
1029.9
1069.4
1103.6
1136.9
1164.8
10.4
20.7
31.1
T5
T6(°C)
300
403.5
455.1
506.7
545.2
583.5
615.1
646.6
673.6
700.3
723.6
746.5
766.5
786.0
802.9
819.3
Note that, at p ≈ 6.9 or t = 6.9 × 2.07 = 14.3s, the mid-point temperature is To ≈ 1000K.
300
300
300
300
300
300
300
300
300
300
300
300
300
300
300
300
<
(b) The steady-state temperature distribution can be obtained by continuing the marching
solution until only small changes in Tm are noted. From the table above, note that at p = 15 or t
= 31s, the temperature distribution is still changing with time. It is likely that at least 15 more
calculation sets are required to see whether steady-state is being approached.
COMMENTS: (1) This problem should be solved with a computer rather than a handcalculator. For such a situation, it would be appropriate to decrease the spatial increment in
order to obtain better estimates of the temperature distribution.
(2) If the rod were very long, the steady-state temperature
distribution would be very flat at the mid-length x = 0.
Performing an energy balance on the small control volume
shown to the right, find
E g − E out =
0
r ∆x
4 =
I2 e − es P∆x To4 − Tsur
0.
Ac
(
)
Substituting numerical values, find To = 2003K. It is unlikely that the present rod would ever
reach this steady-state, maximum temperature. That is, the effect of conduction along the rod
will cause the center temperature to be less than this value.
PROBLEM 5.110
KNOWN: Thin, circular-disc subjected to induction heating causing a uniform heat generation
in a prescribed region; upper surface exposed to convection process.
FIND: (a) Transient finite-difference equation for a node in the region subjected to induction
heating, (b) Sketch the steady-state temperature distribution on T-r coordinates; identify
important features.
SCHEMATIC:
ASSUMPTIONS: (1) Thickness w << ro, such that conduction is one-dimensional in rdirection, (2) In prescribed region, q is uniform, (3) Bottom surface of disc is insulated, (4)
Constant properties.
ANALYSIS: (a) Consider the nodal point arrangement for
the region subjected to induction heating. The size of the
control volume =
is V 2π rm ⋅ ∆r ⋅ w. The energy
conservation requirement for the node m has the form
with
E in − E out + E g =
E st
 =
q a + q b + q conv + qV
E st .
Recognizing that qa and qb are conduction terms and qconv is the convection process,
p
p
p
p
Tm
 
 
∆r   Tm-1 -∆r   Tm+1 Tm
+ k  2  rm +  w 
k  2pp
 rm - 2  w 
∆r
∆r
2 
 

 
)
(
p
T p+1 − Tm
p
+ h [ 2pp
rm ⋅ ∆r ] T∞ − Tm
+ q [ 2 =
rm ⋅ ∆r ⋅ w ] r c p [ 2p rm ⋅ ∆r ⋅ w ] m
.
∆t
Upon regrouping, the finite-difference equation has the form,
2
  ∆r  p
 ∆r  p
 ∆r  T + q ∆r  + 1 - ∆r  p
p+1
+
+
+
Tm= Fo  1 T
1
T
Bi
  2Fo Bi ⋅ Fo   Tm
∞
m-1 
m+1




k  
w
 w 
 2rm 
  2rm 

where
Fo = α∆t/∆r 2
Bi = h∆r/k.
(b) The steady-state temperature distribution has
these features:
1. Zero gradient at r = 0, r0
2. No discontinuity at r1, r2
3. Tmax occurs in region r1 < r < r2
Note also, distribution will not be linear anywhere;
distribution is not parabolic in r1 < r < r2 region.
<
PROBLEM 5.111
KNOWN: An electrical cable experiencing uniform volumetric generation; the lower half is well
insulated while the upper half experiences convection.
FIND: (a) Explicit, finite-difference equations for an interior node (m,n), the center node (0,0), and an
outer surface node (M,n) for the convective and insulated boundaries, and (b) Stability criterion for each
FDE; identify the most restrictive criterion.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional (r,φ), transient conduction, (2) Constant properties, (3) Uniform

q.
ANALYSIS: The explicit, finite-difference equations may be obtained by applying energy balances to
appropriate control volumes about the node of interest. Note the coordinate system defined above where
(r,φ) → (m∆r, n∆φ). The stability criterion is determined from the coefficient associated with the node of
interest.
Interior Node (m,n). The control volume
for an interior node is
V= rm ∆φ ⋅ ∆r ⋅ 
(with rm =
m∆r,  =
1 ) where  is the
length normal to the page. The
conservation of energy requirement is
E in − E out + E g =
E st
( q1 + q 2 )r + ( q3 + q 4 )q
 =
+ qV
r cV
p
p
p+1
p
− Tm,n
Tm,n
∆t
p
p
p
p
Tm-1,n --Tm,n
Tm+1,n Tm,n
Tm,n+1 Tm,n
1
1


k ⋅  m -  ∆r ⋅ ∆φ ⋅
+ k ⋅  m +  ∆r ⋅ ∆φ ⋅
+ k ⋅ ∆r ⋅
2
2
∆r
∆r
( m∆r ) ∆φ


+ k ⋅ ∆r ⋅
p
p
Tm,n-1
- Tm,n
( m∆r ) ∆φ
p+1
p
Tm,n
- Tm,n

+ q ( m∆r ⋅ ∆φ=
) ⋅ ∆r r c ( m∆r ⋅ ∆φ ) ⋅ ∆r ⋅
∆t
(1)
Define the Fourier number as
k ∆t α∆t
Fo =
⋅
=
(2)
r c ∆r 2 ∆r 2
and then regroup the terms of Eq. (1) to obtain the FDE,
 m - 1/2

m + 1/2 p
1
q


p+1
p
p
p
=
+
+ Tm,n-1
+ ∆r 2 
Tm,n
Fo 
Tm-1,n
Tm+1,n +
Tm,n+1
m
k
 m

( m∆φ )2


 
2

p
 + 1 Tm,n
(3) <
+ −Fo  2 +
.
2

m
∆
φ


(
)

 
Continued …
(
)
PROBLEM 5.111 (Cont.)
The stability criterion requires that the last term on the right-hand side in braces be positive. That is, the
p
coefficient of Tm,n must be positive and the stability criterion is
2
Fo ≤ 1/2 1 + 1/ ( m∆φ ) 
(4)


2
Note that, for m >> 1/2 and (m∆φ) >>1, the FDE takes the form of a 1-D cartesian system.
Center Node (0,0). For the control volume,
V=
π ( ∆r/2 ) ⋅1. The energy balance is
2
E ′in − E ′out + E ′g =
E ′st where E ′in =
Σq′n .
p
p
T1,n − To
∆r
∆r
∑ k ⋅  2 ∆φ  ⋅ ∆r + q p  2 
n =0
N
2
p+1
2
p
 ∆r  To − To
(5)
 2 
∆t
where N = (2π/∆φ) - 1, the total number of qn. Using the definition of Fo, find

q
 1 N p
p+1
2
p
=
∆r  + (1 − 4Fo ) To . <
To
4Fo 
T1,n +
∑
+
N
1
4k


n =0
By inspection, the stability criterion is Fo ≤ 1/4.
(7)
Surface Nodes (M,n). The control volume
for the surface node is V = (M - ¼)∆r∆φ⋅∆r/2.1.
From the energy balance,
= r c ⋅p
 = E ′st
E ′in − E ′out + E ′g = ( q1′ + q′2 )r + ( q′3 + q′4 )φ + qV
p
p
(
p
)
p
TM-1,n -TM,n
∆r TM,n+1 TM,n
p
k ⋅ ( M - 1/2 ) ∆r ⋅ ∆φ
+ h ( M∆r ⋅ ∆φ ) T∞ - TM,n +k ⋅ ⋅
∆r
2
( M∆r ) ∆φ
p
p
p+1
p
TM,n
∆r TM,n-1 -∆r 
∆r  TM,n TM,n


+k ⋅
⋅
+ q ( M - 1/4 ) ∆r ⋅ ∆=
φ⋅
r c ( M - 1/4 ) ∆r ⋅ ∆φ ⋅
.




∆t
2
2 
2 
( M∆r ) ∆φ
2
Regrouping and using the definitions for Fo = α∆t/∆r and Bi = h∆r/k,
 M - 1/2

1
q
p
p
p
p+1
=
Tm,n
Fo  2
TM-1,n +
TM,n+1 - TM,n-1 + 2Bi ⋅ T∞ + ∆r 2 
k
( M-1/4 ) M ( ∆φ )2
 M - 1/4


 M-1/2

M
1

 p
(8) <
  TM,n
+ 1 - 2Fo 
+ Bi ⋅
+
.
2
M-1/4
M - 1/4 ( M - 1/4 ) M ( ∆φ )  






1  M − 1/2
M
1
 . (9)
The stability criterion is
+ Bi
+
Fo ≤ 
2  M − 1/4
M − 1/4 ( M − 1/4 ) M ( ∆φ )2 


(
)
To determine which stability criterion is most restrictive, compare Eqs. (4), (7) and (9). The most
restrictive (lowest Fo) has the largest denominator. For small values of m, it is not evident whether Eq.
(7) is more restrictive than Eq. (4); Eq. (4) depends upon magnitude of ∆φ. Likewise, it is not clear
whether Eq. (9) will be more or less restrictive than Eq. (7). Numerical values must be substituted.
PROBLEM 5.112
KNOWN: Initial temperature distribution in two bars that are to be soldered together; interface contact
resistance.
FIND: (a) Explicit FDE for T4,2 in terms of Fo and Bi = ∆x/k R ′′t,c ; stability criterion, (b) T4,2 one
time step after contact is made if Fo = 0.01 and value of ∆t; whether the stability criterion is satisfied.
SCHEMATIC:
PROPERTIES: Table A-1, Steel, AISI 1010 (1000K): k = 31.3 W/m⋅K, c = 1168 J/kg⋅K, r = 7832
3
kh/m .
ASSUMPTIONS: (1) Two-dimensional transient conduction, (2) Constant properties, (3) Interfacial
solder layer has negligible thickness.
ANALYSIS: (a) From an energy balance on
the control volume V = (∆x/2)⋅∆y⋅1.
+ E =
E − E
E
in
out
g
st
p+1
p
T4,2 − T4,2
qa + q b + qc + qd =
ρ cV
.
∆t
Note that q a = ∆T/R ′′t,c A c while the remaining qi are conduction terms,
(
1
R ′′t,c
(
)
)
(
p
p
T3,2 − T4,2 ∆y + k ( ∆x/2 )
(
+ k ( ∆x/2 )
p
p
T4,3 − T4,2
)
∆y
p
p
T4,1 − T4,2
∆y
)
=
+ k ( ∆y
(
)
p
p
T5,2 − T4,2
)
∆x
ρ c [( ∆x / 2 ) ⋅ ∆y ]
p+1
p
T4,2 − T4,2
∆t
.
2
Defining Fo ≡ ( k/ρ c ) ∆t/∆x and Bic ≡ ∆y/R ′′t,c k, regroup to obtain
(
)
T=
Fo T4,3 + 2T5,2 + T4,1 + 2Bi T3,2 + (1 − 4Fo − 2FoBi ) T4,2 .
4,2
p+1
p
p
p
p
p
<
p
The stability criterion requires the coefficient of the T4,2 term be zero or positive,
(1 − 4Fo − 2FoBi ) ≥ 0
Fo ≤ 1/ ( 4 + 2Bi )
or
(
<
)
(b) For
=
Fo 0.01 and
=
Bi 0.020m/ 2 × 10-5 m 2 ⋅ K/W × 31.3W/m
=
⋅ K 31.95,
=
T4,2
0.01 (1000 + 2 × 900 + 1000 + 2 × 31.95 × 700 ) K + (1 − 4 × 0.01 − 2 × 0.01 × 31.95 )1000K
p+1
p+1
<
T4,2 = 485.30K + 321.00K = 806.3K.
With Fo = 0.01, the time step is
∆t= Fo ∆x 2 ( ρ c/k )= 0.01( 0.020m )
2
( 7832kg/m3 ×1168J/kg ⋅ K/31.3W/m ⋅ K )=
With Bi = 31.95 and Fo = 0.01, the stability criterion, Fo ≤ 0.015, is satisfied.
1.17s.
<
<
PROBLEM 5.113
KNOWN: Flue of square cross-section, initially at a uniform temperature, is suddenly exposed to hot
flue gases.
FIND: Temperature distribution in the wall and heat loss rate at 5, 10, 50 and 100 hours after
introduction of gases using the implicit finite-difference method.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional transient conduction, (2) Constant properties.
-7
2
PROPERTIES: Flue (given): k = 0.85 W/m⋅K, α = 5.5 × 10 m /s.
ANALYSIS: The network representing the flue cross-sectional area is shown with ∆x = ∆y = 50 mm.
Initially all nodes are at Ti = 25°C when suddenly the interior and exterior surfaces are exposed to
convection processes, (T∞,i, hi) and (T∞,o, ho), respectively Referring to the network above, note that
there are four types of nodes: interior (02, 03, 06, 07, 10, 11, 14, 15, 17, 18, 20); plane surfaces with
convection (interior – 01, 05, 09); interior corner with convection (13), plane surfaces with convection
(exterior – 04, 08, 12, 16, 19, 21); and, exterior corner with convection. The system of finite-difference
equations representing the network is obtained using IHT|Tools|Finite-difference equations|Twodimensional|Transient. The IHT code is shown in Comment 2 and the results for t = 5, 10, 50 and 100
hour are tabulated below.
p
T17
(T18p+1 + T14p+1 + T18p+1 + T14p+1 ) =
Node 17
p+1
− Fo
(1 + 4Fo ) T17
Node 13
4
1 + 4Fo 1 + 1 Bi  T p+1 − 2 Fo 2T p+1 + T + 2T p+1 + T p+1 =
p
T13 + Bii ⋅ Fo ⋅ T∞ ,i
i
9
13
14
14
9

 3 
3
3

Node 12
Node 22
(
(1 + 2Fo ( 2 + Bio ) ) T12p+1 − Fo ( 2T11p+1 + T16p+1 + T8p+1 ) =
p+1
p+1
p+1
− 2Fo ( T21
+ T21
(1 + 4Fo (1 + Bio ) ) T22
)=
)
p
T12
+ 2Bio ⋅ Fo ⋅ T∞,o
p
T22
+ 4Bio ⋅ Fo ⋅ T∞,o
Numerical values for the relevant parameters are (using a time step of one hour):
α∆t 5.5 ×10−6 m 2 / s × 3600s
=
Fo
=
∆x 2
( 0.050m )2
= 7.92000
h o ∆x 5 W/m 2 ⋅ K × 0.050m
=
Bio =
= 0.29412
k
0.85 W/m ⋅ K
=
Bii
h i ∆x 100 W/m 2 ⋅ K × 0.050m
=
= 5.88235
k
0.85 W/m ⋅ K
The system of FDEs can be represented in matrix notation, [A][T] = [C]. The coefficient matrix [A] and
terms for the right-hand side matrix [C] are given on the following page.
Continued …
PROBLEM 5.113 (Cont.)
For this problem a standard computer program was used to obtain the solution matrix [T]. The
0= 25°C. The results are tabulated below.
initial temperature distribution was Tm
T(m,n) (C)
Node/time (h)
0
5
10
50
100
T01
25
335.00
338.90
340.20
340.20
T02
25
248.00
274.30
282.90
282.90
T03
25
179.50
217.40
229.80
229.80
T04
25
135.80
170.30
181.60
181.60
T05
25
334.50
338.50
339.90
339.90
T06
25
245.30
271.90
280.80
280.80
T07
25
176.50
214.60
227.30
227.30
T08
25
133.40
168.00
179.50
179.50
T09
25
332.20
336.60
338.20
338.20
T10
25
235.40
263.40
273.20
273.20
T11
25
166.40
205.40
219.00
219.00
T12
25
125.40
160.40
172.70
172.70
T13
25
316.40
324.30
327.30
327.30
T14
25
211.00
243.00
254.90
254.90
T15
25
146.90
187.60
202.90
202.90
T16
25
110.90
146.70
160.20
160.20
T17
25
159.80
200.50
216.20
216.20
T18
25
117.40
160.50
177.50
177.50
T19
25
90.97
127.40
141.80
141.80
T20
25
90.62
132.20
149.00
149.00
T21
25
72.43
106.70
120.60
120.60
T22
25
59.47
87.37
98.89
98.89
<
The rate of heat loss (per unit depth) from the exterior of the flue to the environment can be found by
summing the convective heat loss from nodes 04, 08, 12, 16, 19, 21, and 22 (and multiplying by 8 due to
symmetry, see Schematic).
Continued …
PROBLEM 5.113 (Cont.)
1
1

qloss =
8ho ∆x  (T4 − T∞ ,o ) + (T8 − T∞ ,o ) + (T12 − T∞ ,o ) + (T16 − T∞ ,o ) + (T19 − T∞ ,o ) + (T21 − T∞ ,o ) + (T22 − T∞ ,o ) 
2
2

The results for the different times are shown below.
t (hr)
qloss (W/m)
5
961.5
10
1376
50
1530
100
1530
<
COMMENTS: (1) Note that the steady-state condition is reached by t = 50 hours; this can be seen by
comparing the distributions for t = 50 and 100 hours. Within 10 hours, the flue is within a few degrees of
the steady-state condition and the heat loss rate is within 10% of the steady-state value. (2) The IHT code
for performing the numerical solution is shown in its entirety below. Use has been made of symmetry in
writing the FDEs. The tabulated results above were obtained by copying from the IHT Browser and
pasting the desired columns into EXCEL.
// From Tools|Finite-difference equations|Two-dimensional|Transient
// Interior surface nodes, 01, 05, 09, 13
/* Node 01: plane surface node, s-orientation; e, w, n labeled 05, 05, 02 . */
rho * cp * der(T01,t) = fd_2d_psur_s(T01,T05,T05,T02,k,qdot,deltax,deltay,Tinfi,hi,q''a)
q''a = 0
// Applied heat flux, W/m^2; zero flux shown
qdot = 0
rho * cp * der(T05,t) = fd_2d_psur_s(T05,T09,T01,T06,k,qdot,deltax,deltay,Tinfi,hi,q''a)
rho * cp * der(T09,t) = fd_2d_psur_s(T09,T13,T05,T10,k,qdot,deltax,deltay,Tinfi,hi,q''a)
/* Node 13: internal corner node, w-s orientation; e, w, n, s labeled 14, 09, 14, 09. */
rho * cp * der(T13,t) = fd_2d_ic_ws(T13,T14,T09,T14,T09,k,qdot,deltax,deltay,Tinfi,hi,q''a)
// Interior nodes, 02, 03, 06, 07, 10, 11, 14, 15, 18, 20
/* Node 02: interior node; e, w, n, s labeled 06, 06, 03, 01. */
rho * cp * der(T02,t) = fd_2d_int(T02,T06,T06,T03,T01,k,qdot,deltax,deltay)
rho * cp * der(T03,t) = fd_2d_int(T03,T07,T07,T04,T02,k,qdot,deltax,deltay)
rho * cp * der(T06,t) = fd_2d_int(T06,T10,T02,T07,T05,k,qdot,deltax,deltay)
rho * cp * der(T07,t) = fd_2d_int(T07,T11,T03,T08,T06,k,qdot,deltax,deltay)
rho * cp * der(T10,t) = fd_2d_int(T10,T14,T06,T11,T09,k,qdot,deltax,deltay)
rho * cp * der(T11,t) = fd_2d_int(T11,T15,T07,T12,T10,k,qdot,deltax,deltay)
rho * cp * der(T14,t) = fd_2d_int(T14,T17,T10,T15,T13,k,qdot,deltax,deltay)
rho * cp * der(T15,t) = fd_2d_int(T15,T18,T11,T16,T14,k,qdot,deltax,deltay)
rho * cp * der(T17,t) = fd_2d_int(T17,T18,T14,T18,T14,k,qdot,deltax,deltay)
rho * cp * der(T18,t) = fd_2d_int(T18,T20,T15,T19,T17,k,qdot,deltax,deltay)
rho * cp * der(T20,t) = fd_2d_int(T20,T21,T18,T21,T18,k,qdot,deltax,deltay)
// Exterior surface nodes, 04, 08, 12, 16, 19, 21, 22
/* Node 04: plane surface node, n-orientation; e, w, s labeled 08, 08, 03. */
rho * cp * der(T04,t) = fd_2d_psur_n(T04,T08,T08,T03,k,qdot,deltax,deltay,Tinfo,ho,q''a)
rho * cp * der(T08,t) = fd_2d_psur_n(T08,T12,T04,T07,k,qdot,deltax,deltay,Tinfo,ho,q''a)
rho * cp * der(T12,t) = fd_2d_psur_n(T12,T16,T08,T11,k,qdot,deltax,deltay,Tinfo,ho,q''a)
rho * cp * der(T16,t) = fd_2d_psur_n(T16,T19,T12,T15,k,qdot,deltax,deltay,Tinfo,ho,q''a)
rho * cp * der(T19,t) = fd_2d_psur_n(T19,T21,T16,T18,k,qdot,deltax,deltay,Tinfo,ho,q''a)
rho * cp * der(T21,t) = fd_2d_psur_n(T21,T22,T19,T20,k,qdot,deltax,deltay,Tinfo,ho,q''a)
/* Node 22: external corner node, e-n orientation; w, s labeled 21, 21. */
rho * cp * der(T22,t) = fd_2d_ec_en(T22,T21,T21,k,qdot,deltax,deltay,Tinfo,ho,q''a)
// Rate of heat loss from nodes 04, 08, 12, 16, 19, 21, 22
qloss = 8*ho*deltax*((0.5*T04+T08+T12+T16+T19+T21+0.5*T22)-6*Tinfo)
// Input variables
deltax = 0.050
deltay = 0.050
Tinfi = 350
hi = 100
Tinfo = 25
ho = 5
k = 0.85
alpha = 5.55e-7
alpha = k / (rho * cp)
rho = 1000
// arbitrary value
Continued …
PROBLEM 5.113 (Cont.)
(3) The results for t = 50 hour, representing the steady-state condition, are shown below,
arranged according to the coordinate system.
Tmn (C)
x/y (mm)
0
50
100
150
200
250
300
0
181.60
179.50
172.70
160.20
141.80
120.60
98.89
149.00
50
229.80
227.30
219.00
202.90
177.50
100
282.90
280.80
273.20
172.70
216.20
150
340.20
339.90
338.20
327.30
Compared to the steady-state temperature distribution determined using a finer mesh (not shown), ∆x =
∆y = 25 mm, the results for the coarser mesh are slightly higher, within a fraction of 1°C, along the midsection of the flue, but notably higher in the vicinity of the inner corner. (For example, node 13 is 2.6°C
higher with the coarser mesh.)
PROBLEM 5.114
KNOWN: Electrical heating elements embedded in a ceramic plate of known properties and
dimensions. Environmental temperature and heat transfer coefficient. Initially plate is at a
uniform temperature and suddenly heaters are energized with known heat rate.
FIND: Time required for the difference between the surface above a heating element and initial
temperatures to reach 95% of the difference for steady-state conditions using the implicit finitedifference method.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction, (2) Constant properties, (3) No internal
generation except for Node 7, (4) Heating element approximates a line source; wire diameter is
negligible.
ANALYSIS: The grid for the symmetry element above consists of 12 nodes. Nodes 1-3 are
points on a surface experiencing convection; nodes 4-12 are interior nodes; node 7 is a special
case with internal generation and because of symmetry, q′ht = 25 W/m. Their finite-difference
equations are derived as follows.
Surface Node 2. From an energy balance on the prescribed control volume with ∆x/∆y = 3,
T p+1 − T2p
E = E = q′ + q′ + q′ + q′ = ρ cV 2
in
st
a
b
c
d
∆t
p+1
k
∆y T1
2
− T2p+1
∆x
(
+ h∆x T∞ − T2p+1
)
p+1
p+1
p+1
p
T5p+1 − T2p+1
∆y T3 − T2
 ∆y  T2 − T2
+k
+ k∆x
= ρ c  ∆x 
.
2
∆x
∆y
2 
∆t

Continued …
PROBLEM 5.114 (Cont.)
Divide by k, use the following definitions, and regroup to obtain the finite-difference equations.
N ≡=
h∆x/k 100 W/m 2 ⋅ K × 0.006m/2 W/m
=
⋅ K 0.3000
Fo ≡ ( k/ρ c ) ∆t/∆x ⋅ ∆y = α∆t/∆x ⋅ ∆y =
1.5 ×10−6 m 2 / s ×1s/ ( 0.006 × 0.002 ) m 2 =
0.1250
) (
(
)
(
1  ∆y 
1  ∆y 
T1p+1 − T2p+1 + N T∞ − T2p+1 +   T3p+1 − T2p+1


2  ∆x 
2  ∆x 
 ∆x 
1
+   T5p+1 − T2p+1
=
T2p+1 − T2p
2Fo
 ∆y 
)
(
(1)
(2)
)
)
(
1  ∆y  p+1   ∆x 
1  p+1 1  ∆x  p+1
 ∆y 
+   T3
T1 −    + N +   +
 T


2  ∆x 
2  ∆y 
 ∆x  2Fo  2
  ∆y 
 ∆x 
1 p
+   T5p+1 =
− NT∞ −
T .
2Fo 2
 ∆y 
(3)
Substituting numerical values for Fo and N, and using T∞ = 30°C and ∆x/∆y = 3, find
0.16667T1p+1 − 7.63333T2p+1 + 0.16667T3p+1 + 3.00000T5p+1 =
9.0000 − 4.0000T2p .
(4)
By inspection and use of Eq. (3), the FDEs for Nodes 1 and 3 can be inferred.
Interior Node 7. From an energy balance on the
prescribed control volume with ∆x/∆y = 3,
E ′in + E ′g =
E ′st
where E ′g = 2q′ht and E ′in represents the conduction
terms −q′a + q′b + q′c + q′d ,
k∆y
T8p+1 − T7p+1
∆x
+ k∆x
T4p+1 − T7p+1
∆y
+ k∆y
T8p+1 − T7p+1
∆x
p+1
− T7p+1
T10
T p+1 − T7p
′ht ρ c ( ∆x ⋅ ∆y ) 7
+ 2q=
.
∆y
∆t
Using the definition of Fo, Eq. (2), and regrouping, find
1  ∆x  p+1   ∆x   ∆y 
1  p+1
T4 −    +   +
 T7


2  ∆y 
  ∆y   ∆x  2Fo 
+ k∆x
q′
1  ∆x  p+1
1 p
 ∆y 
+   T8p+1 +   T10
=
− ht −
T
2  ∆y 
k 2Fo 7
 ∆x 
(5)
p+1
1.50000T4p+1 − 7.33333T7p+1 + 0.33333T8p+1 + 1.50000T10
=
−12.5000 − 4.0000T7p .
Continued …
(6)
PROBLEM 5.114 (Cont.)
Recognizing the form of Eq. (5), it is a simple matter to infer the FDE for the remaining interior
points for which q ht = 0. In matrix notation [A][T] = [C], the coefficient matrix [A] and RHS
matrix [C] are:
Recall that the problem asks for the time required for the
surface temperature difference (relative to the initial
temperature) above a heating element (Node 1) to reach 95%
of the difference for steady-state conditions. This provides
information on approximately how long it takes for the plate
to come to a steady operating condition. Not knowing the
steady-state temperature distribution, use the implicit FDE in matrix form above to step through
time → ∞ to the steady-state solution; that is, proceed to p → 10,20…100 until the solution
matrix [T] does not change. The results of the analysis are tabulated below. Column 2 labeled
Tm(∞) is the steady-state distribution. The criterion is for Node 1 to satisfy (T2-Ti) = 0.95(T2(∞)Ti), or T2 = 54.5°C. The third column is the temperature distribution at t = to = 248s, Tm(248s);
at this elapsed time, Node 1 has met the criterion. Therefore the time required is 248 s.
<
m
1
2
3
4
5
6
7
8
9
10
11
12
Tm(∞)
55.80
49.93
47.67
59.03
51.72
49.19
63.89
52.98
50.14
62.84
53.35
50.46
Tm(248s)
54.51
48.64
46.38
57.64
50.32
47.79
62.42
51.52
48.68
61.35
51.86
48.98
COMMENTS: Note that at t = 248 s, the only nodes that have met the 95% criterion are Nodes
1, 4, 7, and 10. Can you explain these nodes were the first to reach this limit? Which nodes will
be the last to reach their limits?
PROBLEM 5S.1
PROBLEM 5S.7
PROBLEM 5S.9
KNOWN: Steel (plain carbon) billet of square cross-section initially at a uniform temperature
of 30°C is placed in a soaking oven and subjected to a convection heating process with
prescribed temperature and convection coefficient.
FIND: Time required for billet center temperature to reach 600°C.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in x1 and x2 directions, (2) Constant
properties, (3) Heat transfer to billet is by convection only.
PROPERTIES: Table A-1, Steel, plain carbon (T = (30+600)°C/2 = 588K = ≈ 600K): r =
3
-5 2
7854 kg/m , cp = 559 J/kg⋅K, k = 48.0 W/m⋅K, α =k/rcp = 1.093 × 10 m /s.
ANALYSIS: The billet corresponds to Case (e), Figure 5S.11 (infinite rectangular bar). Hence,
the temperature distribution is of the form
θ ∗ ( x1=
, x 2 , t ) P ( x1, t ) × P ( x 2 , t )
where P(x,t) denotes the distribution corresponding to the plane wall. Because of symmetry in
the x1 and x2 directions, the P functions are identical. Hence,
θ ( 0, 0, t ) θ o ( 0, t ) 
=


θi
 θ i  Plane wall
2
θ= T − T∞

where θ i =
Ti − T∞
and L =
0.15m.
=
θ o T ( 0,t ) − T∞
Substituting numerical values, find
1/ 2
1/ 2 

θ o ( 0, t )  T ( 0,0,t ) − T∞ 
600 − 750 ) C 
(
=

= =
0.46.

θi
 ( 30 − 750 ) C 
 Ti − T∞ 


Consider now the Heisler chart for the plane wall, Figure 5S.1. For the values
θo
θ o∗ =
≈ 0.46
θi
find
∗ Fo
t=
=
αt
L2
k
48.0 W/m ⋅ K
Bi-1 =
3.2
=
=
hL 100 W/m 2 ⋅ K × 0.15m
≈ 3.2.
Hence,
3.2 L2
=
t =
α
3.2 ( 0.15 m )
= 6587
=
s 1.83 h.
1.093 ×10−5 m 2 / s
2
<
PROBLEM 5S.10
KNOWN: Initial temperature of fire clay brick which is cooled by convection.
FIND: Center and corner temperatures after 50 minutes of cooling.
SCHEMATIC:
ASSUMPTIONS: (1) Homogeneous medium with constant properties, (2) Negligible radiation
effects.
3
PROPERTIES: Table A-3, Fire clay brick (900K): r = 2050 kg/m , k = 1.0 W/m⋅K, cp = 960
-6 2
J/kg⋅K. α = 0.51 × 10 m /s.
ANALYSIS: From Figure 5S.11, the center temperature is given by
T ( 0,0,0,t ) − T∞
Ti − T∞
= P1 ( 0, t ) × P2 ( 0, t ) × P3 ( 0, t )
where P1, P2 and P3 must be obtained from Figure 5S.1.
L1 0.03m:
=
Bi
=
1
h L1
= 1.50
k
Fo
=
1
L
=
2 0.045m:
Bi
=
2
h L2
= 2.25
k
Fo
=
2
L
=
3 0.10m:
Bi
=
3
h L3
= 5.0
k
Fo
=
3
αt
= 1.70
L21
αt
= 0.756
L22
αt
= 0.153
L23
Hence from Figure 5S.1,
P1 ( 0, t ) ≈ 0.22
Hence,
P2 ( 0, t ) ≈ 0.50
T ( 0,0,0,t ) − T∞
Ti − T∞
P3 ( 0, t ) ≈ 0.85.
≈ 0.22 × 0.50 × 0.85 =
0.094
and the center temperature is
T ( 0,0,0,t ) ≈ 0.094 (1600 − 313) K + 313 K =
434 K.
<
Continued …
PROBLEM 5S.10 (Cont.)
The corner temperature is given by
T ( L1, L 2 , L3 , t ) − T∞
Ti − T∞
= P ( L1, t ) × P ( L 2 , t ) × P ( L3 , t )
where
P=
( L1, t )
θ ( L1, t )
⋅ P1 ( 0, t ) , etc.
θo
and similar forms can be written for L2 and L3. From Figure 5S.2,
θ ( L1, t )
≈ 0.55
θo
θ ( L2 , t )
≈ 0.43
θo
θ ( L3 , t )
≈ 0.25.
θo
Hence,
P ( L1, t ) ≈ 0.55 × 0.22 =
0.12
P ( L 2 , t ) ≈ 0.43 × 0.50 =
0.22
P ( L3 , t ) ≈ 0.85 × 0.25 =
0.21
and
T ( L1, L 2 , L3 , t ) − T∞
Ti − T∞
≈ 0.12 × 0.22 × 0.21 =
0.0056
or
T ( L1, L 2 , L3 , t ) ≈ 0.0056 (1600 − 313) K + 313 K.
The corner temperature is then
T ( L1, L 2 , L3 , t ) ≈ 320 K.
<
COMMENTS: (1) The foregoing temperatures are overpredicted by ignoring radiation, which
is significant during the early portion of the transient.
(2) Note that, if the time required to reach a certain temperature were to be determined, an
iterative approach would have to be used. The foregoing procedure would be used to compute
the temperature for an assumed value of the time, and the calculation would be repeated until the
specified temperature were obtained.
PROBLEM 5S.11
KNOWN: Cylindrical copper pin, 100 mm long × 50 mm diameter, initially at 20°C; end faces are
subjected to intense heating, suddenly raising them to 500°C; at the same time, the cylindrical surface is
subjected to a convective heating process (T∞,h).
FIND: (a) Temperature at center point of cylinder after a time of 8 seconds from sudden application of
heat, (b) Consider parameters governing transient diffusion and justify simplifying assumptions that could
be applied to this problem.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction, (2) Constant properties and convection heat transfer
coefficient.
(
)
PROPERTIES: Table A-1, Copper, pure T ≈ ( 500 + 20 ) C/2 ≈ 500K : r = 8933 kg/m3, c = 407
J/kg⋅K, k = 386 W/m⋅K, α = k/rc = 386 W/m⋅K/8933 kg/m3 × 407 J/kg⋅K = 1.064 × 10-4m2/s.
ANALYSIS: (1) The pin can be treated as a two-dimensional system comprised of an infinite cylinder
whose surface is exposed to a convection process (T∞,h) and of a plane wall whose surfaces are
maintained at a constant temperature (Te). This configuration corresponds to the short cylinder, Case (i)
of Figure 5S.11,
θ ( r,x,t )
(1)
= C ( r,t ) × P ( x,t ) .
θi
For the infinite cylinder, using Figure 5S.4, with
)
(
2
-3
hro 100 W/m ⋅ K 25 × 10 m
Bi =
6.47 × 10-3 and
=
=
k
385 W/m ⋅ K
m2
1.064 × 10-4
× 8s
at
Fo =
1.36,
= s
=
-3 2
ro2
25 × 10 m
)
(
θ ( 0,8s ) 
 ≈ 1.
θi  cyl
For the infinite plane wall, using Figure 5S.1, with
=
C ( 0,8s )
find
Bi
=
find
hL
k
→∞
P ( 0,8s )
=
or
Bi-1 → 0
and
θ ( 0,8s ) 
≈ 0.5.

θi  wall
Fo
=
(2)
at
=
L2
1.064 × 10-4 m 2 / s × 8s
= 0.34,
(50 ×10-3m )
2
(3)
θ ( 0, 0, 8s ) T ( 0,0,8s ) − T∞
Combining Equations (2) and (3) with Eq. (1), find
=
=
≈ 1 × 0.5 0.5
θi
Ti − T∞
<
T ( 0,0,8s ) =T∞ + 0.5 ( Ti − T∞ ) =500 + 0.5 ( 20 − 500 ) =260 C.
(b) The parameters controlling transient conduction with convective boundary conditions are the Biot and
Fourier numbers. Since Bi << 0.1 for the cylindrical shape, we can assume radial gradients are negligible.
That is, we need only consider conduction in the x-direction.
PROBLEM 5S.12
KNOWN: Cylindrical-shaped meat roast weighing 2.25 kg, initially at 6°C, is placed in an oven
and subjected to convection heating with prescribed (T∞,h).
FIND: Time required for the center to reach a done temperature of 80°C.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in x and r directions, (2) Uniform and
constant properties, (3) Properties approximated as those of water.
(
)

PROPERTIES: Table A-6, Water, liquid T =
(80 + 6 ) C/2 ≈ 315K :/ r = 1/vf = 1/1.009 ×
-3 3
3
-7 2
10 m /kg = 991.1 kg/m , cp,f = 4179 J/kg⋅K, k = 0.634 W/m⋅K, α = k/rc = 1.531 × 10 m /s.
ANALYSIS: The dimensions of the roast are determined from the requirement ro = L and
knowledge of its weight and density,
1/ 3
1/ 3 

 M 
2.25 kg
(1)
=
=0.0712m.
or ro =L =


 2πr 
 2π 991.1 kg/m3 
The roast corresponds to Case (i), Figure 5S.11, and the temperature distribution may be
T ( x,r,t ) − T∞
expressed as the product of one-dimensional solutions,
= P ( x,t ) × C ( r,t ) , where
Ti − T∞
P(x,t) and C(r,t) are defined by Equations 5S.2 and 5S.3, respectively. For the center of the
cylinder,
M =r V =r ⋅ 2L ⋅ π ro2
T ( 0,0,t ) − T∞
=
Ti − T∞

80 − 175 ) C
(=
( 6 − 175) C
(2)
0.56.
In terms of the product solutions,
T ( 0,0,t ) − T∞
T ( 0,t ) − T∞ 
= 0.56
=

Ti − T∞
Ti − T∞ 
T ( 0,t ) − T∞ 

Ti − T∞ 
wall
cylinder
×
For each of these shapes, we need to find values of θo/θi such that their product satisfies
Equation (3). For both shapes,
h ro hL 15 W/m 2 ⋅ K × 0.0712m
=
Bi
= =
= 1.68
k
k
0.634 W/m ⋅ K
or
Bi-1 ≈ 0.6
Fo =
α t/ro2 =
α t/L2 =
1.53 ×10−7 m 2 / s × t/ ( 0.0712m ) =×
3.020 10−5 t.
Continued …
2
(3)
PROBLEM 5S.12 (Cont.)
A trial-and-error solution is necessary. Begin by assuming a value of Fo; obtain the respective
θo/θi values from Figures 5S.1 and 5S.4; test whether their product satisfies Equation (3). Two
trials are shown as follows:
θo 
θo 
Trial Fo
t(hrs) θ o / θ i ) wall
θ o / θ i )cyl
 × 
θ i  w θ i  cyl
1
2
0.4
0.3
3.68
2.75
0.72
0.78
0.50
0.68
0.36
0.53
For Trial 2, the product of 0.53 agrees closely with the value of 0.56 from Equation (2). Hence,
it will take approximately 2 ¾ hours to roast the meat.
PROBLEM 5S.13
KNOWN: A long alumina rod, initially at a uniform temperature of 850 K, is suddenly exposed
to a cooler fluid.
FIND: Temperature of the rod after 30 s, at an exposed end, T(0,0,t), and at an axial distance
6mm from the end, T(0, 6 mm, t).
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in (r,x) directions, (2) Constant properties,
(3) Convection coefficient is same on end and cylindrical surfaces.
PROPERTIES: Table A-2, Alumina, polycrystalline aluminum oxide (assume
3
T ≈ ( 850 + 600 ) K/2 =
725K): r = 3970 kg/m , c = 1154 J/kg⋅K, k = 12.4 W/m⋅K.
ANALYSIS: First, check if system behaves as a lumped capacitance. Find
=
Bi
hLc h ( ro / 2 ) 500 W/m ⋅ K ( 0.010m/2 )
=
=
= 0.202.
k
k
12.4 W/m ⋅ K
Since Bi > 0.1, rod does not behave as spacewise isothermal object. Hence, treat rod as a semiinfinite cylinder, the multi-dimensional system Case (f), Figure 5S.11.
The product solution can be written as
θ ( r,x,t ) θ ( r,t ) θ ( x,t )
θ ∗ ( r,x,t ) =
=
×
=C r∗ , t∗ × S x∗ , t∗
θi
θi
θi
) (
(
)
Infinite cylinder, C(r*,t*). Using the Heisler charts with r* = r = 0 and
-1
 h ro 
=
Bi-1 =

 k 
-1
 500 W/m 2 ⋅ K × 0.01m 
=
2.48.


12.4 W/m ⋅ K


2
−6 2
Evaluate α = k/rc = 2.71 × 10 m /s, find =
Fo α t/r
=
o 2.71× 10 m / s × 30s/(0.01m) =
-6 2
2
-1
0.812. From the Heisler chart, Figure 5S.4, with Bi = 2.48 and Fo = 0.812, read C(0,t*) =
θ(0,t)/θi = 0.61.
Continued …
PROBLEM 5S.13 (Cont.)
Semi-infinite medium, S(x*,t*). Recognize this as Case (3), Figure 5.7. From Equation 5.63,
note that the LHS needs to be transformed as follows,
T − Ti
T − T∞
T − T∞
=
1−
S ( x,t ) =
.
T∞ − Ti
Ti − T∞
Ti − T∞
Thus,

1/ 2 

 

 hx h 2α t   
h (α t )   
x
x





 .


− exp  +
+
S ( x,t ) =
1 − erfc
 erfc
1/ 2  
2
1/
2

k
k




k  

 2 (α t )  
 2 (α t )
  


Evaluating this expression at the surface (x = 0) and 6 mm from the exposed end, find



2 ⋅ K 2 2.71× 10−6 m 2 / s × 30s  
500
W/m




S ( 0,30s ) =
1 − erfc ( 0 ) − exp 0 +

2




(12.4 W/m ⋅ K )


 



)
(



2 ⋅ K 2.71× 10-6 m 2 / s × 30s 1/ 2  
500
W/m




 
erfc 0 +

12.4 W/m ⋅ K





  
)
(
{
}
S ( 0,30s ) =1 − 1 − exp ( 0.1322 )  erfc ( 0.3636 )  =0.693.
Note that Table B.2 was used to evaluate the complementary error function, erfc(w).






0.006m
S ( 6mm,30s ) =
1 -
erfc 
 2 2.71×10-6 m 2 / s × 30s 1/ 2 




)
(


 500 W/m 2 ⋅ K × 0.006m


exp 



+ 0.1322  erfc ( 0.3327 + 0.3636 )   =
0.835.
12.4 W/m ⋅ K


 

The product solution can now be evaluated for each location. At (0,0),
T ( 0,0,30s ) − T∞
=C 0,t∗ × S 0,t∗ =0.61× 0.693 =0.423.
θ ∗ ( 0, 0, t ) =
Ti − T∞
( ) ( )
T ( 0,0,30s ) =T∞ + 0.423 ( Ti − T∞ ) =350K + 0.423 ( 850 − 350 ) K =561 K.
Hence,
At (0,6mm),
( ) (
<
)
θ ∗ ( 0, 6mm,t ) =C 0,t∗ × S 6mm,t∗ =0.61× 0.835 =0.509
T ( 0,6mm,30s ) = 604 K.
<
COMMENTS: Note that the temperature at which the properties were evaluated was a good
estimate.
PROBLEM 5S.14
KNOWN: Stainless steel cylinder of Example 5S.1, 80-mm diameter by 60-mm length, initially at 600
2
K, suddenly quenched in an oil bath at 300 K with h = 500 W/m ⋅K. Use the Transient Conduction,
Plane Wall and Cylinder models of IHT to obtain the following solutions.
FIND: (a) Calculate the temperatures T(r,x,t) after 3 min: at the cylinder center, T(0, 0, 3 min), at the
center of a circular face, T(0, L, 3 min), and at the midheight of the side, T(ro, 0, 3 min); compare your
results with those in the example; (b) Calculate and plot temperature histories at the cylinder center, T(0,
0, t), the mid-height of the side, T(ro, 0, t), for 0 ≤ t ≤ 10 min; comment on the gradients and what effect
they might have on phase transformations and thermal stresses; and (c) For 0 ≤ t ≤ 10 min, calculate and
plot the temperature histories at the cylinder center, T(0, 0, t), for convection coefficients of 500 and 1000
2
W/m ⋅K.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in r- and x-coordinates, (2) Constant properties.
3
PROPERTIES: Stainless steel (Example 5S.1): r = 7900 kg/m , c = 526 J/kg⋅K, k = 17.4 W/m⋅K.
ANALYSIS: The following results were obtained using the Transient Conduction models for the Plane
Wall and Cylinder of IHT. Salient portions of the code are provided in the Comments.
(a) Following the methodology for a product solution outlined in Example 5S.1, the following results
were obtained at t = to = 3 min
(r, x, t)
P(x, t)
C(r, t)
0, 0, to
0, L, to
ro, 0, to
0.6357
0.4365
0.6357
0.5388
0.5388
0.3273
T(r, x, t)-IHT
(K)
402.7
370.5
362.4
T(r, x, t)-Ex
(K)
405
372
365
Continued …
PROBLEM 5S.14 (Cont.)
The temperatures from the one-term series calculations of the Example 5S.1 are systematically higher
than those resulting from the IHT multiple-term series model, which is the more accurate method.
(b) The temperature histories for the center and mid-height of the side locations are shown in the graph
below. Note that at early times, the temperature difference between these locations, and hence the
gradient, is large. Large differences could cause variations in microstructure and hence, mechanical
properties, as well as induce residual thermal stresses.
(c) Effect of doubling the convection coefficient is to increase the quenching rate, but much less
than by a factor of two as can be seen in the graph below.
Quenching with h = 500 W /m^2.K
Effect of increased conv. coeff. on quenching rate
600
500
T(0, 0, t) (C)
T(x, r, t) (C)
600
400
300
500
400
300
0
2
4
6
8
10
0
Time, t (min)
2
4
6
8
10
Time, t (min)
Mid-height of side (0,ro)
Center (0, 0)
h = 500 W/m^2.K
h = 1000 W/m^2.K
COMMENTS: From IHT menu for Transient Conduction | Plane Wall and Cylinder, the models were
combined to solve the product solution. Key portions of the code, less the input variables, are copied
below.
// Plane wall temperature distribution
// The temperature distribution is
T_xtP = T_xt_trans("Plane Wall",xstar,FoP,BiP,Ti,Tinf)
// The dimensionless parameters are
xstar = x / L
BiP = h * L / k
// Eq 5.9
FoP= alpha * t / L^2
// Eq 5.33
alpha = k/ (rho * cp)
// Dimensionless representation, P(x,t)
P_xt = (T_xtP - Tinf ) / (Ti - Tinf)
// Cylinder temperature distribution
// The temperature distribution T(r,t) is
T_rtC = T_xt_trans("Cylinder",rstar,FoC,BiC,Ti,Tinf)
// The dimensionless parameters are
rstar = r / ro
BiC = h * ro / k
FoC= alpha * t / ro^2
// Dimensionless representation, C(r,t)
C_rt= (T_rtC - Tinf ) / (Ti - Tinf)
// Product solution temperature distribution
(T_xrt - Tinf) / (Ti - Tinf) = P_xt * C_rt
// Eq 5.42
// Eq 5.50
PROBLEM 6.1
KNOWN: Temperature distribution at x2 in laminar thermal boundary layer.
FIND: (a) Whether plate is being heated or cooled, (b) Temperature distributions at two other x
locations. Locations of largest and smallest heat fluxes, (c) Temperature distribution at x2 for lower and
higher free stream velocities. Which velocity condition causes the largest heat flux.
SCHEMATIC:
T∞
T∞
Free stream
Thermal
boundary
layer
δt
Ts
x1
x2
x3
x
ASSUMPTIONS: (1) Steady-state conditions, (2) Laminar, incompressible flow.
ANALYSIS: (a) Since the sketch indicates that the free stream temperature is greater than the surface
temperature, the plate is being heated by the fluid. This is consistent with the fact that the surface heat
flux in the positive y-direction is given by Eq. 6.3:
∂T
qs′′ = − k f
∂y y = 0
From the sketch, the temperature gradient is positive, therefore the heat flux is negative. The heat transfer
<
is in the negative y-direction, the plate is being heated by the fluid.
(b) At every location in the boundary layer, the temperature must vary from Ts at the surface to T∞ in the
free stream. This change must occur within the thermal boundary layer thickness, as shown in the sketch
below.
T∞
T∞
T∞
T∞
Ts
x1
x2
x3
x
Continued...
PROBLEM 6.1 (Cont.)
The magnitude of the heat flux is proportional to the temperature gradient at the surface, ∂T / ∂y y =0 ,
which is shown schematically as a dashed line. The temperature gradient is steeper (larger) at x1 where
the thermal boundary layer is thinner and less steep (smaller) at x3 where the thermal boundary layer is
<
thicker. Therefore, the magnitude of the local heat flux is largest at x1 and smallest at x3.
(c) As the free stream velocity increases the boundary layer becomes thinner. Sketches for a low and high
free stream velocity are shown below.
T∞
Free stream
T∞
Thermal
boundary
layer
Ts
x2
Low freestream velocity case.
T∞
T∞
Free stream
Thermal
boundary
layer
Ts
x2
x
High freestream velocity case.
The temperature gradient, shown as the dashed line, is steeper for the higher free stream velocity case.
Therefore the higher free stream velocity case has the higher convective heat flux.
<
COMMENTS: It is important to understand how the temperature gradient at the surface varies as the
thickness of the boundary layer changes.
PROBLEM 6.2
KNOWN: Form of the velocity and temperature profiles for flow over a surface.
FIND: Expressions for the friction and convection coefficients.
SCHEMATIC:
ANALYSIS: The shear stress at the wall is
τs = µ
∂ u
= µ  A + 2By − 3Cy 2 
= Aµ .


 y=0
∂ y  y=0
Hence, the friction coefficient has the form,
τs
2Aµ
=
2 / 2 ρ u2
ρ u∞
∞
=
Cf
Cf =
2Aν
2
u∞
.
<
The convection coefficient is
2

−k f ( ∂ T/∂ y ) y=0 −k f  E + 2Fy − 3Gy  y=0
h =
=
Ts − T∞
D − T∞
h=
−k f E
.
D − T∞
<
COMMENTS: It is a simple matter to obtain the important surface parameters from knowledge
of the corresponding boundary layer profiles. However, it is rarely a simple matter to determine
the form of the profile.
PROBLEM 6.3
KNOWN: Boundary layer temperature distribution.
FIND: Surface heat flux.
SCHEMATIC:
PROPERTIES: Table A-4, Air (Ts = 300K): k = 0.0263 W/m⋅K.
ANALYSIS: Applying Fourier’s law at y = 0, the heat flux is
∂ T
∂ y
u y
 u 

=
−k ( T∞ − Ts )  Pr ∞  exp  −Pr ∞ 
ν  y=0
y=0
 ν 

u
−k ( T∞ − Ts ) Pr ∞
q′′s =
−k
q′′s =
ν
−0.0263 W/m ⋅ K (100K ) 0.7 × 5000 1/m.
q′′s =
q′′s = −9205 W/m 2 .
COMMENTS: (1) Negative flux implies convection heat transfer to the surface.
(2) Note use of k at Ts to evaluate q′′s from Fourier’s law.
<
PROBLEM 6.4
PROBLEM 6.5
PROBLEM 6.6
PROBLEM 6.7
KNOWN: Expression for the local heat transfer coefficient of a circular, hot gas jet at T∞
directed normal to a circular plate at Ts of radius ro.
FIND: Heat transfer rate to the plate by convection.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Flow is axisymmetric about the plate, (3) For
h(r), a and b are constants and n ≠ -2.
ANALYSIS: The convective heat transfer rate to the plate follows from Newton’s law of
cooling
q conv
= ∫ dq conv
= ∫ h ( r ) ⋅ dA ⋅ ( T∞ − Ts ) .
A
A
The local heat transfer coefficient is known to have the form,
h (r=
) a + br n
and the differential area on the plate surface is
dA = 2π r dr.
Hence, the heat rate is
q conv=
ro
∫0
( a + brn ) ⋅ 2π r dr ⋅ ( T∞ − Ts )
r
b n+2  o
a
q conv =2π ( T∞ − Ts )  r 2 +
r

n+2
2
0
b n+2 
a
q conv =
2π  ro2 +
ro  ( T∞ − Ts ) .
n+2
2

<
COMMENTS: Note the importance of the requirement, n ≠ -2. Typically, the radius of the jet
is much smaller than that of the plate.
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:
1 L
hL = ∫ hx dx
L 0
−0.5
For laminar flow, hx = Alam x 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:
hL ,lam
11.0 W/m 2 ⋅ K, L =
0.1 m
=
2
1m
3.48 W/m ⋅ K, L =
<
=
A
3.98 W/m1.8 ⋅ K. Thus,
For turbulent flow, hx = Aturb x −0.2 where
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 6.9
KNOWN: Expression for face-averaged Nusselt numbers on a cylinder of rectangular cross section.
Dimensions of the cylinder.
FIND: Average heat transfer coefficient over the entire cylinder. Plausible explanation for variations in
the face-averaged heat transfer coefficients.
c = 40 mm
SCHEMATIC:
Air
V = 10 m/s
d = 30 mm
T∞ = 300 K
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties.
PROPERTIES: Table A.4, air (300 K): k = 0.0263 W/m⋅K, ν = 1.589 × 10-5 m2/s, Pr = 0.707.
ANALYSIS:
For the square cylinder, c/d = 40 mm/30 mm = 1.33
Vd 10 m/s × 30 × 10-3 m
Red =
=
= 18,880
ν
1.589 × 10-5 m 2 /s
Therefore, for the front face C = 0.674, m = ½. For the sides, C = 0.107, m = 2/3 while for the back C =
0.153, m = 2/3.
Front face:
Nu d,f = 0.674 × 18,8801/2 × 0.7071/3 = 82.44
hf =
kNu d
0.0263 W/m ⋅ K × 82.44
=
= 72.27 W/m 2 ⋅ K
d
30 × 10-3 m
Side faces:
Nu d,s = 0.107 × 18,8802/3 × 0.7071/3 = 67.36
hs =
kNu d,s
d
=
0.0263 W/m ⋅ K × 67.36
-3
30 × 10 m
= 59.05 W/m 2 ⋅ K
Back face:
Nu d,b = 0.153 × 18,8802/3 × 0.7071/3 = 96.43
hb =
kNu d,b
d
=
0.0263 W/m ⋅ K × 96.43
30 × 10-3 m
= 84.54 W/m 2 ⋅ K
Continued…
PROBLEM 6.9 (Cont.)
For the entire square cylinder of unit length,
h=
h f A f + 2h s As + h b A b
A f + 2As + A b
 72.27 W/m 2 ⋅ K × 30 × 10-3 m + 2 × 59.05 W/m 2 ⋅ K × 40 × 10-3 m 


 + 84.54 W/m 2 ⋅ K × 30 × 10-3 m



h=
-3
3
(2 × 30 × 10 m + 2 × 40 × 1 m)
h = 67.35 W/m 2 ⋅ K
<
The face-averaged heat transfer coefficients are largest on the back face and smallest on the side faces.
Plausible explanations for the variations of the face-averaged heat transfer coefficients are complex fluid
flow patterns including vortex shedding on the back face and development of relatively thick boundary
layers along the sides.
COMMENT: See S.Y. Yoo, J.H. Park, C.H. Chung and M.K. Chung, Journal of Heat Transfer, Vol.
125, pp. 1163-1169, 2003 for details.
PROBLEM 6.10
KNOWN: Variation of local heat transfer coefficient around a circular collector tube.
FIND: (a) Estimate the average heat transfer coefficient, (b) Case with highest collector efficiency.
SCHEMATIC:
40
hθ (W/m2·K)
Case 3
Case 1
20
Case 2
0
π
0
2π
θ
ASSUMPTIONS: Solar irradiation is independent of the reflector orientation.
ANALYSIS: (a) From Eq. 6.13,
2π
1
1
1 2π
hdAs
hθ (=
D / 2) Ldθ
hθ dθ
=
=
2π θ 0
As As
π DL θ 0=
where L is the collector tube length. Hence, the average heat transfer coefficient may be estimated as the
average value of the local heat transfer coefficient. Approximate values of the average heat transfer
coefficient are shown in the sketch below.
∫
h
=
∫
∫
hθ (W/m2·K)
40
Case 3
20
Case 1
Case 2
0
0
π
θ
2π
For Case 1, h ≈ 20 W/m2⋅K; for Case 2, h ≈ 10 W/m2⋅K, for Case 3, h ≈ 30 W/m2⋅K
<
(b) The collector tube will be hotter than the ambient air. Hence, convective losses from the collector tube
will diminish the overall collector efficiency. Therefore, Case 2 will have the highest collector efficiency.
<
COMMENTS: (1) For case 2, the parabolic reflector partially “shields” the collector tube from the
wind, resulting in reduced heat transfer coefficients. The flow adjacent to the tube also experiences a
change in direction for case 2, with a large recirculation pattern established behind the reflector. (2) None
of the cases experiences symmetrical flow over the collector tube, as would be expected without the
reflector in place. (3) See Naeeni and Yaghoubi, “Analysis of Wind Flow Around a Parabolic Collector
(2) Heat Transfer from Receiver Tube,” Renewable Energy, Vol. 32, pp. 1259 – 1272, 2007, for
additional discussion.
PROBLEM 6.11
KNOWN: Temperature distribution in boundary layer for helium flow over a flat plate.
FIND: Variation of local convection coefficient along the plate and value of average coefficient.
SCHEMATIC:
T(x,y) = 25 + 60exp(-600xy)
T∞ = 25°C
L=3m
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties.
PROPERTIES: Table A-4, Helium (Ts = 85°C = 358 K), k = 0.173 W/m⋅K.
ANALYSIS: From Eq. 6.5,
k ∂ T ∂ y y =0
k ( 60 × 600x )
h=
−
=
+
( Ts − T∞ )
( Ts − T∞ )
where k is evaluated at y= 0 and Ts = T(x,0) = 85°C. Hence, with Ts - T∞ = 60°C = 60 K,
=
h
0.173 W m ⋅ K ( 36, 000x ) K m
= 104x W m 2 ⋅ K
60 K
(
)
<
and the convection coefficient increases linearly with x.
350
311
0
1
2
3
The average coefficient over the range 0 ≤ x ≤ 3 m is
3
1 L
104 3
104 x 2
=
h =
hdx
=
xdx
= 155 W m 2 ⋅ K
∫
∫
0
0
L
3
3 2
0
<
PROBLEM 6.12
PROBLEM 6.13
PROBLEM 6.13 (Cont.)
PROBLEM 6.14
KNOWN: Local convection coefficient on rotating disk. Radius and surface temperature of disk.
Temperature of stagnant air.
FIND: Local heat flux and total heat rate. Nature of boundary layer.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat transfer from back surface and edge of disk.
ANALYSIS: If the local convection coefficient is independent of radius, the local heat flux at every
point on the disk is
′′ h ( Ts − T∞=
q=
C 600 W / m 2
) 20 W / m2 ⋅ K ( 50 − 20 ) °=
<
Since h is independent of location, h= h= 20 W / m 2 ⋅ K and the total power requirement is
Pelec ==
q hAs ( Ts − T∞ ) =
hπ ro2 ( Ts − T∞ )
Pelec
=
20 ) °C
( 20 W / m2 ⋅ K )π ( 0.1m )2 (50 −=
18.9 W
<
If the convection coefficient is independent of radius, the boundary layer must be of uniform thickness δ.
Within the boundary layer, air flow is principally in the circumferential direction. The circumferential
velocity component uθ corresponds to the rotational velocity of the disk at the surface (y = 0) and
increases with increasing r (uθ = Ωr). The velocity decreases with increasing distance y from the surface,
approaching zero at the outer edge of the boundary layer (y → δ).
PROBLEM 6.15
KNOWN: Dimensions and temperatures of rotating and stationary disks, air gap spacing between disks,
rotational speed. Correlation for the local Nusselt number.
FIND: Value of the average Nusselt number, total heat flux from the disk’s top surface, total power
requirement. Comment on the nature of the flow between the disks.
SCHEMATIC:
Stationary disk, Td = 20°C
ro = 100 mm
Air
T∞ = 20°C
g = 1 mm
Rotating disk
Ts = 50°C
Air
Ω = 150 rad/s
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible viscous
dissipation.
PROPERTIES: Table A-4, air ( T = (50°C + 20°C)/2 = 35°C ≈ 308K): ν = 16.69×10-6 m2/s, k = 0.0269
W/m⋅K.
ANALYSIS: From the problem statement, Nu
=
r
(
)
h( r )r
= 70 1 + e −140G Rer−o 0.456 Rer0.478 . Since
k
Rer = Ωr 2 /ν , the local heat transfer coefficient is

=
h( r ) k 70 1 + e −140G


(
)
 Ωro2 


 ν 
−0.456
Ω
ν 
 
0.478 
 r −0.044


The average heat transfer coefficient may be evaluated from
0.478  ro

2 −0.456
1
2π k 
Ω
−140G  Ωro 
 r × r −0.044 dr
+
h=
h( r )dAs =
70
1
e


 
2
As
ν
ν


π
r
 
o


As

0
(
∫
)
∫
or

1.022k 
=
h
70 1 + e −140G
2

ro

(
)
 Ωro2 


 ν 
−0.456
Ω
 
ν 
0.478 
 ro1.956


Continued…
PROBLEM 6.15 (Cont.)
Substituting values,
h=
1.022 × 0.0269W/m ⋅ K
(0.100m)2

× 70 1 + e −140×0.01


(
)
 150rad/s × ( 0.100m )2 


 16.69 × 10−6 m 2 /s 


−0.456
150rad/s



−6 2 
 16.69 × 10 m /s 
0.478 
 ( 0.100m )1.956


or
=
h 30.8 W/m 2 ⋅ K
<
The average Nusselt number is
Nu=
hD =
/ k 30.8W/m 2 ⋅ K × 0.200m / 0.0269W/m ⋅=
K 229
D
<
The heat flux from the top surface of the disk is
q=
= 924W/m 2
" h(Ts − Td=
) 30.8W/m 2 ⋅ K × (50 − 20)°C
<
Therefore, the total electric power requirement is
2
2
=
=
P q" A
q "π r=
924W/m 2 × π × (0.100m)=
29W
s
o
<
Note that if only conduction were occurring between the two disks, the heat flux would be
=
− Td ) (0.0269W/m ⋅ K / 0.001m) × (50 − =
q " ( k / g )(Ts=
20)°C 807W/m 2 .
The conduction heat flux is slightly less than the calculated quantity with rotation, suggesting that
advection in the cross-gap direction is small, and that heat transfer between the disks is dominated by
conduction. The laminar flow between the disks is characterized by very small velocities in the cross-gap
direction.
COMMENTS: (1) The slight increase in heat transfer rate is due to edge effects where air can exit or
enter the space between the disks, and enhance heat transfer between the disks by mixing. (2) The
Reynolds number is Rero =
Ωro2 /ν =
150rad/s × ( 0.100m ) /16.69 × 10−6 m 2 / s =
90,000. Transition to
2
turbulent flow begins at a Reynolds number of approximately 180,000 for this configuration. (3) See Pelle
and Harmand, “Heat Transfer Measurements in an Opened Rotor-Stator System Air-Gap,” Experimental
Thermal and Fluid Science, ‘Vol. 31, pp. 165 – 180, 2007, for more information.
PROBLEM 6.16
KNOWN: Air flow over a flat plate of known length, location of transition from laminar to turbulent
flow, value of the critical Reynolds number.
FIND: (a) Free stream velocity with properties evaluated at T = 350 K, (b) Expression for the average
convection coefficient, h lam (x) , as a function of the distance x from the leading edge in the laminar
region, (c) Expression for the average convection coefficient h turb (x) , as a function of the distance x from
the leading edge in the turbulent region, (d) Compute and plot the local and average convection
coefficients over the entire plate length.
SCHEMATIC:
hlam=Clamx-0.5
T∞, u∞
hturb=Cturbx-0.2
x
Laminar
Turbulent
xc
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties.
PROPERTIES: Table A.4, air (T = 350 K): k = 0.030 W/m⋅K, ν = 20.92 × 10-6 m2/s, Pr = 0.700.
ANALYSIS:
(a) Using air properties evaluated at 350 K with xc = 0.5 m,
u x
Re x,c = ∞ c = 5 × 105
ν
u ∞ =×
5 105 ν x c =×
5 105 × 20.92 × 10−6 m 2 / s 0.5 m =
20.9 m s
(b) From Eq. 6.13, the average coefficient in the laminar region, 0 ≤ x ≤ xc, is
hlam ( x ) =
1 x
h lam ( x ) dx =
x ∫0
1
x
Clam
x -0.5
∫o x
dx =
1
x
Clam x
0.5
<
= 2Clam x -0.5 = 2h lam (x) (1)
<
(c) The average coefficient in the turbulent region, xc ≤ x ≤ L, is
xc
x 

1  xc
x
x 0.5
x 0.8



h turb ( x ) = ∫ h lam ( x ) dx + ∫ h turb ( x ) dx =Clam
+ C turb


0
x


x
0.5
0.8

c

0
x c 
Continued…
PROBLEM 6.16 (Cont.)
(
)
1
h turb ( x ) =  2Clam x c0.5 + 1.25C turb x 0.8 − x c0.8 

x 
(2)
<
(d) The local and average coefficients, Eqs. (1) and (2) are plotted below as a function of x for the range 0
≤ x ≤ L.
Convection coefficient (W/m^2.K)
150
100
50
0
0
0.5
Distance from leading edge, x (m)
Local - laminar, x <= xc
Local - turbulent, x => xc
Average - laminar, x <= xc
Average - turbulent, x => xc
1
PROBLEM 6.17
PROBLEM 6.18
PROBLEM 6.19
PROBLEM 6.19 (Cont.)
PROBLEM 6.20
KNOWN: Pressure dependence of the dynamic viscosity, thermal conductivity and specific heat.
FIND: (a) Variation of the kinematic viscosity and thermal diffusivity with pressure for an
incompressible liquid and an ideal gas, (b) Value of the thermal diffusivity of air at 350 K for pressures of
1, 5 and 10 atm, (c) Location where transition occurs for air flow over a flat plate with T∞ = 350 K, p = 1,
5 and 10 atm, and u∞ = 2 m/s.
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Transition at Rex,c = 5 × 105,
(4) Ideal gas behavior.
PROPERTIES: Table A.4, air (350 K): µ = 208.2 × 10-7 N⋅s/m2, k = 0.030 W/m⋅K, cp = 1009 J/kg⋅K, ρ =
0.995 kg/m3.
ANALYSIS:
(a) For an ideal gas
p = ρRT or ρ = p/RT
(1)
while for an incompressible liquid, ρ = constant
(2)
The kinematic viscosity is ν = μ/ρ
(3)
Therefore, for an ideal gas
ν = μRT/p or ν ∝ p-1
and for an incompressible liquid
(4)
<
<
ν = μ/ρ or ν is independent of pressure.
The thermal diffusivity is
α= k / ρc
Therefore, for an ideal gas,
α = kRT/pc or α ∝ p-1
(6)
For an incompressible liquid α = k/ρc or α is independent of pressure
<
<
(b) For T = 350 K, p = 1 atm, the thermal diffusivity of air is
α=
0.030 W/m ⋅ K
0.995 kg/m3 × 1009 J/kg ⋅ K
= 29.9 × 10-6 m 2 /s
<
Using Equation 6, at p = 5 atm,
Continued…
PROBLEM 6.20 (Cont.)
α = 29.9 × 10-6 m 2 /s/5 = 5.98 × 10-6 m2/s
<
At p = 10 atm,
α = 29.9 × 10-6 m 2 /s/10 = 2.99 × 10-6 m2/s
<
(c) For transition over a flat plate,
x u
Re x,c = c ∞ = 5 × 105
ν
Therefore
x c = 5 × 105 (ν/u ∞ )
For T∞ = 350 K, p = 1 atm,
ν = μ/ρ = 208.2 × 10-7 N ⋅ s/m 2 0.995 kg/m3 = 20.92 × 10-6 m 2 /s
Using Equation 4, at p = 5 atm
ν = 20.92 × 10-6 m 2 /s 5 = 4.18 × 10-6 m 2 /s
At p = 10 atm,
ν = 20.92 × 10-6 m 2 /s 10 = 2.09 × 10-6 m 2 /s
Therefore, at p = 1 atm
x c = 5 × 105 × 20.92 × 10-6 m 2 /s/(2m/s) = 5.23 m
<
At p = 5 atm,
x c = 5 × 105 × 4.18 × 10-6 m 2 /s/(2m/s) = 1.05 m
<
At p = 10 atm
x c = 5 × 105 × 2.09 × 10-6 m 2 /s/(2m/s) = 0.523 m
<
COMMENT: Note the strong dependence of the transition length upon the pressure for the gas (the
transition length is independent of pressure for the incompressible liquid).
PROBLEM 6.21
PROBLEM 6.21 (Cont.)
PROBLEM 6.22
KNOWN: Laminar boundary layer flow over a flat plate. Atmospheric pressure. Type of fluid, and/or
temperature, and/or δ/δt.
FIND: Missing information for type of fluid, and/or temperature, and or δ/δt.
SCHEMATIC:
Air: T = 300 K, δ/δt = 0.89
Water: T = 350 K
Engine oil: δ/δt = 15.4
Unknown fluid: δ/δt = 45.3
δt (x) or δ(x)
δ (x) or δt(x)
T∞
u∞
p = 1 atm
L
x
ASSUMPTIONS: (1) Steady-state, (2) Laminar boundary layer flow, (3) Properties are constant at the
specified temperature.
PROPERTIES: Table A-4, Air, (T = 300 K): Pr = 0.707. Table A-6, Water, (T = 350 K): Pr = 2.29.
ANALYSIS: For air at 300 K and water at 350 K, the Prandtl number can be found in Appendix A, and
the ratio of boundary layer thicknesses is
δ 
δ 
1/3
1/3
1/3
1/3
=
=
=
Pr
0.707
0.891,  = Pr=
2.29
1.32
 =

air
water
 δ t air
 δ t  water
<
For engine oil, the Prandtl number can be found from the given boundary layer thickness ratio:
3
δ 
3
=
Proil  =
15.4
=
3650

δ
 t oil
<
<
Interpolating in Table A.5 gives Toil = 309 K.
For the unknown fluid:
3
δ 
3
=
=
Pr?  =
93,000
 45.3
δ
 t ?
Searching through Appendix A, we find that the fluid with the largest Prandtl number is glycerin, but the
tabulated values are still not as large as the required value. Assuming that the fluid is glycerin at a
temperature less than 273 K and extrapolating, we find T = 272 K.
<
COMMENTS: Most gases have Pr < 1, and therefore δ < δt. Most liquids have Pr > 1, and therefore δ >
δt. Liquid metals are an exception to this trend. Their Prandtl numbers are very small, and consequently δ
« δt for liquid metals.
PROBLEM 6.23
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 ReL,2 to differ
from ReL,1. However, for the prescribed conditions, this non-constant property effect is small.
PROBLEM 6.24
PROBLEM 6.25
KNOWN: Critical Reynolds number for a cylinder in cross flow. Critical Mach number.
FIND: Critical cylinder diameter below which, if the flow of air at atmospheric pressure and temperature
is turbulent, compressibility effects may be important.
SCHEMATIC:
Air
T∞ = 23°C
V
ASSUMPTIONS: (1) Steady-state conditions. (2) Air behaves as an ideal gas.
PROPERTIES: Table A.4, air (T = 300 K): M = 28.97 kg/kmol, cp = 1.007 kJ/kg∙K, µ = 184.6 × 10-7
N∙s/m2.
ANALYSIS: The density of an ideal gas may be found from the equation of state,
ρ=
p
RT
and the speed of sound for an ideal gas is a = γ RT .
From the definition of the Mach number, the air velocity may be expressed as
V = Ma⋅ a
Substituting the preceding equations into the definition of the Reynolds number yields
Re
=
D
VD VDρ Ma γ RT Dp
Ma
D
=
=
=
ν
µ
µ RT
µ
γ
Dp
RT
Letting Re = Rec and Ma = Mac, the preceding equation can be rearranged to write an expression for the
critical cylinder diameter,
Dc =
Rec
Mac
RT µ
γ p
Before evaluating the critical cylinder diameter, we note that the gas constant for air is
Continued...
PROBLEM 6.25 (Cont.)
=
R
R 8315 J/kmol ⋅ K
=
= 287 J / kg ⋅ K
M 28.97 kg/kmol
and the specific heat at constant volume, cv, is
cv = c p − R = 1007 J / kg ⋅ K − 287 J / kg ⋅ K = 720 J / kg ⋅ K
Therefore, the ratio of specific heats for air is
=
g
c p 1007 J/kg ⋅ K
=
= 1.399
cv
720 J/kg ⋅ K
For the conditions of the problem, the critical cylinder diameter is
Rec RT m 2 × 105 287 J/kg ⋅ K × 300 K 184.6 × 10−7 N ⋅ s/m 2
Dc =
=
×
=
0.030 m =
30 mm <
Mac g p
0.3
1.399
1.0133 × 105 N/m 2
COMMENTS: (1) The expression for the critical Reynolds number (ReD,c = 2 × 105) is plotted using log-log
scales in the figure below. Laminar flow occurs to the left of the sloped line, while turbulent flow occurs to the
right of the sloped line. The velocity associated with the critical Mach number is identified by the horizontal line
that separates regions of incompressible flow (below the horizontal line) and compressible flow (above the
horizontal line). From this plot, it is evident that below the critical cylinder diameter, Dc, if the flow is turbulent,
compressibility effects may be important. If the flow is laminar, the flow may or may not be compressible. In
general, turbulent flow is difficult to achieve in situations involving small length scales.
Mac
Compressible
Log V
Incompressible
ReD,c
Turbulent
Laminar
Dc
Log D
(2) The value of the critical Reynolds number is geometry-dependent. Care must be taken to apply the
correct value of the critical Reynolds number in any calculation involving convection heat transfer.
PROBLEM 6.26
KNOWN: Experimental measurements of the heat transfer coefficient for a square bar in cross
flow.
FIND: (a) h for the condition when L = 1m and V = 15m/s, (b) h for the condition when L =
1m and V = 30m/s, (c) Effect of defining a side as the characteristic length.
SCHEMATIC:
ASSUMPTIONS: (1) Functional form Nu = CRem Pr n applies with C, m, n being constants,
(2) Constant properties.
ANALYSIS: (a) For the experiments and the condition L = 1m and V = 15m/s, it follows that
Pr as well as C, m, and n are constants. Hence
hL ∝
( VL )m .
Using the experimental results, find m. Substituting values
m
m
h1L1  V1L1 
50 × 0.5  20 × 0.5 
=
=


h 2 L 2  V2 L 2 
40 × 0.5  15 × 0.5 
giving m = 0.782. It follows then for L = 1m and V = 15m/s,
L  V⋅L 
h=
h1 1 

L  V1 ⋅ L1 
m
W
0.5  15 ×1.0 
50
=
×


m 2 ⋅ K 1.0  20 × 0.5 
0.782
34.3W/m 2 ⋅ K.
=
<
59.0W/m 2 ⋅ K.
=
<
(b) For the condition L = 1m and V = 30m/s, find
L  V⋅L 
h=
h1 1 

L  V1 ⋅ L1 
m
W
0.5  30 ×1.0 
50
=
×


m 2 ⋅ K 1.0  20 × 0.5 
0.782
(c) If the characteristic length were chosen as a side rather than the diagonal, the value of C
would change. However, the coefficients m and n would not change.
COMMENTS: The foregoing Nusselt number relation is used frequently in heat transfer
analysis, providing appropriate scaling for the effects of length, velocity, and fluid properties on
the heat transfer coefficient.
PROBLEM 6.27
KNOWN: Form of the Nusselt number correlation for forced convection and fluid properties.
FIND: Expression for figure of merit FF and values for air, water and a dielectric liquid.
-5
2
PROPERTIES: Prescribed. Air: k = 0.026 W/m⋅K, ν = 1.6 × 10 m /s, Pr = 0.71. Water: k = 0.600
-6 2
-6 2
W/m⋅K, ν = 10 m /s, Pr = 5.0. Dielectric liquid: k = 0.064 W/m⋅K, ν = 10 m /s, Pr = 25
n
ANALYSIS: With Nu L ~ Rem
L Pr , the convection coefficient may be expressed as
h~
m
m 
n
k  VL 
n ~ V  k Pr
Pr


L n 
L1− m  n m




The figure of merit is therefore
FF =
k Pr n
<
nm
and for the three fluids, with m = 0.80 and n = 0.33,
(
FF W ⋅ s0.8 / m 2.6 ⋅ K
)
Air
167
Water
64, 400
Dielectric
11, 700
<
Water is clearly the superior heat transfer fluid, while air is the least effective.
COMMENTS: The figure of merit indicates that heat transfer is enhanced by fluids of large k, large Pr
and small ν.
PROBLEM 6.28
PROBLEM 6.29
PROBLEM 6.30
PROBLEM 6.31
PROBLEM 6.32
KNOWN: Flow over a flat plate. Velocity and temperature of two fluids. Variation of boundary layer
thickness with x for laminar flow.
FIND: (a) Location where transition to turbulence occurs for each fluid, (b) Plot of velocity boundary
layer thickness for 0 ≤ x ≤ xc for each fluid, (c) Plot of thermal boundary layer thickness over the same
range. Which fluid has the largest local temperature gradient at the surface, Nusselt number, and heat
transfer coefficient.
SCHEMATIC:
u∞ = 2 m/s
T∞ = 300K (air)
T∞ = 380K (oil)
xc
Ts > T∞
x
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible flow, (3) Transition occurs at a critical
Reynolds number of 5 × 105.
PROPERTIES: Table A.4, Air (T = 300 K): ν = 15.89 × 10-6 m2/s, k = 0.0263 W/m⋅K, Pr = 0.707.
Table A.5, Engine Oil (T = 380 K): ν = 16.9 × 10-6 m2/s, k = 0.136 W/m⋅K, Pr = 233.
ANALYSIS: (a) Transition occurs at Rex,c = u∞x/ν = 5 × 105. Therefore, for air
xc =
5 × 105
ν
u∞
=
5 × 105
15.89 × 10−6 m 2 / s
=
4.0 m
2 m/s
<
<
The value for engine oil is 4.2 m.
(b) The velocity boundary layer thickness is given by
=
δ
5
δ
=
. Thus, for air
x
Rex
5x
15.89 × 10−6 m 2 / s × x
= 5 xν=
= 0.0141x1/ 2
/ u∞ 5
2
m
/
s
Rex
<
where x and δ are expressed in meters. The corresponding result for engine oil is δ = 0.0145x1/ 2 .
<
These results are plotted below.
Continued…
PROBLEM 6.32 (Cont.)
0.03
Engine oil
Air
delta (m)
0.02
0.01
0
0
0.5
1
1.5
2
2.5
x (m)
3
3.5
4
4.5
<
(c) From Eq. 6.55 with n = 1/3, δt = δ⋅Pr-1/3. Thus, for air
δt =
δ Pr −1/3 =
0.0141x1/ 2 × (0.707) −1/3 =
0.0158 x1/ 2
Similarly for engine oil δ t = 0.00586 x1/ 2 .
<
<
The results for the two fluids are shown below.
0.03
Air
deltat (m)
0.02
0.01
Engine oil
0
0
0.5
1
1.5
2
2.5
x (m)
3
3.5
4
4.5
<
Continued…
PROBLEM 6.32 (Cont.)
The two fluids are subjected to the same temperature difference between the surface and the free stream.
Since the thermal boundary layer thickness is the distance over which the temperature varies from the
surface temperature to the free stream temperature, the fluid with the smaller value of δt must have a
larger temperature gradient, −∂T/∂yy = 0.
Therefore, engine oil has the larger temperature gradient at the surface.
<
The local Nusselt number is given by Nu = hx/k, where h is defined in Equation 6.5. Therefore,
Nu =
−∂T / ∂y y =0
Ts − T∞
x
At a given x location, since Ts – T∞ is the same for both fluids, the fluid with the larger temperature
gradient has the larger local Nusselt number.
<
Engine oil has the larger local Nusselt number.
The heat transfer coefficient is given by Equation 6.5:
h=
− k f ∂T / ∂y
y =0
Ts − T∞
Since engine oil has a larger temperature gradient and a larger thermal conductivity, it is associated with a
larger heat transfer coefficient.
<
COMMENTS: (1) Since the kinematic viscosity of the two fluids is nearly the same, their local
Reynolds numbers, transition locations, and velocity boundary layer thicknesses are comparable. (2) The
much higher Prandtl number of the engine oil results in a much thinner thermal boundary layer and
consequently a larger temperature gradient at the surface and higher heat transfer coefficient.
PROBLEM 6.33
KNOWN: Expression for the local heat transfer coefficient of air at prescribed velocity and temperature
flowing over electronic elements on a circuit board and heat dissipation rate for a 4 × 4 mm chip located
120mm from the leading edge.
FIND: Surface temperature of the chip surface, Ts.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Power dissipated within chip is lost by convection
across the upper surface only, (3) Chip surface is isothermal, (4) The average heat transfer coefficient for
the chip surface is equivalent to the local value at x = L, (5) Negligible radiation.
PROPERTIES: Table A-4, Air (assume Ts = 45°C, Tf = (45 + 25)/2 = 35°C = 308K, 1atm): ν = 16.69 ×
-6 2
-3
10 m /s, k = 26.9 × 10 W/m⋅K, Pr = 0.703.
ANALYSIS: From an energy balance on the chip (see above),
q conv
= E=
g 30W.
(1)
Newton’s law of cooling for the upper chip surface can be written as
T
=
s T∞ + q conv / h A chip
(2)
( )
where A chip =  2 . Assume that the average heat transfer coefficient h over the chip surface is
equivalent to the local coefficient evaluated at x = L. That is, h chip ≈ h x ( L ) where the local coefficient
can be evaluated from the special correlation for this situation,
0.85
hxx
 Vx 
Nu
=
= 0.04 
Pr1/ 3
x

ν 
k
and substituting numerical values with x = L, find
0.85
k  VL 
h x = 0.04 
Pr1/ 3

L ν 
 0.0269 W/m ⋅ K   10 m/s × 0.120 m 
=
h x 0.04 

 
0.120 m

16.69 ×10-6 m 2 / s 
0.85
=
( 0.703
)1/ 3
107 W/m 2 ⋅ K.
The surface temperature of the chip is from Eq. (2),
2
-3
2


T=
42.5 C.
s 25 C + 30 × 10 W/ 107 W/m ⋅ K × ( 0.004m ) =



<
COMMENTS: (1) Note that the estimated value for Tf used to evaluate the air properties was
reasonable. (2) Alternatively, we could have evaluated h chip by performing the integration of the local
value, h(x).
PROBLEM 6.34
KNOWN: Expression for the local heat transfer coefficient of air at prescribed velocity and temperature
flowing over electronic elements on a circuit board and heat dissipation rate for a 4 × 4 mm chip located
120 mm from the leading edge. Atmospheric pressure in Mexico City.
FIND: (a) Surface temperature of chip, (b) Air velocity required for chip temperature to be the same at
sea level.
SCHEMATIC:
p = 76.5 kPa
ASSUMPTIONS: (1) Steady-state conditions, (2) Power dissipated in chip is lost bey convection across
the upper surface only, (3) Chip surface is isothermal, (4) The average heat transfer coefficient for the
chip surface is equivalent to the local value at x = L, (5) Negligible radiation, (6) Ideal gas behavior.
PROPERTIES: Table A.4, air (p = 1 atm, assume Ts = 45 °C, Tf = (45 °C + 25 °C)/2 = 35 °C): k =
0.0269 W/m⋅K, ν = 16.69 × 10-6 m2/s, Pr = 0.706.
ANALYSIS:
(a) From an energy balance on the chip (see above),
q conv
= E=
g 30W.
(1)
Newton’s law of cooling for the upper chip surface can be written as
T
=
s T∞ + q conv / h A chip
(2)
where A chip =  2 . From Assumption 4, h chip ≈ h x ( L ) where the local coefficient can be evaluated
from the correlation provided in Problem 6.33.
0.85
hxx
 Vx 
Nu
=
=
0.04
Pr1/ 3
x


k
ν 
The kinematic viscosity is
μ
ν=
ρ
while for an ideal gas,
p
ρ=
RT
(3)
(4)
(5)
Combining Equations 4 and 5 yields
ν ∝ p-1
(6)
Continued…
PROBLEM 6.34 (Cont.)
The Prandtl number is
ν
μρc
μc
Pr =
=
=
α
ρk
k
(7)
which is independent of pressure.
Therefore, at sea level (p = 1 atm)
k = 0.0269 W/m ⋅ K, ν = 16.69 × 10-6 m 2 /s, Pr = 0.706
h x = 0.04
k  VL 
L  ν 
0.85
Pr1/3
 0.0269 W/m ⋅ K   10 m/s × 0.120 m 
h x = 0.04 


0.120 m

 16.69 × 10-6 m 2 /s 
-3
Ts = 25°C + 30 × 10 W
0.85
(0.706)1/3 = 107 W/m 2 ⋅ K
107 W/m 2 ⋅ K × (0.004 m) 2
= 42.5°C
In Mexico City (p = 76.5 kPa)
101.3lPa 
-6
2
ν = 16.69 × 10-6 m 2 /s × 
 = 22.10 × 10 m /s
 76.5kPa 
k = 0.0269 W/m ⋅ K, Pr = 0.706
 0.0269 W/m ⋅ K   10 m/s × 0.120 m 
h x = 0.04 


0.120 m

  22.10 × 10-6 m 2 /s 
-3
Ts = 25°C + 30 × 10 W
0.85
84.5 W/m 2 ⋅ K × (0.004 m) 2
(0.706)1/3 = 84.5 W/m 2 ⋅ K
<
= 47.2°C
(b) For the same chip temperature, it is required that hx = 107 W/m2∙K. Therefore
 0.0269 W/m×K   V × 0.120 m 
h x = 107 W/m ⋅ K = 0.04 


0.120 m

  22.10 × 10-6 m 2 /s 
2
From which we may find V = 13.2 m/s
0.85
(0.706)1/3
<
COMMENTS: (1) In Part (a), the chip surface temperature increased from 42.4 °C to 47.2 °C. This is
considered to be significant and the electronics packaging engineer needs to consider the effect of large
changes in atmospheric pressure on the efficacy of the electronics cooling scheme. (2) Careful
consideration needs to be given to the effect changes in the atmospheric pressure on the kinematic
viscosity and, in turn, on changes in transition lengths which might affect local convective heat transfer
coefficients.
PROBLEM 6.35
KNOWN: Location and dimensions of computer chip on a circuit board. Form of the convection
correlation. Maximum allowable chip temperature and surface emissivity. Temperature of cooling air
and surroundings.
FIND: Effect of air velocity on maximum power dissipation, first without and then with consideration of
radiation effects.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible temperature variations in chip, (3) Heat transfer
exclusively from the top surface of the chip, (4) The local heat transfer coefficient at x = L provides a
good approximation to the average heat transfer coefficient for the chip surface.
( T∞ + Tc )
PROPERTIES: Table A.4, air =
(T
Pr = 0.703.
2 = 328 K): ν = 18.71 × 10-6 m2/s, k = 0.0284 W/m⋅K,
ANALYSIS: Performing an energy balance for a control surface about the chip, we obtain Pc = qconv +
(
)
2
2
εs ( Tc + Tsur ) Tc + Tsur . With
qrad, where qconv = hA s ( Tc − T∞ ) , qrad = h r A s ( Tc − Tsur ) , and h r =
h ≈ h L , the convection coefficient may be determined from the correlation provided in Problem 6.33
1/3
(NuL = 0.04 Re0.85
L Pr ). Hence,
(
1/ 3
2
Pc  2  0.04 ( k L ) Re0.85
=
( Tc − T∞ ) + es ( Tc + Tsur ) Tc2 + Tsur
L Pr

) ( Tc − Tsur )
where ReL = VL/ν. Computing the right side of this expression for ε = 0 and ε = 0.85, we obtain the
following results.
0.3
Chip power, Pc(W)
0.25
0.2
0.15
0.1
0.05
0
0
5
10
15
20
25
Velocity, V(m/s)
epsilon = 0.85
epsilon = 0
Since hL increases as V0.85, the chip power must increase with V in the same manner. Radiation exchange
increases Pc by a fixed, but small (6 mW) amount. While hL varies from 14.5 to 223 W/m2⋅K over the
prescribed velocity range, hr = 6.5 W/m2⋅K is a constant, independent of V.
COMMENTS: Alternatively, h could have been evaluated by integrating hx over the range 118 ≤ x ≤
122 mm to obtain the appropriate average. However, the value would be extremely close to hx=L.
PROBLEM 6.36
PROBLEM 6.37
KNOWN: Characteristic length of a microscale chemical detector, free stream velocity and temperature,
hydrogen wind tunnel pressure and free stream velocity.
FIND: Model length scale and hydrogen temperature needed for similarity.
SCHEMATIC:
Air
Hydrogen
Hydrogen
T∞ =T 25°C
=?
T∞ = ?
∞
V = 0.5 m/s
Lm = ?
= 0.5
m/s
V = V10
m/s
Lm = ?
Ls = 80 µm
p = 0.5 atm
p = 0.5 atm
p = 1 atm
Heated model
Heated sensor
model
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible microscale or
nanoscale effects, (4) Ideal gas behavior.
PROPERTIES: Table A.4, air (T = 25 °C): Prs = 0.707, νs = 15.71 × 10-6 m2/s, hydrogen (250 K) Pr =
0.707, ν = 81.4 × 10-6 m2/s.
ANALYSIS: For similarity we require Rem = Res and Prm = Prs. For the sensor,
Vs Ls
10 m/s × 80 × 10-6 m
=
= 50.93
νs
1.571 × 10-5 m 2 /s
Prs = 0.707
Res =
For the model, Prm = Prs = 0.707.
From Table A.4, we note Prs = 0.707, ν = 81.4 ×10-6 m2/s at T∞ = 250 K and p = 1 atm.
<
The value of the Prandtl number is independent of pressure for an ideal gas. The kinematic viscosity is
pressure-dependent. Hence,
μ
μ
ρ(at 1.0 atm)
ν(at 0.5 atm) =
=
×
ρ(at 0.5 atm)
ρ(at 1.0 atm)
ρ(at 0.5 atm)
For an ideal gas,
1.0 atm
ν(at 0.5 atm) = ν(at 1.0 atm)×
= 2ν(at 1.0 atm)
0.5 atm
Therefore,
νm = 81.4 × 10-6 m2/s × 2 = 163 × 10-6 m2/s
For similarity,
Continued…
PROBLEM 6.37 (Cont.)
Rem = Res = 50.93 =
or
Vm L m
0.5 m/s × L m
=
νm
163 × 10-6 m 2 /s
Lm = 16.6 × 10-3 m = 16.6 mm
<
COMMENTS: (1) From Section 2.2.1, we know that the mean free path of air at room conditions is
approximately 80 nm. Since Ls is three orders of magnitude greater than the mean free path, the air may
be treated as a continuum. (2) Hydrogen can leak from enclosures easily. By keeping the wind tunnel
pressure below atmospheric, we avoid possible leakage of flammable hydrogen into the lab. Also, if leaks
occur, air must enter the wind tunnel. It is much easier to seal against air leaks than hydrogen leaks. (3)
Prm = 0.707 at 100 K also. However, the operation of the hydrogen wind tunnel at such a low temperature
would be much more difficult than at 250 K.
PROBLEM 6.38
KNOWN: Drag force and air flow conditions associated with a flat plate.
FIND: Rate of heat transfer from the plate.
SCHEMATIC:
ASSUMPTIONS: (1) Chilton-Colburn analogy is applicable.
3
PROPERTIES: Table A-4, Air (70°C,1 atm): ρ = 1.018 kg/m , cp = 1009 J/kg⋅K, Pr = 0.70, ν
-6 2
= 20.22 × 10 m /s.
ANALYSIS: The rate of heat transfer from the plate is
( ) ( Ts − T∞ )
q 2h L2
=
where h may be obtained from the Chilton-Colburn analogy,
Cf
h
Pr 2 / 3
= St Pr 2/3
=
2
r u ∞ cp
2
Cf 1 t s
1 ( 0.075 N/2 ) / ( 0.2m )
=
=
= 5.76 ×10−4.
2
2
3
2
2 r u ∞ / 2 2 1.018 kg/m ( 40 m/s ) / 2
jH
=
Hence,
C
h = f r u ∞ cp Pr -2/3
2
-2 / 3
h=
5.76 ×10-4 1.018kg/m3 40m/s (1009J/kg ⋅ K ) ( 0.70 )
)
(
h 30 W/m 2 ⋅ K.
=
The heat rate is
(
q=
2 30 W/m 2 ⋅ K
) ( 0.2m )2 (120 − 20) C
<
q = 240 W.
COMMENTS: Although the flow is laminar over the entire surface (ReL = u∞L/ν = 40 m/s ×
-6 2
5
0.2m/20.22 × 10 m /s = 4.0 × 10 ), the pressure gradient is zero and the Chilton-Colburn
analogy is applicable to average, as well as local, surface conditions. Note that the only
contribution to the drag force is made by the surface shear stress.
PROBLEM 6.39
PROBLEM 6.40
PROBLEM 6.41
KNOWN: Dimensions and temperature of a thi n, rough plate. Velocity of air flow parallel to plate (at
an angle of 45° to a side). Heat transfer rate from plate to air.
FIND: Drag force on plate.
SCHEMATIC:
L = 0.2 m
Air
T∞ = 20°C
u∞ = 30 m/s
Ts = 80°C
ASSUMPTIONS: (1) The modified Reynolds analogy holds, (2) Constant properties.
PROPERTIES: Table A-4, Air (50°C = 323 K): cp = 1008 J/kg∙K, Pr = 0.704.
ANALYSIS: The modified Reynolds analogy, Equation 6.70, combined with the definition of the
Stanton number, Equation 6.67, yields
(1)
Cf /2= (Nu/Re)Pr -1/3
The drag force is related to the friction coefficient according to
FD = τs As = Cf ⋅ ρu 2∞ As / 2
(2)
Combining Equations (1) and (2)
Nu -1/3 2
FD =
Pr ρu ∞ As
Re
Substituting the definitions of Nu and Re, we find
hLc ν
h ν -1/3
h
FD =
Pr -1/3ρu 2∞ As =
Pr u ∞ As = Pr 2/3u ∞ As
k u ∞ Lc
cp α
cp
Where Lc is a characteristic length used to define Nu and Re. With hAs = q/ ΔT we have
FD =
qu ∞ Pr 2/3
2000 W × 30 m/s × (0.704) 2/3
= 0.785 N
=
c pΔT
1008 J/kg ⋅ K × 60 K
<
COMMENTS: (1) Heat transfer or friction coefficient correlations for this simple configuration
apparently do not exist. (2) Experiments to measure the drag force would be relatively simple to
implement and measured drag forces could be used to determine the heat transfer coefficients using the
Reynolds analogy. (3) The solution demonstrates advantages associated with working the problem
symbolically and only introducing numbers at the end. First, the length scale in Nu and Re did not have
to be defined because it cancelled out. Second, the properties k, ν, and ρ also cancelled out.
PROBLEM 6.42
KNOWN: Nominal operating conditions of aircraft and characteristic length and average friction
coefficient of wing.
FIND: Average heat flux needed to maintain prescribed surface temperature of wing.
SCHEMATIC:
ASSUMPTIONS: (1) Applicability of modified Reynolds analogy, (2) Constant properties.
-6
2
PROPERTIES: Prescribed, Air: ν = 16.3 × 10 m /s, k = 0.022 W/m⋅K, Pr = 0.72.
ANALYSIS: The average heat flux that must be maintained over the surface of the air foil is
q ′′ h ( Ts − T∞ ) , where the average convection coefficient may be obtained from the modified Reynolds
=
analogy.
Nu L
Nu L
Cf
=
St=
Pr 2 / 3
=
Pr 2 / 3
2
ReL Pr
ReL Pr1/ 3
−6 2
7
Hence, with Re L =
VL / ν =
100 m / s ( 2m ) / 16.3 × 10 m / s =
1.23 × 10 ,
(
)
0.0025
1/ 3
Nu L =
1.23 ×107 ( 0.72 )
=
13, 780
2
=
h
k
0.022 W / m ⋅ K
=
Nu L
=
780 ) 152 W / m 2 ⋅ K
(13,
L
2m
=
q′′ 152 W / m 2 ⋅ K 5 − ( −
=
23)  °C 4260 W / m 2
<
COMMENTS: If the flow is turbulent over the entire airfoil, the modified Reynolds analogy provides a
good measure of the relationship between surface friction and heat transfer. The relation becomes more
approximate with increasing laminar boundary layer development on the surface and increasing values of
the magnitude of the pressure gradient.
PROBLEM 6.43
PROBLEM 6.44
PROBLEM 6.45
KNOWN: Evaporation rate from pan of water of prescribed diameter. Water temperature. Air
temperature and relative humidity.
FIND: (a) Convection mass transfer coefficient, (b) Evaporation rate for increased relative humidity, (c)
Evaporation rate for increased temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Water vapor is saturated at liquid interface and may be approximated as a perfect
gas.
-1 (49.4 m3/kg)-1 = 0.0202
PROPERTIES: Table A-6, Saturated water vapor (Ts = 296K): ρ A,sat
= v=
g
-1
-1 13.98 m3 / kg =
kg/m3; (Ts = 320 K): ρ A,sat
= v=
0.0715 kg/m3.
g
)
(
ANALYSIS: (a) Since evaporation is a convection mass transfer process, the rate equation has the form
(
)
 evap h m A ρ A,s − ρ A,∞ and the mass transfer coefficient is
=
m
=
hm
 evap
m
=
p D2 / 4 ρ A,s − ρ A,∞
)(
(
)
1.5 ×10−5 kg/s
= 0.0179 m/s
(p /4 )( 0.23 m )2 0.0202 kg/m3
<
with Ts = T∞ = 23°C and φ∞ = 0.
(b) If the relative humidity of the ambient air is increased to 50%, the ratio of the evaporation rates is
 evap (φ∞ = 0.5 )
m
 evap (φ∞ = 0 )
m
h m A  ρ A,sat ( Ts ) − φ∞ ρ A,sat ( T∞ ) 
=
h m A ρ A,sat ( Ts )
= 1 − φ∞
ρ A,sat ( T∞ )
.
ρ A,sat ( Ts )

0.0202 kg/m3 
 evap (φ∞ =
Hence, m
0.5 ) =
1.5 ×10−5 kg/s 1 − 0.5
0.75 ×10−5 kg/s.
=
3

0.0202 kg/m 
<
(c) If the temperature of the ambient air is increased from 23°C to 47°C, with φ∞ = 0 for both cases, the
ratio of the evaporation rates is
aaa
 evap T
m
=
h m Aρ A,sat 47 C
ρ A,sat 47 C
s T=
∞ 47 C
)
(
( )
)
(
( )
0.0715 kg/m
T =
47 C ) =
1.5 × 10 kg/s
(T =
0.0202 kg/m
(
(
)
)
.
= =
aaa
 evap T
m
=
T
=
23
C
h
A
ρ
23
C
ρ
23
C
s
∞
m A,sat
A,sat
 evap
Hence, m
s
∞
a
−5
3
3
=
5.31 × 10−5 kg/s.
<
COMMENTS: Note the highly nonlinear dependence of the evaporation rate on the water temperature.
 evap increases by 350%.
For a 24°C rise in Ts , m
PROBLEM 6.46
KNOWN: Number, diameter, and length of cylindrical rollers. Temperature and evaporation rate of
water film on 135° arc of cylinders exposed to high-velocity air. Air temperature and relative humidity.
FIND: Area-averaged mass transfer coefficient for exposed roller surfaces.
SCHEMATIC:
Paper approaching roller
Paper leaving roller
Water (A) film
Ts = 90°C
T∞=35°C
Φ∞= 0.95
135°
+
N = 30 rollers, L = 3 m
nA = 80 kg/s
D = 1.5 m
Hot roller
ASSUMPTIONS: (1) Steady-state, (2) Uniform properties, (3) Water vapor is saturated at liquid
interface.
−1
PROPERTIES: Table A-6, Saturated water vapor (363 K), ρ=
υ=
0.4193 kg/m3 ; Saturated
g
g
−1
water vapor (308 K), ρ=
υ=
0.03898 kg/m3 .
g
g
ANALYSIS: The mass transfer coefficient can be found from the known evaporation rate, area, and
properties according to Equation 6.19:
= n=
m evap
hm As ( ρ A, s − ρ A,∞ )
A
=
ρ A, s ρ=
0.4193 kg/m3 , ρ A,∞ =
φ∞ ρ A,sat (T∞ ) =
0.95 × 0.03898 kg/m3 =
0.0370 kg/m3 , and
Here
A,sat (Ts )
A=
N π DL(135° / 360°=
) 30π × 1.5 m × 3 m(135° / 360°=
) 159 m 2 . Thus,
s
hm
=
m evap
80 kg/s
=
= 1.32 m/s
2
As ( ρ A, s − ρ A,∞ ) 159 m (0.4193 − 0.0370) kg/m3
<
COMMENTS: (1) The freestream air is nearly saturated. Care must be taken to avoid saturated
conditions (and, in turn, possible condensation of water vapor) by forcibly expelling the moist air from
the plant. (2) Evaporation from the paper that is in contact with the roller occurs at a high rate, and is
driven by a complex heat transfer process involving conduction in the hot roller, a contact resistance at
the roller-paper interface, and conduction in the moist paper.
PROBLEM 6.47
PROBLEM 6.48
KNOWN: Species concentration profile, CA(y), in a boundary layer at a particular location for
flow over a surface.
FIND: Expression for the mass transfer coefficient, hm, in terms of the profile constants, CA,∞
and DAB. Expression for the molar convection flux, N A
′′ .
SCHEMATIC:
ASSUMPTIONS: (1) Parameters D, E, and F are constants at any location x, (2) DAB, the mass
diffusion coefficient of A through B, is known.
ANALYSIS: The convection mass transfer coefficient is defined in terms of the concentration
gradient at the wall,
h m ( x ) = −D AB
∂ CA / ∂ y y=0
( CA,s − CA,∞ )
.
The gradient at the surface follows from the profile, CA(y),
)
(
∂ CA
∂
= Dy 2 + Ey + F
=
+ E.
∂ y y=0 ∂ y
y=0
Hence,
−D ABE
D ABE
−
=
hm ( x ) =
.
CA,s − CA,∞
F − CA,∞
(
) (
<
)
The molar flux follows from the rate equation,
(
)
′′A h m CA,s − CA,∞=
N=
N′′A = −D ABE.
−D ABE
( CA,s − CA,∞ )
(
)
⋅ CA,s − CA,∞ .
<
COMMENTS: It is important to recognize that the influence of species B is present in the
property DAB. Otherwise, all the parameters relate to species A.
PROBLEM 6.49
PROBLEM 6.50
PROBLEM 6.50 (Cont.)
PROBLEM 6.51
PROBLEM 6.52
KNOWN: Average convection mass transfer coefficient for air flow over a surface coated with species
A (water, naphthalene, or acetone). Characteristic length. Free stream velocity. Air temperature.
FIND: Values of Sc, Sh L , and Re L for each case. Values of C, m, and n for Sherwood number
correlation. Area-averaged mass transfer coefficient for evaporation of benzene from the same object.
SCHEMATIC:
(A)
—
Water (hm=1 2× 1 0- 3m/s)
—
Naphthalene (hm=3. 6× 1 0- 3m/s)
—
Acetone (hm=8× 1 0- 3m/s)
(B)
Air
T=300K
V=2m/s (Water)
=1 . 5m/s (Naphthalene)
=2. 5m/s (Acetone)
=3m/s (Benzene)
ASSUMPTIONS: (1) Steady-state, (2) Uniform properties, (3) Species A is saturated at the air-coating
interface, (4) Concentration of species A is small.
PROPERTIES: For the fluids at 300 K, using the temperature correction DAB ~ T3/2:
ν(m2/s)×10-6
Table Fluid(s)
A-4
A-8
A-8
A-8
A-8
Air
Water vapor – Air
Naphthalene - Air
Acetone - Air
Benzene – Air
DAB(m2/s)
15.89
0.263 × 10-4
0.620 × 10-5
0.127 × 10-4
0.889 × 10-5
ANALYSIS: Water (νapor) - Air:
=
Sh
L
=
Re
L
h m L ( 0.012 m / s )1 m
=
= 457
D AB 0.263 × 10−4 m 2 / s
VL
=
( 2 m / s )1 m
<
= 1.26 × 105
ν air 15.89 × 10−6 m 2 / s
ν air 15.89 × 10−6 m 2 / s
=
Sc =
= 0.605
D AB 0.263 × 10−4 m 2 / s
Naphthalene (νapor) - Air:
Sh L =
581 Re L =×
9.44 104
Sc =
2.56
<
Sh L =
631 Re L =
1.57 × 105
Sc =
1.25
<
Acetone (vapor) - air:
Continued…
PROBLEM 6.52 (Cont.)
The coefficients C, m, and n can be found by solving the three simultaneous equations:
m
n
m
n
n
=
Sh L,water CRe
=
Sh L,naph CRe
=
Sh L,ace CRe mL,aceScace
L,water Sc water ,
L,naph Sc naph ,
Solving these using IHT yields:
C = 0.829, m = 0.549, n = 0.275
<
For benzene evaporating in air flowing at 3 m/s,
( 3 m / s )1 m
ν air 15.89 × 10−6 m 2 / s
VL
Re L = =
=
1.89 × 105 , Sc =
=
=
1.79
−6
2
ν air 15.89 × 10 m / s
D AB 0.889 × 10−5 m 2 / s
h m,water =
(CRe mL,water Sc nwater )D AB
L
0.829 × (1.89 × 105 )0.549 (1.79)0.275 × 0.889 × 10−5 m 2 / s
=
= 6.8 × 10−3 m/s
1m
<
COMMENTS: Note that ν should be evaluated for the mixture of A and B. It is assumed that species A
is present in small concentration, so that ν is associated solely with the freestream fluid B.
PROBLEM 6.53
KNOWN: Convection heat transfer correlation for flow over a contoured surface.
FIND: (a) Evaporation rate from a water film on the surface, (b) Steady-state film temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (b) Constant properties, (c) Negligible radiation, (d) Heat
and mass transfer analogy is applicable.
-6 2
PROPERTIES: Table A-4, Air (300K, 1 atm): k = 0.0263 W/m⋅K, ν = 15.89×10 m /s, Pr = 0.707;
3
Table A-6, Water (Ts ≈ 280K): vg = 130.4 m /kg, hfg = 2485 kJ/kg; Table A-8, Water-air (T ≈ 298K):
-4 2
DAB = 0.26×10 m /s.
ANALYSIS: (a) The mass evaporation rate is
 evap =
m
nA =
h m A  ρ A,sat ( Ts ) − φ∞ ρ A,sat ( T∞ )  =
h m A ρ A,sat ( Ts ) .
From the heat and mass transfer analogy:
(10 m/s ) 1m =
VL
=
ReL =
6.29 ×105
-6
2
ν
15.89 ×10 m / s
(
Sh L =
0.43 6.29 ×105
)
0.58
Sh L = 0.43 Re0.58
Sc0.4
L
ν
15.89 ×10-6 m 2 / s
=
Sc ==
0.61
D AB
26 ×10-6 m 2 / s
814
( 0.61)0.4 =
D AB
0.26 ×10-4 m 2 / s
=
=
hm =
Sh L
(814 ) 0.0212 m/s
L
1m
-1
ρ A,sat
=
0.0077 kg/m3.
( Ts ) v=
g ( Ts )
Hence,
 evap =
m
0.0212m/s × 1m 2 × 0.0077kg/m3 =
1.63 ×10−4 kg/s.
<
(b) From a surface energy balance, q′′conv = q′′evap , or
 ′′evap h fg
h L ( T∞ − Ts ) = m
With
(
Nu L =
0.43 6.29 ×105
=
hL
Ts = T∞ −
)
0.58
( m ′′evaphfg ) .
hL
864
( 0.707 )0.4 =
k
0.0263 W/m ⋅ K
=
=
Nu L
864 22.7 W/m 2 ⋅ K.
L
1m
(
)
1.63 ×10-4 kg/s ⋅ m 2 2.485 ×106 J/kg
Hence,
<
=
Ts =
300K 282.2K.
22.7 W/m 2 ⋅ K
COMMENTS: The saturated vapor density, ρA,sat, is strongly temperature dependent, and if the initial
guess of Ts needed for its evaluation differed from the above result by more than a few degrees, the
density would have to be evaluated at the new temperature and the calculations repeated.
PROBLEM 6.54
KNOWN: Dimensions of rectangular naphthalene rod. Velocity and temperature of air flow. Molecular
weight and saturation pressure of naphthalene.
FIND: Mass loss after 30 minutes.
SCHEMATIC:
pA,sat
ASSUMPTIONS: (1) Constant properties, (2) Mass loss is small, so dimensions remain unchanged, (3)
Viscosity of air-naphthalene mixture is approximately that of air.
PROPERTIES: Table A-4, Air (300 K): ν = 15.89 × 10-6 m2/s. Table A-8, Naphthalene in air, (300 K):
DAB = 0.62 × 10-5 m2/s, Sc = ν/D AB = 2.56.
ANALYSIS: We will use the heat and mass transfer analogy, with the Nusselt number correlation known
from Problem 6.9 to be of the form
Nu d = CRedm Pr1/3
Then invoking Equation 6.59,
m 1/ 3
=
Sh d CRe
=
h md / D AB
d Sc
Now Red = Vd/ν = 10 m/s × 0.03 m/15.89×10-6 m 2 /s = 18,880 . We find the values of C and m from
Problem 6.9 with c/d = 0.33, for the front, sides, and back of the rod:
C
0.674
0.153
0.174
front
sides
back
m
1/2
2/3
2/3
Shd
126.7
148.5
168.8
hm(m/s)
0.0262
0.0307
0.0349
The average mass transfer coefficient is
h m = (h m,front d + 2h m,sidec + h m,back d)/(2d + 2c)
0.0262 m / s × 0.03 m + 2 × 0.0307 m / s × 0.01 m + 0.0349 m / s × 0.03 m
2 × 0.03 m + 2 × 0.01 m
= 0.0306 m/s
=
Then the mass loss can be found from
Δm = n A Δt = h m A tot (ρ A,s - ρ A,∞ )Δt
Continued…
PROBLEM 6.54 (Cont.)
Here ρA,∞ = 0 and ρA,s can be found from the saturation pressure, using the ideal gas law:
ρ A,s =
p A,sat
R iTs
=
p A,satM A
RTs
-4
=
1.33 × 10 bar × 128.16 kg/kmol
8.314 × 10-2 m3 ⋅ bar/kmol ⋅ K × 300 K
= 6.83 × 10-4 kg/m3
Thus, finally,
Δm = 0.0306 m/s × (2 × 0.03 m + 2 × 0.01 m) × 0.5 m
× (6.83 × 10-4 - 0) kg/m3 × 30 min × 60 s/min
= 1.50 × 10-3 kg
<
COMMENTS: The average depth of surface recession is given by δ = h m (ρ A,s - ρ A,∞ )Δt/ρ A,sol where
ρ A,sol is the density of solid naphthalene, ρ A,sol = 1025 kg/m3. Thus δ = 37 μm and the assumption that
the dimensions remain unchanged is good.
PROBLEM 6.55
PROBLEM 6.56
PROBLEM 6.57
PROBLEM 6.58
PROBLEM 6.59
KNOWN: Thin layer of water on concrete surface experiences evaporation, convection with ambient
air, and radiation exchange with the sky. Average convection coefficient estimated as h = 53 W/m2⋅K.
FIND: (a) Heat fluxes associated with convection, q′′conv , evaporation, q′′evap , and radiation exchange
with the sky, q′′rad , (b) Use results to explain why the concrete is wet instead of dry, and (c) Direction
of heat flow and the heat flux by conduction into or out of the concrete.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor behaves as an ideal gas, (3) Constant
properties, (4) Water surface is small compare to large, isothermal surroundings (sky), and (4)
Applicability of the heat-mass transfer analogy.
PROPERTIES: Table A.4, Air (Tf = (T∞ + Ts)/2 = 282.5 K): ρ = 1.243 kg/m3, cp = 1007 J/kg⋅K, α =
2.019 ×105 m2/s; Table A.8, Water-air (Tf = 282.5 K): DAB = 0.26 × 10-4 m2/s (282.5/298)3/2 = 0.24 ×
10-4 m2/s; Table A.6, Water (Ts = 275 K): ρA,s = ρg = 1/νg = 1/181.7 = 0.0055 kg/m3, hfg = 2497 kJ/kg;
Table A.6, Water ( T∞ = 290 K): ρA,s = 1/69.7 = 0.0143 kg/m3.
ANALYSIS: (a) The heat fluxes associated with the processes shown on the schematic are
Convection:
+795 W m 2
q′′conv =
h ( T∞ − Ts ) =
53 W m 2 ⋅ K ( 290 − 275 ) K =
Radiation Exchange:
(
)
(
)
<
4
=
+131W m 2
εs Ts4 − Tsky
q′′rad =
0.96 × 5.76 × 10−8 W m 2 ⋅ K 4 2754 − 2404 K 4 =
<
Evaporation:
q′′evap =
n ′′A h fg =
−2.255 ×10−4 kg s ⋅ m 2 × 2497 ×103 J kg =
−563.1W m 2
<
where the evaporation rate from the surface is
(
)
n ′′A =
h m ρ A,s − ρ A, ∞ =
0.050 m s ( 0.0055 − 0.7 × 0.0143) kg m3 =
−2.255 × 10−4 kg s ⋅ m 2
Continued...
PROBLEM 6.59 (Cont.)
and where the mass transfer coefficient is evaluated from the heat-mass transfer analogy, Eq. 6.60, with n
= 1/3,
 α 
h
= ρ cp Le2 / 3 = ρ cp 

hm
 DAB 
2/3
 2.019 ×10−5 

= 1.243kg m3 ×1007 J kg ⋅ K 
 0.26 ×10−4 


2/3
h
= 1058 J m3 ⋅ K
hm
3 ⋅ K 0.050 m s
=
h m 53 W m 2 ⋅ K 1058 J m
=
(b) From the foregoing evaporation calculations, note that water vapor from the air is condensing on the
liquid water layer. That is, vapor is being transported to the surface, explaining why the concrete surface
is wet, even without rain.
(c) From an overall energy balance on the water film considering conduction in the concrete as shown in
the schematic,
E in − E out =
0
q′′conv − q′′evap − q′′rad − q′′cond =
0
q′′cond = q′′conv − q′′evap − q′′rad
(
)(
)
=
q′′cond 1795 W m3 − −563.1W m 2 − +131W m 2 = 1227 W m 2
The heat flux by conduction is into the concrete.
<
PROBLEM 6.60
PROBLEM 6.61
KNOWN: Dry air at 32°C flows over a wetted plate of width 1 m maintained at a surface temperature of
27°C by an embedded heater supplying 432 W.
FIND: (a) The evaporation rate of water from the plate, nA (kg/h) and (b) The plate temperature Ts when
all the water is evaporated, but the heater power remains the same.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor behaves as an ideal gas, (3) Constant
properties, and (4) Applicability of the heat-mass transfer analogy.
PROPERTIES: Table A.4, Air (Tf = (32 + 27)°C/2 = 302.5 K): ρ = 1.153 kg/m3, cp = 1007 J/kg⋅K, α =
2.287 ×105 m2/s; k = 0.02648 W/m⋅K; ν = 1.614 × 10-5 m2/s; Pr = 0.704. Air (Tf = (86 + 27)°C/2 = 332
K); k = 0.02867 W/m⋅K; ν = 1.911 × 10-5 m2/s; Pr = 0.703, Table A.8, Water-air (Tf ≈ 300 K): DAB = 0.26
× 10-4 m2/s; Table A.6, Water (Ts = 27°C = 300 K): ρA,s = 1/νg = 1/39.13 = 0.0256 kg/m3, hfg = 2438
kJ/kg.
ANALYSIS: (a) Perform an energy balance on the wetted plate to obtain the evaporation rate, nA.
E in −
=
E out 0
Pe + q conv −=
q evap 0
Pe + hAs ( T∞ − Ts ) − n A h fg =
0
(1)
In order to find h , invoke the heat-mass transfer analogy, Eq. (6.60) with n = 1/3,
h
hm
= ρ c p Le
2/3
 α 
= ρ cp 

 D AB 
The evaporation rate equation
(
=
n A h m As ρ A,s − ρ A,∞
2/3
 2.287 × 10−5 
= 1.153 kg m × 1007 J kg ⋅ K 

 0.26 × 10−4 


2/3
3
= 1066 J m ⋅ K (2)
)
Substituting Eqs. (2) and (3) into Eq. (1), find h m
)
(
3
(3)
(
)
Pe + 1066 J m3 ⋅ K h m As ( T∞ − Ts ) − h m As ρ A,s − ρ A,∞ h fg =
0
(4)
432 W + 1066 J m3 ⋅ K ( 32 − 27 ) K − ( 0.0256 − 0 ) kg m3 × 2438 × 103 J kg  ( 0.200 × 1) m 2 ⋅ h m =
0


432 + [5330 - 62,413] × 0.20 h m = 0
h m = 0.0378 m/s
Using Eq. (3), find
nA =
0.0378 m s ( 0.200 ×1) m 2 ( 0.0256 − 0 ) kg m3 =
1.94 ×10−4 kg s =
0.70 kg h
(b) When the plate is dry, all the power must be removed by convection,
Pe = qconv = h d As(Ts - T∞ )
<
(5)
1/3
where h d is the convection heat transfer coefficient under dry conditions. Since Nu L = C Re1/2
L Pr ,
Continued…
PROBLEM 6.61 (Cont.)
k
hd = h  d
 k
1/2
 ν 
 ν 
 d 
1/3
 Prd 
 Pr 


(6)
Combining Eqs. (2), (5), and (6) yields
432 W
Ts = 32°C +
1/2
0.2 m 2 × h m ×1066
kd

  1.614 ×10−5 m 2 / s 

×
×

νd
m3 ⋅ K  0.02648 W/m ⋅ K  

J
1/3
 Pr 
× d 
 0.707 
which may be solved iteratively or with IHT to yield
Ts= 86°C
<
PROBLEM 6.62
PROBLEM 6.63
PROBLEM 6.64
KNOWN: Initial temperature and droplet diameter of water mist. Expression for Nusselt number.
Temperature and relative humidity of air stream.
FIND: (a) Initial convection heat transfer rate, evaporative heat loss rate, and rate of change of droplet
temperature, (b) Steady-state droplet temperature.
SCHEMATIC:
Ti = 10°C
Air
T∞ = 32°C
f∞ = 0.20 or 0.95
D = 100 µm
Water
ASSUMPTIONS: (1) The properties of the air-water vapor mixture can be approximated by pure air
properties, (2) Air properties and DAB evaluated at Tam = (Ts + T∞)/2.
PROPERTIES: Table A-4, Air (T = 294 K): k = 0.0258 W/m⋅K; Table A-8, H2O in Air (T = 294 K):
DAB = 0.26 × 10-4 m2/s (294/298)3/2 = 0.255 × 10-4 m2/s; Table A-6, Water (T = 283 K): ρA,sat(Ts) = 1/111.8
m3/kg = 0.00894 kg/m3, hfg = 2478 kJ/kg, ρw = 1000 kg/m3, cp,w = 4193 J/kg⋅K; (T = 305 K): ρA,sat(T∞) =
1/29.74 m3/kg = 0.0336 kg/m3.
ANALYSIS: (a) The convection heat transfer coefficient is
h=
Nu D k / D =
2k / D =
2 × 0.0258 W/m ⋅ K / 100 × 10−6 m =
516 W/m 2 ⋅ K
At the initial time, when the droplet temperature is Ti = 10°C, the convection heat transfer rate to the
droplet is given by
qconv= hAs (T∞ − Ti )= 516 W/m 2 ⋅ K × π (100 × 10−6 m) 2 × (32 − 10)°C= 357 × 10−6 W
<
Referring to Eq. 6.59, since Nu is independent of Pr, n = 0, and we have
Nu ==
Sh 2
or
hm =
2 DAB / D =
2 × 0.255 × 10−4 m 2 /s / 100 × 10−6 m =
0.510 m/s
Thus the rate of evaporative heat loss can be expressed as
=
qevap m=
hm As ( ρ A,s − ρ A,∞ )h fg
evap h fg
For φ∞ = 0.20, ρA,∞ = φ∞ρA,sat(T∞) = 0.2 × 0.0336 kg/m3 = 0.00672 kg/m3. Thus,
=
qevap m=
hm As ( ρ A,s − ρ A,∞ )h fg
evap h fg
= 0.510 m/s × p (100 × 10−6 m) 2 × (0.00894 − 0.00672) kg/m3 × 2478 × 103 J/kg
= 88.1 × 10−6 W
<
Continued…
PROBLEM 6.64 (Cont.)
Similarly, for φ∞ = 0.95, ρA,∞ = φ∞ρA,sat(T∞) = 0.95 × 0.0336 kg/m3 = 0.0319 kg/m3
−912 × 10−6 W
qevap =
<
The rate of change of temperature of the droplet can be found from an energy balance on the droplet:
(ρc V )
p
For φ∞ = 0.20,
w
dT
= qconv − qevap
dt
dT qconv − qevap
(357 − 88.1) × 10−6 W
= =
= 122 K/s
dt
( ρ c pV )w 1000 kg/m3 × 4193 J/kg ⋅ K × p × (100 × 10−6 m)3 / 6
<
(357 + 912) × 10−6 W
dT qconv − qevap
= =
= 578 K/s
dt
( ρ c pV )w 1000 kg/m3 × 4193 J/kg ⋅ K × p × (100 × 10−6 m)3 / 6
<
And for φ∞ = 0.95,
For φ∞ = 0.20, the evaporative heat loss is positive; the droplet is evaporating. Convection from the warm
air is warming the droplet, while evaporation is cooling it. The net effect is to warm the droplet. For φ∞ =
0.95, the evaporative heat loss is negative; water from the humid environment is condensing on the
droplet. Both convection and condensation cause warming of the droplet, so the rate of temperature
increase is higher in this case. Eventually, the droplet will become warm enough so that condensation
<
ceases and evaporation will begin.
(b) At steady-state, an energy balance requires qconv = qevap. Eq. 6.64 applies, with hm / h = DAB / k .
Thus, the following implicit equation must be solved for Ts:
T∞ − Ts h fg
=
DAB
[ ρ A,sat (Ts ) − ρ A,∞ ]
k
where the properties DAB and k should be evaluated at Tam = (Ts + T∞)/2, and hfg should be evaluated at Ts.
This equation is most easily solved in IHT, which yields
Ts = 289 K for φ∞ = 0.20
Ts = 304 K for φ∞ = 0.95
<
<
COMMENTS: (1) As expected, the steady-state droplet temperature is lower for the low humidity
conditions. (2) The steady-state temperature does not depend on the droplet diameter because the areas for
convection and evaporation are the same and the heat and mass transfer coefficients have the same
dependence on diameter. (3) Based on the initial rate of temperature increase, it appears that the droplet
will take less than 1 s to reach its steady-state temperature. A complete analysis of the transient problem
shows that the steady-state temperature is reached in approximately 0.17 and 0.15 s for the φ∞ = 0.20 and
0.95 cases, respectively. In this time period, the change in diameter of the droplet is less than 1% in both
cases. Of course, the droplet will eventually evaporate.
PROBLEM 6.65
PROBLEM 6.65 (Cont.)
PROBLEM 6.66
PROBLEM 6.66 (Cont.)
PROBLEM 6.67
PROBLEM 6.67 (Cont.)
PROBLEM 6.68
KNOWN: Initial plate temperature Tp (0) and saturated air temperature (T∞) in a dishwasher at the start
2
of the dry cycle. Thermal mass per unit area of the plate Mc/As = 1600 J/m ⋅K.
FIND: (a) Differential equation to predict plate temperature as a function of time during the dry cycle
and (b) Rate of change in plate temperature
at the start of the dry cycle assuming the average convection
heat transfer coefficient is 3.5 W/m2⋅K.
SCHEMATIC:
ASSUMPTIONS: (1) Plate is spacewise isothermal, (2) Negligible thermal resistance of water film on
plate, (3) Heat-mass transfer analogy applies.
3
PROPERTIES: Table A-4, Air ( T =(55 + 65)°C/2 = 333 K, 1 atm): ρ = 1.0516 kg/m , cp = 1008
-6 2
J/kg⋅K, Pr = 0.703, ν = 19.24× 10 m /s; Table A-6, Saturated water vapor, (Ts = 65°C = 338 K): ρA =
3
3
1/vg = 0.1592 kg/m , hfg = 2347 kJ/kg; (Ts = 55°C = 328 K): ρA = 1/vg = 0.1029 kg/m ; Table A-8, Air-4 2
3/2
4 2
water vapor (Ts = 65°C = 338 K, 1 atm): DAB = 0.26 × 10 m /s (338/298) = 0.314 × 10- m /s.
ANALYSIS: (a) Perform an energy balance on a rate basis on the plate,
E in
=
− E out E st
q′′conv=
− q′′evap
( Mc/As ) ( dTp / dt ) .
Using the rate equations for the heat and mass transfer fluxes, find
h T∞ − Tp ( t )  − h m  ρ A,s ( Ts ) − ρ A,∞ ( T∞ )  h fg =
( Mc/As )( dT/dt ) .
<
(b) To evaluate the change in plate temperature at t = 0, the start of the drying process when Tp (0) =
65°C and T∞ = 55°C, evaluate h m from knowledge
of h 3.5 W/m 2 ⋅ K using the heat-mass transfer
=
analogy, Eq. 6.60, with n = 1/3,
2/3
2/3
h
 Sc 
 ν / D AB 
2/3
= rrr
=
cp Le
cp=
cp 
 

hm
 Pr 
 Pr 
and evaluating thermophysical properties at their appropriate temperatures, find
 19.24 ×10-6 m 2 / s/0.314 ×10-4 m 2 / s 
3.5 W/m 2 ⋅ K

1.0516 kg/m3 ×1008 J/kg ⋅ K 
=


hm
0.703


2/3
Substituting numerical values into the conservation expression of part (a), find

hm =
3.619 ×10-3 m/s.
(
)
<
3.5 W/m 2 ⋅ K ( 55 -65 ) C 3.619 ×10-3m/s ( 0.1592 - 0.1029 ) kg/m3 × 2347 ×103 J/kg
= 1600 J/m 2 ⋅ K dTp / dt
dTp / dt =
− [35.0 + 478.2] W/m 2 ⋅ K/1600 J/m 2 ⋅ K =
−0.32 K/s.
COMMENTS: This rate of temperature change will not be sustained for long, since, as the plate cools,
the rate of evaporation (which dominates the cooling process) will diminish.
PROBLEM 6S.1
KNOWN: Two-dimensional flow conditions for which v = 0 and T = T(y).
FIND: (a) Verify that u = u(y), (b) Derive the x-momentum equation, (c) Derive the energy equation.
SCHEMATIC:
Pressure & shear forces
Energy fluxes
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)
Negligible body forces, (4) v = 0, (5) T = T(y) or ∂T/∂x = 0, (6) Thermal energy generation occurs only
by viscous dissipation.
ANALYSIS: (a) From the mass continuity equation, it follows from the prescribed conditions that ∂u/∂x
= 0. Hence u = u(y).
(b) From Newton’s second law of motion, ΣFx =
(Rate of increase of fluid momentum)x,
 p −  p + ∂ p dx  dy ⋅ 1 +  −τ + τ + ∂ τ dy =
dx ⋅ 1

 ∂ y 
  ∂ x 


Hence, with τ = µ ( ∂ u/∂ y ) , it follows that
−
{(
ρ u)u +
∂
∂ x
[( ρ u ) u ] dx
}
dy ⋅ 1 − ( ρ u ) u dy ⋅ 1
∂ p
∂ 2u
= µ
.
∂ x
∂ y2
∂ p ∂ τ ∂
( ρ=
=
+
u ) u  0
∂ x ∂ y ∂ x
<
(c) From the conservation of energy requirement and the prescribed conditions, it follows that
E in − E out =
0, or
 pu + ρ u e + u 2 / 2  dy ⋅1 +  −k ∂ T + τ u + ∂ (τ u ) dy  dx ⋅1




∂ y
 ∂ y

)
(
{
− pu +
or,
(
)
(
∂ (τ u ) ∂
∂
−
( pu ) −  ρ u e + u 2 / 2
∂ y
∂ x
∂ x
(
τ
)
}

∂
∂ 
∂ T ∂  ∂ T 
+
−k
ρ u e + u 2 / 2  dx dy ⋅ 1 − τ u − k
dy dx ⋅ 1 =0
( pu ) dx + ρ u e + u 2 / 2 +




∂ x
∂ x
∂ y ∂ y  ∂ y  

0
) + ∂∂ y k ∂∂ Ty  =
∂ u
∂ τ
∂ p
∂ 2T
0.
+u
−u
+k
=
∂ y
∂ y
∂ x
∂ y2
Noting that the second and third terms cancel from the momentum equation,
2
 ∂ 2T 
∂ u 
µ
+
k
0.

=

 ∂ y 2 
∂ y 
<
PROBLEM 6S.2
KNOWN: Oil properties, journal and bearing temperatures, and journal speed for a lightly
loaded journal bearing.
FIND: Maximum oil temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant
properties, (3) Clearance is much less than journal radius and flow is Couette.
ANALYSIS: The temperature distribution corresponds to the result obtained in the text
Example on Couette flow,
µ 2  y  y 2 
T(y) =
T0 +
U  −   .
2k
 L  L  
The position of maximum temperature is obtained from
dT
µ 2  1 2y 
U  − 
= 0=
dy
2k
 L L2 
or,
y = L/2.
The temperature is a maximum at this point since d 2T/dy 2 < 0. Hence,
Tmax = T ( L/2 ) = T0 +
mm
U2
2 1 1
2k
U  −  = T0 +
8k
2 4
2
10-2 kg/s ⋅ m (10m/s )
a
T=
max 40 C +
8 × 0.15 W/m ⋅ K
Tmax = 40.83a C.
<
COMMENTS: Note that Tmax increases with increasing µ and U, decreases with increasing k,
and is independent of L.
PROBLEM 6S.3
KNOWN: Diameter, clearance, rotational speed and fluid properties of a lightly loaded journal bearing.
Temperature of bearing.
FIND: (a) Temperature distribution in the fluid, (b) Rate of heat transfer from bearing and operating
power.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)
Couette flow.
3
-5 2
-3
PROPERTIES: Oil (Given): ρ = 800 kg/m , ν = 10 m /s, k = 0.13 W/m⋅K; µ = ρν = 8 × 10 kg/s⋅m.
ANALYSIS: (a) For Couette flow, the velocity distribution is linear, u(y) = U(y/L), and the energy
equation and general form of the temperature distribution are
2
2
 du 
d 2T
µ  U 2
C
U
k
T=
=
−µ   =
−µ  
−   y 2 + 1 y + C2 .
2
L
 dy 
dy
2k  L 
k
2
Considering the boundary conditions dT/dy)y=L = 0 and T(0) = T0, find C2 = T0 and C1 = µU /L.
Hence,
2
T=
T0 + µ U 2 / k ( y/L ) − 1/ 2 ( y/L )  .


( )

<

(b) Applying Fourier’s law at y = 0, the rate of heat transfer per unit length to the bearing is
− k (π D )
q′ =
dT 
dy  y=0
=
− (π D )
m U2
L
(
=
− π × 75 × 10−3 m
) 8 ×10
−3
kg/s ⋅ m (14.14 m/s )
0.25 × 10−3 m
2
=
−1507.5 W/m
where the velocity is determined as
U=
0.0375m × 3600 rev/min ( 2π rad/rev ) / ( 60 s/min ) =
14.14 m/s.
( D/2 ) ω =
The journal power requirement is
=
P′ F(′ y=L )=
U τ s( y=L ) ⋅ π D ⋅ U
)
(
=
P′ 452.5kg/s 2 ⋅ m π × 75 ×10-3m 14.14m/s
= 1507.5kg ⋅ =
m/s3 1507.5W/m
<
where the shear stress at y = L is
 14.14 m/s 
452.5 kg/s 2 ⋅ m.
=
-3
 0.25 ×10 m 
τ s( y=L ) =
m ( ∂ u/∂ y ) y=L =
m =
8 ×10-3 kg/s ⋅ m 
L
U
COMMENTS: Note that q′ = P′, which is consistent with the energy conservation requirement.
PROBLEM 6S.4
KNOWN: Conditions associated with the Couette flow of air or water.
FIND: (a) Force and power requirements per unit surface area, (b) Viscous dissipation, (c) Maximum
fluid temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Fully-developed Couette flow, (2) Incompressible fluid with constant properties.
-7
2
-3
PROPERTIES: Table A-4, Air (300K): µ = 184.6 × 10 N⋅s/m , k = 26.3 × 10 W/m⋅K; Table A-6,
-6
2
Water (300K): µ = 855 × 10 N⋅s/m , k = 0.613 W/m⋅K.
ANALYSIS: (a) The force per unit area is associated with the shear stress. Hence, with the linear
velocity profile for Couette
flow, τ µ=
=
( du/dy ) µ ( U/L ) .
τ air = 184.6 ×10−7 N ⋅ s/m 2 ×
200 m/s
= 0.738 N/m 2
0.005 m
200 m/s
Water:
=34.2 N/m 2 .
t water =855 ×10−6 N ⋅ s/m 2 ×
0.005 m
With the required power given by P/A = τ ⋅ U,
<
Air:
=
( P/A )air
<
Air:
0.738 N/m 2 ) 200 m/s 147.6 W/m 2
(=
Water:=
34.2 N/m 2 ) 200 m/s 6840 W/m 2 .
( P/A )water (=
=
(b) The viscous dissipation
is µΦ µ=
( du/dy )2 µ ( U/L )2 . Hence,
184.6 ×10−7
2
N ⋅ s  200 m/s 
4
3
 0.005 m  = 2.95 ×10 W/m
2
m
Air:
( mΦ )air =
Water:
855 ×10−6
( mΦ )water =
<
2
N ⋅ s  200 m/s 
1.37 ×106 W/m3.
 0.005 m  =
2

m 
(c) From the solution to Part 4 of Example 6S.1, the location of the maximum temperature corresponds to
ymax = L/2. Hence, Tmax
= T0 + m U 2 / 8k and
2
184.6 ×10-7 N ⋅ s/m 2 ( 200 m/s )
aa
Air:
27 C +
=
30.5 C
( Tmax )air =
8 × 0.0263 W/m ⋅ K
<
855 ×10-6 N ⋅ s/m 2 ( 200 m/s )
aa
T
=
27
C
+
=
34.0 C.
( max )water
8 × 0.613 W/m ⋅ K
COMMENTS: (1) The viscous dissipation associated with the entire fluid layer, µΦ ( LA ) , must equal
2
Water:
b g
b g
Hence, Tmax,water ≈ Tmax,air .
the power, P. (2) Although µΦ
water >> µΦ air , k water >> k air .
PROBLEM 6S.5
KNOWN: Velocity and temperature difference of plates maintaining Couette flow. Mean
temperature of air, water or oil between the plates.
FIND: (a) Pr⋅Ec product for each fluid, (b) Pr⋅Ec product for air with plate at sonic velocity.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Couette flow, (3) Air is at 1 atm.
PROPERTIES: Table A-4, Air (300K, 1atm), cp = 1007 J/kg⋅K, Pr = 0.707, γ = 1.4, R= 287.02
J/kg⋅K; Table A-6, Water (300K): cp = 4179 J/kg⋅K, Pr = 5.83; Table A-5, Engine oil (300K), cp
= 1909 J/kg⋅K, Pr = 6400.
ANALYSIS: The product of the Prandtl and Eckert numbers is dimensionless,
Pr=
⋅ Ec Pr
U2
m2 / s2
m2 / s2
∩
∩
.
cp ∆T ( J/kg ⋅ K ) K
kg ⋅ m 2 / s 2 / kg
)
(
Substituting numerical values, find
Air
Pr⋅Ec 0.0028
Water
0.0056
Oil
13.41
<
(b) For an ideal gas, the speed of sound is
c = (γ R T )
1/ 2
where R, the gas constant for air, is Ru/ M = 8.315 kJ/kmol⋅K/(28.97 kg/kmol) = 287.02 J/kg⋅K.
Hence, at 300K for air,
U = c = (1.4 × 287.02 J/kg ⋅ K × 300K )
1/ 2
= 347.2 m/s.
For sonic velocities, it follows that
=
Pr ⋅ Ec
2
347.2 m/s )
(
0.707
=
1007J / kg ⋅ K × 25K
3.38.
<
COMMENTS: From the above results it follows that viscous dissipation effects must be
considered in the high speed flow of gases and in oil flows at moderate speeds. For Pr⋅Ec to be
less than 0.1 in air with ∆T = 25°C, U should be < 60 m/s.
~
PROBLEM 6S.6
KNOWN: Couette flow with moving plate isothermal and stationary plate insulated.
FIND: Temperature of stationary plate and heat flux at the moving plate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)
Couette flow.
ANALYSIS: The energy equation is given by
2
 ∂ 2T 
∂ u 
=
0 k
+µ

 ∂ y 2 
∂ y 
Integrating twice find the general form of the temperature distribution,
2
2
2
∂ T
µ U
∂ T
µ U
=
−  
=
−   y + C1
∂ y
k L
k L
∂ y2
µ  U 2 2
−
T ( y) =
y + C1y + C2 .
2k  L 
Consider the boundary conditions to evaluate the constants,
∂ T/∂ y y=0 =0 → C1 =0 and T ( L ) =TL → C2 =TL +
µ
2k
U2.
Hence, the temperature distribution is
 µ U2 
T ( y) =
TL + 

 2k 
  y 2 
1 −    .
L
   
The temperature of the lower plate (y = 0) is
 µ U2 
T ( 0=
.
) TL + 
<
The heat flux to the upper plate (y = L) is
∂ T
µ U2
−k
=
q′′ ( L ) =
.
y=L
<
2k 


∂ y
L
COMMENTS: The heat flux at the top surface may also be obtained by integrating the viscous
dissipation over the fluid layer height. For a control volume about a unit area of the fluid layer,
E ′′g =
E ′′out
2
L ∂ u 
∫ µ
q′′ ( L )
 dy =
0
∂ y 
µ U2
q′′ ( L ) = .
L
PROBLEM 6S.7
KNOWN: Couette flow with heat transfer. Lower (insulated) plate moves with speed U and upper plate
is stationary with prescribed thermal conductivity and thickness. Outer surface of upper plate maintained
at constant temperature, Tsp = 40°C.
FIND: (a) On T-y coordinates, sketch the temperature distribution in the oil and the stationary plate, and
(b) An expression for the temperature at the lower surface of the oil film, T(0) = To, in terms of the plate
speed U, the stationary plate parameters (Tsp, ksp, Lsp) and the oil parameters (µ, ko, Lo). Determine this
temperature for the prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed Couette flow and (3) Incompressible
fluid with constant properties.
ANALYSIS: (a) The temperature distribution is shown above with these key features: linear in plate,
parabolic in oil film, discontinuity in slope at plate-oil interface, and zero gradient at lower plate surface.
(b) From Example 6S.1, the general solution to the conservation equations for the temperature distribution
in the oil film is
2
µ  U 
2
To ( y ) =
where
A= 
−Ay + C3 y + C4

2k o  Lo 
and the boundary conditions are,
At y = 0, insulated boundary
dTo 
= 0;

dy  y = 0
At y = Lo, heat fluxes in oil and plate are equal,
C3 = 0
q′′o ( Lo ) = q′′sp ( Lo )
Continued...
PROBLEM 6S.7 (Cont.)
To ( Lo ) − Tsp
dT 
−k o o 
=
dy  y = L
R sp
o
 dTo 
= −2ALo


dy


y = Lo
R =
−AL2o + C4
To ( Lo ) =
 sp Lsp k sp

k Lsp 
C4 =
Tsp + AL2o 1 + 2 o

Lo k sp 

Hence, the temperature distribution at the lower surface is
To ( 0 ) =−A ⋅ 0 + C4
k Lsp 
µ 2
To ( 0 ) =
Tsp +
U 1 + 2 o

2k o
Lo k sp 

<
Substituting numerical values, find
To ( 0 ) = 40 C +
0.799 N ⋅ s m 2
0.145 3 
×  = 116.9 C
( 5 m s )2 1 + 2
2 × 0.145 W m ⋅ K
5
1.5 

<
COMMENTS: (1) Give a physical explanation about why the maximum temperature occurs at the lower
surface.
(2) Sketch the temperature distribution if the upper plate moved with a speed U while the lower plate is
stationary and all other conditions remain the same.
PROBLEM 6S.8
KNOWN: Shaft of diameter 100 mm rotating at 9000 rpm in a journal bearing of 70 mm length.
Uniform gap of 1 mm separates the shaft and bearing filled with lubricant. Outer surface of bearing is
water-cooled and maintained at Twc = 30°C.
FIND: (a) Viscous dissipation in the lubricant, µΦ(W/m3), (b) Heat transfer rate from the lubricant,
assuming no heat lost through the shaft, and (c) Temperatures of the bearing and shaft, Tb and Ts.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed Couette flow, (3) Incompressible fluid
with constant properties, and (4) Negligible heat lost through the shaft.
ANALYSIS: (a) The viscous dissipation, µΦ, Eq. 6S.20, for Couette flow from Example 6S.1, is
2
2
2
 47.1m s 
 du 
U
0.03 N ⋅ s m 2 
6.656 ×107 W m3
=
mmm
Φ
=

 =

 =

L
 dy 
 0.001m 
<
where the velocity distribution is linear and the tangential velocity of the shaft is
U=
pp
DN =( 0.100 m ) × 9000 rpm × ( min 60s ) =
47.1m s .
(b) The heat transfer rate from the lubricant volume ∀ through the bearing is
q = mm
Φ ⋅ ∀ = Φ (π D ⋅ L ⋅  ) = 6.65 × 107 W m3 (π × 0.100 m × 0.001m × 0.070 m ) = 1462 W
<
where  = 70 mm is the length of the bearing normal to the page.
Continued...
PROBLEM 6S.8 (Cont.)
(c) From Fourier’s law, the heat rate through the bearing material of inner and outer diameters, Di and Do,
and thermal conductivity kb is, from Eq. (3.32),
qr =
2π k b ( Tb − Twc )
ln ( Do Di )
q ln ( Do Di )
Tb Twc + r
=
2π k b
1462 W ln ( 200 100 )
Tb =
30 C +
=
81.2 C
2π × 0.070 m × 45 W m ⋅ K
<
To determine the temperature of the shaft, T(0) = Ts, first the temperature distribution must be found
beginning with the general solution, Example 6S.1,
2
µ U
−   y 2 + C3 y + C4
T ( y) =
2k  L 
The boundary conditions are, at y = 0, the surface is adiabatic
dT 
=
0=
C3 0

dy  y = 0
and at y = L, the temperature is that of the bearing, Tb
µ  U 2 2
T ( L ) =Tb =−   L + 0 + C4
2k  L 
C4 =Tb +
µ
2k
U2
Hence, the temperature distribution is
 y2 
2

T ( y) =
Tb +
U 1 −
 L2 
2k
µ


and the temperature at the shaft, y = 0, is
Ts =
T ( 0 ) =+
Tb
m
2k
0.03 N ⋅ s m 2
2

U =
81.3 C +
2 × 0.15 W m ⋅ K
303 C
( 47.1m s )2 =
<
PROBLEM 6S.9
KNOWN: Couette flow with heat transfer.
FIND: (a) Dimensionless form of temperature distribution, (b) Conditions for which top plate is
adiabatic, (c) Expression for heat transfer to lower plate when top plate is adiabatic.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) incompressible fluid with constant properties, (3)
Negligible body forces, (4) Couette flow.
ANALYSIS: (a) From Example 6.4, the temperature distribution is
µ 2  y  y 2 
y
T =T0 +
U  −    + ( TL − T0 )
2k
 L  L  
L
 y  y 2  y
T − T0
µ U2
 −   +
=
TL − T0 2k ( TL − T0 )  L  L   L


or, with
θ ≡ ( T − T0 ) TL − T0 ,
η ≡ y L,
Ec ≡ U 2 cp ( TL − T0 )
Pr⋅ Ec
 1

θ=
η − η 2 + η= η 1 + Pr⋅ Ec (1 − η ) 
2
 2

Pr ≡ cp µ k ,
(
)
(1)
<
(2)
<
(b) For there to be zero heat transfer at the top plate, (dT/dy)y=L = 0. Hence, (dθ/dη)η=1 = 0.
dθ 
Pr⋅ Ec
Pr⋅ Ec
=
+ 1 =0
(1 − 2η ) η =1 + 1 =−

dη η =1
2
2
There is no heat transfer at the top plate if,
Ec⋅Pr = 2.
(c) The heat transfer rate to the lower plate (per unit area) is
( T − T ) dq
dT
q′′0 =
−k
=
−k L 0
dy y 0=
L
dη η 0
=
T − T  Pr⋅ Ec
q′′o =
−k L 0 
(1 − 2η ) η =0 + 1
L  2

T − T  Pr⋅ Ec 
q′′0 =
−k L 0 
+ 1 =
−2k ( TL − T0 ) L
L  2

<
Continued...
PROBLEM 6S.9 (Cont.)
(d) Using Eq. (1), the dimensionless temperature distribution is plotted as a function of dimensionless
distance, η = y/L. When Pr⋅Ec = 0, there is no dissipation and the temperature distribution is linear, so
that heat transfer is by conduction only. As Pr ⋅Ec increases, viscous dissipation becomes more
important. When Pr⋅Ec = 2, heat transfer to the upper plate is zero. When Pr⋅Ec > 2, the heat rate is out
of the oil film at both surfaces.
theta = (T(y)-T0)/(TL-T0)
2
1.5
1
0.5
0
0
0.25
0.5
eta = y/L
Pr*Ec = 0, conduction
Pr*Ec = 1
Pr*Ec = 2, adiabatic at y=L
Pr*Ec = 4
Pr*Ec = 10
0.75
1
PROBLEM 6S.10
KNOWN: Steady, incompressible, laminar flow between infinite parallel plates at different
temperatures.
FIND: (a) Form of continuity equation, (b) Form of momentum equations and velocity profile.
Relationship of pressure gradient to maximum velocity, (c) Form of energy equation and temperature
distribution. Heat flux at top surface.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional flow (no variations in z) between infinite, parallel plates, (2)
Negligible body forces, (3) No internal energy generation, (4) Incompressible fluid with constant
properties.
ANALYSIS: (a) For two-dimensional, steady conditions, the continuity equation is
∂ ( ρ u ) ∂ ( ρ v)
0.
+
=
∂ x
∂ y
Hence, for an incompressible fluid (constant ρ) in parallel flow (v = 0),
∂ u
= 0.
∂ x
<
The flow is fully developed in the sense that, irrespective of y, u is independent of x.
(b) With the above result and the prescribed conditions, the momentum equations reduce to
∂ p
∂ 2u
∂ p
−
0=
∂ x
+µ
∂ y2
−
0=
∂ y
<
Since p is independent of y, ∂p/∂x = dp/dx is independent of y and
∂ 2u
d 2 u dp
µ = µ=
.
∂ y2
dy 2 dx
Since the left-hand side can, at most, depend only on y and the right-hand side is independent of y, both
sides must equal the same constant C. That is,
µ
d 2u
dy 2
= C.
Hence, the velocity distribution has the form
u ( y )=
C 2
y + C1y + C2 .
2µ
Using the boundary conditions to evaluate the constants,
u ( 0 ) =0
→
C2 =0 and u ( L ) =0
→
C1 =−CL/2µ .
Continued …..
PROBLEM 6S.10 (Cont.)
u ( y)
=
The velocity profile is
)
(
C 2
y − Ly .
2µ
The profile is symmetric about the midplane, in which case the maximum velocity exists at y = L/2.
Hence,
C  L2 
L2 dp
u ( L/2 ) =
u max =
or
u max =
.
−
− 
<
2mm
8 dx
 4 
(c) For fully developed thermal conditions, (∂T/∂x) = 0 and temperature depends only on y. Hence with v
= 0, ∂u/∂x = 0, and the prescribed assumptions, the energy equation becomes
2
 du 
d 2T
dp
∂ i
ρu
∂ x
= k
With i = e + p/ρ,
+u
+µ  .
dx
 dy 
dy 2
∂ i ∂ e 1 dp
where
=+
∂ x ∂ x r dx
∂ e ∂ e∂ T ∂ e∂ r
0.
=
+
=
∂ x ∂ T∂ x ∂ r∂ x
2
d 2T
 du 
=
+µ  .
0 k
 dy 
dy 2
Hence, the energy equation becomes
<
With du/dy = (C/2µ) (2y - L), it follows that
d 2T
C2
=
−
4y 2 − 4Ly + L2 .
2
4kµ
dy
)
(
Integrating twice,
C2  y 4 2Ly3 L2 y 2 
T ( y) =
−
+
 −
 + C3 y + C4
4kµ  3
3
2 


Using the boundary conditions to evaluate the constants,
T ( 0 ) =T2
→
C4 =T2
and
T ( L ) =T1
C2
y
T ( y ) =T2 +   ( T1 − T2 ) −
4kµ
L
Hence,
→
C3 =
C2 L3 ( T1 − T2 )
.
+
24kµ
L
 y 4 2Ly3 L2 y 2 L3 y 
+
−
 −
.
3
2
6 
 3

<
From Fourier’s law,
q′′ ( L ) =
−k
q′′ ( L=
)
∂ T
k
C2
=( T2 − T1 ) +
∂ y y=L L
4µ
k
C2 L3
.
( T2 − T1 ) +
L
24 µ
4 3
L3 
 L − 2L3 + L3 − 
6
 3

<
COMMENTS: The third and second terms on the right-hand sides of the temperature distribution and
heat flux, respectively, represents the effects of viscous dissipation. If C is large (due to large µ or umax),
viscous dissipation is significant. If C is small, conduction effects dominate.
PROBLEM 6S.11
KNOWN: Steady, incompressible flow of binary mixture between infinite parallel plates with
different species concentrations.
FIND: Form of species continuity equation and concentration distribution. Species flux at
upper surface.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional flow, (2) No chemical reactions, (3) Constant
properties.
ANALYSIS: For fully developed conditions, ∂CA/∂x = 0. Hence with v = 0, the species
conservation equation reduces to
d 2CA
dy 2
<
= 0.
Integrating twice, the general form of the species concentration distribution is
CA (=
y ) C1y + C2 .
Using appropriate boundary conditions and evaluating the constants,
=
CA ( 0 ) CA,2
CA ( L ) = CA,1
→
→
C2 =CA,2
C1 = CA,1 − CA,2 / L,
(
)
the concentration distribution is
(
)
CA ( y ) =
CA,2 + ( y/L ) CA,1 − CA,2 .
<
From Fick’s law, the species flux is
N′′A ( L ) = −D AB
N′′A ( L )
=
(
dCA
dy y=L
)
D AB
CA,2 − CA,1 .
L
<
COMMENTS: An analogy between heat and mass transfer exists if viscous dissipation is
2
2
negligible. The energy equation is then d T/dy = 0. Hence, both heat and species transfer are
influenced only by diffusion. Expressions for T(y) and q′′ ( L ) are analogous to those for CA(y)
and N′′A ( L ) .
PROBLEM 6S.12
KNOWN: Flow conditions between two parallel plates, across which vapor transfer occurs.
FIND: (a) Variation of vapor molar concentration between the plates and mass rate of water
production per unit area, (b) Heat required to sustain the process.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed, incompressible flow with
constant properties, (3) Negligible body forces, (4) No chemical reactions, (5) All work
interactions, including viscous dissipation, are negligible.
ANALYSIS: (a) The flow will be fully developed in terms of the vapor concentration field, as
well as the velocity and temperature fields. Hence
∂ CA
0=
or
CA ( x,y ) CA ( y ) .
=
∂ x
 = 0, v = 0 and constant DAB, the species conservation equation
Also, with ∂CA/∂t = 0, N
A
reduces to
d 2CA
dy 2
= 0.
Separating and integrating twice,
CA
=
( y ) C1 ( y ) + C2 .
Applying the boundary conditions,
CA ( 0 )= CA,0
→
CA ( L ) =→
CA,L
C2= CA,0
CA,L =
C1L + C2
CA,0 − CA,L
C1 =
−
L
find the species concentration distribution,
(
CA ( y ) =CA,0 − CA,0 − CA,L
) ( y/L ) .
<
From Fick’s law, Eq. 6.7, the species transfer rate is
CA,0 − CA,L
∂ CA 
N′′A =
N′′A,s =
−D AB
=
D AB
.

∂ y  y=0
L
Continued …..
PROBLEM 6S.12 (Cont.)
Multiplying by the molecular weight of water vapor, MA, the mass rate of water production per
unit area is
CA,0 − CA,L
′′
=
M
n ′′A M=
N
D
.
A A
A AB
L
<
(b) Heat must be supplied to the bottom surface in an amount equal to the latent and sensible heat
transfer from the surface,
=
q′′ q′′lat + q′′sen
 dT 
=
q′′ n ′′A,s h fg +  −k 
.
 dy  y=0
The temperature distribution may be obtained by solving the energy equation, which, for the
prescribed conditions, reduces to
d 2T
dy 2
= 0.
Separating and integrating twice,
T (=
y ) C1y + C2 .
Applying the boundary conditions,
T ( 0 ) = T0
T (L) =
TL
→
→
C2 = T0
C1 =
( T1 − T0 ) / L
find the temperature distribution,
T ( y ) =T0 − ( T0 − TL ) y/L.
Hence,
−k
(T − T )
dT 
k 0 L .
=

dy  y=0
L
Accordingly,
q′′ M A D AB
CA,0 − CA,L
L
h fg + k
( T0 − TL ) .
L
<
COMMENTS: Despite the existence of the flow, species and energy transfer across the air are
uninfluenced by advection and transfer is only by diffusion. If the flow were not fully
developed, advection would have a significant influence on the species concentration and
temperature fields and hence on the rate of species and energy transfer. The foregoing results
would, of course, apply in the case of no air flow. The physical condition is an example of
Poiseuille flow with heat and mass transfer.
PROBLEM 6S.13
KNOWN: The conservation equations, Eqs. 6S.24 and 6S.31.
FIND: (a) Describe physical significance of terms in these equations, (b) Identify
approximations and special conditions used to reduce these equations to the boundary layer
equations, Eqs. 6.29 and 6.30, (c) Identify the conditions under which these two boundary layer
equations have the same form and, hence, an analogy will exist.
ANALYSIS: (a) The energy conservation equation, Eq. 6S.24, has the form
∂ i
∂ i
∂  ∂ T ∂  ∂ T  ∂ p
∂ p

ρu =
+ρ v
k
+
k
+ u
+v
+ µΦ + q.




∂ x
∂ y ∂ x  ∂ x ∂ y ∂ y  ∂ x
∂ y 
1a
1b
2a
2b
3
4 5
The terms, as identified, have the following physical significance:
1. Change of enthalpy (thermal + flow work) advected in x and y directions,
2. Change of conduction flux in x and y directions,
3. Work done by static pressure forces,
4. Word done by viscous stresses,
5. Rate of energy generation.
The species mass conservation equation for a constant total concentration has the form
∂ CA
∂ CA
∂ CA  ∂ 
∂ CA  
∂ 
u
D
D
=
+v
+
+ NA
AB
AB

∂ x
∂ y ∂ x 
∂ x  ∂ y 
∂ y 
1a
1b
2a
2b
3
1. Change in species transport due to advection in x and y directions,
2. Change in species transport by diffusion in x and y directions, and
3. Rate of species generation.
<
<
(b) The special conditions used to reduce the above equations to the boundary layer equations
 = 0 ) , without internal
are: constant properties, incompressible flow, non-reacting species ( N
A
heat generation ( q = 0 ) , species diffusion has negligible effect on the thermal boundary layer,
u(∂ p/∂ x) is negligible. The approximations are,

∂ u
∂ u ∂ v
>>
,
,
u >> v
∂ y
∂ x ∂ y

∂ T
∂ T
>>
Concentration b.1.:
Thermal b.1.:

∂ x
∂ y
The resulting simplified boundary layer equations are
Velocity boundary layer
∂ T
∂ T
∂ 2T ν  ∂ u 
+v
= a
+ 
u

∂ x
∂ y
∂ y2 c  ∂ y 
2
∂ v
∂ x
 ∂ CA
∂ CA
>>
.

∂ x
 ∂ y
∂ CA
∂ CA
∂ 2CA
+v
= D AB
u
∂ x
∂ y
∂ y2
<
1a
1b
2a
3
1c
1d
2b
where the terms are: 1. Advective transport, 2. Diffusion, and 3. Viscous dissipation.
(c) When viscous dissipation effects are negligible, the two boundary layer equations have
identical form. If the boundary conditions for each equation are of the same form, an analogy
between heat and mass (species) transfer exists.
PROBLEM 6S.14
KNOWN: Thickness and inclination of a liquid film. Mass density of gas in solution at free surface of
liquid.
FIND: (a) Liquid momentum equation and velocity distribution for the x-direction. Maximum velocity,
(b) Continuity equation and density distribution of the gas in the liquid, (c) Expression for the local
Sherwood number, (d) Total gas absorption rate for the film, (e) Mass rate of NH3 removal by a water
film for prescribed conditions.
SCHEMATIC:
NH3 (A) – Water (B)
L = 2m
δ = 1 mm
D = 0.05m
W = πD = 0.157m
3
ρA,o = 25 kg/m
-9 2
DAB = 2 × 10 m /s
φ = 0°
ASSUMPTIONS: (1) Steady-state conditions, (2) The film is in fully developed, laminar flow, (3)
Negligible shear stress at the liquid-gas interface, (4) Constant properties, (5) Negligible gas
concentration at x = 0 and y = δ, (6) No chemical reactions in the liquid, (7) Total mass density is
constant, (8) Liquid may be approximated as semi-infinite to gas transport.
3
-6
2
PROPERTIES: Table A-6, Water, liquid (300K): ρf = 1/vf = 997 kg/m , µ = 855 × 10 N⋅s/m , ν =
-6 2
µ/ρf = 0.855 × 10 m /s.
ANALYSIS: (a) For fully developed flow (v = w = 0, ∂u/∂x = 0), the x-momentum equation is
∂τ yx / ∂ y + X where τ yx =
µ ( ∂ u/∂ y ) and X =
0=
( r g ) cos φ .
That is, the momentum equation reduces to a balance between gravitational and shear forces. Hence,
)
(
µ ∂ 2 u/∂ y 2 = − ( ρ g ) cos φ .
Integrating,
∂ u/∂ y =
− ( g cos φ /ν ) y + C1
Applying the boundary conditions,
→
∂ u/∂ y ) y=0 =0
C1 =0
u=
(δ ) 0
Hence,
=
u
=
C2 g cos φ
→
g cos φ
(
)
δ 2=
− y2
2ν
and the maximum velocity exists at y = 0,
u=
( 0)
max u=
− ( g cos φ /2ν ) y 2 + C1y + C2 .
u=
δ2
.
2ν
g cos φ δ 2 
2
1 − ( y/δ ) 



2ν
<
( g cos φ δ 2 ) / 2ν .
<
(b) Species transport within the liquid is influenced by diffusion in the y-direction and advection in the xdirection. Hence, the species continuity equation with u assumed equal to umax throughout the region of
gas penetration is
Continued …..
PROBLEM 6S.14 (Cont.)
2
∂ρ
∂ ρA
∂ 2 ρ A u max ∂ρ A
=
u A D=
.
AB
∂ x
D AB ∂ x
∂ y2
∂ y2
Appropriate boundary conditions are: ρA(x,0) = ρA,o and ρA(x,∞) = 0 and the entrance condition is:
ρA(0,y) = 0. The problem is therefore analogous to transient conduction in a semi-infinite medium due to
a sudden change in surface temperature. From Section 5.7, the solution is then
rr
y
y
A − A,o
= erf
=
rr
erfc
A
A,o
1/ 2
1/ 2
0 − r A,o
2 ( D AB x/u max )
2 ( D AB x/u max )
(c) The Sherwood number is defined as
<
Sh x
=
n ′′A,x − D AB∂r A / ∂ y y=0
where=
h m,x ≡
h m,x x
rr
A,o
A,o
D AB
 y u max 
2
1
 u

=
− ρ A,o
exp  −
=
− ρ A,o  max 

∂ y y=0
 p D AB x 
(p )1/ 2  4 D AB x  2 ( D AB x/u max )1/ 2 y=0
1/ 2
2
∂ρ A
.
Hence,
1/ 2
1/ 2
1/ 2
1/ 2
u max D AB 
u max x 
1
1
 u max x 
 ν 


h m,x
Sh x
=
=




 =



1/
2
1/
2
π x

(π )  D AB 
(π )  ν   D AB 
and with Rex ≡ umax x/ν,
1/ 2
1/2
1/2
=
Sh x =
1/ (π )  Re1/2
0.564 Re1/2
x Sc
x Sc .

<

(d) The total gas absorption rate may be expressed as
=
n A h m,x ( W ⋅ L ) ρ A,o
where the average mass transfer convection coefficient is
1/ 2
1 L
1  u max D AB 
∫ h m,x dx =
h m,x =

π
L 0
L 
Hence, the absorption rate per unit width is
1/ 2
 4u max D AB 
=


0 x1/2
π L
∫L
dx
.
n A / W = ( 4u max D AB L / π )
ρ A,o .
(e) From the foregoing results, it follows that the ammonia absorption rate is
1/ 2
1/ 2
nA
 4u max D AB L 
W ρ A,o
=

π
<
1/ 2
 4 g cosφδ 2 D L 
AB


2πn


W ρ A,o .
Substituting numerical values,
(
) (
)
1/ 2
2


2
-3
-9 2
2 × 10 m /s 2m 
 4 × 9.8 m/s × 1 × 10 m
n=
A 

-6 2
2π × 0.855 × 10 m / s




3
=
( 0.157m ) 25 kg/m
6.71 × 10
-4
kg/s.
<
COMMENTS: Note that ρA,o ≠ ρA,∞, where ρA,∞ is the mass density of the gas phase. The value of
ρA,o depends upon the pressure of the gas and the solubility of the gas in the liquid.
PROBLEM 7.1
PROBLEM 7.2
PROBLEM 7.3
PROBLEM 7.4
PROBLEM 7.5
PROBLEM 7.5 (Cont.)
PROBLEM 7.6
PROBLEM 7.7
PROBLEM 7.8
PROBLEM 7.8 (Cont.)
PROBLEM 7.9
PROBLEM 7.9 (Cont.)
PROBLEM 7.10
PROBLEM 7.10 (Cont.)
PROBLEM 7.11
PROBLEM 7.11 (Cont.)
PROBLEM 7.12
PROBLEM 7.12 (Cont.)
PROBLEM 7.13
PROBLEM 7.13 (Cont.)
PROBLEM 7.14
PROBLEM 7.15
PROBLEM 7.15 (Cont.)
PROBLEM 7.15 (Cont.)
PROBLEM 7.15 (Cont.)
PROBLEM 7.16
PROBLEM 7.16 (Cont.)
PROBLEM 7.17
KNOWN: Surface characteristics of a flat plate in an air stream.
FIND: Orientation which minimizes convection heat transfer.
SCHEMATIC:
ASSUMPTIONS: (1) Surface B is sufficiently rough to trip the boundary layer when in the upstream
position (Configuration 2).
PROPERTIES: Table A-4, Air (Tf = 343K, 1 atm): v = 20.2 × 10−6 m2/s, k = 29.5 × 10−3 W/m⋅K,
Pr = 0.7.
ANALYSIS: Since Configuration (2) results in a turbulent boundary layer over the entire surface, the
lowest heat transfer is associated with Configuration (1). Find
Re
=
L
u∞L
20 m/s ×1m
=
= 0.99 × 106.
v
20.2 × 10−6 m 2 /s
Hence in Configuration (1), transition will occur just before the rough surface ( x c = 0.48m).
Note that
1/3
0.037 ( 0.99 × 106 )4/5 − 871 0.7
Nu
=
=
1284
L,1


Nu L,2 =
0.037 ( 0.99 × 106 )
4/5
(0.7)1/3 =
2056 > Nu L,1 .
h L,1L
k
=
1284
For Configuration (1): = Nu
L,1
Hence
And
h L,1 = 1284 ( 29.5 × 10−3 W/m ⋅ K ) /1m = 37.9W/m 2 ⋅ K
q1 = h L,1A ( TS − T∞ ) = 37.9 W/m 2 ⋅ K (0.5m × 1m)(110 − 30)K
q1 = 1515 W.
<
PROBLEM 7.18
KNOWN: Dimensions and orientation of a flat plate placed in atmospheric airstream, airstream velocity
and temperature, plate temperature.
FIND: The average convection coefficient for the entire surface.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Boundary layer on rough
plate is tripped at leading edge, (4) Transition occurs at Rex,c = 500,000.
PROPERTIES: Table A.4, air (Tf = 60°C = 343K, p = 1 atm): v = 20.2 × 10−6 m2/s, Pr = 0.701,
k = 0.0295 W/m⋅K
ANALYSIS: For both the smooth and rough sections of the plate,
Re
=
L
u∞ L 20 m/s × 0.5m
=
= 500,000
v
20.2 × 10−6 m 2 /s
Therefore, flow over portion A of the plate is nearly all laminar except near the trailing edge.
For portion A, Eq. 7.38 yields
hA =
k
0.0295 W/m ⋅ K
0.037 ReL4/5 − 871 Pr1/3 =
0.037 × 500,0004/5 − 871 0.7011/3 = 24.6 W/m 2 ⋅ K
L
0.5m
For portion B, Eq. 7.38 yields
0.0295W/m ⋅ K
k
0.037 ReL4/5  Pr1/3 =
0.037 × 500,0004/5  0.7011/3 =
70.3W/m 2 ⋅ K
hB =
0.5m
L
Therefore, for the entire plate consisting of half portion A and Half portion B,
=
h
h A + hB
= 47.45 W/m 2 ⋅ K
2
<
COMMENTS: (1) The value of the average coefficient for orientations 1 and 2 of Problem 7.17 are
h1 = 37.9 W/m 2 ⋅ K and h 2 = 60.7 W/m 2 ⋅ K, respectively. The average heat transfer coefficient found
here is approximately midway between the values associated with the other two orientations. This is
coincidental, since the nature of the flow is very different for the three different orientations. (2) In reality,
boundary layer development cannot proceed on the smooth and rough parts of the plate independent of
the other part of the plate, as implied in the solution above. Rather, there will be an intermediate region
that will be characterized by complex flow and heat transfer phenomena. The solution should be viewed
as a first approximation of actual heat transfer conditions.
PROBLEM 7.19
KNOWN: Plate dimensions and initial temperature. Velocity and temperature of air in parallel flow over
plates.
FIND: Initial rate of heat transfer from plate. Rate of change of plate temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible radiation, (2) Negligible effect of conveyor velocity on boundary layer
development, (3) Plates are isothermal, (4) Negligible heat transfer from sides of plate, (5)
Rex,c = 5 × 105, (6) Constant properties.
PROPERTIES: Table A-1, AISI 1010 steel (583K): kp = 48.7 W/m⋅K, c = 553 J/kg⋅K, p =7832 kg/m3.
Table A-4, Air (p = 1 atm, Tf = 443K): v = 31.6 × 10−6 m2/s, k = 0.0368 W/m⋅K, Pr = 0.687.
ANALYSIS: The initial rate of heat transfer from a plate is
q=
2 h AS ( Ti − T∞ ) = 2h L2 (Ti − T∞ )
With Re L =
u ∞ L/v =
10m/s × 1m/31.6 × 10−6 m 2 /s =
3.16 × 105 , flow is laminar over the entire surface and
Nu L =
0.664Re1/L 2 Pr1/3 =
0.664 ( 3.16 × 105 ) (0.687)1/3 =
336
1/ 2
h =(k/L)Nu L =(0.0368W/m ⋅ K/1m)336 =12.4W/m 2 ⋅ K
Hence,
q=
2 × 12.4 W/m 2 ⋅ K(1m) 2 (310 − 30)º C =
6944 W
<
Performing an energy balance at an instant of time for a control surface about the plate, − E out =
E St , we
obtain,
ρd L2 c
dT
=
− h 2 L2 (Ti − T∞ )
dt i
2 (12.4W/m 2 ⋅ K ) (310 − 30)º C
dT
=
−
=
−0.27º C/s
dt i
7832 kg/m3 × 0.006 m × 553J/kg ⋅ K
<
= h(δ /2)/k=
7.6 × 10−4 use of the lumped capacitance method is
COMMENTS: (1) With Bi
p
appropriate. (2) Despite the large plate temperature and the small convection coefficient, if adjoining
plates are in close proximity, radiation exchange with the surroundings will be small and the assumption
of negligible radiation is justifiable.
PROBLEM 7.20
PROBLEM 7.20 (Cont.)
PROBLEM 7.20 (Cont.)
PROBLEM 7.21
PROBLEM 7.22
PROBLEM 7.22 (Cont.)
PROBLEM 7.23
PROBLEM 7.23 (Cont.)
PROBLEM 7.24
PROBLEM 7.24 (Cont.)
PROBLEM 7.24 (Cont.)
PROBLEM 7.24 (Cont.)
PROBLEM 7.24 (Cont.)
PROBLEM 7.25
PROBLEM 7.25 (Cont.)
PROBLEM 7.26
PROBLEM 7.27
PROBLEM 7.27 (Cont.)
PROBLEM 7.27 (Cont.)
PROBLEM 7.28
PROBLEM 7.29
PROBLEM 7.30
PROBLEM 7.30 (Cont.)
PROBLEM 7.31
PROBLEM 7.32
PROBLEM 7.32 (Cont.)
PROBLEM 7.33
PROBLEM 7.34
PROBLEM 7.35
PROBLEM 7.36
PROBLEM 7.36 (Cont.)
PROBLEM 7.37
PROBLEM 7.38
PROBLEM 7.38 (Cont.)
PROBLEM 7.39
PROBLEM 7.39 (Cont.)
PROBLEM 7.40
PROBLEM 7.40 (Cont.)
PROBLEM 7.41
PROBLEM 7.42
PROBLEM 7.42 (Cont.)
PROBLEM 7.43
PROBLEM 7.43 (Cont.)
PROBLEM 7.44
PROBLEM 7.45
PROBLEM 7.45 (Cont.)
PROBLEM 7.46
PROBLEM 7.46 (Cont.)
PROBLEM 7.47
PROBLEM 7.48
PROBLEM 7.48 (Cont.)
PROBLEM 7.49
PROBLEM 7.50
PROBLEM 7.50 (Cont.)
PROBLEM 7.51
PROBLEM 7.52
PROBLEM 7.52 (Cont.)
PROBLEM 7.52 (Cont.)
PROBLEM 7.53
PROBLEM 7.53 (Cont.)
PROBLEM 7.54
PROBLEM 7.54 (Cont.)
PROBLEM 7.55
PROBLEM 7.56
PROBLEM 7.56 (Cont.)
PROBLEM 7.57
PROBLEM 7.57 (Cont.)
PROBLEM 7.57 (Cont.)
PROBLEM 7.58
PROBLEM 7.58 (Cont.)
PROBLEM 7.59
PROBLEM 7.60
PROBLEM 7.60 (Cont.)
PROBLEM 7.61
PROBLEM 7.61 (Cont.)
PROBLEM 7.62
PROBLEM 7.62 (Cont.)
PROBLEM 7.62 (Cont.)
PROBLEM 7.63
PROBLEM 7.63 (Cont.)
PROBLEM 7.64
PROBLEM 7.65
PROBLEM 7.66
PROBLEM 7.66 (Cont.)
PROBLEM 7.67
PROBLEM 7.68
PROBLEM 7.68 (Cont.)
PROBLEM 7.69
PROBLEM 7.69 (Cont.)
PROBLEM 7.70
PROBLEM 7.70 (Cont.)
PROBLEM 7.71
PROBLEM 7.72
PROBLEM 7.72 (Cont.)
PROBLEM 7.73
PROBLEM 7.74
PROBLEM 7.75
PROBLEM 7.76
PROBLEM 7.76 (Cont.)
PROBLEM 7.76 (Cont.)
PROBLEM 7.77
PROBLEM 7.77 (Cont.)
PROBLEM 7.78
PROBLEM 7.78 (Cont.)
PROBLEM 7.79
PROBLEM 7.79 (Cont.)
PROBLEM 7.80
PROBLEM 7.81
PROBLEM 7.82
PROBLEM 7.83
PROBLEM 7.83 (Cont.)
PROBLEM 7.84
PROBLEM 7.84 (Cont.)
PROBLEM 7.85
PROBLEM 7.85 (Cont.)
PROBLEM 7.86
PROBLEM 7.86 (Cont.)
PROBLEM 7.87
PROBLEM 7.88
PROBLEM 7.88 (Cont.)
PROBLEM 7.89
PROBLEM 7.89 (Cont.)
PROBLEM 7.89 (Cont.)
PROBLEM 7.90
PROBLEM 7.90 (Cont.)
PROBLEM 7.91
PROBLEM 7.91 (Cont.)
PROBLEM 7.91 (Cont.)
PROBLEM 7.92
PROBLEM 7.92 (Cont.)
PROBLEM 7.93
PROBLEM 7.93 (Cont.)
PROBLEM 7.93 (Cont.)
PROBLEM 7.94
PROBLEM 7.95
PROBLEM 7.96
PROBLEM 7.97
PROBLEM 7.97 (Cont.)
PROBLEM 7.98
PROBLEM 7.99
PROBLEM 7.99 (Cont.)
PROBLEM 7.100
PROBLEM 7.101
PROBLEM 7.101 (Cont.)
PROBLEM 7.102
PROBLEM 7.102 (Cont.)
PROBLEM 7.103
PROBLEM 7.103 (Cont.)
PROBLEM 7.103 (Cont.)
PROBLEM 7.103 (Cont.)
PROBLEM 7.104
PROBLEM 7.105
PROBLEM 7.106
PROBLEM 7.106 (Cont.)
PROBLEM 7.107
PROBLEM 7.108
PROBLEM 7.108 (Cont.)
PROBLEM 7.109
PROBLEM 7.109 (Cont.)
PROBLEM 7.110
PROBLEM 7.110 (Cont.)
PROBLEM 7.111
PROBLEM 7.111 (Cont.)
PROBLEM 7.112
PROBLEM 7.112 (Cont.)
PROBLEM 7.113
PROBLEM 7.113 (Cont.)
PROBLEM 7.114
PROBLEM 7.114 (Cont.)
PROBLEM 7.115
PROBLEM 7.116
PROBLEM 7.117
PROBLEM 7.117 (Cont.)
PROBLEM 7.118
PROBLEM 7.119
PROBLEM 7.119 (Cont.)
PROBLEM 8.1
PROBLEM 8.2
PROBLEM 8.2 (Cont.)
PROBLEM 8.3
PROBLEM 8.4
PROBLEM 8.4 (Cont.)
PROBLEM 8.5
PROBLEM 8.5 (Cont.)
PROBLEM 8.5 (Cont.)
PROBLEM 8.6
PROBLEM 8.6 (Cont.)
PROBLEM 8.7
PROBLEM 8.7 (Cont.)
PROBLEM 8.8
PROBLEM 8.9
PROBLEM 8.9 (Cont.)
PROBLEM 8.10
PROBLEM 8.10 (Cont.)
PROBLEM 8.11
PROBLEM 8.11 (Cont.)
PROBLEM 8.12
PROBLEM 8.12 (Cont.)
PROBLEM 8.12 (Cont.)
PROBLEM 8.13
PROBLEM 8.14
PROBLEM 8.15
PROBLEM 8.15 (Cont.)
PROBLEM 8.16
PROBLEM 8.17
PROBLEM 8.18
PROBLEM 8.18 (Cont.)
PROBLEM 8.19
PROBLEM 8.19 (Cont.)
PROBLEM 8.20
PROBLEM 8.21
PROBLEM 8.21 (Cont.)
PROBLEM 8.21 (Cont.)
PROBLEM 8.22
PROBLEM 8.22 (Cont.)
PROBLEM 8.23
PROBLEM 8.24
PROBLEM 8.24 (Cont.)
PROBLEM 8.25
PROBLEM 8.25 (Cont.)
PROBLEM 8.26
PROBLEM 8.27
PROBLEM 8.27 (Cont.)
PROBLEM 8.27 (Cont.)
PROBLEM 8.28
PROBLEM 8.28 (Cont.)
PROBLEM 8.29
PROBLEM 8.29 (Cont.)
PROBLEM 8.30
PROBLEM 8.31
PROBLEM 8.31 (Cont.)
PROBLEM 8.32
PROBLEM 8.33
PROBLEM 8.33 (Cont.)
PROBLEM 8.34
PROBLEM 8.35
PROBLEM 8.35 (Cont.)
PROBLEM 8.36
PROBLEM 8.37
PROBLEM 8.37 (Cont.)
PROBLEM 8.38
PROBLEM 8.38 (Cont.)
PROBLEM 8.39
PROBLEM 8.39 (Cont.)
PROBLEM 8.40
PROBLEM 8.40 (Cont.)
PROBLEM 8.41
PROBLEM 8.41 (Cont.)
PROBLEM 8.42
PROBLEM 8.42 (Cont.)
PROBLEM 8.43
PROBLEM 8.43 (Cont.)
PROBLEM 8.44
PROBLEM 8.45
PROBLEM 8.46
PROBLEM 8.47
PROBLEM 8.47 (Cont.)
PROBLEM 8.48
PROBLEM 8.48 (Cont.)
PROBLEM 8.48 (Cont.)
PROBLEM 8.49
PROBLEM 8.49 (Cont.)
PROBLEM 8.49 (Cont.)
PROBLEM 8.50
PROBLEM 8.50 (Cont.)
PROBLEM 8.51
PROBLEM 8.51 (Cont.)
PROBLEM 8.52
PROBLEM 8.53
PROBLEM 8.53 (Cont.)
PROBLEM 8.53 (Cont.)
PROBLEM 8.54
PROBLEM 8.54 (Cont.)
PROBLEM 8.55
PROBLEM 8.56
PROBLEM 8.56 (Cont.)
PROBLEM 8.57
PROBLEM 8.58
PROBLEM 8.58 (Cont.)
PROBLEM 8.59
PROBLEM 8.59 (Cont.)
PROBLEM 8.60
PROBLEM 8.60 (Cont.)
PROBLEM 8.61
PROBLEM 8.62
PROBLEM 8.62 (Cont.)
PROBLEM 8.63
PROBLEM 8.63 (Cont.)
PROBLEM 8.64
PROBLEM 8.65
PROBLEM 8.66
PROBLEM 8.67
PROBLEM 8.67 (Cont.)
PROBLEM 8.68
PROBLEM 8.68 (Cont.)
PROBLEM 8.69
PROBLEM 8.69 (Cont.)
PROBLEM 8.70
PROBLEM 8.70 (Cont.)
PROBLEM 8.71 (a), (b)
Continued…
PROBLEM 8.71 (c)
PROBLEM 8.72
PROBLEM 8.72 (Cont.)
PROBLEM 8.73
PROBLEM 8.74
PROBLEM 8.75
PROBLEM 8.75 (Cont.)
PROBLEM 8.76
PROBLEM 8.76 (Cont.)
PROBLEM 8.76 (Cont.)
PROBLEM 8.77
PROBLEM 8.78
PROBLEM 8.78 (Cont.)
PROBLEM 8.79
PROBLEM 8.79 (Cont.)
PROBLEM 8.79 (Cont.)
PROBLEM 8.80
PROBLEM 8.80 (Cont.)
PROBLEM 8.80 (Cont.)
PROBLEM 8.80 (Cont.)
PROBLEM 8.81
PROBLEM 8.81 (Cont.)
PROBLEM 8.81 (Cont.)
PROBLEM 8.82
PROBLEM 8.83
PROBLEM 8.83 (Cont.)
PROBLEM 8.84
PROBLEM 8.85
PROBLEM 8.85 (Cont.)
PROBLEM 8.86
PROBLEM 8.86 (Cont.)
PROBLEM 8.87
PROBLEM 8.87 (Cont.)
PROBLEM 8.88
PROBLEM 8.89
PROBLEM 8.89 (Cont.)
PROBLEM 8.89 (Cont.)
PROBLEM 8.90
PROBLEM 8.90 (Cont.)
PROBLEM 8.91
PROBLEM 8.91 (Cont.)
PROBLEM 8.91 (Cont.)
PROBLEM 8.92
PROBLEM 8.93
PROBLEM 8.94
PROBLEM 8.95
PROBLEM 8.95 (Cont.)
PROBLEM 8.96
PROBLEM 8.96 (Cont.)
PROBLEM 8.96 (Cont.)
PROBLEM 8.97
PROBLEM 8.98
PROBLEM 8.98 (Cont.)
PROBLEM 8.99
PROBLEM 8.100
PROBLEM 8.100 (Cont.)
PROBLEM 8.101
PROBLEM 8.102
PROBLEM 8.102 (Cont.)
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.)