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.