Download 3-30 An exposed hot surface of an industrial natural gas furnace is

3-30 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through
that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also,
the length of time it will take for the insulation to pay for itself from the energy it saves will be determined.
Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant.
3 The furnace operates continuously. 4 The given heat transfer coefficient accounts for the radiation effects.
Properties The thermal conductivity of the glass wool insulation is given to be k = 0.038 W/m°C.
Analysis The rate of heat transfer without insulation is
A  (2 m)(1.5 m)  3 m2
Insulation
Q  hA(Ts  T )  (10 W / m2 . C)(3 m2 )(80  30) C  1500 W
In order to reduce heat loss by 90%, the new heat transfer rate and thermal
resistance must be
Q  010
.  1500 W  150 W
T
T (80  30) C
Q 

 Rtotal 

 0.333  C / W
Rtotal
150 W
Q
Rinsulation
Ro
T
Ts
L
and in order to have this thermal resistance, the thickness of insulation must be
1
L
Rtotal  Rconv  Rinsulation 

hA kA
1
L


 0.333 C/W
2
2
(10 W/m .C)(3 m ) (0.038 W/m. C)(3 m 2 )
L  0.034 m  3.4 cm
Noting that heat is saved at a rate of 0.91500 = 1350 W and the furnace operates continuously and thus 36524 =
8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year is
Q
t
(1.350 kJ/s)(8760 h)  3600 s  1 therm 
Energy Saved  saved

  517 .4 therms


Efficiency
0.78
 1 h  105,500 kJ 
The money saved is
Money saved  ( Energy Saved)(Cost of energy)  (517.4 therms)($0.55 / therm)  $284.5 (per year)
The insulation will pay for its cost of $250 in
Money spent
$250
Payback period 

 0.88 yr
Money saved $284.5 / yr
which is less than one year.
3-55 A wall is constructed of two layers of sheetrock spaced by 5 cm  12 cm wood studs. The
space between the studs is filled with fiberglass insulation. The thermal resistance of the wall and
the rate of heat transfer through the wall are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat
transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat
transfer coefficients account for the radiation heat transfer.
Properties The thermal conductivities are given to be k = 0.17 W/m°C for sheetrock, k = 0.11
W/m°C for wood studs, and k = 0.034 W/m°C for fiberglass insulation.
Analysis (a) The representative surface area is A  1  0.65  0.65 m2 . The thermal resistance
network and the individual thermal resistances are
Ri
T1
R2
R1
R3
R4
R5
T2
1
1

 0.185 C/W
2
hi A (8.3 W/m .C) (0.65 m 2 )
L
0.01 m
R1  R 4  R sheetrock 

 0.090 C/W
kA (0.17 W/m. C) (0.65 m 2 )
Ri 
L
0.12 m

 21 .818 C/W
kA (0.11 W/m. C) (0.05 m 2 )
L
0.12 m
R3  R fiberglass 

 5.882 C/W
kA (0.034 W/m. C) (0.60 m 2 )
R 2  R stud 
(
1
1

 0.045 C/W
2
o
ho A (34 W/m . C) (0.65 m 2 )
1
1
1
1





 R mid  4.633 C/W
R 2 R3 21 .818 5.882
Ro 
1
R mid
Rtotal  Ri  R1  R mid  R 4  Ro  0.185  0.090  4.633  0.090  0.045  4.858 C/W (for a 1 m  0.65 m section)
T T
[20  (5)]C
Q  1  2 
 5.15 W
Rtotal
4.858 C/W
b) Then steady rate of heat transfer through entire wall becomes
(12 m) (5 m)
Q total  (5.15 W)
 475 W
0.65 m 2
3-41C An interface acts like a very thin layer of insulation, and thus the thermal contact
resistance has significance only for highly conducting materials like metals. Therefore, the
thermal contact resistance can be ignored for two layers of insulation pressed against each other.
3-42C An interface acts like a very thin layer of insulation, and thus the thermal contact
resistance is significant for highly conducting materials like metals. Therefore, the thermal
contact resistance must be considered for two layers of metals pressed against each other.
3-48 A thin copper plate is sandwiched between two epoxy boards. The error involved in the
total thermal resistance of the plate if the thermal contact conductances are ignored is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the
plate is large. 3 Thermal conductivities are constant.
Properties The thermal conductivities are given to be k = 386 W/m°C for copper plates and k =
0.26 W/m°C for epoxy boards. The contact conductance at the interface of copper-epoxy layers
is given to be hc = 6000 W/m2C.
Analysis The thermal resistances of different layers for unit
surface area of 1 m2 are
Copp
er
1
1
Rcontact 

 0.00017 C/W
Epox
2
2
plate Epox
hc Ac (6000 W/m .C)(1 m )
y
y
Rplate 
L
0.001 m

 2.6  106  C / W
kA (386 W / m.  C)(1 m2 )
Repoxy 
L
0.005 m

 0.01923  C / W
kA (0.26 W / m.  C)(1 m2 )
Q
5 mm 5 mm
The total thermal resistance is
R total  2 Rcontact  R plate  2 Repoxy
 2  0.00017  2.6  10 6  2  0.01923  0.03914 C/W
Then the percent error involved in the total thermal resistance of the plate if the thermal
contact resistances are ignored is determined to be
%Error 
2Rcontact
2  0.00017
100 
100  0.87%
Rtotal
0.03914
which is negligible.
3-68 A steam pipe covered with 3-cm thick glass wool insulation is subjected to convection on its surfaces. The rate
of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2
Heat transfer is one-dimensional since there is thermal symmetry about the center line and no
variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact
resistance at the interface is negligible.
Properties The thermal conductivities are given to be k = 15 W/m°C for steel and k = 0.038
W/m°C for glass wool insulation
Analysis The inner and the outer surface areas of the insulated pipe per
unit length are
Ai  Di L   (0.05 m)(1 m)  0157
.
m2
Ao  Do L   (0.055  0.06 m)(1 m)  0.361 m2
Ri
The individual thermal resistances are
R1
R2
T1
Ro
T2
1
1
Ri 

 0.08 C/W
2
hi Ai (80 W/m .C) (0.157 m 2 )
ln( r2 / r1 )
ln( 2.75 / 2.5)
R1  R pipe 

 0.00101 C/W
2k1 L
2 (15 W/m. C) (1 m)
R 2  Rinsulation 
ln( r3 / r2 )
ln(5.75 / 2.75 )

 3.089 C/W
2k 2 L
2 (0.038 W/m. C) (1 m)
1
1

 0.1847 C/W
ho Ao (15 W/m 2 .C) (0.361 m 2 )
 Ri  R1  R 2  Ro  0.08  0.00101  3.089  0.1847  3.355 C/W
Ro 
Rtotal
Then the steady rate of heat loss from the steam per m. pipe length becomes
T T
(320  5) C
Q  1 2 
 93.9 W
Rtotal
3.355  C / W
The temperature drops across the pipe and the insulation are

Tpipe  QR
pipe  (93.9 W)(0.00101  C / W)  0.095 C

Tinsulation  QR
insulation  (93.9 W)(3.089  C / W)  290 C
3-108 A commercially available heat sink is to be selected to keep the case temperature of a transistor below 90C
in an environment at 20 C .
Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 90 C . 3 The contact
resistance between the transistor and the heat sink is negligible.
Analysis The thermal resistance between the transistor attached to the sink and
the ambient air is determined to be
R
Ts
T
T
 T (90  20) C
T
Q 

 Rcase  ambient  transistor

 1.75  C / W
Rcase  ambient
40 W
Q
The thermal resistance of the heat sink must be below 175
. o C / W . Table 3-4 reveals that HS6071 in vertical
position, HS5030 and HS6115 in both horizontal and vertical position can be selected.
3-111E The handle of a stainless steel spoon partially immersed in boiling water extends 7 in. in the air from the
free surface of the water. The temperature difference across the exposed surface of the spoon handle is to be
determined.
Assumptions 1 The temperature of the submerged portion of the spoon is equal to the water temperature. 2 The
temperature in the spoon varies in the axial direction only (along the spoon), T(x). 3 The heat transfer from the tip of
the spoon is negligible. 4 The heat transfer coefficient is constant and uniform over the entire spoon surface. 5 The
thermal properties of the spoon are constant. 6 The heat transfer coefficient accounts for the effect of radiation from
the spoon..
Properties The thermal conductivity of the spoon is given to be k = 8.7 Btu/hft°F.
Analysis Noting that the cross-sectional area of the spoon is constant and measuring x from the free surface of water,
the variation of temperature along the spoon can be expressed as
T ( x)  T cosh a ( L  x)

Tb  T
cosh aL
h, T
D
where
p  2(0.5 / 12 ft  0.08 / 12 ft )  0.0967 ft
Tb
Ac  (0.5 / 12 ft)(0.08 / 12 ft )  0.000278 ft 2
a
hp

kAc
(3 Btu / h.ft 2 .  F)(0.0967 ft )
(8.7 Btu / h.ft.  F)(0.000278 ft 2 )
L = 7 in
 10.95 ft -1
Noting that x = L = 7/12=0.583 ft at the tip and substituting, the tip temperature of the
spoon is determined to be
0
.
00
. 8
5
i
in
n
cosh a( L  L)
cosh aL
cosh 0
1
= 75 F + (200  75)
= 75 F + (200  75)
= 75.4 F
cosh(10 .95  0.583 )
296
T ( L)  T  (Tb  T )
Therefore, the temperature difference across the exposed section of the spoon handle is
T  Tb  Ttip  (200  75.4)F  124.6F
3-114 A circuit board houses 80 logic chips on one side, dissipating 0.04 W each through the back side of the board
to the surrounding medium. The temperatures on the two sides of the circuit board are to be determined for the cases
of no fins and 864 aluminum pin fins on the back surface.
Assumptions 1 Steady operating conditions exist. 2 The temperature in the board and along the fins varies in one
direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and
is dissipated from the back side of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer
coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7
The heat transfer coefficient accounts for the effect of radiation from the fins.
Properties The thermal conductivities are given to be k = 20 W/m°C for the circuit board, k = 237 W/m°C for the
aluminum plate and fins, and k = 1.8 W/m°C for the epoxy adhesive.
Analysis (a) The total rate of heat transfer dissipated by the chips is
Q  80  (0.04 W)  3.2 W
2 cm
The individual resistances are
Repoxy
Rboard
RAluminum
Rconv
T1
T2
T2
A  (012
. m)(018
. m)  0.0216 m2
L
0.003 m

 0.00694  C / W
kA (20 W / m.  C)(0.0216 m2 )
1
1


 0.9259  C / W
hA (50 W / m2 .  C)(0.0216 m2 )
Rboard 
Rconv
Rtotal  Rboard  Rconv  0.00694  0.9259  0.93284  C / W
The temperatures on the two sides of the circuit board are
T T

Q  1 2 
 T1  T2  QR
total  40 C  (3.2 W)(0.93284  C / W)  43.0 C
Rtotal
T T

Q  1 2 
 T2  T1  QR
board  43.0 C  (3.2 W)(0.00694  C / W)  40.5  0.02  43.0 C
Rboard
Therefore, the board is nearly isothermal.
(b) Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined
to be
hp

kAc
a
hD
kD / 4
2

4h

kD
4(50 W / m2 .  C)
 18.37 m-1
(237 W / m.  C)(0.0025 m)
tanh aL tanh(18.37 m-1  0.02 m)

 0.957
aL
18.37 m-1  0.02 m
The fins can be assumed to be at base temperature provided that the fin area is modified by multiplying it by 0.957
(or alternately use the method in parenthesis below). Then the various thermal resistances are
 fin 
L
0.0002 m

 0.0051  C / W
kA (18
. W / m.  C)(0.0216 m2 )
L
0.002 m


 0.00039  C / W
kA (237 W / m.  C)(0.0216 m2 )
Repoxy 
RAl
Afinned   fin nDL  0.957  864 (0.0025 m)(0.02 m)  0130
.
m2
Aunfinned  0.0216  864
D 2
4
Atotal,with fins  Afinned  Aunfinned
Rconv 
(or use Rcon v 
Rtotal
 0.0216  864 
 (0.0025) 2
4
 0130
.
 0.017  0147
.
m2
 0.0174 m2
1
1

 01361
.
C / W
hAtotal,with fins (50 W / m2 .  C)(0147
.
m2 )
1
and overall 
NAf
Qtotal
 1
1   f  ; you should get the same answer)
hAt b
At
overall hAt
 Rboard  Repoxy  Raluminum  Rconv
 0.00694  0.0051  0.00039  01361
.
 01484
.
C / W
Then the temperatures on the two sides of the circuit board becomes
T T

Q  1 2 
 T1  T2  QR
.
 C / W)  40.5 C
total  40 C  (3.2 W)( 01484
Rtotal
T T

Q  1 2 
 T2  T1  QR
board  40.5 C  (3.2 W)(0.00694  C / W)  40.5  0.02  40.5 C
Rboard
3-124 Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between
the pipes is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial
direction). 3 Thermal conductivity of the concrete is constant.
T1 = 60C
Properties The thermal conductivity of concrete is given to be k =
0.75 W/m°C.
T2 = 15C
Analysis The shape factor for this configuration is given in Table 3-5
to be
S

2L
 4 z  D1 2  D 2 2 

cosh 1 


2
D
D
1
2


2(8 m)
2
z = 40 cm
 9.078 m
 4(0.4 m) 2  (0.05 m) 2  (0.05 m) 2 

cosh


2(0.05 m)( 0.05 m)


Then the steady rate of heat transfer between the pipes becomes
Q  Sk (T  T )  (9.078 m)(0.75 W/m.C)(60  15)C  306 W
1 
1
2
D = 5 cm
L=8m