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PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
Fourth Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2011
Chapter 3
STEADY HEAT CONDUCTION
PROPRIETARY AND CONFIDENTIAL
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re, mperature.
Steady Heat Conduction in Plane Walls
3-1C The temperature distribution in a plane wall will be a straight line during steady and one dimensional heat transfer with constant wall thermal conductivity.
3-2C In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out of it. Also, the temperature at any point in the wall remains constant. Therefore, the energy content of the wall does not change during steady heat conduction. However, the temperature along the wall and thus the energy content of the wall will change during transient conduction.
3-3C Convection heat transfer through the wall is expressed as = hA (^) s ( Ts − T ∞). In steady heat transfer, heat transfer rate
to the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer coefficient three times that of the inner surface will experience three times smaller temperature drop compared to the inner surface. Therefo at the outer surface, the temperature will be closer to the surrounding air te
Q
3-4C The new design introduces the thermal resistance of the copper layer in addition to the thermal resistance of the aluminum which has the same value for both designs. Therefore, the new design will be a poorer conductor of heat.
3-5C (a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod is the bottom or
the top surface area of the rod,. (b) If the top and the bottom surfaces of the rod are insulated, the heat transfer
area of the rod is the lateral surface area of the rod,
As = π D^2 / 4
A = π DL.
3-6C The thermal resistance of a medium represents the resistance of that medium against heat transfer.
3-7C The combined heat transfer coefficient represents the combined effects of radiation and convection heat transfers on a surface, and is defined as h combined = h convection + h radiation. It offers the convenience of incorporating the effects of radiation in the convection heat transfer coefficient, and to ignore radiation in heat transfer calculations.
3-8C Yes. The convection resistance can be defined as the inverse of the convection heat transfer coefficient per unit surface area since it is defined as R (^) conv = 1 /( hA ).
3-9C The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat transfers occur simultaneously.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
A
3-17 The two surfaces of a wall are maintained at specified temperatures. The rate of heat loss through the wall is to be determined.
Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant.
Wall
5 °C
Q &
L = 0.25 m
Properties The thermal conductivity is given to be k = 0.8 W/m⋅°C.
Analysis The surface area of the wall and the rate of heat loss through the wall are (^14) °C
= ( 3 m)×( 6 m)= 18 m^2
= 518 W
- 25 m
( 14 5 )C
(^12) ( 0. 8 W/m C)(18m (^2) ) L
T T
Q & kA
3-18 Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface temperature of the bottom of the pan is given. The boiling heat transfer coefficient and the outer surface temperature of the bottom of the pan are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the thickness of the bottom of the pan is small relative to its diameter. 3 The thermal conductivity of the pan is constant.
Properties The thermal conductivity of the aluminum pan is given to be k = 237 W/m⋅°C.
Analysis ( a ) The boiling heat transfer coefficient is
2
2 2
- 0491 m 4
( 0. 25 m) 4
π D π
As 95 °C 108 °C
800 W
= 1254 W/m^2. ° C A (^) s ( Ts − T ∞ ) ( 0. 0491 m^2 )( 108 − 95 ) °C
800 W
Q
h
Q hAsTs T &
The ou r surface temperature of the bottom of the pan is
0.5 cm
( b ) te
= 108.3 ° C
( 237 W/m C)(0.0491m )
( 800 W)( 0. 005 m) , , 1 108 C+ 2
, ,
kA
QL
T T
L
T T
Q kA
souter sinner
souter sinner
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3-19 The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through the window and the inner surface temperature are to be determined.
Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant. 4 Heat transfer by radiation is negligible.
Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C.
Analysis The area of the window and the individual resistances are
A =( 1. 5 m)×( 2. 4 m)= 3. 6 m^2 L
Glass
T 1
Q &
0. 02778 0. 00214 0. 01111 0. 04103 C/W
0. 01111 C/W
2 (^25 W/m^2 .C)(^3.^6 m^2 )
o , 2 = ° °
= (^) conv = = hA
R R
000214 C/W
( 0. 78 W/m.C)( 3. 6 m )
- 006 m
0. 02778 C/W
, 1 , 2
2 1
glass
Rtal Rconv Rglass R conv
kA
L
R
R R
he steady rate of heat transfer through window glass is then
1 (^10 W/m^2 .C)(^3.^6 m^2 )
i , 1 ° conv (^) hA
to
Ri R glass Ro T (^) T ∞ 1 T ∞ 2
= 707 W
0. 04103 C/W
1 2 [^24 (^5 )]C
R total
T T
Q &
The inner surface temperature of the window glass can be determined from
⎯⎯→ = − = ° − ° = 4.4 ° C
= ∞^1 ∞ 1 , 1 24 C ( 707 W)(0.0277 8 C/W)
, 1
1 1 conv conv
T T QR
R
T T
Q & &
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3-21 A double-pane window consists of two layers of glass separated by an evacuated space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined.
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of the glass is constant. 4 Heat transfer by radiation is negligible.
Properties The thermal conductivity of the glass is given to be k glass = 0.78 W/m⋅°C.
Analysis Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum is zero (no medium to conduct heat) and thus its thermal resistance is zero. Therefore, if radiation is disregarded, the heat transfer through the window will be zero. Then the answer of this problem is zero since the problem states to disregard radiation.
Vacuum
Discussion In reality, heat will be transferred between the glasses by radiation. We do not know the inner surface temperatures of windows. In order to determine radiation heat resistance we assume them to be 5°C and 15°C, respectively, and take the emissivity to be 1. Then individual resistances are
A =( 1. 5 m)×( 2. 4 m)= 3. 6 m^2
0. 09505 C/W
0. 01111 C/W
( 25 W/m. C)( 3. 6 m)
0. 05402 C/W
1 ( 5. 67 108 W/m^2 .K^4 )( 3. 6 m^2 )[ 2882 2782 ][ 288 278 ]^3
× − + +
R R
K
0. 00107 C/W
( 0. 78 W/m.C)( 3. 6 m )
- 003 m
0. 02778 C/W
, 1 1 , 2
o 2 o 2 2
o , 2
2 1
1 1 3 glass
i , 1
total conv rad conv
conv
conv
R R R R R
hA
kA
L
R R R
R
The steady rate of heat transfer through window glass then becomes
Ri R 1 R rad R 3 Ro R T ∞^1 T ∞^2 h 1 A ( 10 W/m^2 .°C)( 3. 6 m^2 )
( 2 +^2 )( + )
s surr s surr
rad AT T T T
R
= 274 W
0. 09505 C/W
1 2 [^21 (^5 )]C
R total
T T
Q &
The inner surface temperature of the window glass can be determined from
⎯⎯→ = − = ° − ° = 13.4 ° C
= ∞^1 ∞ 1 , 1 21 C ( 274 W)(0.0277 8 C/W)
, 1
1 1 conv conv
T T QR
R
T T
Q & &
Similarly, the inner surface temperatures of the glasses are calculated to be 13.1 and -1.7°C (we had assumed them to be 15 and 5°C when determining the radiation resistance). We can improve the result obtained by reevaluating the radiation resistance and repeating the calculations.
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3-22 Prob. 3-20 is reconsidered. The rate of heat transfer through the window as a function of the width of air space is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
A=1.5*2.4 [m^2] L_glass=3 [mm] k_glass=0.78 [W/m-C] L_air=12 [mm] T_infinity_1=21 [C] T_infinity_2=-5 [C] h_1=10 [W/m^2-C] h_2=25 [W/m^2-C]
"PROPERTIES" k_air=conductivity(Air,T=25)
"ANALYSIS" R_conv_1=1/(h_1A) R_glass=(L_glassConvert(mm, m))/(k_glassA) R_air=(L_airConvert(mm, m))/(k_airA) R_conv_2=1/(h_2A) R_total=R_conv_1+2*R_glass+R_air+R_conv_ Q_dot=(T_infinity_1-T_infinity_2)/R_total
Lair [mm]
Q
[W]
2 4 6 8 10 12 14 16 18 20
100
150
200
250
300
350
400
450
L (^) air [mm]
Q
[W]
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
3-25 A very thin transparent heating element is attached to the inner surface of an automobile window for defogging purposes, the inside surface temperature of the window is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 Thermal resistance of the thin heating element is negligible.
Properties Thermal conductivity of the window is given to be k = 1.2 W/m · °C.
Analysis The thermal resistances are
h A
R
i
i
hA
R
o
o
= and kA
L
R (^) win=
From energy balance and using the thermal resistance concept, the following equation is expressed:
o
o h i
∞, i − R R
T T
qA R
T T
win
or 1 /( ) /( ) 1 /( )
, 1 1 , L kA h A
T T
qA hA
T T
o
o h i
i
∞ −^ ∞
o
o h i
i Lk h
T T
q h
T T
, 1 1 ,
∞ −^ ∞
1 / 15 W/m^2 ⋅°C ( 0. 005 m/ 1. 2 W/m⋅°C)+( 1 / 100 W/m C)
( 5 C )
1300 W/m
22 C 1 2 1 − − °
° − T T
equation: (2 -T_1)/(1/15)+1300=(T_1-(-5))/(0.005/1.2+1/100)
Discussion In actuality, the ambient temperature and the convective heat transfer coefficient outside the automobile vary with weather conditions and the automobile speed. To maintain the inner surface temperature of the window, it is necessary to vary the heat flux to the heating element according to the outside condition.
Copy the following line and paste on a blank EES screen to solve the above
2
Solving by EES software, the inside surface temperature of the window is
T 1 = 14.9 ° C
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
3-26 A process of bonding a transparent film onto a solid plate is taking place inside a heated chamber. The temperatures inside the heated chamber and on the transparent film surface are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 Thermal contact resistance is negligible.
Properties The thermal conductivities of the transparent film and the solid plate are given to be 0.05 W/m · °C and 1.2 W/m · °C, respectively.
Analysis The thermal resistances are
hA
R
conv =
k A
L
R
f
f f =
and k (^) sA s
Using the thermal resistance conc
L
R = s
ept, the following e uation is expressed: q
s
b f
b R
T T
R R
T T 2
conv
R
inside the cham
earranging and solving for the temperature ber yields
( ) (^) b f
f Rs Ls / k s ⎝
b f b b (^) T k
L
h
T T
R R T
T T
T ⎟⎟+
conv 2 ∞
⎟+ ° = 127 ° C
∞ =^70 C
05 W/m C
001 m 70 W/m C
- 013 m/ 1. 2 W/m C
( 70 52 ) C
T 2
T he surfac etemperature of the transparent film is
s
b f
b R
T T
R
T 1 T − 2
b f
f s s
b f b s
b (^) T k
L
L k
T T
R T
R
T T
T ⎟⎟+
2 2 1
⎟+ ° =^103 °^ C
= 70 C
05 W/m C
001 m
013 m/ 1. 2 W/m C
( 70 52 ) C
T 1
Discussion If a thicker transparent film were to be bonded on the solid plate, then the inside temperature of the heated chamber would have to be higher to maintain the temperature of the bond at 70 °C.
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3-29 A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding air and surfaces. For a given deep body temperature, the outer skin temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat transfer coefficient is constant and uniform over the entire exposed surface of the person. 3 The surrounding surfaces are at the same temperature as the indoor air temperature. 4 Heat generation within the 0.5-cm thick outer layer of the tissue is negligible.
Q conv
T skin
Q rad
Properties The thermal conductivity of the tissue near the skin is given to be k = 0.3 W/m⋅°C.
Analysis The skin temperature can be determined directly from
= 35.5 ° C
( 0. 3 W/m C)(1.7m )
( 150 W)( 0. 005 m) 1 37 C 2
1
kA
QL
T T
L
T T
Q kA
skin
skin
&
3-30 A double-pane window is considered. The rate of heat loss through the window and the temperature difference across
Properties The thermal conductivities of glass and air are given to be 0.78 W/m⋅K and 0.025 W/m⋅K, respectively.
Analysis ( a ) The rate of heat transfer through the window is determined to be
the largest thermal resistance are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer coefficients are constant.
[ ]
[ ] = 210 W
× °
40 W/m ⋅ °C^0.^78 W/m⋅°C^0.^025 W/m⋅°C^0.^78 W/m⋅°C 20 W/m ⋅ °C 2
2 2
× °
( 1 1. 5 m ) 20 - (-20) C
1 0. 004 m 0. 005 m 0. 004 m 1
( 1 1. 5 m ) 20 - (-20) C
2
g o
g a
a g
g i k h
L
k
L
k
L
h
A T
Q &
( b ) Noting that the largest resistance is through the air gap, the temperature difference across the air gap is determined from
= 28 ° C
⋅° ×
( 0. 025 W/m C)( 1 1. 5 m )
- 005 m ( 210 W) 2 k A
L
T QR Q
a
a a & a &
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3-31E A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal resistance of the wall and its R-value of insulation are to be determined.
Assumptions 1 Heat transfer through the wall is one-dimensional. 2 Thermal conductivities are constant.
Properties The thermal conductivities are given to be k sheetrock = 0.10 Btu/h⋅ft⋅°F and k insulation = 0.020 Btu/h⋅ft⋅°F.
L 1 L 2 L 3
R 1 R 2 R 3
Analysis ( a ) The surface area of the wall is not given and thus we consider a unit surface area ( A = 1 ft^2 ). Then the R -value of insulation of the wall becomes equivalent to its thermal resistance, which is determined from.
= + = × + = 30.17 ft^2. ° F.h/Btu
- 17 ft .F.h/Btu ( 0. 020 Btu/h.ft.F)
7 / 12 ft
500 ft .F.h/Btu ( 0. 10 Btu/h.ft.F)
6 / 12 ft
1 2
2 2
2 2
2 1
1 1 3
R R R
k
L
R R
k
L
R R R
total
fiberglass
sheetrock
( b ) Therefore, this is approximately a R -30 wall in English units.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
3-33 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
Insulation
L
Ts
R insulation Ro
A =( 2 m)(1.5m)= 3 m^2
Q^ & = hA ( T (^) s − T ∞)=( 10 W/m^2 ⋅°C)( 3 m^2 )( 110 − 32 )°C= 2340 W
In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be (^) T ∞
0. 333 C/W
( 110 32 )C
T
R
T
Q^ & (^) total 234 W
0. 10 × 2340 W= 234 W
R Q
Q
total &
and in order to have this thermal resistance, the thickness of insulation must be
= = 3.4 cm
- 034 m
( 10 W/m^2 C)(3m^2 ) ( 0. 038 W/m.C)(3m^2 ) L
0. 333 C/W
conv L
kA
L
hA
R (^) total R Rinsulation
oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h furnace efficiency is 78%, the amount of natural gas saved per year is
N t h per year, and that the
- 1 therms 105,500kJ
1 therm 1 h
3600 s
( 2. 106 kJ/s)(8760h) Energy Saved ⎟⎟= ⎠
Efficiency
Q^ & (^) saved t
he money aved is gy = =
he insulation will pay for its cost of $250 in
T s
Money saved= ( Ener Saved)(Costofenergy) ( 807. 1 therms)($1.10/therm) $ 887. 8 (peryear)
T
= = = 0.282 yr $887.8/yr
Moneysaved
Moneyspent Paybackperiod
which is equal to 3.4 months.
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Q hATs T ∞
3-34 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 coefficients accounts for the radiation effects.
Properties The thermal conductivity of the expanded perlite insulation is given to be k = 0.052 W/m⋅°C.
Analysis The rate of heat transfer without insulation is
Insulation
L
Ts
R insulation Ro
A =( 2 m)(1.5m)= 3 m^2
= ( − )=( 10 W/m^2 ⋅°C)( 3 m^2 )( 110 − 32 )°C= 2340 W
In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be T ∞
0. 333 C/W
( 110 32 )C
T
R
T
Q^ & (^) total 234 W
0. 10 × 2340 W= 234 W
R Q
Q
total &
and in order to have this thermal resistance, the thickness of insulation must be
= = 4.7 cm
- 047 m
( 10 W/m^2 C)(3m^2 ) ( 0. 052 W/m C)(3m^2 ) L
0. 333 C/W
conv L
kA
L
hA
R (^) total R Rinsulation
oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h furnace efficiency is 78%, the amount of natural gas saved per year is
N t h per year, and that the
- 1 therms 105,500kJ
1 therm 1 h
3600 s
( 2. 106 kJ/s)(8760h) Energy Saved ⎟⎟= ⎠
Efficiency
Q^ & (^) saved t
oney
gy
The m saved is
Money saved= ( Ener Saved)(Costofenergy)=( 807. 1 therms)($1.10/therm)=$ 887. 8 (peryear)
T he insulation will pay for its cost of $250 in
= = = 0.282 yr $887.8/yr
Moneysaved
Moneyspent Paybackperiod
which is equal to 3.4 months.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
3-36 Two of the walls of a house have no windows while the other two walls have single- or double-pane windows. The average rate of heat transfer through each wall, and the amount of money this household will save per heating season by converting the single pane windows to double pane windows are to be determined.
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is disregarded.
Properties The thermal conductivities are given to be k = 0.026 W/m⋅°C for air, and 0.78 W/m⋅°C for glass.
Analysis The rate of heat transfer through each wall can be determined by applying thermal resistance network. The convection resistances at the inner and outer surfaces are common in all cases.
Walls without windows: Wall
0. 003571 0. 05775 0. 001389 0. 06271 C/W
0. 001389 C/W
( 18 W/m^2 C)( 10 4 m )
o
R (^) o (^) h A
0. 05775 C/W
( 10 4 m )
- 31 m C/W
0. 003571 C/W
( 7 W/m C)( 10 4 m )
tal wall
2
2
2 wall wall
2 2
⋅° ×
×
⋅° ×
i o
i
i
R R R R
A
R value kA
L
R
hA
R
Q &
L
to
Then
Ri R wall Ro = 255.1 W °
0. 06271 C/W
1 2 (^248 )C
R total
T T
Q &
Wall with single pane windows:
0. 001786 0. 000583 0. 000694 0. 003063 C/W
0. 000694 C/W
( 18 W/m C)( 20 4 m )
0. 000583 C/W
0. 002968 C/W
( 0. 78 W/m C)( 1. 2 1. 8 )m
- 005 m
0. 033382 C/W
( 20 4 ) 5 ( 1. 2 1. 8 )m
- 31 m C/W
0. 001786 C/W
( 7 W/m C)( 20 4 m )
total eqv
2 2
o eqv wall glass
2 o 2
glass glass
2
2 wall wall
2 2
⋅° ×
⋅ ×
× − ×
⋅° ×
i o
o
o
eqv
i
i
R R R R
h A
R
R
R R R
kA
L
R
A
R value kA
L
R
hA
R
R i R wall R o
R glass
Then
= 5224 W
0. 003063 C/W
( 24 8 )C
total
1 2 R
T T
Q &
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4th wall with double pane windows:
R wall
R glass R air R glass
R i R o
0. 001786 0. 020717 0. 000694 0. 023197 C/W
0. 020717 C/W
eqv
= + = + ⎯⎯→ R = °
R 0. 033382
2 2 0. 002968 0. 267094 0. 27303 C/W
0. 002968 C/W
( 0. 78 W/m C)( 1. 2 1 )m
- 005 m
0. 033382 C/W
( 20 4 ) 5 ( 1. 2 1. 8 )m
- 31 m C/W
total eqv
eqv wall window
glass air
2 2
glass glass
2
2 wall wall
= + = × + = °
⋅° ×
× − ×
R Ri R R o
R R
R R R
kA
L
R
A
R value kA
L
R
0. 267094 C/W
( 0. 026 W/m C)( 1. 2 1. 8 )m
- 015 m 2 o 2
air air = ° ⋅ ×
kA
L
R
window
Then
= 690 W
0. 023197 C/W
( 24 8 )C
total
1 2 R
T T
Q &
The rate of heat transfer which will be saved if the single pane windows are converted to double pane windows is
The amount of energy and money saved during a 7-month long heating season by switching from single pane to double pane windows become
Money savings = (Energy saved)(Unit cost of energy) = (22,851 kWh)($0.08/kWh) = $
5224 690 4534 W
pane
double pane
Q &save (^) = Q &single− Q & = − =
Q (^) save = Q & save ∆ t =( 4. 534 kW)(7× 30 × 24 h)=22,851 kWh
