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 =hAs(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,
4 / 2 D As =π DL A=π .
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 hcombined = hconvection + hradiation. 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 Rconv =1 hA/( ).
3-9C The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat
r coefficient is he
ce will be
3-10C For a surface of A at which the convection and radiation heat transfer coefficients are conv and hrad, the single
equivalent heat transfe qv =hconv+hrad when the medium and the surrounding surfaces are at the same
temperature. Then the equivalent thermal resistan
h
) /(
1 h A
Reqv= eqv .
3-11C The thermal resistance network associated with a five-layer composite wall involves five single-layer resistances
connected in series.
3-12C Once the rate of heat transfer is known, the temperature drop across any layer can be determined by multiplying heat transfer rate by the thermal resistance across that layer,
Q&
layer layer QR
T = & ∆
3-13C The temperature of each surface in this case can be determined from
) ( / ) ( ) ( / ) ( 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 ∞ − ∞ ∞ − ∞ − ∞ ∞ − ∞ ∞ + = ⎯→ ⎯ − = − = ⎯→ ⎯ − = s s s s s s s s R Q T T R T T Q R Q T T R T T Q & & & &
where R∞−i is the thermal resistance between the environment ∞ and surface i.
3-14C Yes, it is.
window, and thus the heat sfer rate will be smaller relative to the one which consists of a single 8 mm thick glass sheet.
nsfer and slow down the heat gain of the drink wrapped in a lanket. Therefore, the drink left on a table will warm up faster.
3-15C The window glass which consists of two 4 mm thick glass sheets pressed tightly against each other will probably have
thermal contact resistance which serves as an additional thermal resistance to heat transfer through tran
3-16C The blanket will introduce additional resistance to heat tra
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 =(3m)×(6m)=18m2 W 518 = ° − ° ⋅ = − = m 25 . 0 C ) 5 14 ( ) m C)(18 W/m 8 . 0 ( 2 2 1 L T T kA Q&
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 m 0491 . 0 4 m) 25 . 0 ( 4 = = =πD π As 95°C 108°C 800 W C . W/m 1254 2 ° = ° − − ∞) (0.0491m )(108 95) C (T T 2 As s = = − = ∞ W 800 ) ( Q h T T hA Q s s & &
The ou r surface temperature of the bottom of the pan is
0.5 cm (b) te C 108.3° = ° ⋅ ° = + = − = ) m C)(0.0491 W/m 237 ( m) 005 . 0 W)( 800 ( + C 108 2 1 , , , , kA L Q T T L T T kA Q inner s outer s inner s outer s & &
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
2 m 6 . 3 m) 4 . 2 ( m) 5 . 1 ( × = = A L Glass T1
Q&
C/W 04103 . 0 01111 . 0 00214 . 0 02778 . 0 C/W 01111 . 0 ) m 6 . 3 ( C) . W/m 25 ( 2 2 2 2 , o = ° ° = = = conv A h R R 1 1 C/W 00214 0 ) m 6 . 3 ( C) W/m. 78 . 0 ( m 006 . 0 C/W 02778 . 0 1 1 2 , 1 , 2 1 glass ° = + + = + + = ° = ° = = ° = = = = conv glass conv tal R R R R A k L R R Rhe steady rate of heat transfer through window glass is then ) m 6 . 3 ( C) . W/m 10 ( 2 2 1 1 , i ° conv h A . to Rglass Ri Ro T T∞1 T∞2 W 707 = ° ° − − = − = ∞ ∞ C/W 04103 . 0 C )] 5 ( 24 [ 2 1 total R T T Q&
The inner surface temperature of the window glass can be determined from
C 4.4° = ° − ° = − = ⎯→ ⎯ − = ∞ 1 ∞1 ,1 24 C (707 W)(0.02778 C/W) 1 , 1 1 conv conv R Q T T R T T Q& &
3-20 A double-pane window consists of two layers of glass separated by a stagnant air 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 conductivities of the glass and air are constant. 4 Heat transfer by radiation is negligible.
Air
Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/m⋅°C and kair = 0.026 W/m⋅°C.
Analysis The area of the window and the individual resistances are A=(1.5m)×(2.4m)=3.6m2 C/W 16924 . 0 01111 . 0 12821 . 0 ) 00107 . 0 ( 2 02778 . 0 2 C/W 01111 . 0 ) m 6 . 3 ( C) . W/m 25 ( 1 1 C/W 12821 . 0 ) m 6 . 3 ( C) W/m. 026 . 0 ( 2 2 2 2 = ° ° = = = air A k R R 0.012m C/W 00107 . 0 ) 6 . 3 ( C) W/m. 78 . 0 ( m 003 . 0 C/W 02778 . 0 1 1 2 , 2 1 1 , o 2 o 2 2 2 , o 2 1 1 glass 3 1 1 , i ° = + + + = + + + = = = = = ° = ° = = = = ° = = = = conv conv total conv conv R R R R R A h R R L A k L R R R R
The steady rate of heat transfer through window glass then becomes
R1 R2 R3 Ri Ro T∞1 T∞2 R ) m 6 . 3 ( C) . W/m 10 ( 2 2 1A ° h m W 154 = ° ° − − = − = ∞ ∞ C/W 16924 . 0 C )] 5 ( 21 [ 2 1 total R T T Q&
The inner surface temperature of the window glass can be determined from
C 16.7° ° − ° = − = ⎯→ ⎯ − = ∞ 1 ∞1 ,1 21C (154 W)(0.02778 C/W)= 1 , 1 1 conv conv R Q T T R T T Q& &
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 kglass = 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 2 m 6 . 3 m) 4 . 2 ( m) 5 . 1 ( × = = A C/W 09505 . 0 01111 . 0 05402 . 0 ) 00107 . 0 ( 2 02778 . 0 2 C/W 01111 . 0 ) m 6 . 3 ( C) . W/m 25 ( 1 1 C/W 05402 . 0 ] 278 288 ][ 278 288 )[ m 6 . 3 ( ) .K W/m 10 67 . 5 ( 1 8 2 4 2 2 2 3 = = = ° = + + × − R R K 1 1 C/W 00107 . 0 ) m 6 . 3 ( C) W/m. 78 . 0 ( m 003 . 0 C/W 02778 . 0 1 1 2 , 1 1 , o 2 o 2 2 2 , o 2 1 1 glass 3 1 1 , i ° = + + + = + + + = = = ° = ° = = = = ° = = = = conv rad conv total conv conv R R R R R A h A k L R R R R
The steady rate of heat transfer through window glass then becomes
R1 Rrad R3 Ri Ro T∞1 T∞2 R ) m 6 . 3 ( C) . W/m 10 ( 2 2 1A ° h ) )( ( 2 + 2 + = surr s surr s rad T T T T A R εσ W 274 = ° ° − − = − = ∞ ∞ C/W 09505 . 0 C )] 5 ( 21 [ 2 1 total R T T Q&
The inner surface temperature of the window glass can be determined from
C 13.4° = ° − ° = − = ⎯→ ⎯ − = ∞ 1 ∞1 ,1 21 C (274 W)(0.02778C/W) 1 , 1 1 conv conv R Q T T 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.
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_1*A) R_glass=(L_glass*Convert(mm, m))/(k_glass*A) R_air=(L_air*Convert(mm, m))/(k_air*A) R_conv_2=1/(h_2*A) R_total=R_conv_1+2*R_glass+R_air+R_conv_2 Q_dot=(T_infinity_1-T_infinity_2)/R_total Lair [mm] Q [W] 2 4 6 8 10 12 14 16 18 20 414 307.4 244.5 202.9 173.4 151.4 134.4 120.8 109.7 100.5 2 4 6 8 10 12 14 16 18 20 100 150 200 250 300 350 400 450 Lair [mm] Q [W ]
A
3-23E The inner and outer surfaces of the walls of an electrically heated house remain at specified temperatures during a
winter day. The amount of heat lost from the house that day and its cost are to be determined.
Assumptions 1 Heat transfer through the walls is steady since the surface temperatures of the walls remain constant at the
specified values during the time period considered. 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 walls is constant.
Properties The thermal conductivity of the brick wall is given to be k = 0.40 Btu/h⋅ft⋅°F. Analysis We consider heat loss through the walls only. The total heat transfer area is
=2(50×9+35×9)=1530ft2
Wall
T2
Q&
LThe rate of heat loss during the daytime is
6120Btu/h ft 1 F ) 45 55 ( ) ft F)(1530 ft Btu/h 40 . 0 ( 2 2 1 day = ° − ° ⋅ ⋅ = − = L T T kA Q&
The rate of heat loss during nighttime is
T1 Btu/h 240 , 12 ft 1 C ) 35 55 ( ) ft F)(1530 ft Btu/h 40 . 0 ( 2 2 1 night L = ° − ° ⋅ ⋅ = − =kAT T Q&
The amount of heat loss from the house that night will be
Btu 232,560 = + = + = ∆ = ⎯→ ⎯ = Q Q Q t 10Qday 14Qnight
Q& & & &
∆ ) Btu/h 240 , 12 ( h) 14 ( ) Btu/h 6120 ( h) 10 ( t
hen the cost of this heat loss for that day becomes T
$6.13
= =(232,560/3412kWh)($0.09/kWh)
Cost
3-24 A cylindrical resistor on a circuit board dissipates 0.15 W of power steadily in
heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resist
a specified environment. The amount of or are to be determined.
ssumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the resistor.
pates during a 24-hour period is h) W)(24 15 . 0 ( t Q Q & A
Analysis (a) The amount of heat this resistor dissi
Resistor 0.15 W
Q&
Wh 3.6 = = ∆ =(b) The heat flux on the surface of the resistor is
2 2 2 m 000127 . 0 m m)(0.012 003 . 0 ( 4 m) 003 . 0 ( 2 4 2 + = + = D πDL π As π π )= = = =1179 W/m2 2 m 000127 . 0 W 15 . 0 s A Q q& &
(c) The surface temperature of the resistor can be determined from
W 15 . 0
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
A h R i i = 1 R h A o o = 1 and kA L Rwin=
From energy balance and using the thermal resistance concept, the following equation is expressed:
o o h i i− ∞, R R T T A q R T T + − = + ∞ win , 1 1 & or ) /( 1 ) /( ) /( 1 , 1 1 , A h kA L T T A q A h T T o o h i i + − = + − ∞ ∞ & o o h i i h k L T T q h T T / 1 / / 1 , 1 1 , + − = + − ∞ ∞ & C) W/m 100 / 1 ( C) W/m 2 . 1 / m 005 . 0 ( C W/m 15 / 1 2⋅° ⋅° + ) C 5 ( W/m 1300 C 22° −T1 + 2= T1− − ° 2⋅° 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
C 14.9°
=
1
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 Rconv = 1 A k L R f f f = and A ks s
Using the thermal resistance conc
L R = s
ept, the llowing e uation is expressed:
fo 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 s s s L k R / ⎝ b b f b T k L h T T T R R T T T ⎟⎟+ ⎠ ⎞ ⎜ ⎜ ⎛ + − = + + − = 2 1 conv 2 ∞ C 127° = ° + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° ⋅ + ° ⋅ ° ⋅ ° − = ∞ 70 C C W/m 05 . 0 m 001 . 0 C W/m 70 1 C W/m 2 . 1 / m 013 . 0 C ) 52 70 ( 2 The surfac temperature of the transparent film is
T e s b f b R T T R T T1− = − 2 b f f s s b b f s b T k L k L T T T R R T T T ⎟⎟+ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = + − = / 2 2 1 C 103° = ° + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° ⋅ ° ⋅ ° − = 70 C C W/m 05 . 0 m 001 . 0 C W/m 2 . 1 / m 013 . 0 C ) 52 70 ( 1 T
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.
3-27 A power transistor dissipates 0.15 W of power steadily in a specified environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the transistor.
Analysis (a) The amount of heat this transistor dissipates during a 24-hour period is Air, 30°C kWh 0.0036 = = = ∆ =Q t (0.15 W)(24h) 3.6 Wh Q &
(b) The heat flux on the surface of the transistor is
2 2 2 m 0001021 . 0 m) m)(0.004 005 . 0 ( 4 m) 005 . 0 ( 2 4 2 = + = + = π π π π DL D As Power Transistor 0.15 W 2 W/m 1469 = = = 2 m 0001021 . 0 s A q& Q& 0.15 W
(c) The surface temperature of the transistor can be determined from
C 111.6° = ° ⋅ + ° = + = ⎯→ ⎯ − = ∞ ∞ ) m 21 C)(0.00010 W/m (18 W 15 . 0 C 30 ) ( 2 2 s s s s T T T T hAQ hA Q& &
3-28 A circuit board houses 100 chips, each dissipating 0.06 W. The surface heat flux, the surface temperature of the chips, and the thermal resistance between the surface of the board and the cooling medium are to be determined.
1 Steady operating conditions exist. 2
Assumptions
transferred uniformly from the entire front surface.
Heat transfer from the back surface of the board is negligible. 2 Heat is
nalysis (a The heat flux on the surface of the circuit board is
A ) Chips Ts
Q&
T∞ 2 m 0216 . 0 m) m)(0.18 12 . 0 ( = = s A 2 W 278 = = = 2 m 0216 . 0 s A q& Q& (100×0.06) W /me sur ce temperature of the chips is
(b) Th fa − =hAs(Ts T∞) Q& C 67.8° = ° ⋅ × ° = + = ∞ ) m 0216 . 0 )( C W/m 10 ( W ) 06 . 0 100 ( + C 40 2 2 s s hA Q T T &
(c) The thermal resistance is
C/W 4.63° = ° ⋅ = = ) m 0216 . 0 ( C) W/m 10 ( 1 1 2 2 s conv hA R
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.
Qconv
Tskin
Qrad
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
C 35.5° = ° ⋅ − ° = − = − = ) m C)(1.7 W/m 3 . 0 ( m) 005 . 0 W)( 150 ( C 37 2 1 1 kA L Q T T L T T kA Q skin skin & &
3-30 A double-pane window is considered. The rate of heat loss through the window and the temperature difference across
PropertiesThe 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.
[
]
[
]
W 210 = + + + + ° × ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ C 0.78 W/m C 0.025 W/m C 0.78 W/m C 20 W/m C W/m 40 2 2 2 = + + + + ° × = + + + + ∆ = 05 . 0 000513 . 0 2 . 0 000513 . 0 025 . 0 C (-20) -20 ) m 5 . 1 1 ( 1 m 004 . 0 m 005 . 0 m 004 . 0 1 C (-20) -20 ) m 5 . 1 1 ( 1 1 2 o g g a a g g i k h L k L k L h T A Q&b) Noting that the largest resistance is through the air gap, the temperature difference across the air gap is determined from
( C 28° = × ° ⋅ = = = ∆ ) m 5 . 1 1 ( C) W/m 025 . 0 ( m 005 . 0 W) 210 ( 2 A k L Q R Q T a a a a & &
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
ksheetrock = 0.10 Btu/h⋅ft⋅°F and kinsulation = 0.020 Btu/h⋅ft⋅°F.
L1 L2 L3
R1 R2 R3
Analysis (a) The surface area of the wall is not given and thus we consider a unit surface area (A = 1 ft2). Then the R-value of insulation of the wall becomes equivalent to its thermal resistance, which is determined from.
F.h/Btu . ft 30.17 2 ° = + × = + = ° = ° = = = ° = ° = = = = 17 . 29 500 . 0 2 2 F.h/Btu . ft 17 . 29 F) Btu/h.ft. 020 . 0 ( ft 12 / 7 F.h/Btu . ft 500 . 0 F) Btu/h.ft. 10 . 0 ( ft 12 / 6 . 0 2 1 2 2 2 2 2 1 1 3 1 R R R k L R R k L R R R total fiberglass sheetrock
3-32 The roof of a house with a gas furnace consists of a concrete that is losing heat to the outdoors by radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The emissivity
and thermal conductivity of the roof are constant.
Q&
Tin=20°C
Tsky = 100 K
Tair =10°C
Properties The thermal conductivity of the concrete is given to be
k = 2 W/m⋅°C. The emissivity of both surfaces of the roof is given
to be 0.9. L=15 cm
Analysis When the surrounding surface temperature is different than the ambient temperature, the thermal resistances network approach becomes cumbersome in problems that involve radiation. Therefore, we will use a different but intuitive approach.
In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal to the heat transfer from the roof to the surroundings (by convection and radiation), that must be equal to the heat transfer through the roof by conduction. That is,
rad + conv gs, surroundin to roof cond roof, rad + conv roof, to room Q Q Q
Q&= & = & = &
Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out , respectively, the quantities above can be expressed as
[
4]
, 4 4 2 8 2 , 2 2 4 , 4 , rad + conv roof, to room K) 273 ( K) 273 20 ( ) K W/m 10 67 . 5 )( m 300 )( 9 . 0 ( C ) )(20 m C)(300 W/m 5 ( ) ( ) ( + − + ⋅ × + ° − ° ⋅ = − + − = − in s in s in s room in s room i T T T T A T T A h Q& ε σ m 15 . 0 ) m 300 )( C W/m 2 ( 2 , , , , cond roof, out s in s out s in s T T L T T kA Q& = − = ⋅° −[
4 4]
, 4 2 8 2 m 300 )( 9 . 0 ( + , 2 2 4 4 , , rad + conv surr, to roof K) 100 ( K) 273 ( ) K W/m 10 67 . 5 )( C ) 10 )( m C)(300 W/m 12 ( ) ( ) ( − + ⋅ × ° − ° ⋅ = − + − = − out s out s surr out s surr out s o T T T T A T T A h Q& ε σlving the quations above simultaneously gives
he total amount of natural gas consumption during a 14-hour period is
So e
Q&=37,440 W ,Ts,in =7.3°C ,andTs,out =−2.1°C T therms 36 . 22 kJ 10 80 . 0 80 . 0 80 . 0 = = ⎜⎜⎝ = Qgas total 5,500 therm 1 ) s 3600 14 )( kJ/s 440 . 37 ( = ⎟⎟ ⎠ ⎞ ⎛ × ∆t Q Q &
inally, the money lost through the roof during that period is F $26.8 = =(22.36 therms)($1.20/therm) lost Money
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 Rinsulation Ro A=(2m)(1.5m)=3m2 Q&=hA(Ts−T∞)=(10 W/m2⋅°C)(3m2)(110−32)°C=2340 W In order to reduce heat loss by 90%, the new heat transfer rate and
thermal resistance must be T∞
C/W 333 . 0 C ) 32 110 ( − ° = ° = ∆ = ⎯→ ⎯ ∆ = T R T Q& total W 234 W 234 W 2340 10 . 0 × = Q R Q total & &
and in order to have this thermal resistance, the thickness of insulation must be = cm 3.4 = = ° ° ⋅ m 034 . 0 ) m C)(3 W/m. 038 . 0 ( ) m C)(3 W/m 10 ( 2 2 2 L ° = + = + = + = C/W 333 . 0 1 1 conv L kA L hA R R Rtotal insulation
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
therms 1 . 807 kJ 105,500 therm 1 h 1 s 3600 0.78 h) kJ/s)(8760 106 . 2 ( Saved Energy ⎟⎟= ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∆ = Efficiency t Q&saved he money aved is gy = =
he insulation will pay for its cost of $250 in
T s
Ener ( saved
Money = Saved)(Cost ofenergy) (807.1 therms)($1.10/therm) $887.8(per year) T yr 0.282 = = = $887.8/yr $250 saved Money spent Money period Payback
∞
T T hA Q& s
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 Rinsulation Ro A=(2m)(1.5m)=3m2 = ( − )=(10 W/m2⋅°C)(3m2)(110−32)°C=2340 W In order to reduce heat loss by 90%, the new heat transfer rate and thermal
resistance must be T∞ C/W 333 . 0 C ) 32 110 ( − ° = ° = ∆ = ⎯→ ⎯ ∆ = T R T Q& total W 234 W 234 W 2340 10 . 0 × = Q R Q total & &
and in order to have this thermal resistance, the thickness of insulation must be = cm 4.7 = = ° ⋅ ° ⋅ m 047 . 0 ) m C)(3 W/m 052 . 0 ( ) m C)(3 W/m 10 ( 2 2 2 L ° = + = + = + = C/W 333 . 0 1 1 conv L kA L hA R R Rtotal insulation
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
therms 1 . 807 kJ 105,500 therm 1 h 1 s 3600 0.78 h) kJ/s)(8760 106 . 2 ( Saved Energy ⎟⎟= ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∆ = Efficiency t Q&saved oney gy The m saved is Ener ( saved
Money = Saved)(Cost ofenergy)=(807.1 therms)($1.10/therm)=$887.8(per year) he insulation will pay for its cost of $250 in
T yr 0.282 = = = $887.8/yr $250 saved Money spent Money period Payback
3-35 Prob. 3-33 is reconsidered. The effect of thermal conductivity on the required insulation thickness is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN" A=2*1.5 [m^2] T_s=110 [C] T_infinity=32 [C] h=10 [W/m^2-C] k_ins=0.038 [W/m-C] f_reduce=0.90 "ANALYSIS" Q_dot_old=h*A*(T_s-T_infinity) Q_dot_new=(1-f_reduce)*Q_dot_old Q_dot_new=(T_s-T_infinity)/R_total R_total=R_conv+R_ins R_conv=1/(h*A)
R_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm"
kins [W/m.C] Lins [cm] 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 1.8 2.25 2.7 3.15 3.6 4.05 4.5 4.95 5.4 5.85 6.3 6.75 7.2 0.02 0.03 0.04 0.05 0.06 0.07 0.08 1 2 3 4 5 6 7 8 kins [W/m-C] Lin s [ c m ]
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 C/W 06271 . 0 001389 . 0 05775 . 0 003571 . 0 C/W 001389 . 0 ) m 4 10 ( C) W/m 18 ( 2 = = o o h A R 1 1 C/W 05775 . 0 ) m 4 10 ( C/W m 31 . 2 C/W 003571 . 0 ) m 4 10 ( C) W/m 7 ( 1 1 wall tal 2 2 2 wall wall 2 2 ° = + + = + + = ° = × ° ⋅ ° = × ° ⋅ = − = = ° = × ° ⋅ = = o i i i R R R R A value R kA L R A h R
Q&
L to Then Rwall Ri Ro W 255.1 = ° ° − = − = ∞ ∞ C/W 06271 . 0 C ) 8 24 ( 2 1 total R T T Q&Wall with single pane windows:
C/W 003063 . 0 000694 . 0 000583 . 0 001786 . 0 C/W 000694 . 0 ) m 4 20 ( C) W/m 18 ( 1 1 C/W 000583 . 0 002968 . 0 1 5 033382 . 0 1 1 5 1 1 C/W 002968 . 0 m ) 8 . 1 2 . 1 )( C W/m 78 . 0 ( m 005 . 0 C/W 033382 . 0 m ) 8 . 1 2 . 1 ( 5 ) 4 20 ( C/W m 31 . 2 C/W 001786 . 0 ) m 4 20 ( C) W/m 7 ( 1 1 eqv total 2 2 o glass wall eqv 2 o 2 glass glass 2 2 wall wall 2 2 ° = + + = + + = ° = × ° ⋅ = = = → + = + = ° = × ⋅ = = ° = × − × ° ⋅ = − = = ° = × ° ⋅ = = o i o o eqv i i R R R R A h R R R R R kA L R A value R kA L R A h R Ri Rwall Ro Rglass Then W 5224 = ° ° − = − = ∞ ∞ C/W 003063 . 0 C ) 8 24 ( total 2 1 R T T Q&
4th wall with double pane windows:
Rwall
Rglass Rair Rglass
Ri Ro C/W 023197 . 0 000694 . 0 020717 . 0 001786 . 0 C/W 020717 . 0 27303 . 0 1 5 1 1 5 1 1 eqv ° = ⎯→ ⎯ + = + = R R 0.033382 C/W 27303 . 0 267094 . 0 002968 . 0 2 2 C/W 002968 . 0 m ) 1 2 . 1 )( C W/m 78 . 0 ( m 005 . 0 C/W 033382 . 0 m ) 8 . 1 2 . 1 ( 5 ) 4 20 ( C/W m 31 . 2 eqv total eqv window wall air glass 2 2 glass glass 2 2 wall wall ° = + + = + + = ° = + × = + = ° = × ° ⋅ = = ° = × − × ° ⋅ = − = = o i R R R R R R R R R kA L R A value R kA L R 8 . C/W 267094 . 0 m ) 8 . 1 2 . 1 )( C W/m 026 . 0 ( m 015 . 0 2 o 2 air air = ° × ⋅ = = kA L R window Then W 690 = ° ° − = − = ∞ ∞ C/W 023197 . 0 C ) 8 24 ( total 2 1 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) = $1828
W 4534 690 5224 pane double pane single save =Q −Q = − =
Q& & &
kWh 22,851 = h) 24 30 kW)(7 534 . 4 ( × × = ∆ =Q t
, − sout room oAT T h Q&
room and the refrigerated space can be expressed as
3-37 The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of sheet metal. The
minimum thickness of insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is to be determined.
Assumptions 1 Heat transfer through the refrigerator walls is steady since the temperatures of the food compartment and the
kitchen air remain constant at the specified values. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation effects.
Properties The thermal conductivities are given to be k = 15.1 W/m⋅°C for sheet metal and 0.035 W/m⋅°C for fiberglass
insulation.
Analysis The minimum thickness of insulation can be determined by assuming the outer surface temperature of the refrigerator to be 20°C. In steady operation, the rate of heat transfer through the refrigerator wall is constant, and thus heat transfer between the room and the refrigerated space is equal to the heat transfer between the room and the outer surface of the refrigerator. Considering a unit surface area, 1 mm L 1 mm insulation ) ( = W 36 = C ) 20 24 )( m 1 ( C) W/m 9 ( 2⋅° 2 − ° =
Using the thermal resistance network, heat transfer between the
Ri R1 Rins R3 Ro i insulation metal o refrig room total refrig room h k L k L h T T A Q R T T Q 1 2 1 / + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = − = &
& Troom Trefrig
Substitutin g, C W/m 4 1 C W/m 035 . 0 C W/m 15.1 m 001 . 0 2 C W/m 9 1 + C ) 2 24 ( W/m 36 2= − ° 2 2 2 2 ⋅° + ⋅° + ⋅° × ° ⋅ L
Solv ing for L, the minimum thickness of insulation is determined to be
3-38 Prob. 3-37 is reconsidered. The effects of the thermal conductivities of the insulation material and the sheet
metal on the thickness of the insulation is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN" k_ins=0.035 [W/m-C] L_metal=0.001 [m] k_metal=15.1 [W/m-C] T_refrig=2 [C] T_kitchen=24 [C] h_i=4 [W/m^2-C] h_o=9 [W/m^2-C] T_s_out=20 [C] "ANALYSIS"
A=1 [m^2] “a unit surface area is considered"
Q_dot=h_o*A*(T_kitchen-T_s_out) Q_dot=(T_kitchen-T_refrig)/R_total
R_total=R_conv_i+2*R_metal+R_ins+R_conv_o R_conv_i=1/(h_i*A)
R_metal=L_metal/(k_metal*A)
R_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm"
R_conv_o=1/(h_o*A) 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 kins [W/m-C] Lin s [ c m ] kins [W/m.C] Lins [cm] 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.4997 0.6247 0.7496 0.8745 0.9995 1.124 1.249 1.374 1.499 1.624 1.749 1.874 1.999
kmetal [W/m.C] Lins [cm] 10 30.53 51.05 71.58 92.11 112.6 133.2 153.7 174.2 194.7 215.3 235.8 256.3 276.8 297.4 317.9 338.4 358.9 379.5 400 0.8743 0.8748 0.8749 0.8749 0.8749 0.8749 0.8749 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0 50 100 150 200 250 300 350 400 0.8743 0.8744 0.8745 0.8746 0.8747 0.8748 0.8749 0.875 kme tal [W/m-C] Lin s [ c m ]
3-39 Heat is to be conducted along a circuit board with a copper layer on one side. The percentages of heat conduction
along the copper and epoxy layers as well as the effective thermal conductivity of the board are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional
since heat transfer from the side surfaces is disregarded 3 Thermal conductivities are
constant. Copper
Epoxy
Ts t epoxy
tcopper
Properties The thermal conductivities are given to be k = 386 W/m⋅°C for copper and
0.26 W/m⋅°C for epoxy layers.
Analysis We take the length in the direction of heat transfer to be L and the width of the board to be w. Then heat conduction along this two-layer board can be expressed as
[
]
LT T kA T kA Q Q Q + =⎜⎛ ∆ ⎟⎞ +⎜⎛ ∆ ⎟⎞ epoxy copper & & &eat conduction along an “equivalent” board of thickness t = tcopper + tepoxy and thermal conductivity k can be expressed as
w kt kt L L ∆ + = ⎠ ⎝ ⎠ ⎝ copper epoxy epoxy copper ) ( ) ( = H eff Q L T w t t k L T kA ⎜ ⎝ = Q ⎟ = + ∆ ⎠ ⎞ ⎛ ∆ ) (copper epoxy eff board &
Setting the two relations above equal to each other and solving for the effective conductivity gives
epoxy copper epoxy copper epoxy copper epoxy copper ) ( ) ( ) ( ) ( ) ( t t kt kt k kt kt t t keff eff + + = ⎯→ ⎯ + = +
Note that heat conduction is proportional to kt. Substituting, the fractions of heat onducted along the copper and epoxy layers as well as the effective thermal conductivity of the board are determined to be
c 99.2% 0.8% = = = = = = = = ° = + = + = ° = ° ⋅ = ° = ° ⋅ = 992 . 0 038912 . 0 0386 . 0 ) ( ) ( 008 . 0 038912 . 0 000312 . 0 ) ( ) ( C W/ 038912 . 0 000312 . 0 0386 . 0 ) ( ) ( ) ( C W/ 000312 . 0 m) C)(0.0012 W/m 26 . 0 ( ) ( C W/ 0386 . 0 m) C)(0.0001 W/m 386 ( ) ( total copper copper total epoxy epoxy f epoxy copper total epoxy copper kt kt f kt kt kt kt kt kt kt and C W/m. 29.9 ° = + ° × + × = m ) 0012 . 0 0001 . 0 ( C W/ ) 0012 . 0 26 . 0 0001 . 0 386 ( eff k
3-40E A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermal conductivity of the
board along its 9 in long side and the fraction of the heat conducted through copper along that side are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is
one-dimensional since heat transfer from the side surfaces are disregarded 3
Thermal conductivities are constant. Copper
Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for
copper and 0.15 Btu/h⋅ft⋅°F for epoxy layers.
Analysis We take the length in the direction of heat transfer to be L and the width of the board to be w. Then heat conduction along this two-layer plate can be expressed as (we treat the two layers of epoxy as a single layer that is twice as thick)
[
]
L T w kt kt L T kA L T kA Q Q Q ∆ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∆ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∆ = + = epoxy copper epoxy copper epoxy copper ) ( ) ( & & & Epoxy Ts ½ tepoxy tcopper ½ tepoxy EpoxyHeat conduction along an “equivalent” plate of thick ness t = tcopper + tepoxy
and thermal conductivity keff can be expressed as Q
L L ⎠
⎝ eff copper epoxy board T w t t k T kA Q& ⎜⎛ ∆ ⎟⎞ = ( + ) ∆
Setting the two relations above equal to each other and solving for the effective conductivity gives = epoxy copper epoxy copper epoxy copper eff t t eff +
Note that heat conduction is proportional to kt. Substituting, the
epoxy copper ( ) ) ( ) ( ) ( ) (t t kt kt k kt kt k + = + ⎯⎯→ = +
fraction of heat conducted along the copper layer and the effective th rmal conductivity of the plate are determined to be
and e F Btu/h. 93292 . 0 00375 . 0 9292 . 0 ) ( ) ( ) ( F Btu/h. 00375 . 0 ft) F)(0.15/12 Btu/h.ft. 15 . 0 ( 2 ) ( F Btu/h. 9292 . 0 ft) F)(0.05/12 Btu/h.ft. 223 ( ) ( epoxy copper total epoxy copper ° = + = + = ° = ° = ° = ° = kt kt kt kt kt F . Btu/h.ft 32.0 2 ° = + ° = + + = ft )] 12 / 15 . 0 ( 2 ) 12 / 05 . 0 [( F Btu/h. 93292 . 0 t ) ( ) ( epoxy copper epoxy copper t kt kt keff 99.6% = = = = 0.996 93292 . 0 9292 . 0 ) ( ) ( total copper copper kt kt f
3-41 Warm air blowing over the inner surface of an automobile windshield is used for defrosting ice accumulated on the outer surface. The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the windshield is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 The automobile is operating at 1 atm.
Properties Thermal conductivity of the windshield is given to be k = 1.4 W/m · °C.
Analysis The thermal resistances are
A h R i i = 1 A h R o o 1 = and kA win
From energy balance and using the thermal res
L
R =
istance oncept, th following equation is expressed:
c e i i o o R R T T R T T + − = − ∞ ∞ win , 1 1 , win 1 ,o ∞ or , 1 R R T T T T R i o − − − = ∞ i k L T T i − ⎟ ⎞ ⎜ ⎛ − = ∞ 1 1 1 , h T T h ∞,o − 1⎜⎝ o ⎟⎠ ection i
For the ice to begin melting, the outer surface temperature of the windshield (
T
1) should be at least 0 °C. The conv heat transfer coefficient for the warm air isC W/m 112 2⋅° = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ° ⋅ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° ⋅ ° − − ° − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = − −1 ∞ ∞ 1 2 1 , , 1 C W/m 4 . 1 m 005 . 0 C W/m 200 1 C ) 0 10 ( C ) 25 0 ( 1 k L h T T T T h o o i i
Discussion In practical situations, the ambient temperature and the convective heat transfer coefficient outside the
automobile vary with weather conditions and the automobile speed. Therefore the convection heat transfer coefficient of the warm air necessary to melt the ice should be varied as well. This is done by adjusting the warm air flow rate and temperature.
3-42 The thermal contact conductance for an aluminum plate attached on a copper plate, that is heated electrically, is 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.
Properties The thermal conductivity of the aluminum plate is given to be 235 W/m · °C.
Analysis The thermal resistances are
kA L Rcond =
and
hA
From energy balance and using the thermal res
R = 1
istance concept, the following equation is expressed:
conv conv cond 1 elec /A R R R T T A q c + + − = ∞ & or ) /( 1 ) /( / 1 elec hA kA L A R T T A q + + − = ∞ & c
Rearranging the equation and solving for the contact resistance yields
C/W m 10 258 6 C W/m 67 1 C W/m 235 025 . 0 C ) 20 100 ( 2 − ° − = m W/m 5300 1 2 5 2 elec 1 ° ⋅ × = ° ⋅ − ° ⋅ − − − = − ∞ . h k L q T T Rc &
herma contact conductance is
Discussion By comparing the value of the thermal contact conductance with the values listed in Table 3-2, the surface conditions of the plates appear to be milled.
The t l C W/m 16000 2⋅° = = c c R h 1/
Thermal Contact Resistance
3-43C The resistance that an interface offers to heat transfer per unit interface area is called thermal contact resistance, . The inverse of thermal contact resistance is called the thermal contact conductance.
c
R
3-44C The thermal contact resistance will be greater for rough surfaces because an interface with rough surfaces will contain
more air gaps whose thermal conductivity is low.
3-45C 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.
3-46C 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.
3-47C Heat transfer through the voids at an interface is by conduction and radiation. Evacuating the interface eliminates heat
transfer by conduction, and thus increases the thermal contact resistance.
3-48C Thermal contact resistance can be minimized by (1) applying a thermally conducting liquid on the surfaces before
they are pressed against each other, (2) by replacing the air at the interface by a better conducting gas such as helium or hydrogen, (3) by increasing the interface pressure, and (4) by inserting a soft metallic foil such as tin, silver, copper, nickel, or aluminum between the two surfaces.
3-49 The thickness of copper plate whose thermal resistance is equal to the thermal contact resistance is to be determined.
Properties The thermal conductivity of copper is k = 386 W/m⋅°C.
Analysis Noting that thermal contact resistance is the inverse of thermal contact conductance, the thermal contact resistance
is determined to be C/W . m 10 143 . 7 C . W/m 000 , 14 1 1 5 2 2 c
For a unit surface area, the therm
° × = ° = = − h R
al resistance of a flat plate is defined as
c
k
R= L where L is the thickness of the plate and k is e thermal conductivity. Setting he equivalent thickness is determined from t e relation above to be
copper. Note that e thermal contact resistance in this case is greater than the sum of the thermal resistances of both plates.
ror involved in the total thermal resistance of the ting conditions exist. 2 Heat transfer is one-dimensional since the plate is large. 3 Thermal
ers is given to be hc = 6000 W/m2⋅°C. rmal resistances of different layers for unit surface
a of 1 m th R=Rc, t h cm 2.76 = = ° ⋅ × ° ⋅ = = = (386 W/m C)(7.143 10−5m2 C/W) 0.0276m c kR kR L
Therefore, the interface between the two plates offers as much resistance to heat transfer as a 2.76 cm thick th
3-50 A thin copper plate is sandwiched between two epoxy boards. The er
plate if the thermal contact conductances are ignored is to be determined.
Assumptions 1 Steady opera
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 lay
Analysis The the
are 2 are Epoxy 7 mm Epoxy 7 mm Copper plate
Q&
C ° 0.00017 /W ) m C)(1 W/m 6000 ( 1 1 2 2 c contact = ° ⋅ = = c A h R C/W 10 6 . 2 ) m C)(1 W/ (386 kA m m 001 . 0 6 2 ate = × ° ° ⋅ = = L − R pl C/W 02692 . 0 ) m C)(1 W/m (0.26 m 007 . 0 2 epoxy = ° ° ⋅ = = kA L RThe total thermal resistance is 2 2 contact plate epoxy
total= R +R + R R C/W 05419 . 0 02692 . 0 2 10 6 . 2 00017 . 0 2× + × 6 + × = ° = −
nvolved in the total thermal resistance of the ntact resistances are ignored is determined to be
Then the percent error i plate if the thermal co
Rcontact Rcontact Repoxy T1 T2 Repoxy Rplate 0.63% = × × = × = 100 05419 . 0 00017 . 0 2 100 2 Error % total contact R R
3-51 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulation sleeve. For
specified top and bottom surface temperatures, the rate of heat transfer along the cylinders and the temperature drop at the interface are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is
one-dimensional in the axial direction since the lateral surfaces of both cylinders are well-insulated. 3 Thermal conductivities are
constant.
Bar Bar Interface
Properties The thermal conductivity of aluminum bars is given to be
k = 176 W/m⋅°C. The contact conductance at the interface of aluminum-aluminum plates for the case of ground surfaces and of 20 atm ≈ 2 MPa pressure is hc = 11,400 W/m2⋅°C (Table 3-2).
Ri Rglass Ro
Analysis (a) The thermal resistance network in this case consists of two conduction resistance and the contact resistance, and they are
determined to be T1 T2 C/W 0447 . 0 /4] m) (0.05 C)[ W/m 400 , 11 ( 1 1 2 2 c contact = ° ° ⋅ = = π c A h R C/W 4341 . 0 /4] m) (0.05 C)[ W/m (176 m 15 . 0 plate= = L R 2 = ° ° ⋅ π kA
hen the rate of heat transfer is determined to be T W 142.4 = ° × + +2 bar (0.0447 2 0.4341) contact total R R R ° − = ∆ = ∆ = T T (150 20) C Q& C/W
e rate of heat transfer through the bars is 142.4 W. ) The temperature drop at the interface is determined to be Therefore, th (b C 6.4° = ° = =
Generalized Thermal Resistance Networks
3-52C Two approaches used in development of the thermal resistance network in the x-direction for multi-dimensional
problems are (1) to assume any plane wall normal to the axis to be isothermal and (2) to assume any plane parallel to the x-axis to be adiabatic.
3-53C The thermal resistance network approach will give adequate results for multi-dimensional heat transfer problems if
heat transfer occurs predominantly in one direction.
3-54C Parallel resistances indicate simultaneous heat transfer (such as convection and radiation on a surface). Series
resistances indicate sequential heat transfer (such as two homogeneous layers of a wall).
3-55 A typical section of a building wall is considered. The average heat flux through the wall is to be determined.
Assumptions 1 Steady operating conditions exist.
Properties The thermal conductivities are given to be k23b = 50 W/m⋅K, k23a = 0.03 W/m⋅K, k12 = 0.5 W/m⋅K, k34 = 1.0 W/m⋅K.
Analysis We consider 1 m2 of wall area. The thermal resistances are
C/W m 1 . 0 C) W/m 0 . 1 ( m 1 . 0 C/W m 10 32 . 1 0.005) C)(0.6 W/m 50 ( m 005 . 0 m) 08 . 0 ( ) ( C/W m 645 . 2 0.005) C)(0.6 W/m 03 . 0 ( m 6 . 0 m) 08 . 0 ( ) ( C/W m 02 . 0 C) W/m 5 . 0 ( m 01 . 0 2 34 34 34 2 5 23b b 23 23 2 23a a 23 23 2 12 12 12 ° ⋅ = ° ⋅ = = ° ⋅ × = + ° ⋅ = + = ° ⋅ = + ° ⋅ = + = ° ⋅ = ° ⋅ = = − k t R L L k L t R L L k L t R k t R b a b b a a
The total thermal resistance and the rate of heat transfer are
C/W m 120 . 0 1 . 0 10 32 . 1 645 . 2 10 32 . 1 645 . 2 02 . 0 2 5 5 34 23 23 23 23 12 total ° ⋅ = + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × + × + = + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + = − − R R R R R R R b a b a
3-56 A wall consists of horizontal bricks separated by plaster layers. There are also plaster layers on each side of the wall, and a rigid foam on the inner side of the wall. The rate of heat transfer through the wall is 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 by radiation is disregarded.
Properties The thermal conductivities are given to be k = 0.72 W/m⋅°C for bricks, k = 0.22 W/m⋅°C for plaster layers, and k
= 0.026 W/m⋅°C for the rigid foam.
Analysis We consider 1 m deep and 0.28 m high portion of wall which is representative of the entire wall. The thermal resistance network and individual resistances are
R7 R6 R5 R4 R3 R2 R1 Ri T∞1 T∞2 C/W 737 . 4 179 . 0 804 . 0 ) 325 . 0 ( 2 747 . 2 357 . 0 2 C/W 804 . 0 45 . 45 1 833 . 0 1 45 . 45 1 1 1 1 1 C/W 179 . 0 ) m 1 28 . 0 ( C) W/m 20 ( 1 1 C/W 833 . 0 ) m 1 25 . 0 ( C) W/m 72 . 0 ( m 15 . 0 C/W 45 . 45 ) m 1 015 . 0 ( C) W/m 22 . 0 ( m 15 . 0 C/W 325 . 0 ) m 1 28 . 0 ( C) W/m 22 . 0 ( m 02 . 0 = = = =R R L R C/W 747 . 2 ) m 1 28 . 0 ( C) W/m 026 . 0 ( C/W 357 . 0 ) m 1 1 1 2 1 5 4 3 2 2 2 , o 2 4 2 5 3 2 6 2 2 1 2 2 1 , ° = + + + + = + + + + = ° = ⎯→ ⎯ + + = + + = ° = × ° ⋅ = = = ° = × ° ⋅ = = = ° = × ° ⋅ = = = = ° = × ° ⋅ ° = × ° ⋅ = = = ° = × = = = o mid i total mid mid conv brick o centerplaster sideplaster foam conv i R R R R R R R R R R R A h R R kA L R R A h L R R R kA kA R R R R
The steady rate of heat transfer through the wall per is m 02 . 0 28 . 0 ( C) W/m 10 ( 1 ⋅° L A h 2 m 28 . 0 W 49 . 5 C/W 737 . 4 C )] 4 ( 22 [( 2 1 = ° ° − − = − = ∞ ∞ total R T T Q&
Then steady rate of heat transfer through the entire wall becomes
W 470 = × = 22 m 28 . 0 m ) 6 4 ( W) 49 . 5 ( total Q&
3-57 Prob. 3-56 is reconsidered. The rate of heat transfer through the wall as a function of the thickness of the rigid
nalysis The problem is solved using EES, and the solution is given below.
] ] -C] [m^2] ter/(k_plaster*A_3) side+R_mid+R_conv_2 ity_2)/R_total _dot_total=Q_dot*A/A_1 ] foam is to be plotted. A "GIVEN" A=4*6 [m^2] L_brick=0.15 [m] L_plaster_center=0.15 [m L_plaster_side=0.02 [m "L_foam=2 [cm]" k_brick=0.72 [W/m-C] k_plaster=0.22 [W/m-C] k_foam=0.026 [W/m T_infinity_1=22 [C] T_infinity_2=-4 [C] h_1=10 [W/m^2-C] h_2=20 [W/m^2-C] A_1=0.28*1 [m^2] A_2=0.25*1 [m^2] A_3=0.015*1 "ANALYSIS" R_conv_1=1/(h_1*A_1)
R_foam=(L_foam*Convert(cm, m))/(k_foam*A_1) "L_foam is in cm"
R_plaster_side=L_plaster_side/(k_plaster*A_1) R_plaster_center=L_plaster_cen R_brick=L_brick/(k_brick*A_2) R_conv_2=1/(h_2*A_1) 1/R_mid=2*1/R_plaster_center+1/R_brick R_total=R_conv_1+R_foam+2*R_plaster_ Q_dot=(T_infinity_1-T_infin Q Lfo [cm am Qtotal [W] 1 2 3 4 5 6 7 8 9 10 141.7 662.8 470.5 364.8 297.8 251.6 217.8 192 171.7 155.3 1 2 3 4 5 6 7 8 9 10 100 200 300 400 500 600 700 Qto ta l [ W ] Lfoam [cm]
3-58 A wall is constructed of two layers of sheetrock spaced by 5 cm × 16 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 . The thermal resistance network and the individual thermal resistances are
2 m 65 . 0 65 . 0 1× = = A R5 R4 R3 R2 R1 Ri T∞1 T∞2 W 40 . 4 C/W 588 . 6 C )] 9 ( 20 [ section) m 0.65 m 1 a (for 045 . 0 090 . 0 178 . 6 090 . 0 185 . 0 C/W 178 . 6 843 . 7 1 091 . 29 1 1 1 1 C/W 045 . 0 ) m 65 . 0 ( C) W/m 34 ( 1 1 C/W 843 . 7 ) m 60 . 0 ( C) W/m 034 . 0 ( m 16 . 0 C/W 091 . 29 ) m 05 . 0 ( C) W/m 11 . 0 ( m 16 . 0 C/W 090 . 0 ) m 65 . 0 ( C) W/m 17 . 0 ( m 01 . 0 C/W 185 . 0 1 1 ) m 65 . 0 ( C) W/m 3 . 8 ( 2⋅° 2 iA h 2 1 4 1 3 2 2 o 2 2 3 2 2 2 4 1 = ° ° − − = − = × ° = + + + + = + + + + = ° = ⎯→ ⎯ + = + = ° = ⋅ = = ° = ° ⋅ = = = ° = ° ⋅ = = = ° = ° ⋅ = = = = ° = = = ∞ ∞ total o mid i total mid mid o o fiberglass stud sheetrock i R T T Q R R R R R R R R R R A h R kA L R R kA L R R kA L R R R R & C/W 6.588
(b) Then steady rate of heat transfer through entire wall becomes
W 406 = = 2 m 65 . 0 m) 5 ( m) 12 ( W) 40 . 4 ( total Q&
