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SI Heat 4e SM Chap02 - 솔루션
기계요소설계 (국립한국해양대학교)
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2-1
Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
Fourth Edition in SI Units
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2011
Chapter 2
HEAT CONDUCTION EQUATION
PROPRIETARY AND CONFIDENTIAL
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2-2
Introduction
2-1C The heat transfer process from the kitchen air to the refrigerated space is
transient in nature since the thermal conditions in the kitchen and the
refrigerator, in general, change with time. However, we would analyze this
problem as a steady heat transfer problem under the worst anticipated conditions
such as the lowest thermostat setting for the refrigerated space, and the
anticipated highest temperature in the kitchen (the so-called design conditions).
If the compressor is large enough to keep the refrigerated space at the desired
temperature setting under the presumed worst conditions, then it is large enough
to do so under all conditions by cycling on and off. Heat transfer into the
refrigerated space is three-dimensional in nature since heat will be entering
through all six sides of the refrigerator. However, heat transfer through any wall
or floor takes place in the direction normal to the surface, and thus it can be
analyzed as being one-dimensional. Therefore, this problem can be simplified
greatly by considering the heat transfer to be onedimensional at each of the four
sides as well as the top and bottom sections, and then by adding the calculated
values of heat transfer at each surface.
2-2C The term steady implies no change with time at any point within the medium while transient implies variation with
time or time dependence. Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer
through a medium at any location although both quantities may vary from one location to another. During transient heat
transfer, the temperature and heat flux may vary with time as well as location. Heat transfer is one-dimensional if it occurs
primarily in one direction. It is two-dimensional if heat tranfer in the third dimension is negligible.
2-3C Heat transfer through the walls, door, and the top and bottom sections of an oven is transient in nature since the thermal
conditions in the kitchen and the oven, in general, change with time. However, we would analyze this problem as a steady
heat transfer problem under the worst anticipated conditions such as the highest temperature setting for the oven, and the
anticipated lowest temperature in the kitchen (the so called “design” conditions). If the heating element of the oven is large
enough to keep the oven at the desired temperature setting under the presumed worst conditions, then it is large enough to do
so under all conditions by cycling on and off.
Heat transfer from the oven is three-dimensional in nature since heat will be entering through all six sides of the
oven. However, heat transfer through any wall or floor takes place in the direction normal to the surface, and thus it can be
analyzed as being one-dimensional. Therefore, this problem can be simplified greatly by considering the heat transfer as
being one- dimensional at each of the four sides as well as the top and bottom sections, and then by adding the calculated
values of heat transfers at each surface.
2-4C Heat transfer to a potato in an oven can be modeled as one-dimensional since temperature differences (and thus heat
transfer) will exist in the radial direction only because of symmetry about the center point. This would be a transient heat
transfer process since the temperature at any point within the potato will change with time during cooking. Also, we would
use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body can be described
by a constant value of the radius in spherical coordinates. We would place the origin at the center of the potato.
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2-3
2-5C Assuming the egg to be round, heat transfer to an egg in boiling water can be modeled as one-dimensional since
temperature differences (and thus heat transfer) will primarily exist in the radial direction only because of symmetry about
the center point. This would be a transient heat transfer process since the temperature at any point within the egg will change
with time during cooking. Also, we would use the spherical coordinate system to solve this problem since the entire outer
surface of a spherical body can be described by a constant value of the radius in spherical coordinates. We would place the
origin at the center of the egg.
2-6C Heat transfer to a hot dog can be modeled as two-dimensional since temperature differences (and thus heat transfer)
will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the
azimuthal direction. This would be a transient heat transfer process since the temperature at any point within the hot dog will
change with time during cooking. Also, we would use the cylindrical coordinate system to solve this problem since a
cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly
at the center of the hot dog. Heat transfer in a very long hot dog could be considered to be one-dimensional in preliminary
calculations.
2-7C Heat transfer to a roast beef in an oven would be transient since the temperature at any point within the roast will
change with time during cooking. Also, by approximating the roast as a spherical object, this heat transfer process can be
modeled as one-dimensional since temperature differences (and thus heat transfer) will primarily exist in the radial direction
because of symmetry about the center point.
2-8C Heat loss from a hot water tank in a house to the surrounding medium can be considered to be a steady heat transfer
problem. Also, it can be considered to be two-dimensional since temperature differences (and thus heat transfer) will exist in
the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the azimuthal
direction.)
2-9C Yes, the heat flux vector at a point P on an isothermal surface of a medium has to be perpendicular to the surface at that
point.
2-10C Isotropic materials have the same properties in all directions, and we do not need to be concerned about the variation
of properties with direction for such materials. The properties of anisotropic materials such as the fibrous or composite
materials, however, may change with direction.
2-11C In heat conduction analysis, the conversion of electrical, chemical, or nuclear energy into heat (or thermal) energy in
solids is called heat generation.
2-12C The phrase “thermal energy generation” is equivalent to “heat generation,” and they are used interchangeably. They
imply the conversion of some other form of energy into thermal energy. The phrase “energy generation,” however, is vague
since the form of energy generated is not clear.
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2-4
2-13C Heat transfer to a canned drink can be modeled as two-dimensional since temperature differences (and thus heat
transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in
the azimuthal direction. This would be a transient heat transfer process since the temperature at any point within the drink
will change with time during heating. Also, we would use the cylindrical coordinate system to solve this problem since a
cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly
at the center of the bottom surface.
2-14 A certain thermopile used for heat flux meters is considered. The minimum heat flux this meter can detect is to be
determined.
Assumptions 1 Steady operating conditions exist.
Properties The thermal conductivity of kapton is given to be 0.345 W/m⋅K.
Analysis The minimum heat flux can be determined from
q& = k
Δt
0.1°C
= (0.345 W/m ⋅ °C)
= 17.3 W/m 2
0.002 m
L
2-15 The rate of heat generation per unit volume in the uranium rods is given. The total rate of heat generation in each rod is
to be determined.
g = 2×108 W/m3
Assumptions Heat is generated uniformly in the uranium rods.
Analysis The total rate of heat generation in the rod is determined
by multiplying the rate of heat generation per unit volume by the
volume of the rod
D = 5 cm
L=1m
E& gen = e&genV rod = e&gen (πD 2 / 4) L = (2 × 10 8 W/m 3 )[π (0.05 m) 2 / 4](1 m) = 3.93 × 10 5 W = 393 kW
2-16 The variation of the absorption of solar energy in a solar pond with depth is given. A relation for the total rate of heat
generation in a water layer at the top of the pond is to be determined.
Assumptions Absorption of solar radiation by water is modeled as heat generation.
Analysis The total rate of heat generation in a water layer of surface area A and thickness L at the top of the pond is
determined by integration to be
E& gen =
∫V
e& gen dV =
∫
L
x =0
e& 0 e −bx ( Adx) = Ae& 0
e −bx
−b
L
=
0
Ae& 0 (1 − e −bL )
b
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2-5
2-17 The rate of heat generation per unit volume in a stainless steel plate is given. The heat flux on the surface of the plate is
to be determined.
Assumptions Heat is generated uniformly in steel plate.
Analysis We consider a unit surface area of 1 m2. The total rate of
heat generation in this section of the plate is
E& gen = e&genV plate = e&gen ( A × L) = (5 × 10 6 W/m 3 )(1 m 2 )(0.03 m) = 1.5 ×10 5 W
e
L
Noting that this heat will be dissipated from both sides of the plate, the heat flux on
either surface of the plate becomes
q& =
E& gen
Aplate
=
1.5 × 10 5 W
2 ×1 m
2
= 75,000 W/m 2 = 75 kW/m 2
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2-6
Heat Conduction Equation
2-18C The one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and heat
∂ 2 T e& gen 1 ∂T
. Here T is the temperature, x is the space variable, e&gen is the heat generation per unit
+
=
generation is
k
α ∂t
∂x 2
volume, k is the thermal conductivity, α is the thermal diffusivity, and t is the time.
2-19C The one-dimensional transient heat conduction equation for a long cylinder with constant thermal conductivity and
1 ∂ ⎛ ∂T ⎞ e&gen 1 ∂T
heat generation is
. Here T is the temperature, r is the space variable, g is the heat generation per
=
⎟+
⎜r
r ∂r ⎝ ∂r ⎠
α ∂t
k
unit volume, k is the thermal conductivity, α is the thermal diffusivity, and t is the time.
2-20 We consider a thin element of thickness Δx in a large plane wall (see Fig. 2-12 in the text). The density of the wall is ρ,
the specific heat is c, and the area of the wall normal to the direction of heat transfer is A. In the absence of any heat
generation, an energy balance on this thin element of thickness Δx during a small time interval Δt can be expressed as
ΔE element
Q& x − Q& x + Δx =
Δt
where
ΔE element = E t + Δt − E t = mc(Tt + Δt − Tt ) = ρcAΔx(Tt + Δt − Tt )
Substituting,
T
− Tt
Q& x − Q& x + Δx = ρcAΔx t + Δt
Δt
Dividing by AΔx gives
−
T
− Tt
1 Q& x + Δx − Q& x
= ρc t + Δt
A
Δx
Δt
Taking the limit as Δx → 0 and Δt → 0 yields
1 ∂ ⎛ ∂T ⎞
∂T
⎟ = ρc
⎜ kA
A ∂x ⎝
∂t
∂x ⎠
since from the definition of the derivative and Fourier’s law of heat conduction,
Q& x + Δx − Q& x ∂Q ∂ ⎛
∂T ⎞
=
=
⎜ − kA
⎟
Δx →0
∂x ⎠
∂x ∂x ⎝
Δx
lim
Noting that the area A of a plane wall is constant, the one-dimensional transient heat conduction equation in a plane wall with
constant thermal conductivity k becomes
∂ 2T
∂x
2
=
1 ∂T
α ∂t
where the property α = k / ρc is the thermal diffusivity of the material.
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2-7
2-21 We consider a thin cylindrical shell element of thickness Δr in a long cylinder (see Fig. 2-14 in the text). The density of
the cylinder is ρ, the specific heat is c, and the length is L. The area of the cylinder normal to the direction of heat transfer at
any location is A = 2πrL where r is the value of the radius at that location. Note that the heat transfer area A depends on r in
this case, and thus it varies with location. An energy balance on this thin cylindrical shell element of thickness Δr during a
small time interval Δt can be expressed as
ΔE element
Q& r − Q& r + Δr + E& element =
Δt
where
ΔE element = E t + Δt − E t = mc(Tt + Δt − Tt ) = ρcAΔr (Tt + Δt − Tt )
E& element = e& genV element = e& gen AΔr
Substituting,
− Tt
T
Q& r − Q& r + Δr + e& gen AΔr = ρcAΔr t + Δt
Δt
where A = 2πrL . Dividing the equation above by AΔr gives
−
− Tt
T
1 Q& r + Δr − Q& r
+ e& gen = ρc t + Δt
Δr
A
Δt
Taking the limit as Δr → 0 and Δt → 0 yields
∂T
1 ∂ ⎛ ∂T ⎞
⎟ + e& gen = ρc
⎜ kA
A ∂r ⎝
∂t
∂r ⎠
since, from the definition of the derivative and Fourier’s law of heat conduction,
Q& r + Δr − Q& r ∂Q ∂ ⎛
∂T ⎞
=
=
⎜ − kA
⎟
Δr →0
∂r ⎠
∂r ∂r ⎝
Δr
lim
Noting that the heat transfer area in this case is A = 2πrL and the thermal conductivity is constant, the one-dimensional
transient heat conduction equation in a cylinder becomes
1 ∂ ⎛ ∂T ⎞
1 ∂T
⎜r
⎟ + e& gen =
r ∂r ⎝ ∂r ⎠
α ∂t
where α = k / ρc is the thermal diffusivity of the material.
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2-8
2-22 We consider a thin spherical shell element of thickness Δr in a sphere (see Fig. 2-16 in the text). The density of the
sphere is ρ, the specific heat is c, and the length is L. The area of the sphere normal to the direction of heat transfer at any
location is A = 4πr 2 where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this
case, and thus it varies with location. When there is no heat generation, an energy balance on this thin spherical shell
element of thickness Δr during a small time interval Δt can be expressed as
ΔE element
Q& r − Q& r + Δr =
Δt
where
ΔE element = E t + Δt − E t = mc(Tt + Δt − Tt ) = ρcAΔr (Tt + Δt − Tt )
Substituting,
T
−T
Q& r − Q& r + Δr = ρcAΔr t + Δt t
Δt
where A = 4πr 2 . Dividing the equation above by AΔr gives
−
T
− Tt
1 Q& r + Δr − Q& r
= ρc t + Δt
Δt
A
Δr
Taking the limit as Δr → 0 and Δt → 0 yields
∂T
1 ∂ ⎛ ∂T ⎞
⎟ = ρc
⎜ kA
A ∂r ⎝
∂t
∂r ⎠
since, from the definition of the derivative and Fourier’s law of heat conduction,
Q& r + Δr − Q& r ∂Q ∂ ⎛
∂T ⎞
=
=
⎜ − kA
⎟
Δr →0
∂r ⎠
∂r ∂r ⎝
Δr
lim
Noting that the heat transfer area in this case is A = 4πr 2 and the thermal conductivity k is constant, the one-dimensional
transient heat conduction equation in a sphere becomes
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎟=
⎜r
∂r ⎠ α ∂t
r 2 ∂r ⎝
where α = k / ρc is the thermal diffusivity of the material.
2-23 For a medium in which the heat conduction equation is given in its simplest by
∂ 2T
∂x
2
=
1 ∂T
:
α ∂t
(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is
constant.
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2-9
2-24 For a medium in which the heat conduction equation is given by
∂ 2T
∂x
+
2
∂ 2T
∂y
2
=
1 ∂T
:
α ∂t
(a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is
constant.
2-25 For a medium in which the heat conduction equation is given by
1 ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞
⎟ + e& gen = 0 :
⎟ + ⎜k
⎜ kr
r ∂r ⎝ ∂r ⎠ ∂z ⎝ ∂z ⎠
(a) Heat transfer is steady, (b) it is two-dimensional, (c) there is heat generation, and (d) the thermal conductivity is variable.
2-26 For a medium in which the heat conduction equation is given in its simplest by
1 d ⎛ dT ⎞
⎜ rk
⎟ + e&gen = 0 :
r dr ⎝ dr ⎠
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is heat generation, and (d) the thermal conductivity is variable.
2-27 For a medium in which the heat conduction equation is given by
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎟=
⎜r
∂r ⎠ α ∂t
r 2 ∂r ⎝
(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is
constant.
2-28 For a medium in which the heat conduction equation is given in its simplest by r
d 2T dT
+
=0:
dr 2 dr
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is
constant.
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2-10
2-29 We consider a small rectangular element of length Δx, width Δy, and height Δz = 1 (similar to the one in Fig. 2-20).
The density of the body is ρ and the specific heat is c. Noting that heat conduction is two-dimensional and assuming no heat
generation, an energy balance on this element during a small time interval Δt can be expressed as
Rate of heat ⎞ ⎛ Rate of heat conduction ⎞ ⎛ Rate of change of ⎞
⎛
⎟
⎟ ⎜
⎟ ⎜
⎜
at the surfaces at
⎟ = ⎜ the energy content ⎟
⎜ conduction at the ⎟ − ⎜
⎟ ⎜ of the element ⎟
⎜ surfaces at x and y ⎟ ⎜ x + Δx and y + Δy
⎠
⎠ ⎝
⎠ ⎝
⎝
or
ΔE element
Q& x + Q& y − Q& x + Δx − Q& y + Δy =
Δt
Noting that the volume of the element is V element = ΔxΔyΔz = ΔxΔy × 1 , the change in the energy content of the element can
be expressed as
ΔE element = E t + Δt − E t = mc(Tt + Δt − Tt ) = ρcΔxΔy (Tt + Δt − Tt )
T
− Tt
Q& x + Q& y − Q& x + Δx − Q& y + Δy = ρcΔxΔy t + Δt
Δt
Substituting,
Dividing by ΔxΔy gives
−
&
&
− Tt
T
1 Q& x + Δx − Q& x
1 Q y + Δy − Q y
−
= ρc t + Δt
Δy
Δx
Δx
Δy
Δt
Taking the thermal conductivity k to be constant and noting that the heat transfer surface areas of the element for heat
conduction in the x and y directions are Ax = Δy × 1 and A y = Δx × 1, respectively, and taking the limit as Δx, Δy, and Δt → 0
yields
∂ 2T
∂x
2
+
∂ 2T
∂y
2
=
1 ∂T
α ∂t
since, from the definition of the derivative and Fourier’s law of heat conduction,
∂T ⎞
∂ ⎛ ∂T ⎞
∂ 2T
1 Q& x + Δx − Q& x
1 ∂Q x
1 ∂ ⎛
=
=
⎜ − kΔyΔz
⎟ = − ⎜k
⎟ = −k 2
Δx →0 ΔyΔz
Δx
ΔyΔz ∂x
ΔyΔz ∂x ⎝
∂x ⎠
∂x ⎝ ∂x ⎠
∂x
lim
&
&
∂T ⎞
∂ ⎛ ∂T ⎞
1 Q y + Δy − Q y
1 ∂Q y
1 ∂ ⎛
∂ 2T
⎜⎜ − kΔxΔz
⎟⎟ = − ⎜⎜ k
⎟⎟ = −k
=
=
Δy →0 ΔxΔz
Δy
ΔxΔz ∂y
ΔxΔz ∂y ⎝
∂y ⎠
∂y ⎝ ∂y ⎠
∂y 2
lim
Here the property α = k / ρc is the thermal diffusivity of the material.
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2-11
2-30 We consider a thin ring shaped volume element of width Δz and thickness Δr in a cylinder. The density of the cylinder
is ρ and the specific heat is c. In general, an energy balance on this ring element during a small time interval Δt can be
expressed as
ΔE element
(Q& r − Q& r + Δr ) + (Q& z − Q& z + Δz ) =
Δt
Δz
But the change in the energy content of the element can be expressed as
ΔE element = E t + Δt − E t = mc(Tt + Δt − Tt ) = ρc( 2πrΔr ) Δz (Tt + Δt − Tt )
rr
r+Δr
Substituting,
− Tt
T
(Q& r − Q& r + Δr ) + (Q& z − Q& z + Δz ) = ρc(2πrΔr )Δz t + Δt
Δt
Dividing the equation above by ( 2πrΔr )Δz gives
−
T
− Tt
1 Q& r + Δr − Q& r
1 Q& z + Δz − Q& z
−
= ρc t + Δt
2πrΔz
2πrΔr
Δr
Δz
Δt
Noting that the heat transfer surface areas of the element for heat conduction in the r and z directions are
Ar = 2πrΔz and Az = 2πrΔr , respectively, and taking the limit as Δr , Δz and Δt → 0 yields
∂T
1 ∂ ⎛ ∂T ⎞ 1 ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞
⎟ + ⎜k
⎜k
⎟+
⎜ kr
⎟ = ρc
r ∂r ⎝ ∂r ⎠ r 2 ∂φ ⎜⎝ ∂φ ⎟⎠ ∂z ⎝ ∂z ⎠
∂t
since, from the definition of the derivative and Fourier’s law of heat conduction,
∂ ⎛
∂T ⎞
1 Q& r + Δr − Q& r
1 ∂Q
1
1 ∂ ⎛ ∂T ⎞
=
=
⎜ − k (2πrΔz )
⎟=−
⎜ kr
⎟
Δr →0 2πrΔz
Δr
∂r ⎠
r ∂r ⎝ ∂r ⎠
2πrΔz ∂r 2πrΔz ∂r ⎝
lim
∂ ⎛
∂T ⎞
1 Q& z + Δz − Q& z
1 ∂Qz
1
∂ ⎛ ∂T ⎞
=
=
⎜ − k (2πrΔr )
⎟ = − ⎜k
⎟
Δz →0 2πrΔr
Δz
∂z ⎠
2πrΔr ∂z
2πrΔr ∂z ⎝
∂z ⎝ ∂z ⎠
lim
For the case of constant thermal conductivity the equation above reduces to
1 ∂ ⎛ ∂T ⎞ ∂ 2 T 1 ∂T
=
⎜r
⎟+
r ∂r ⎝ ∂r ⎠ ∂z 2 α ∂t
where α = k / ρc is the thermal diffusivity of the material. For the case of steady heat conduction with no heat generation it
reduces to
1 ∂ ⎛ ∂T ⎞ ∂ 2 T
=0
⎜r
⎟+
r ∂r ⎝ ∂r ⎠ ∂z 2
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2-12
2-31 Consider a thin disk element of thickness Δz and diameter D in a long cylinder (Fig. P2-31). The density of the cylinder
is ρ, the specific heat is c, and the area of the cylinder normal to the direction of heat transfer is A = πD 2 / 4 , which is
constant. An energy balance on this thin element of thickness Δz during a small time interval Δt can be expressed as
⎛ Rate of heat ⎞ ⎛ Rate of heat
⎞ ⎛ Rate of heat ⎞ ⎛ Rate of change of ⎞
⎜
⎟ ⎜
⎟ ⎜
⎟ ⎜
⎟
conduction
at
conduction
at
the
−
⎜
⎟ ⎜
⎟ + ⎜ generation inside ⎟ = ⎜ the energy content ⎟
⎜ the surface at z ⎟ ⎜ surface at z + Δz ⎟ ⎜ the element ⎟ ⎜ of the element ⎟
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
or,
ΔE element
Q& z − Q& z + Δz + E& element =
Δt
But the change in the energy content of the element and the rate of heat generation within the element can be expressed as
ΔE element = E t + Δt − E t = mc(Tt + Δt − Tt ) = ρcAΔz (Tt + Δt − Tt )
and
E& element = e& genV element = e& gen AΔz
Substituting,
T
− Tt
Q& z − Q& z + Δz + e& gen AΔz = ρcAΔz t + Δt
Δt
Dividing by AΔz gives
−
T
− Tt
1 Q& z + Δz − Q& z
+ e& gen = ρc t + Δt
A
Δt
Δz
Taking the limit as Δz → 0 and Δt → 0 yields
∂T
1 ∂ ⎛ ∂T ⎞
⎟ + e& gen = ρc
⎜ kA
A ∂z ⎝
∂t
∂z ⎠
since, from the definition of the derivative and Fourier’s law of heat conduction,
Q& z + Δz − Q& z ∂Q ∂ ⎛
∂T ⎞
=
=
⎟
⎜ − kA
Δz → 0
∂z ⎠
∂z ∂z ⎝
Δz
lim
Noting that the area A and the thermal conductivity k are constant, the one-dimensional transient heat conduction equation in
the axial direction in a long cylinder becomes
∂ 2T
e& gen
∂z
k
+
2
=
1 ∂T
α ∂t
where the property α = k / ρc is the thermal diffusivity of the material.
2-32 For a medium in which the heat conduction equation is given by
∂ 2T 1 ∂T
1 ∂ ⎛ 2 ∂T ⎞
1
=
⎟+ 2
⎜r
2 ∂r
2
∂r ⎠ r sin θ ∂φ 2 α ∂t
r
⎝
(a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is
constant.
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2-13
Boundary and Initial Conditions; Formulation of Heat Conduction Problems
2-33C The mathematical expressions of the thermal conditions at the boundaries are called the boundary conditions. To
describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate
system along which heat transfer is significant. Therefore, we need to specify four boundary conditions for two-dimensional
problems.
2-34C The mathematical expression for the temperature distribution of the medium initially is called the initial condition.
We need only one initial condition for a heat conduction problem regardless of the dimension since the conduction equation
is first order in time (it involves the first derivative of temperature with respect to time). Therefore, we need only 1 initial
condition for a two-dimensional problem.
2-35C A heat transfer problem that is symmetric about a plane, line, or point is said to have thermal symmetry about that
plane, line, or point. The thermal symmetry boundary condition is a mathematical expression of this thermal symmetry. It is
equivalent to insulation or zero heat flux boundary condition, and is expressed at a point x0 as ∂T ( x 0 , t ) / ∂x = 0 .
2-36C The boundary condition at a perfectly insulated surface (at x = 0, for example) can be expressed as
−k
∂T (0, t )
=0
∂x
or
∂T (0, t )
= 0 which indicates zero heat flux.
∂x
2-37C Yes, the temperature profile in a medium must be perpendicular to an insulated surface since the slope ∂T / ∂x = 0 at
that surface.
2-38C We try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linear expression that
causes mathematical difficulties while solving the problem; often making it impossible to obtain analytical solutions.
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2-14
2-39 Heat conduction through the bottom section of an aluminum pan that is used to cook stew on top of an electric range is
considered (Fig. P2-45). Assuming variable thermal conductivity and one-dimensional heat transfer, the mathematical
formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for
steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be variable.
3 There is no heat generation in the medium. 4 The top surface at x = L is subjected to specified temperature and the bottom
surface at x = 0 is subjected to uniform heat flux.
Analysis The heat flux at the bottom of the pan is
q& s =
E& gen
Q& s
0.90 × (900 W)
=
=
= 31,831 W/m 2
2
As πD / 4 π (0.18 m) 2 / 4
Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as
d ⎛ dT ⎞
⎟=0
⎜k
dx ⎝ dx ⎠
−k
dT (0)
= q& s = 31,831 W/m 2
dx
T ( L) = T L = 108°C
2-40 A spherical container of inner radius r1 , outer radius r2 , and thermal conductivity k is
given. The boundary condition on the inner surface of the container for steady one-dimensional
conduction is to be expressed for the following cases:
(a) Specified temperature of 50°C: T (r1 ) = 50°C
(b) Specified heat flux of 45 W/m2 towards the center: k
r1
r2
dT (r1 )
= 45 W/m 2
dr
(c) Convection to a medium at T∞ with a heat transfer coefficient of h: k
dT (r1 )
= h[T (r1 ) − T∞ ]
dr
2-41 Heat is generated in a long wire of radius ro covered with a plastic insulation layer at a constant rate of e&gen . The heat
flux boundary condition at the interface (radius ro) in terms of the heat generated is to be expressed. The total heat generated
in the wire and the heat flux at the interface are
E& gen = e& genV wire = e& gen (πro2 L)
E& gen e& gen (πro2 L) e& gen ro
Q&
=
=
q& s = s =
A
A
(2πro ) L
2
D
egen
L
Assuming steady one-dimensional conduction in the radial direction, the heat flux boundary condition can be expressed as
−k
dT (ro ) e& gen ro
=
dr
2
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2-15
2-42 A long pipe of inner radius r1, outer radius r2, and thermal conductivity
k is considered. The outer surface of the pipe is subjected to convection to a
medium at T∞ with a heat transfer coefficient of h. Assuming steady onedimensional conduction in the radial direction, the convection boundary
condition on the outer surface of the pipe can be expressed as
−k
r1
r2
dT (r2 )
= h[T (r2 ) − T∞ ]
dr
2-43 A spherical shell of inner radius r1, outer radius r2, and thermal
conductivity k is considered. The outer surface of the shell is subjected to
radiation to surrounding surfaces at Tsurr . Assuming no convection and
steady one-dimensional conduction in the radial direction, the radiation
boundary condition on the outer surface of the shell can be expressed as
−k
h, T∞
[
dT (r2 )
4
= εσ T (r2 ) 4 − Tsurr
dr
ε
k
r1
Tsurr
r2
]
2-44 A spherical container consists of two spherical layers A and B that are at
perfect contact. The radius of the interface is ro. Assuming transient onedimensional conduction in the radial direction, the boundary conditions at the
interface can be expressed as
ro
T A (ro , t ) = T B (ro , t )
and
−kA
∂T A (ro , t )
∂T B (ro , t )
= −k B
∂r
∂r
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2-16
2-45 Heat conduction through the bottom section of a steel pan that is used to boil water on top of an electric range is
considered. Assuming constant thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the
differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant.
3 There is no heat generation in the medium. 4 The top surface at x = L is subjected to convection and the bottom surface at
x = 0 is subjected to uniform heat flux.
Analysis The heat flux at the bottom of the pan is
q& s =
E& gen
Q& s
0.85 × (1250 W)
=
=
= 33,820 W/m 2
2
As πD / 4 π (0.20 m) 2 / 4
Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as
d 2T
=0
dx 2
dT (0)
= q& s = 33,280 W/m 2
dx
dT ( L)
−k
= h[T ( L) − T∞ ]
dx
−k
2-46 A 2-kW resistance heater wire is used for space heating. Assuming constant thermal conductivity and one-dimensional
heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction
problem is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant.
3 Heat is generated uniformly in the wire.
Analysis The heat flux at the surface of the wire is
E& gen
Q&
2000 W
=
= 53.05 W/cm 2
q& s = s =
As 2πro L 2π (0.15 cm)(40 cm)
Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer surface, the differential
equation and the boundary conditions for this heat conduction problem can be expressed as
1 d ⎛ dT ⎞ e& gen
=0
⎟+
⎜r
k
r dr ⎝ dr ⎠
2 kW
dT (0)
=0
dr
dT (ro )
−k
= q& s = 53.05 W/cm 2
dr
D = 0.3 cm
L = 40 cm
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2-17
2-47 The outer surface of the East wall of a house exchanges heat with both convection and radiation., while the interior
surface is subjected to convection only. Assuming the heat transfer through the wall to be steady and one-dimensional, the
mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem
is to be obtained.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal
conductivity is given to be constant. 3 There is no heat generation in the medium.
4 The outer surface at x = L is subjected to convection and radiation while the inner
surface at x = 0 is subjected to convection only.
Analysis Expressing all the temperatures in Kelvin, the differential equation and the
boundary conditions for this heat conduction problem can be expressed as
Tsky
T∞1
h1
T∞2
h2
d 2T
=0
dx 2
−k
dT (0)
= h1[T∞1 − T (0)]
dx
−k
dT ( L)
4
= h1 [T ( L) − T∞ 2 ] + ε 2σ T ( L) 4 − Tsky
dx
L
[
x
]
2-48 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is dropped into a large body of water
at T∞ where it is cooled by convection. Assuming constant thermal conductivity and transient one-dimensional heat transfer,
the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction
problem is to be obtained.
Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given to be constant.
3 There is no heat generation in the medium. 4 The outer surface at r = r0 is subjected to convection.
Analysis Noting that there is thermal symmetry about the midpoint and convection at the outer surface, the differential
equation and the boundary conditions for this heat conduction problem can be expressed as
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎟=
⎜r
∂r ⎠ α ∂t
r 2 ∂r ⎝
∂T (0, t )
=0
∂r
∂T (ro , t )
−k
= h[T (ro ) − T∞ ]
∂r
T (r ,0) = Ti
k
r2
T∞
h
Ti
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2-18
2-49 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is allowed to cool in ambient air at T∞
by convection and radiation. Assuming constant thermal conductivity and transient one-dimensional heat transfer, the
mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem
is to be obtained.
Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given to be variable.
3 There is no heat generation in the medium. 4 The outer surface at r = ro is subjected to convection and radiation.
Analysis Noting that there is thermal symmetry about the midpoint and convection and radiation at the outer surface and
expressing all temperatures in Rankine, the differential equation and the boundary conditions for this heat conduction
problem can be expressed as
ε
∂T
1 ∂ ⎛ 2 ∂T ⎞
⎟ = ρc
⎜ kr
2 ∂r
r
∂t
∂
r
⎠
⎝
Tsurr
∂T (0, t )
=0
∂r
∂T (ro , t )
4
]
−k
= h[T (ro ) − T∞ ] + εσ[T ( ro ) 4 − Tsurr
∂r
T ( r ,0) = Ti
k
T∞
h
r2
Ti
2-50 Water flows through a pipe whose outer surface is wrapped with a thin electric heater that consumes 400 W per m
length of the pipe. The exposed surface of the heater is heavily insulated so that the entire heat generated in the heater is
transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection. Assuming constant
thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the
boundary conditions) of the heat conduction in the pipe is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant.
3 There is no heat generation in the medium. 4 The outer surface at r = r2 is subjected to uniform heat flux and the inner
surface at r = r1 is subjected to convection.
Analysis The heat flux at the outer surface of the pipe is
q& s =
Q& s
Q& s
400 W
=
=
= 979.4 W/m 2
As 2πr2 L 2π (0.065 cm)(1 m)
Noting that there is thermal symmetry about the center line and there is
uniform heat flux at the outer surface, the differential equation and the
boundary conditions for this heat conduction problem can be expressed as
Q = 400 W
h
T∞
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
r1
r2
dT (r1 )
= h[T (ri ) − T∞ ] = 85[T (ri ) − 90]
dr
dT (r2 )
= q& s = 734.6 W/m 2
k
dr
k
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2-19
Solution of Steady One-Dimensional Heat Conduction Problems
2-51C Yes, this claim is reasonable since no heat is entering the cylinder and thus there can be no heat transfer from the
cylinder in steady operation. This condition will be satisfied only when there are no temperature differences within the
cylinder and the outer surface temperature of the cylinder is the equal to the temperature of the surrounding medium.
2-52C Yes, the temperature in a plane wall with constant thermal conductivity and no heat generation will vary linearly
during steady one-dimensional heat conduction even when the wall loses heat by radiation from its surfaces. This is because
the steady heat conduction equation in a plane wall is d 2 T / dx 2 = 0 whose solution is T ( x) = C1 x + C 2 regardless of the
boundary conditions. The solution function represents a straight line whose slope is C1.
2-53C Yes, in the case of constant thermal conductivity and no heat generation, the temperature in a solid cylindrical rod
whose ends are maintained at constant but different temperatures while the side surface is perfectly insulated will vary
linearly during steady one-dimensional heat conduction. This is because the steady heat conduction equation in this case is
d 2T / dx 2 = 0 whose solution is T ( x) = C1 x + C 2 which represents a straight line whose slope is C1.
2-54C Yes, this claim is reasonable since in the absence of any heat generation the rate of heat transfer through a plain wall
in steady operation must be constant. But the value of this constant must be zero since one side of the wall is perfectly
insulated. Therefore, there can be no temperature difference between different parts of the wall; that is, the temperature in a
plane wall must be uniform in steady operation.
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2-20
2-55 A 20-mm thick draw batch furnace front is subjected to uniform
heat flux on the inside surface, while the outside surface is subjected
to convection and radiation heat transfer. The inside surface
temperature of the furnace front is to be determined.
Assumptions 1 Heat conduction is steady. 2 One dimensional heat
conduction across the furnace front thickness. 3 Thermal properties
are constant. 4 Inside and outside surface temperatures are constant.
Properties Emissivity and thermal conductivity are given to be 0.30
and 25 W/m · K, respectively
Analysis The uniform heat flux subjected on the inside
surface is equal to the sum of heat fluxes transferred by
convection and radiation on the outside surface:
4
q& 0 = h(TL − T∞ ) + εσ (TL4 − Tsurr
)
5000 W/m 2 = (10 W/m 2 ⋅ K )[TL − (20 + 273)] K
+ (0.30)(5.67 × 10 −8 W/m 2 ⋅ K 4 )[TL4 − (20 + 273) 4 ] K 4
Copy the following line and paste on a blank EES screen to solve the above equation:
5000=10*(T_L-(20+273))+0.30*5.67e-8*(T_L^4-(20+273)^4)
Solving by EES software, the outside surface temperature of the furnace front is
TL = 594 K
For steady heat conduction, the Fourier’s law of heat conduction can be expressed as
q& 0 = −k
dT
dx
Knowing that the heat flux and thermal conductivity are constant, integrating the differential equation once with respect to x
yields
T ( x) = −
q& 0
x + C1
k
Applying the boundary condition gives
x = L:
T ( L ) = TL = −
q& 0
L + C1
k
→
C1 =
q& 0
L + TL
k
Substituting C1 into the general solution, the variation of temperature in the furnace front is determined to be
T ( x) =
q& 0
( L − x ) + TL
k
The inside surface temperature of the furnace front is
T (0) = T0 =
q& 0
5000 W/m 2
L + TL =
(0.020 m) + 594 K = 598 K
k
25 W/m ⋅ K
Discussion By insulating the furnace front, heat loss from the outer surface can be reduced.
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2-21
2-56 A large plane wall is subjected to specified heat flux and temperature on the left surface and no conditions on the right
surface. The mathematical formulation, the variation of temperature in the plate, and the right surface temperature are to be
determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal
conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the
wall.
Properties The thermal conductivity is given to be k =2.5 W/m⋅°C.
Analysis (a) Taking the direction normal to the surface of the wall to
be the x direction with x = 0 at the left surface, the mathematical
formulation of this problem can be expressed as
d T
=0
dx 2
and
k
q=700 W/m2
T1=80°C
2
L=0.3 m
dT (0)
−k
= q& 0 = 700 W/m 2
dx
T (0) = T1 = 80°C
x
(b) Integrating the differential equation twice with respect to x yields
dT
dx
= C1
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
q&0
k
Heat flux at x = 0:
− kC1 = q& 0 → C1 = −
Temperature at x = 0:
T (0) = C1 × 0 + C2 = T1 → C2 = T1
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T ( x) = −
q& 0
700 W/m 2
x + T1 = −
x + 80°C = −280 x + 80
2.5 W/m ⋅ °C
k
(c) The temperature at x = L (the right surface of the wall) is
T (L) = −280 × (0.3 m) + 80 = -4°C
Note that the right surface temperature is lower as expected.
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2-22
2-57 A large plane wall is subjected to specified heat flux and temperature on the left surface and no conditions on the right
surface. The mathematical formulation, the variation of temperature in the plate, and the right surface temperature are to be
determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal
conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the
wall.
Properties The thermal conductivity is given to be k =2.5 W/m⋅°C.
Analysis (a) Taking the direction normal to the surface of the wall to
be the x direction with x = 0 at the left surface, the mathematical
formulation of this problem can be expressed as
k
d 2T
=0
dx 2
q=1050 W/m2
T1=90°C
and
L=0.3 m
dT (0)
−k
= q& 0 = 1050 W/m 2
dx
T (0) = T1 = 90°C
x
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
q& 0
k
Heat flux at x = 0:
− kC1 = q& 0 → C1 = −
Temperature at x = 0:
T (0) = C1 × 0 + C 2 = T1 → C 2 = T1
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T ( x) = −
q& 0
1050 W/m 2
x + T1 = −
x + 90°C = −420 x + 90
k
2.5 W/m ⋅ °C
(c) The temperature at x = L (the right surface of the wall) is
T (L) = −420 × (0.3 m) + 90 = -36°C
Note that the right surface temperature is lower as expected.
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2-23
2-58 A large plane wall is subjected to specified temperature on the left surface and convection on the right surface. The
mathematical formulation, the variation of temperature, and the rate of heat transfer are to be determined for steady onedimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat
generation.
Properties The thermal conductivity is given to be k = 1.8 W/m⋅°C.
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the
mathematical formulation of this problem can be expressed as
d 2T
dx 2
=0
and
k
T1=90°C
A=30 m2
T (0) = T1 = 90°C
−k
dT ( L)
= h[T ( L) − T∞ ]
dx
L=0.4 m
T∞ =25°C
h=24 W/m2.°C
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
x
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
x = 0:
T (0) = C1 × 0 + C 2 → C 2 = T1
x = L:
− kC1 = h[(C1 L + C 2 ) − T∞ ] → C1 = −
h(C 2 − T∞ )
h(T1 − T∞ )
→ C1 = −
k + hL
k + hL
Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be
T ( x) = −
=−
h(T1 − T∞ )
x + T1
k + hL
(24 W/m 2 ⋅ °C)(90 − 25)°C
(1.8 W/m ⋅ °C) + (24 W/m 2 ⋅ °C)(0.4 m)
= 90 − 90.3 x
x + 90°C
(c) The rate of heat conduction through the wall is
h(T1 − T∞ )
dT
= −kAC1 = kA
Q& wall = −kA
dx
k + hL
(24 W/m 2 ⋅ °C)(90 − 25)°C
= (1.8 W/m ⋅ °C)(30 m 2 )
(1.8 W/m ⋅ °C) + (24 W/m 2 ⋅ °C)(0.4 m)
= 7389 W
Note that under steady conditions the rate of heat conduction through a plain wall is constant.
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2-24
2-59 The top and bottom surfaces of a solid cylindrical rod are maintained at constant temperatures of 20°C and 95°C while
the side surface is perfectly insulated. The rate of heat transfer through the rod is to be determined for the cases of copper,
steel, and granite rod.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat
generation.
Properties The thermal conductivities are given to be k = 380 W/m⋅°C for copper, k = 18 W/m⋅°C for steel, and k = 1.2
W/m⋅°C for granite.
Analysis Noting that the heat transfer area (the area normal to
the direction of heat transfer) is constant, the rate of heat
transfer along the rod is determined from
T − T2
Q& = kA 1
L
Insulated
D = 0.05 m
T1=25°C
T2=95°C
where L = 0.15 m and the heat transfer area A is
A = πD 2 / 4 = π (0.05 m) 2 / 4 = 1.964 × 10 −3 m 2
L=0.15 m
Then the heat transfer rate for each case is determined as follows:
(a) Copper:
T − T2
(95 − 20)°C
= (380 W/m ⋅ °C)(1.964 × 10 −3 m 2 )
= 373.1 W
Q& = kA 1
0.15 m
L
(b) Steel:
T − T2
(95 − 20)°C
= (18 W/m ⋅ °C)(1.964 × 10 −3 m 2 )
= 17.7 W
Q& = kA 1
L
0.15 m
(c) Granite:
T − T2
(95 − 20)°C
= (1.2 W/m ⋅ °C)(1.964 × 10 −3 m 2 )
= 1.2 W
Q& = kA 1
0.15 m
L
Discussion: The steady rate of heat conduction can differ by orders of magnitude, depending on the thermal conductivity of
the material.
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2-25
2-60
plotted.
Prob. 2-59 is reconsidered. The rate of heat transfer as a function of the thermal conductivity of the rod is to be
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
L=0.15 [m]
D=0.05 [m]
T_1=20 [C]
T_2=95 [C]
k=1.2 [W/m-C]
"ANALYSIS"
A=pi*D^2/4
Q_dot=k*A*(T_2-T_1)/L
Q
[W]
0.9817
21.6
42.22
62.83
83.45
104.1
124.7
145.3
165.9
186.5
207.1
227.8
248.4
269
289.6
310.2
330.8
351.5
372.1
392.7
400
350
300
250
Q [W ]
k
[W/m.C]
1
22
43
64
85
106
127
148
169
190
211
232
253
274
295
316
337
358
379
400
200
150
100
50
0
0
50
100
150
200
250
300
350
400
k [W /m -C]
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2-26
2-61 The base plate of a household iron is subjected to specified heat flux on the left surface and to specified temperature on
the right surface. The mathematical formulation, the variation of temperature in the plate, and the inner surface temperature
are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is large relative to its
thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is
no heat generation in the plate. 4 Heat loss through the upper part of the iron is negligible.
Properties The thermal conductivity is given to be k = 60 W/m⋅°C.
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the resistance wires
is transferred to the base plate, the heat flux through the inner surface is determined to be
q& 0 =
Q& 0
800 W
=
= 50,000 W/m 2
Abase 160 × 10 − 4 m 2
Taking the direction normal to the surface of the wall to be the x
direction with x = 0 at the left surface, the mathematical formulation
of this problem can be expressed as
Q =800 W
A=160 cm2
d 2T
=0
dx 2
and
−k
k
T2 =112°C
L=0.6 cm
dT (0)
= q& 0 = 50,000 W/m 2
dx
T ( L) = T2 = 112°C
x
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
q& 0
k
x = 0:
− kC1 = q& 0 → C1 = −
x = L:
T ( L) = C1 L + C 2 = T2 → C 2 = T2 − C1 L → C 2 = T2 +
q& 0 L
k
Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be
q& 0
q& L q& ( L − x)
x + T2 + 0 = 0
+ T2
k
k
k
(50,000 W/m 2 )(0.006 − x)m
=
+ 112°C
60 W/m ⋅ °C
= 833.3(0.006 − x) + 112
T ( x) = −
(c) The temperature at x = 0 (the inner surface of the plate) is
T (0) = 833.3(0.006 − 0) + 112 = 117°C
Note that the inner surface temperature is higher than the exposed surface temperature, as expected.
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2-27
2-62 The base plate of a household iron is subjected to specified heat flux on the left surface and to specified temperature on
the right surface. The mathematical formulation, the variation of temperature in the plate, and the inner surface temperature
are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is large relative to its
thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is
no heat generation in the plate. 4 Heat loss through the upper part of the iron is negligible.
Properties The thermal conductivity is given to be k = 60 W/m⋅°C.
Analysis (a) Noting that the upper part of the iron is well insulated and thus
the entire heat generated in the resistance wires is transferred to the base
plate, the heat flux through the inner surface is determined to be
Q=1200 W
A=160 cm2
Q&
1200 W
= 75,000 W/m 2
q& 0 = 0 =
Abase 160 × 10 − 4 m 2
k
T2 =112°C
L=0.6 cm
Taking the direction normal to the surface of the wall to be the x direction with
x = 0 at the left surface, the mathematical formulation of this problem can be
expressed as
d 2T
=0
dx 2
and
−k
x
dT (0)
= q& 0 = 75,000 W/m 2
dx
T ( L) = T2 = 112°C
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
q& 0
k
x = 0:
− kC1 = q& 0 → C1 = −
x = L:
T ( L) = C1 L + C 2 = T2 → C 2 = T2 − C1 L → C 2 = T2 +
q& 0 L
k
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T ( x) = −
q& L q& ( L − x)
q& 0
+ T2
x + T2 + 0 = 0
k
k
k
(75,000 W/m 2 )(0.006 − x)m
+ 112°C
60 W/m ⋅ °C
= 1250(0.006 − x) + 112
=
(c) The temperature at x = 0 (the inner surface of the plate) is
T (0) = 1250(0.006 − 0) + 112 = 119.5°C
Note that the inner surface temperature is higher than the exposed surface temperature, as expected.
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2-28
2-63
Prob. 2-61 is reconsidered. The temperature as a function of the distance is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
Q_dot=800 [W]
L=0.006 [m]
A_base=160E-4 [m^2]
k=60 [W/m-C]
T_2=112 [C]
"ANALYSIS"
q_dot_0=Q_dot/A_base
T=q_dot_0*(L-x)/k+T_2 "Variation of temperature"
"x is the parameter to be varied"
T
[C]
117
116.5
116
115.5
115
114.5
114
113.5
113
112.5
112
117
116
115
T [C]
x
[m]
0
0.0006
0.0012
0.0018
0.0024
0.003
0.0036
0.0042
0.0048
0.0054
0.006
114
113
112
0
0.001
0.002
0.003
0.004
0.005
0.006
x [m]
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2-29
2-64 Chilled water flows in a pipe that is well insulated from outside. The mathematical formulation and the variation of
temperature in the pipe are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is
thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe.
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this
problem can be expressed as
Insulated
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
and
−k
dT (r1 )
= h[T f − T ( r1 )]
dr
Water
Tf
r2
r1
dT (r2 )
=0
dr
L
(b) Integrating the differential equation once with respect to r gives
r
dT
= C1
dr
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,
dT C1
=
dr
r
T (r ) = C1 ln r + C 2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
r = r2:
r = r1:
C1
= 0 → C1 = 0
r2
−k
C1
= h[T f − (C1 ln r1 + C 2 )]
r1
0 = h(T f − C 2 ) → C 2 = T f
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T (r ) = T f
This result is not surprising since steady operating conditions exist.
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2-30
2-65 The convection heat transfer coefficient between the surface of a pipe carrying superheated vapor and the surrounding
air is to be determined.
Assumptions 1 Heat conduction is steady and one-dimensional and there is thermal symmetry about the centerline. 2
Thermal properties are constant. 3 There is no heat generation in the pipe. 4 Heat transfer by radiation is negligible.
Properties The constant pressure specific heat of vapor is given to be 2190 J/kg · °C and the pipe thermal conductivity is 17
W/m · °C.
Analysis The inner and outer radii of the pipe are
r1 = 0.05 m / 2 = 0.025 m
r2 = 0.025 m + 0.006 m = 0.031 m
The rate of heat loss from the vapor in the pipe can be determined from
Q& loss = m& c p (Tin − Tout ) = (0.3 kg/s)(2190 J/kg ⋅ °C)(7) °C = 4599 W
For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as
d ⎛ dT ⎞
⎟=0
⎜r
dr ⎝ dr ⎠
and
−k
Q&
dT ( r1 ) Q& loss
=
= loss
2π r1 L
dr
A
T (r1 ) = 120 °C
(heat flux at the inner pipe surface)
(inner pipe surface temperature)
Integrating the differential equation once with respect to r gives
dT C1
=
dr
r
Integrating with respect to r again gives
T (r ) = C1 ln r + C 2
where C1 and C 2 are arbitrary constants. Applying the boundary conditions gives
r = r1 :
dT (r1 )
C
1 Q& loss
=−
= 1
dr
k 2π r1 L r1
r = r1 :
T (r1 ) = −
1 Q& loss
ln r1 + C 2
2π kL
→
C1 = −
1 Q& loss
2π kL
→
C2 =
1 Q& loss
ln r1 + T (r1 )
2π kL
Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be
1 Q& loss
1 Q& loss
ln r1 + T (r1 )
ln r +
2π kL
2π kL
1 Q& loss
=−
ln(r / r1 ) + T (r1 )
2π kL
T (r ) = −
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2-31
The outer pipe surface temperature is
1 Q& loss
ln(r2 / r1 ) + T (r1 )
2π kL
1
4599 W
⎛ 0.031 ⎞
ln⎜
=−
⎟ + 120 °C
2π (17 W/m ⋅ °C)(10 m) ⎝ 0.025 ⎠
= 119.1 °C
T (r2 ) = −
From Newton’s law of cooling, the rate of heat loss at the outer pipe surface by convection is
Q& loss = h(2π r2 L)[T (r2 ) − T∞ ]
Rearranging and the convection heat transfer coefficient is determined to be
h=
Q& loss
4599 W
=
= 25.1 W/m 2 ⋅ °C
2π r2 L[T (r2 ) − T∞ ] 2π (0.031 m)(10 m)(119.1 − 25) °C
Discussion If the pipe wall is thicker, the temperature difference between the inner and outer pipe surfaces will be greater. If
the pipe has very high thermal conductivity or the pipe wall thickness is very small, then the temperature difference between
the inner and outer pipe surfaces may be negligible.
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2-32
2-66 A subsea pipeline is transporting liquid hydrocarbon. The temperature variation in the pipeline wall, the inner surface
temperature of the pipeline, the mathematical expression for the rate of heat loss from the liquid hydrocarbon, and the heat
flux through the outer pipeline surface are to be determined.
Assumptions 1 Heat conduction is steady and one-dimensional and there is thermal symmetry about the centerline. 2
Thermal properties are constant. 3 There is no heat generation in the pipeline.
Properties The pipeline thermal conductivity is given to be 60 W/m · °C.
Analysis The inner and outer radii of the pipeline are
r1 = 0.5 m / 2 = 0.25 m
r2 = 0.25 m + 0.008 m = 0.258 m
(a) For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as
d ⎛ dT ⎞
⎟=0
⎜r
dr ⎝ dr ⎠
and
−k
dT (r1 )
= h1[T∞,1 − T (r1 )]
dr
(convection at the inner pipeline surface)
−k
dT (r2 )
= h2 [T (r2 ) − T∞ , 2 ]
dr
(convection at the outer pipeline surface)
Integrating the differential equation once with respect to r gives
dT C1
=
dr
r
Integrating with respect to r again gives
T (r ) = C1 ln r + C 2
where C1 and C 2 are arbitrary constants. Applying the boundary conditions gives
r = r1 :
−k
C
dT(r1 )
= − k 1 = h1 (T∞,1 − C1 ln r1 − C 2 )
r1
dr
r = r2 :
−k
dT (r2 )
C
= − k 1 = h2 (C1 ln r2 + C 2 − T∞, 2 )
dr
r2
C1 and C 2 can be expressed explicitly as
C1 = −
T∞,1 − T∞, 2
k /(r1h1 ) + ln(r2 / r1 ) + k /( r2 h2 )
C 2 = T∞,1 −
⎛ k
⎞
⎜⎜
− ln r1 ⎟⎟
k /( r1 h1 ) + ln(r2 / r1 ) + k /( r2 h2 ) ⎝ r1 h1
⎠
T∞,1 − T∞,2
Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be
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2-33
T (r ) = −
⎡ k
⎤
+ ln(r / r1 )⎥ + T∞,1
⎢
k /(r1 h1 ) + ln(r2 / r1 ) + k /(r2 h2 ) ⎣ r1 h1
⎦
T∞,1 − T∞, 2
(b) The inner surface temperature of the pipeline is
⎤
⎡ k
+ ln(r1 / r1 )⎥ + T∞,1
⎢
k /( r1 h1 ) + ln(r2 / r1 ) + k /( r2 h2 ) ⎣ r1h1
⎦
⎡
⎤
60 W/m ⋅ °C
(70 − 5) °C ⎢
⎥
2
⎣⎢ (0.25 m)(250 W/m ⋅ °C) ⎦⎥
+ 70 °C
=−
60 W/m ⋅ °C
60 W/m ⋅ °C
⎛ 0.258 ⎞
+ ln⎜
⎟+
(0.25 m)(250 W/m 2 ⋅ °C)
⎝ 0.25 ⎠ (0.258 m)(150 W/m 2 ⋅ °C)
T (r1 ) = −
T∞,1 − T∞ , 2
= 45.5 °C
(c) The mathematical expression for the rate of heat loss through the pipeline can be determined from Fourier’s law to be
dT
Q& loss = −kA
dr
dT (r2 )
= −2πLkC1
dr
T∞,1 − T∞ ,2
=
ln(r2 / r1 )
1
1
+
+
2π r1 Lh1
2πLk
2π r2 Lh2
= −k ( 2π r2 L)
(d) Again from Fourier’s law, the heat flux through the outer pipeline surface is
q& 2 = − k
=
=
dT (r2 )
C
dT
= −k
= −k 1
dr
dr
r2
T∞,1 − T∞ , 2
k
k /( r1 h1 ) + ln(r2 / r1 ) + k /( r2 h2 ) r2
(70 − 5) °C
⎛ 60 W/m ⋅ °C ⎞
⎟
⎜
60 W/m ⋅ °C
60 W/m ⋅ °C
⎛ 0.258 ⎞
⎝ 0.258 m ⎠
+
ln
+
⎟
⎜
(0.25 m)(250 W/m 2 ⋅ °C)
⎝ 0.25 ⎠ (0.258 m)(150 W/m 2 ⋅ °C)
= 5947 W/m 2
Discussion Knowledge of the inner pipeline surface temperature can be used to control wax deposition blockages in the
pipeline.
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2-34
2-67 A steam pipe is subjected to convection on the inner surface and to specified temperature on the outer surface.
The mathematical formulation, the variation of temperature in the pipe, and the rate of heat loss are to be determined for
steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is
thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe.
Properties The thermal conductivity is given to be k = 12.5 W/m⋅°C.
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this
problem can be expressed as
T =80°C
d ⎛ dT ⎞
⎟=0
⎜r
dr ⎝ dr ⎠
Steam
dT (r1 )
−k
= h[T∞ − T ( r1 )]
and
150°C
dr
h=70
T (r2 ) = T2 = 70°C
L=9m
(b) Integrating the differential equation once with respect to r gives
dT
r
= C1
dr
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,
dT C1
=
dr
r
T (r ) = C1 ln r + C 2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
C
r = r1:
− k 1 = h[T∞ − (C1 ln r1 + C 2 )]
r1
r = r2:
T ( r2 ) = C1 ln r2 + C 2 = T2
Solving for C1 and C2 simultaneously gives
T − T∞
T 2 − T∞
and C 2 = T2 − C1 ln r2 = T2 − 2
ln r2
C1 =
r2
r2
k
k
ln +
ln +
r1 hr1
r1 hr1
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T − T∞
r
ln + T2
T (r ) = C1 ln r + T2 − C1 ln r2 = C1 (ln r − ln r2 ) + T2 = 2
r2
k
r2
ln +
r1 hr1
=
(80 − 150)°C
r
r
ln
+ 80°C = −18.65 ln
+ 80°C
6
12.5 W/m ⋅ °C
6 cm
6 cm
ln +
5 (70 W/m 2 ⋅ °C)(0.05 m)
(c) The rate of heat conduction through the pipe is
T − T∞
C
dT
= − k (2πrL) 1 = −2πLk 2
Q& = − kA
r
k
r
dr
ln 2 +
r1 hr1
= −2π (9 m)(12.5 W/m ⋅ °C)
(80 − 150)°C
= 13,180 W
6
12.5 W/m ⋅ °C
ln +
5 (70 W/m 2 ⋅ °C)(0.05 m)
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2-35
2-68 A spherical container is subjected to specified temperature on the inner surface and convection on the outer surface.
The mathematical formulation, the variation of temperature, and the rate of heat transfer are to be determined for steady onedimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal
symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation.
Properties The thermal conductivity is given to be k = 30 W/m⋅°C.
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this
problem can be expressed as
d ⎛ 2 dT ⎞
⎜r
⎟=0
dr ⎝
dr ⎠
and
r1
T ( r1 ) = T1 = 0°C
−k
T1
k
r2
T∞
h
dT (r2 )
= h[T ( r2 ) − T∞ ]
dr
(b) Integrating the differential equation once with respect to r gives
dT
r2
= C1
dr
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,
dT C1
=
dr r 2
C1
+ C2
r
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
T (r ) = −
r = r1:
T ( r1 ) = −
r = r2:
−k
C1
+ C 2 = T1
r1
⎛ C
⎞
C1
= h⎜⎜ − 1 + C2 − T∞ ⎟⎟
2
r2
⎝ r2
⎠
Solving for C1 and C2 simultaneously gives
C1 =
r2 (T1 − T∞ )
r
k
1− 2 −
r1 hr2
and
C 2 = T1 +
C1
= T1 +
r1
T1 − T∞ r2
r
k r1
1− 2 −
r1 hr2
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T (r ) = −
⎛ 1 1⎞
C
C1
+ T1 + 1 = C1 ⎜⎜ − ⎟⎟ + T1 =
r1
r
⎝ r1 r ⎠
=
(0 − 25)°C
30 W/m ⋅ °C
1−
2.1
−
2 (18 W/m 2 ⋅ °C)(2.1 m)
T1 − T∞ ⎛ r2 r2 ⎞
⎜ − ⎟⎟ + T1
r2
k ⎜⎝ r1
r ⎠
1− −
r1 hr2
⎛ 2.1 2.1 ⎞
−
⎟ + 0°C = 29.63(1.05 − 2.1 / r )
⎜
r ⎠
⎝ 2
(c) The rate of heat conduction through the wall is
r (T − T )
C
dT
Q& = −kA
= −k (4πr 2 ) 21 = −4πkC1 = −4πk 2 1 ∞
r
k
dr
r
1− 2 −
r1 hr2
= −4π (30 W/m ⋅ °C)
(2.1 m)(0 − 25)°C
= 23,460 W
2.1
30 W/m ⋅ °C
1−
−
2 (18 W/m 2 ⋅ °C)(2.1 m)
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2-36
2-69 A spherical container is subjected to uniform heat flux on the outer surface and specified temperature on the inner
surface. The mathematical formulation, the variation of temperature in the pipe, and the outer surface temperature, and the
maximum rate of hot water supply are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal
symmetry about the mid point. 2 Thermal conductivity is constant. 3 There is no heat generation in the container.
Properties The thermal conductivity is given to be k = 1.5 W/m⋅°C. The specific heat of water at the average temperature of
(100+20)/2 = 60°C is 4.185 kJ/kg⋅°C (Table A-9).
Analysis (a) Noting that the 90% of the 800 W generated by the strip heater is transferred to the container, the heat flux
through the outer surface is determined to be
Q&
Q&
0.90 × 800 W
q& s = s = s 2 =
= 340.8 W/m 2
2
A2 4πr2
4π (0.41 m)
Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r direction, the
mathematical formulation of this problem can be expressed as
d ⎛ 2 dT ⎞
⎟=0
⎜r
dr ⎝
dr ⎠
and
Insulation
T1
k
T (r1 ) = T1 = 120°C
Heater
dT ( r2 )
k
= q& s
dr
r1
r2
r
(b) Integrating the differential equation once with respect to r gives
r2
dT
= C1
dr
Dividing both sides of the equation above by r2 and then integrating,
dT C1
=
dr r 2
C1
+ C2
r
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
T (r ) = −
C1
= q& s → C1 =
2
r = r2:
k
r = r1:
T (r1 ) = T1 = −
r2
q& s r22
k
q& r 2
C1
C
+ C 2 → C 2 = T1 + 1 = T1 + s 2
r1
r1
kr1
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T (r ) = −
⎛ 1 1 ⎞ q& r 2
⎛ 1 1⎞
C1
C
C
+ C 2 = − 1 + T1 + 1 = T1 + ⎜⎜ − ⎟⎟C1 = T1 + ⎜⎜ − ⎟⎟ s 2
r
r
r1
⎝ r1 r ⎠ k
⎝ r1 r ⎠
1⎞
1 ⎞ (340.8 W/m 2 )(0.41 m) 2
⎛ 1
⎛
= 120°C + ⎜
− ⎟
= 120 + 38.19⎜ 2.5 − ⎟
r⎠
1.5 W/m ⋅ °C
⎝ 0.40 m r ⎠
⎝
(c) The outer surface temperature is determined by direct substitution to be
⎛
1⎞
1 ⎞
⎛
Outer surface (r = r2): T (r2 ) = 120 + 38.19⎜⎜ 2.5 − ⎟⎟ = 120 + 38.19⎜ 2.5 −
⎟ = 122.3°C
0.41 ⎠
r2 ⎠
⎝
⎝
Noting that the maximum rate of heat supply to the water is 0.9 × 800 W = 720 W, water can be heated from 20 to 100°C at a
rate of
Q&
0.720 kJ/s
Q& = m& c p ΔT → m& =
=
= 0.002151 kg/s = 7.74 kg/h
c p ΔT (4.185 kJ/kg ⋅ °C)(100 − 20)°C
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2-37
2-70
Prob. 2-69 is reconsidered. The temperature as a function of the radius is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
r_1=0.40 [m]
r_2=0.41 [m]
k=1.5 [W/m-C]
T_1=120 [C]
Q_dot=800 [W]
f_loss=0.10
"ANALYSIS"
q_dot_s=((1-f_loss)*Q_dot)/A
A=4*pi*r_2^2
T=T_1+(1/r_1-1/r)*(q_dot_s*r_2^2)/k "Variation of temperature"
T
[C]
120
120.3
120.5
120.8
121
121.3
121.6
121.8
122.1
122.3
122.5
122
121.5
T [C]
r
[m]
0.4
0.4011
0.4022
0.4033
0.4044
0.4056
0.4067
0.4078
0.4089
0.41
121
120.5
120
0.4
0.402
0.404
0.406
0.408
0.41
r [m]
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2-38
Heat Generation in a Solid
2-71C The cylinder will have a higher center temperature since the cylinder has less surface area to lose heat from per unit
volume than the sphere.
2-72C Heat generation in a solid is simply conversion of some form of energy into sensible heat energy. Some examples of
heat generations are resistance heating in wires, exothermic chemical reactions in a solid, and nuclear reactions in nuclear
fuel rods.
2-73C The rate of heat generation inside an iron becomes equal to the rate of heat loss from the iron when steady operating
conditions are reached and the temperature of the iron stabilizes.
2-74C No, it is not possible since the highest temperature in the plate will occur at its center, and heat cannot flow “uphill.”
2-75C No. Heat generation in a solid is simply the conversion of some form of energy into sensible heat energy. For example
resistance heating in wires is conversion of electrical energy to heat.
2-76 Heat is generated uniformly in a large brass plate. One side of the plate is insulated while the other side is subjected to
convection. The location and values of the highest and the lowest temperatures in the plate are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since the plate is large relative to its thickness, and there is thermal symmetry about the center plane 3 Thermal
conductivity is constant. 4 Heat generation is uniform.
Properties The thermal conductivity is given to be k =111 W/m⋅°C.
Analysis This insulated plate whose thickness is L is equivalent to one-half of
an uninsulated plate whose thickness is 2L since the midplane of the
uninsulated plate can be treated as insulated surface. The highest temperature
will occur at the insulated surface while the lowest temperature will occur at
the surface which is exposed to the environment. Note that L in the following
relations is the full thickness of the given plate since the insulated side
represents the center surface of a plate whose thickness is doubled. The
desired values are determined directly from
Ts = T∞ +
To = Ts +
e& gen L
h
e&gen L2
2k
= 25°C +
(2 × 10 5 W/m 3 )(0.05 m)
44 W/m 2 ⋅ °C
= 252.3°C +
k
egen
Insulated
L=5 cm
T∞ =25°C
h=44 W/m2.°C
= 252.3°C
(2 ×10 5 W/m 3 )(0.05 m) 2
= 254.6°C
2(111 W/m ⋅ °C)
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2-39
2-77
Prob. 2-76 is reconsidered. The effect of the heat transfer coefficient on the highest and lowest temperatures in
the plate is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
L=0.05 [m]
k=111 [W/m-C]
g_dot=2E5 [W/m^3]
T_infinity=25 [C]
h=44 [W/m^2-C]
"ANALYSIS"
T_min=T_infinity+(g_dot*L)/h
T_max=T_min+(g_dot*L^2)/(2*k)
Tmin
[C]
525
425
358.3
310.7
275
247.2
225
206.8
191.7
178.8
167.9
158.3
150
142.6
136.1
130.3
125
550
Tmax
[C]
527.3
427.3
360.6
313
277.3
249.5
227.3
209.1
193.9
181.1
170.1
160.6
152.3
144.9
138.4
132.5
127.3
500
450
400
T m in [C]
h
[W/m2.C]
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
350
300
250
200
150
100
20
30
40
50
60
70
80
90
100
2
h [W /m -C]
550
500
450
T m ax [C]
400
350
300
250
200
150
100
20
30
40
50
60
70
80
90
100
2
h [W /m -C]
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2-40
2-78 A cylindrical nuclear fuel rod is cooled by water flowing through its encased concentric tube. The average temperature
of the cooling water is to be determined.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat generation in the
fuel rod is uniform.
Properties The thermal conductivity is given to be 30 W/m · °C.
Analysis The rate of heat transfer by convection at the fuel rod surface is equal to that of the concentric tube surface:
h1 As ,1 (Ts ,rod − T∞ ) = h2 As , 2 (T∞ − Ts ,tube )
h1 (2π r1 L)(Ts ,rod − T∞ ) = h2 (2π r2 L)(T∞ − Ts , tube )
Ts ,rod =
h2 r2
(T∞ − Ts , tube ) + T∞
h1r1
(a)
The average temperature of the cooling water can be determined by applying Eq. 2-68:
Ts ,rod = T∞ +
e&gen r1
(b)
2h1
Substituting Eq. (a) into Eq. (b) and solving for the average temperature of the cooling water gives
e&gen r1
h2 r2
(T∞ − Ts ,tube ) + T∞ = T∞ +
h1r1
2h1
T∞ =
=
r1 e&gen r1
+ Ts ,tube
r2 2h2
0.005 m ⎡ (50 × 10 6 W/m 3 )(0.005 m) ⎤
⎥ + 40 °C
⎢
0.010 m ⎣⎢
2( 2000 W/m 2 ⋅ °C)
⎦⎥
= 71.25 °C
Discussion The given information is not sufficient for one to determine the fuel rod surface temperature. The convection heat
transfer coefficient for the fuel rod surface (h1) or the centerline temperature of the fuel rod (T0) is needed to determine the
fuel rod surface temperature.
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2-41
2-79 A spherical communication satellite orbiting in space absorbs solar radiation while losing heat to deep space by thermal
radiation. The heat generation rate and the surface temperature of the satellite are to be determined.
Assumptions 1 Heat transfer is steady and one-dimensional. 2 Heat generation is uniform. 3 Thermal properties are constant.
Properties The properties of the satellite are given to be ε = 0.75, α = 0.10, and k = 5 W/m · K.
Analysis For steady one-dimensional heat conduction in sphere, the differential equation is
1 d ⎛ 2 dT ⎞ e&gen
=0
⎟+
⎜r
k
r 2 dr ⎝ dr ⎠
and
T (0) = T0 = 273 K
dT (0)
=0
dr
(midpoint temperature of the satellite)
(thermal symmetry about the midpoint)
Multiply both sides of the differential equation by r 2 and rearranging gives
e&gen 2
d ⎛ 2 dT ⎞
r
⎟=−
⎜r
k
dr ⎝ dr ⎠
Integrating with respect to r gives
r2
e&gen r 3
dT
=−
+ C1
dr
k 3
(a)
Applying the boundary condition at the midpoint (thermal symmetry about the midpoint),
0×
r = 0:
e&gen
dT (0)
=−
× 0 + C1
k
dr
→
C1 = 0
Dividing both sides of Eq. (a) by r 2 and integrating,
e&gen
dT
r
=−
3k
dr
and
T (r ) = −
e&gen
6k
r 2 + C2
(b)
Applying the boundary condition at the midpoint (midpoint temperature of the satellite),
T0 = −
r = 0:
e&gen
6k
→
× 0 + C2
C 2 = T0
Substituting C 2 into Eq. (b), the variation of temperature is determined to be
T (r ) = −
e&gen
6k
r 2 + T0
At the satellite surface ( r = ro ), the temperature is
Ts = −
e&gen
6k
ro2 + T0
(c)
Also, the rate of heat transfer at the surface of the satellite can be expressed as
⎞
⎛4
4
e&gen ⎜ π ro3 ⎟ = As εσ (Ts4 − Tspace
) − As α s q& solar
⎠
⎝3
where
Tspace = 0
The surface temperature of the satellite can be explicitly expressed as
⎡ 1 ⎛4
⎞⎤
3
Ts = ⎢
⎜ π ro e&gen + Asα s q& solar ⎟⎥
⎠⎦
⎣ As εσ ⎝ 3
1/ 4
1/ 4
⎛ e&gen ro / 3 + α s q& solar ⎞
⎟
= ⎜⎜
⎟
εσ
⎠
⎝
(d)
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2-42
Substituting Eq. (c) into Eq. (d)
1/ 4
⎛ e&gen ro / 3 + α s q& solar ⎞
⎟
⎜
⎟
⎜
εσ
⎠
⎝
=−
e&gen
6k
ro2 + T0
⎡ e&gen (1.25 m) / 3 + (0.10)(1000 W/m 2 ) ⎤
⎢
⎥
⎢⎣ (0.75)(5.67 × 10 −8 W/m 2 ⋅ K 4 ) ⎥⎦
1/ 4
=−
e&gen (1.25 m) 2
6(5 W/m ⋅ K )
+ 273 K
Copy the following line and paste on a blank EES screen to solve the above equation:
((e_gen*1.25/3+0.10*1000)/(0.75*5.67e-8))^(1/4)=-e_gen*1.25^2/(6*5)+273
Solving by EES software, the heat generation rate is
e&gen = 233 W/m 3
Using Eq. (c), the surface temperature of the satellite is determined to be
Ts = −
( 233 W/m 3 )
(1.25 m) 2 + 273 K = 261 K
6(5 W/m ⋅ K )
Discussion The surface temperature of the satellite in space is well below freezing point of water.
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2-43
2-80 A 2-kW resistance heater wire with a specified surface temperature is used to boil water. The center temperature of the
wire is to be determined.
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer
is one-dimensional since there is thermal symmetry about the center line and no change in
the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is
uniform.
Properties The thermal conductivity is given to be k = 20 W/m⋅°C.
230°C
Analysis The resistance heater converts electric energy into heat at a rate of 2 kW. The
rate of heat generation per unit volume of the wire is
e& gen =
E& gen
V wire
=
E& gen
πro2 L
r
=
2000 W
π (0.002 m) 2 (0.9 m)
8
= 1.768 × 10 W/m
3
D
The center temperature of the wire is then determined from Eq. 2-71 to be
To = Ts +
e&gen ro2
4k
= 230°C +
(1.768 × 10 8 W/m 3 )(0.002 m) 2
= 238.8°C
4(20 W/m.°C)
2-81 Heat is generated in a long solid cylinder with a specified surface temperature. The variation of temperature in the
cylinder is given by
T (r ) =
2
e& gen ro2 ⎡ ⎛ r ⎞ ⎤
⎢1 − ⎜ ⎟ ⎥ + T s
k ⎢ ⎜⎝ ro ⎟⎠ ⎥
⎦
⎣
(a) Heat conduction is steady since there is no time t variable involved.
80°C
(b) Heat conduction is a one-dimensional.
(c) Using Eq. (1), the heat flux on the surface of the cylinder at r = ro
is determined from its definition to be
q& s = − k
⎡ e& gen ro2 ⎛
dT (ro )
= −k ⎢
dr
⎢⎣
k
⎞⎤
⎜ − 2r ⎟⎥
⎜ r 2 ⎟⎥
⎝ o ⎠⎦ r = r0
k
r
D
⎡ e& gen ro2 ⎛ 2r ⎞⎤
⎜ − o ⎟⎥ = 2e& gen ro = 2(35 W/cm 3 )(4 cm) = 280 W/cm 2
= −k ⎢
⎢⎣ k ⎜⎝ ro2 ⎟⎠⎥⎦
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2-44
2-82
Prob. 2-81 is reconsidered. The temperature as a function of the radius is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
r_0=0.04 [m]
k=25 [W/m-C]
g_dot_0=35E+6 [W/m^3]
T_s=80 [C]
"ANALYSIS"
T=(g_dot_0*r_0^2)/k*(1-(r/r_0)^2)+T_s "Variation of temperature"
T [C]
2320
2292
2209
2071
1878
1629
1324
964.9
550.1
80
2500
2000
T [C]
r [m]
0
0.004444
0.008889
0.01333
0.01778
0.02222
0.02667
0.03111
0.03556
0.04
1500
1000
500
0
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
r [m ]
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2-45
2-83 Heat is generated in a large plane wall whose one side is insulated while the other side is subjected to convection.
The mathematical formulation, the variation of temperature in the wall, the relation for the surface temperature, and the
relation for the maximum temperature rise in the plate are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform.
Analysis (a) Noting that heat transfer is steady and one-dimensional in x direction, the mathematical formulation of this
problem can be expressed as
d 2T
dx
and
2
+
e& gen
=0
k
dT (0)
= 0 (insulated surface at x = 0)
dx
−k
k
egen
dT ( L)
= h[T ( L) − T∞ ]
dx
T∞
h
Insulated
(b) Rearranging the differential equation and integrating,
d 2T
dx
2
=−
e& gen
e& gen
dT
x + C1
=−
dx
k
→
k
L
x
Integrating one more time,
T ( x) =
− e& gen x 2
2k
+ C1 x + C 2
(1)
Applying the boundary conditions:
−e& gen
dT (0)
=0→
(0) + C1 = 0 → C1 = 0
dx
k
B.C. at x = 0:
⎞
⎛ − e& gen L2
⎛ − e& gen ⎞
L ⎟⎟ = h⎜
− k ⎜⎜
+ C 2 − T∞ ⎟
⎟
⎜ 2k
⎝ k
⎠
⎠
⎝
B. C. at x = L:
e& gen L =
C2 =
Dividing by h:
e& gen L
h
+
− he& gen L2
2k
e& gen L2
2k
− hT∞ + C 2 → C 2 = e& gen L +
he& gen L2
2k
+ hT∞
+ T∞
Substituting the C1 and C2 relations into Eq. (1) and rearranging give
T ( x) =
− e& gen x 2
2k
+
e& gen L
h
+
e& gen L2
2k
+ T∞ =
e& gen
2k
( L2 − x 2 ) +
e& gen L
h
+ T∞
which is the desired solution for the temperature distribution in the wall as a function of x.
(c) The temperatures at two surfaces and the temperature difference between these surfaces are
T (0) =
T ( L) =
e& gen L2
2k
e& gen L
h
+
e& gen L
h
+ T∞
+ T∞
ΔTmax = T (0) − T ( L) =
e& gen L2
2k
Discussion These relations are obtained without using differential equations in the text (see Eqs. 2-67 and 2-73).
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2-46
2-84 A long homogeneous resistance heater wire with specified convection conditions at the surface is used to boil water.
The mathematical formulation, the variation of temperature in the wire, and the temperature at the centerline of the wire are
to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal
conductivity is constant. 4 Heat generation in the wire is uniform.
Properties The thermal conductivity is given to be k = 15.2 W/m⋅K.
Analysis Noting that heat transfer is steady and one-dimensional in
the radial r direction, the mathematical formulation of this problem
can be expressed as
r
T∞
h
Water
ro
1 d ⎛ dT ⎞ e& gen
=0
⎟+
⎜r
r dr ⎝ dr ⎠
k
and
−k
0
dT (ro )
= h[T ( ro ) − T∞ ] (convection at the outer surface)
dr
Heater
dT (0)
= 0 (thermal symmetry about the centerline)
dr
Multiplying both sides of the differential equation by r and rearranging gives
e& gen
d ⎛ dT ⎞
r
⎜r
⎟=−
dr ⎝ dr ⎠
k
Integrating with respect to r gives
r
e& gen r 2
dT
=−
+ C1
dr
k 2
(a)
It is convenient at this point to apply the second boundary condition since it is related to the first derivative of the
temperature by replacing all occurrences of r and dT/dr in the equation above by zero. It yields
B.C. at r = 0:
e& gen
dT (0)
=−
× 0 + C1
2k
dr
0×
→ C1 = 0
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating,
e& gen
dT
=−
r
dr
2k
and
T (r ) = −
e& gen
r 2 + C2
4k
(b)
Applying the second boundary condition at r = ro ,
B. C. at r = ro :
k
e& gen ro
2k
e& gen ro e& gen 2
⎛ e& gen 2
⎞
+
ro
ro + C 2 − T∞ ⎟⎟ → C 2 = T∞ +
= h⎜⎜ −
2h
4k
⎝ 4k
⎠
Substituting this C 2 relation into Eq. (b) and rearranging give
T (r ) = T∞ +
e& gen
4k
(ro2 − r 2 ) +
e& gen ro
2h
which is the desired solution for the temperature distribution in the wire as a function of r. Then the temperature at the center
line (r = 0) is determined by substituting the known quantities to be
e&gen 2 e&gen ro
T (0) = T∞ +
ro +
4k
2h
(16.4 × 10 6 W/m 3 )(0.006 m) 2 (16.4 × 10 6 W/m 3 )(0.006 m)
= 100°C +
+
= 125°C
4 × (15.2 W/m ⋅ K)
2 × (3200 W/m 2 ⋅ K)
Thus the centerline temperature will be 25°C above the temperature of the surface of the wire.
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2-47
2-85
Prob. 2-84 is reconsidered. The temperature at the centerline of the wire as a function of the heat generation is to
be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
r_0=0.006 [m]
k=15.2 [W/m-K]
e_dot=16.4 [W/cm^3]
T_infinity=100 [C]
h=3200 [W/m^2-K]
"ANALYSIS"
T_0=T_infinity+e_dot*Convert(W/cm^3,W/m^3)/(4*k)*(r_0^2-r^2) +e_dot*Convert(W/cm^3,W/m^3)*r_0/(2*h)
"Variation of temperature"
r=0 "for centerline temperature"
T0
[C]
115.3
130.6
145.9
161.2
176.5
191.8
207.1
222.4
237.7
253
260
240
220
200
T0 [C]
e
[W/cm3]
10
20
30
40
50
60
70
80
90
100
180
160
140
120
100
10
20
30
40
50
60
70
80
90
100
3
e [W/cm ]
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2-48
2-86 A nuclear fuel rod with a specified surface temperature is used as the fuel in a nuclear reactor. The center temperature
of the rod is 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 change in the axial direction.
3 Thermal conductivity is constant. 4 Heat generation in the rod is uniform.
220°C
egen
Properties The thermal conductivity is given to be k = 29.5 W/m⋅°C.
Analysis The center temperature of the rod is determined from
To = Ts +
e& gen ro2
4k
= 220°C +
Uranium rod
(4 × 10 7 W/m 3 )(0.005 m) 2
= 228°C
4(29.5 W/m.°C)
2-87 Both sides of a large stainless steel plate in which heat is generated uniformly are exposed to convection with the
environment. The location and values of the highest and the lowest temperatures in the plate are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since the plate is large relative to its thickness, and there is thermal symmetry about the center plane 3 Thermal
conductivity is constant. 4 Heat generation is uniform.
Properties The thermal conductivity is given to be k =15.1 W/m⋅°C.
Analysis The lowest temperature will occur at surfaces of plate
while the highest temperature will occur at the midplane. Their
values are determined directly from
Ts = T∞ +
To = Ts +
e& gen L
h
e& gen L2
2k
= 30°C +
(5 × 10 5 W/m 3 )(0.015 m)
= 155°C +
60 W/m 2 ⋅ °C
k
egen
T∞ =30°C
h=60 W/m2⋅°C
2L=3 cm
T∞ =30°C
h=60 W/m2.°C
= 155°C
(5 × 10 5 W/m 3 )(0.015 m) 2
= 158.7°C
2(15.1 W/m ⋅ °C)
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2-49
2-88 A long resistance heater wire is subjected to convection at its outer surface. The surface temperature of the wire is to be
determined using the applicable relations directly and by solving the applicable differential equation.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal
conductivity is constant. 4 Heat generation in the wire is uniform.
Properties The thermal conductivity is given to be k = 15.1 W/m⋅°C.
Analysis (a) The heat generation per unit volume of the wire is
e&gen =
E& gen
V wire
=
E& gen
=
2
πro L
3000 W
π (0.001 m) 2 (6 m)
= 1.592 × 10 8 W/m 3
T∞
h
The surface temperature of the wire is then (Eq. 2-68)
Ts = T∞ +
e&gen ro
2h
= 20°C +
(1.592 × 10 8 W/m 3 )(0.001 m)
2
2(175 W/m ⋅ °C)
T∞
h
k
egen
0
= 475°C
ro
r
(b) The mathematical formulation of this problem can be expressed as
1 d ⎛ dT ⎞ e& gen
=0
⎜r
⎟+
r dr ⎝ dr ⎠
k
and
−k
dT (ro )
= h[T ( ro ) − T∞ ] (convection at the outer surface)
dr
dT (0)
= 0 (thermal symmetry about the centerline)
dr
Multiplying both sides of the differential equation by r and integrating gives
e& gen
d ⎛ dT ⎞
r
⎜r
⎟=−
dr ⎝ dr ⎠
k
e& gen r 2
dT
=−
+ C1
dr
k 2
→r
(a)
Applying the boundary condition at the center line,
B.C. at r = 0:
0×
e& gen
dT (0)
=−
× 0 + C1
dr
2k
→ C1 = 0
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating,
e& gen
dT
=−
r
dr
2k
→
T (r ) = −
e& gen
4k
r 2 + C2
(b)
Applying the boundary condition at r = ro ,
B. C. at r = ro :
−k
e& gen ro
2k
e& gen ro e& gen 2
⎛ e& gen 2
⎞
= h⎜⎜ −
+
ro + C 2 − T∞ ⎟⎟ → C 2 = T∞ +
ro
2h
4k
⎝ 4k
⎠
Substituting this C2 relation into Eq. (b) and rearranging give
T (r ) = T∞ +
e& gen
4k
(ro2 − r 2 ) +
e& gen ro
2h
which is the temperature distribution in the wire as a function of r. Then the temperature of the wire at the surface (r = ro ) is
determined by substituting the known quantities to be
T (r0 ) = T∞ +
e&gen
4k
(ro2 − ro2 ) +
e&gen r0
2h
= T∞ +
e&gen ro
2h
= 20°C +
(1.592 × 10 8 W/m 3 )(0.001 m)
2(175 W/m 2 ⋅ °C)
= 475°C
Note that both approaches give the same result.
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2-50
2-89 Heat is generated uniformly in a resistance heater wire. The temperature difference between the center and the surface
of the wire is to be determined.
Assumptions 1 Heat transfer is steady since there is no change with
time. 2 Heat transfer is one-dimensional since there is thermal
symmetry about the center line and no change in the axial direction. 3
Thermal conductivity is constant. 4 Heat generation in the heater is
uniform.
Properties The thermal conductivity is given to be k = 10 W/m⋅°C.
Analysis The resistance heater converts electric energy into heat at a rate
of 10 kW. The rate of heat generation per unit length of the wire is
E& gen
E& gen
10,000 W
e& gen =
= 2 =
= 3.183 × 10 9 W/m 3
V wire πro L π (0.001 m) 2 (1 m)
r
Ts
ro
0
Heater
Then the temperature difference between the centerline and the surface becomes
ΔTmax =
e& gen ro2
4k
=
(3.183 × 10 9 W/m 3 )(0.001 m) 2
= 79.6°C
4(10 W/m ⋅ °C)
2-90 Heat is generated uniformly in a resistance heater wire. The temperature difference between the center and the surface
of the wire is to be determined.
Assumptions 1 Heat transfer is steady since there is no change with
time. 2 Heat transfer is one-dimensional since there is thermal
symmetry about the center line and no change in the axial direction. 3
Thermal conductivity is constant. 4 Heat generation in the heater is
uniform.
Properties The thermal conductivity is given to be k = 7.8 W/m⋅°C.
Analysis The resistance heater converts electric energy into heat at a rate
of 10 kW. The rate of heat generation per unit volume of the wire is
E& gen
E& gen
10,000 W
e& gen =
= 2 =
= 3.183 × 10 9 W/m 3
V wire πro L π (0.001 m) 2 (1 m)
r
Ts
ro
0
Heater
Then the temperature difference between the centerline and the surface becomes
ΔTmax =
e& gen ro2
4k
=
(3.183 × 10 9 W/m 3 )(0.001 m) 2
= 102.0°C
4(7.8 W/m ⋅ °C)
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2-51
2-91 Heat is generated uniformly in a spherical radioactive material with specified surface temperature. The mathematical
formulation, the variation of temperature in the sphere, and the center temperature are to be determined for steady onedimensional heat transfer.
Assumptions 1 Heat transfer is steady since there is no indication of any changes with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the mid point. 3 Thermal conductivity is constant. 4 Heat generation is
uniform.
Properties The thermal conductivity is given to be k = 15 W/m⋅°C.
Analysis (a) Noting that heat transfer is steady and one-dimensional in the radial
r direction, the mathematical formulation of this problem can be expressed as
1 d ⎛ 2 dT ⎞ e& gen
=0
⎟+
⎜r
dr ⎠
k
r 2 dr ⎝
and
Ts=110°C
k
with e& gen = constant
egen
T (ro ) = Ts = 110°C (specified surface temperature)
0
ro
r
dT (0)
= 0 (thermal symmetry about the mid point)
dr
(b) Multiplying both sides of the differential equation by r2 and rearranging gives
e& gen 2
d ⎛ 2 dT ⎞
r
⎜r
⎟=−
dr ⎝
dr ⎠
k
Integrating with respect to r gives
r2
e&gen r 3
dT
=−
+ C1
dr
k 3
(a)
Applying the boundary condition at the mid point,
B.C. at r = 0:
e& gen
dT (0)
=−
× 0 + C1
dr
3k
0×
→ C1 = 0
Dividing both sides of Eq. (a) by r2 to bring it to a readily integrable form and integrating,
e& gen
dT
=−
r
dr
3k
and
T (r ) = −
e& gen
6k
r 2 + C2
(b)
Applying the other boundary condition at r = r0 ,
B. C. at r = ro :
Ts = −
e& gen
6k
ro2 + C 2
→ C 2 = Ts +
e& gen
6k
ro2
Substituting this C 2 relation into Eq. (b) and rearranging give
T ( r ) = Ts +
e& gen
6k
(ro2 − r 2 )
which is the desired solution for the temperature distribution in the wire as a function of r.
(c) The temperature at the center of the sphere (r = 0) is determined by substituting the known quantities to be
T (0) = Ts +
e&gen
6k
(ro2 − 0 2 ) = Ts +
e&gen ro2
6k
= 110°C +
(5 × 10 7 W/m 3 )(0.04 m) 2
= 999°C
6 × (15 W/ m ⋅ °C)
Thus the temperature at center will be 999°C above the temperature of the outer surface of the sphere.
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2-52
2-92
Prob. 2-91 is reconsidered. The temperature as a function of the radius is to be plotted. Also, the center
temperature of the sphere as a function of the thermal conductivity is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
r_0=0.04 [m]
g_dot=5E7 [W/m^3]
T_s=110 [C]
k=15 [W/m-C]
r=0 [m]
"ANALYSIS"
T=T_s+g_dot/(6*k)*(r_0^2-r^2) "Temperature distribution as a function of r"
T_0=T_s+g_dot/(6*k)*r_0^2 "Temperature at the center (r=0)"
T
[C]
998.9
996.4
989
976.7
959.5
937.3
910.2
878.2
841.3
799.4
752.7
701
644.3
582.8
516.3
444.9
368.5
287.3
201.1
110
1000
900
800
700
T [C]
r
[m]
0
0.002105
0.004211
0.006316
0.008421
0.01053
0.01263
0.01474
0.01684
0.01895
0.02105
0.02316
0.02526
0.02737
0.02947
0.03158
0.03368
0.03579
0.03789
0.04
600
500
400
300
200
100
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
r [m]
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2-53
T0
[C]
1443
546.8
371.2
296.3
254.8
228.4
210.1
196.8
186.5
178.5
171.9
166.5
162
158.2
154.8
151.9
149.4
147.1
145.1
143.3
1600
1400
1200
1000
T0 [C]
k
[W/m.C]
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
800
600
400
200
0
0
50
100
150
200
250
300
350
400
k [W/m-C]
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2-54
2-93 A long homogeneous resistance heater wire with specified surface temperature is used to heat the air. The temperature
of the wire 3.5 mm from the center is to be determined in steady operation.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal
conductivity is constant. 4 Heat generation in the wire is uniform.
Properties The thermal conductivity is given to be k = 6 W/m⋅°C.
Analysis Noting that heat transfer is steady and one-dimensional in the radial r
direction, the mathematical formulation of this problem can be expressed as
r
1 d ⎛ dT ⎞ e& gen
=0
⎜r
⎟+
r dr ⎝ dr ⎠
k
and
180°C
ro
T (ro ) = Ts = 180°C (specified surface temperature)
e&gen
dT (0)
= 0 (thermal symmetry about the centerline)
dr
Resistance wire
Multiplying both sides of the differential equation by r and rearranging gives
e& gen
d ⎛ dT ⎞
r
⎜r
⎟=−
dr ⎝ dr ⎠
k
Integrating with respect to r gives
r
e& gen r 2
dT
=−
+ C1
dr
k 2
(a)
It is convenient at this point to apply the boundary condition at the center since it is related to the first derivative of the
temperature. It yields
B.C. at r = 0:
e& gen
dT (0)
=−
× 0 + C1
2k
dr
0×
→ C1 = 0
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating,
e& gen
dT
=−
r
dr
2k
and
T (r ) = −
e& gen
4k
r 2 + C2
(b)
Applying the other boundary condition at r = ro ,
B. C. at r = ro :
Ts = −
e& gen
4k
ro2 + C 2
→ C 2 = Ts +
e& gen
4k
ro2
Substituting this C2 relation into Eq. (b) and rearranging give
T ( r ) = Ts +
e& gen
4k
(ro2 − r 2 )
which is the desired solution for the temperature distribution in the wire as a function of r. The temperature 3.5 mm from the
center line (r = 0.0035 m) is determined by substituting the known quantities to be
T (0.0035 m) = Ts +
e&gen
4k
(ro2 − r 2 ) = 180°C +
5 × 10 7 W/m 3
[(0.005 m) 2 − (0.0035 m) 2 ] = 207°C
4 × (6 W/ m ⋅ °C)
Thus the temperature at that location will be about 20°C above the temperature of the outer surface of the wire.
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2-55
2-94 Heat is generated in a large plane wall whose one side is insulated while the other side is maintained at a specified
temperature. The mathematical formulation, the variation of temperature in the wall, and the temperature of the insulated
surface are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since the wall is large relative to its thickness, and there is thermal symmetry about the center plane. 3 Thermal
conductivity is constant. 4 Heat generation varies with location in the x direction.
Properties The thermal conductivity is given to be k = 30 W/m⋅°C.
Analysis (a) Noting that heat transfer is steady and one-dimensional in x
direction, the mathematical formulation of this problem can be expressed as
d 2T
dx
2
+
e& gen ( x)
k
=0
k
e&gen
where
e&gen = e&0 e −0.5 x / L
and
dT (0)
= 0 (insulated surface at x = 0)
dx
and e&0 = 8×106 W/m3
Insulated
T2 =30°C
L
T ( L) = T2 = 30°C (specified surface temperature)
x
(b) Rearranging the differential equation and integrating,
d 2T
dx
=−
2
e&0 −0.5 x / L
→
e
k
e& e −0.5 x / L
dT
=− 0
+ C1 →
dx
k − 0.5 / L
dT 2e&0 L −0.5 x / L
=
+ C1
e
dx
k
Integrating one more time,
T ( x) =
2e& 0 L e −0.5 x / L
4e& L2
+ C1 x + C 2 → T ( x) = − 0 e −0.5 x / L + C1 x + C 2
k − 0.5 / L
k
(1)
Applying the boundary conditions:
B.C. at x = 0:
2e& L
2e& L
dT (0) 2e&0 L −0.5×0 / L
=
+ C1 → 0 = 0 + C1 → C1 = − 0
e
k
k
dx
k
B. C. at x = L:
T ( L ) = T2 = −
4e&0 L2 −0.5 L / L
4e& L2
2e& L2
e
+ C1 L + C 2 → C 2 = T2 + 0 e −0.5 + 0
k
k
k
Substituting the C1 and C2 relations into Eq. (1) and rearranging give
T ( x ) = T2 +
e&0 L2
[ 4(e −0.5 − e −0.5 x / L ) + 2(1 − x / L)]
k
which is the desired solution for the temperature distribution in the wall as a function of x.
(c) The temperature at the insulate surface (x = 0) is determined by substituting the known quantities to be
T ( 0) = T 2 +
e& 0 L2
[4(e −0.5 − e 0 ) + (2 − 0 / L)]
k
= 30°C +
(8 × 10 6 W/m 3 )(0.05 m) 2
[4(e −0.5 − 1) + (2 − 0)] = 314°C
(30 W/m ⋅ °C)
Therefore, there is a temperature difference of almost 300°C between the two sides of the plate.
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2-56
2-95
Prob. 2-94 is reconsidered. The heat generation as a function of the distance is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
L=0.05 [m]
T_s=30 [C]
k=30 [W/m-C]
e_dot_0=8E6 [W/m^3]
"ANALYSIS"
e_dot=e_dot_0*exp((-0.5*x)/L) "Heat generation as a function of x"
"x is the parameter to be varied"
8.000x106
7.500x106
7.000x106
3
e
[W/m3]
8.000E+06
7.610E+06
7.239E+06
6.886E+06
6.550E+06
6.230E+06
5.927E+06
5.638E+06
5.363E+06
5.101E+06
4.852E+06
e [W/m ]
x
[m]
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
6.500x106
6.000x106
5.500x106
5.000x106
4.500x106
0
0.01
0.02
0.03
0.04
0.05
x [m]
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2-57
Variable Thermal Conductivity, k(T)
2-96C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature.
2-97C The thermal conductivity of a medium, in general, varies with temperature.
2-98C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly, the
error involved in heat transfer calculation by assuming constant thermal conductivity at the average temperature is (a) none.
2-99C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with constant thermal
conductivity and no heat generation, the temperature in only the plane wall will vary linearly.
2-100C Yes, when the thermal conductivity of a medium varies linearly with temperature, the average thermal conductivity
is always equivalent to the conductivity value at the average temperature.
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2-101 A silicon wafer with variable thermal conductivity is subjected to uniform heat flux at the lower surface. The
maximum allowable heat flux such that the temperature difference across the wafer thickness does not exceed 2 °C is to be
determined.
Assumptions 1 Heat conduction is steady and one-dimensional.
2 There is no heat generation. 3 Thermal conductivity varies
with temperature.
Properties The thermal conductivity is given to be k(T) = (a +
bT + cT2) W/m · K.
Analysis For steady heat transfer, the Fourier’s law of heat
conduction can be expressed as
q& = − k (T )
dT
dT
= −(a + bT + cT 2 )
dx
dx
Separating variable and integrating from x = 0 where T (0) = T1 to x = L where T ( L) = T2 , we obtain
L
T2
0
T1
∫ q&dx = −∫ (a + bT + cT )dT
2
Performing the integration gives
b
c
⎤
⎡
q&L = − ⎢a (T2 − T1 ) + (T22 − T12 ) + (T23 − T13 )⎥
2
3
⎦
⎣
The maximum allowable heat flux such that the temperature difference across the wafer thickness does not exceeding 2 °C is
0.00111
1.29
⎡
⎤
2
2
(598 3 − 600 3 )⎥ W/m
⎢437(598 − 600) − 2 (598 − 600 ) +
3
⎦
q& = − ⎣
(925 × 10 −6 m)
= 1.35 × 10 5 W/m 2
Discussion For heat flux less than 135 kW/m2, the temperature difference across the silicon wafer thickness will be
maintained below 2 °C.
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2-102 A plate with variable conductivity is subjected to specified
temperatures on both sides. The rate of heat transfer through the plate is to be
determined.
k(T)
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2
Thermal conductivity varies quadratically. 3 There is no heat generation.
T2
T1
Properties The thermal conductivity is given to be k (T ) = k 0 (1 + β T 2 ) .
Analysis When the variation of thermal conductivity with temperature k(T) is
known, the average value of the thermal conductivity in the temperature
range between T1 and T2 can be determined from
β
3⎞
⎡
L
β
x
3 ⎤
1
k ⎢(T − T ) + (T − T )⎥
k (T )dT ∫ k (1 + β T ) dT k ⎜ T + 3 T ⎟
∫
⎝
⎠
3
⎦
⎣
=
=
=
=
T2
k avg
⎛
T2
T2
T1
T1
T2 − T1
2
0
0
T1
T2 − T1
T2 − T1
0
2
1
3
2
T2 − T1
⎡ β
⎤
= k 0 ⎢1 + T22 + T1T2 + T12 ⎥
⎣ 3
⎦
(
)
This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal
conductivity kavg equals the rate of heat transfer through the same medium with variable conductivity k(T). Then the rate of
heat conduction through the plate can be determined to be
T − T2
⎡ β
⎤ T − T2
= k 0 ⎢1 + T22 + T1T2 + T12 ⎥ A 1
Q& = k avg A 1
L
L
⎣ 3
⎦
(
)
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-76, and
performed the indicated integration.
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2-103 A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. The variation of
temperature and the rate of heat transfer through the shell are to be determined.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is
no heat generation.
Properties The thermal conductivity is given to be k (T ) = k 0 (1 + β T ) .
Analysis (a) The rate of heat transfer through the shell is expressed as
k(T)
T1
T − T2
Q& cylinder = 2πk avg L 1
ln(r2 / r1 )
T2
where L is the length of the cylinder, r1 is the inner radius, and
r2 is the outer radius, and
r1
r2
r
T +T ⎞
⎛
kavg = k (Tavg ) = k0 ⎜1 + β 2 1 ⎟
2 ⎠
⎝
is the average thermal conductivity.
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat conduction expressed as
dT
Q& = −k (T ) A
dr
where the rate of conduction heat transfer Q& is constant and the heat conduction area A = 2πrL is variable. Separating the
variables in the above equation and integrating from r = r1 where T (r1 ) = T1 to any r where T (r ) = T , we get
Q&
r dr
T
r1
T1
∫ r = −2πL ∫ k (T )dT
Substituting k (T ) = k 0 (1 + β T ) and performing the integrations gives
r
Q& ln = −2πLk 0 [(T − T1 ) + β (T 2 − T12 ) / 2]
r1
Substituting the Q& expression from part (a) and rearranging give
T2 +
2
β
T+
2k avg ln(r / r1 )
2
(T1 − T2 ) − T12 − T1 = 0
βk 0 ln(r2 / r1 )
β
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature distribution T(r)
in the cylindrical shell is determined to be
T (r ) = −
1
β
±
1
β
2
−
2k avg ln(r / r1 )
2
(T1 − T2 ) + T12 + T1
βk 0 ln(r2 / r1 )
β
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the temperature at any
point within the medium must remain between T1 and T2 .
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2-104 A spherical shell with variable conductivity is subjected to specified temperatures on both sides. The variation of
temperature and the rate of heat transfer through the shell are to be determined.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is
no heat generation.
Properties The thermal conductivity is given to be k (T ) = k 0 (1 + β T ) .
Analysis (a) The rate of heat transfer through the shell is expressed as
T − T2
Q& sphere = 4πk avg r1 r2 1
r2 − r1
T2
k(T)
r1
where r1 is the inner radius, r2 is the outer radius, and
T1
r2
r
T +T ⎞
⎛
kavg = k (Tavg ) = k0 ⎜1 + β 2 1 ⎟
2 ⎠
⎝
is the average thermal conductivity.
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat conduction expressed as
dT
Q& = −k (T ) A
dr
where the rate of conduction heat transfer Q& is constant and the heat conduction area A = 4πr2 is variable. Separating the
variables in the above equation and integrating from r = r1 where T (r1 ) = T1 to any r where T (r ) = T , we get
Q&
r dr
T
2
T1
∫ r = −4π ∫ k (T )dT
r1
Substituting k (T ) = k 0 (1 + β T ) and performing the integrations gives
⎛ 1 1⎞
Q& ⎜⎜ − ⎟⎟ = −4πk 0 [(T − T1 ) + β (T 2 − T12 ) / 2]
⎝ r1 r ⎠
Substituting the Q& expression from part (a) and rearranging give
T2 +
2
β
T+
2k avg r2 (r − r1 )
βk 0 r (r2 − r1 )
(T1 − T2 ) − T12 −
2
β
T1 = 0
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature distribution T(r)
in the cylindrical shell is determined to be
T (r ) = −
1
β
±
1
β
2
−
2k avg r2 ( r − r1 )
βk 0 r (r2 − r1 )
(T1 − T2 ) + T12 +
2
β
T1
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the temperature at any
point within the medium must remain between T1 and T2 .
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2-105 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer
through the plate is to be determined.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is
no heat generation.
Properties The thermal conductivity is given to be k (T ) = k 0 (1 + β T ) .
Analysis The average thermal conductivity of the medium in this case is simply
the conductivity value at the average temperature since the thermal conductivity
varies linearly with temperature, and is determined to be
T + T1 ⎞
⎛
k ave = k (Tavg ) = k 0 ⎜1 + β 2
⎟
2 ⎠
⎝
(500 + 350) K ⎞
⎛
= (18 W/m ⋅ K)⎜1 + (8.7 × 10 -4 K -1 )
⎟
2
⎠
⎝
= 24.66 W/m ⋅ K
k(T)
T1
T2
L
Then the rate of heat conduction through the plate becomes
T − T2
(500 − 350)K
Q& = k avg A 1
= ( 24.66 W/m ⋅ K)(1.5 m × 0.6 m)
= 22,190 W = 22.2 kW
L
0.15 m
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq, 2-76, and
performed the indicated integration.
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2-106
Prob. 2-105 is reconsidered. The rate of heat conduction through the plate as a function of the temperature of
the hot side of the plate is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
A=1.5*0.6 [m^2]
L=0.15 [m]
T_1=500 [K]
T_2=350 [K]
k_0=18 [W/m-K]
beta=8.7E-4 [1/K]
"ANALYSIS"
k=k_0*(1+beta*T)
T=1/2*(T_1+T_2)
Q_dot=k*A*(T_1-T_2)/L
Q
[W]
7162
10831
14558
18345
22190
26094
30056
34078
38158
42297
46494
50750
55065
60000
50000
40000
Q [W]
T1
[W]
400
425
450
475
500
525
550
575
600
625
650
675
700
30000
20000
10000
0
400
450
500
550
600
650
700
T1 [K]
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2-64
Special Topic: Review of Differential equations
2-107C We utilize appropriate simplifying assumptions when deriving differential equations to obtain an equation that we
can deal with and solve.
2-108C A variable is a quantity which may assume various values during a study. A variable whose value can be changed
arbitrarily is called an independent variable (or argument). A variable whose value depends on the value of other variables
and thus cannot be varied independently is called a dependent variable (or a function).
2-109C A differential equation may involve more than one dependent or independent variable. For example, the equation
∂ 2 T ( x, t ) e& gen 1 ∂T ( x, t )
has one dependent (T) and 2 independent variables (x and t). the equation
+
=
α
∂t
k
∂x 2
∂ 2 T ( x, t )
∂x
2
+
∂W ( x, t ) 1 ∂T ( x, t ) 1 ∂W ( x, t )
has 2 dependent (T and W) and 2 independent variables (x and t).
=
+
∂x
α
∂t
α
∂t
2-110C Geometrically, the derivative of a function y(x) at a point represents the slope of the tangent line to the graph of the
function at that point. The derivative of a function that depends on two or more independent variables with respect to one
variable while holding the other variables constant is called the partial derivative. Ordinary and partial derivatives are
equivalent for functions that depend on a single independent variable.
2-111C The order of a derivative represents the number of times a function is differentiated, whereas the degree of a
derivative represents how many times a derivative is multiplied by itself. For example, y ′′′ is the third order derivative of y,
whereas ( y ′) 3 is the third degree of the first derivative of y.
2-112C For a function f ( x, y ) , the partial derivative ∂f / ∂x will be equal to the ordinary derivative df / dx when f does not
depend on y or this dependence is negligible.
2-113C For a function f (x) , the derivative df / dx does not have to be a function of x. The derivative will be a constant
when the f is a linear function of x.
2-114C Integration is the inverse of derivation. Derivation increases the order of a derivative by one, integration reduces it
by one.
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2-115C A differential equation involves derivatives, an algebraic equation does not.
2-116C A differential equation that involves only ordinary derivatives is called an ordinary differential equation, and a
differential equation that involves partial derivatives is called a partial differential equation.
2-117C The order of a differential equation is the order of the highest order derivative in the equation.
2-118C A differential equation is said to be linear if the dependent variable and all of its derivatives are of the first degree,
and their coefficients depend on the independent variable only. In other words, a differential equation is linear if it can be
written in a form which does not involve (1) any powers of the dependent variable or its derivatives such as y 3 or ( y ′) 2 , (2)
any products of the dependent variable or its derivatives such as yy ′ or y ′y ′′′ , and (3) any other nonlinear functions of the
dependent variable such as sin y or e y . Otherwise, it is nonlinear.
2-119C A linear homogeneous differential equation of order n is expressed in the most general form as
y ( n ) + f 1 ( x) y ( n −1) + L + f n −1 ( x) y ′ + f n ( x) y = 0
Each term in a linear homogeneous equation contains the dependent variable or one of its derivatives after the equation is
cleared of any common factors. The equation y ′′ − 4 x 2 y = 0 is linear and homogeneous since each term is linear in y, and
contains the dependent variable or one of its derivatives.
2-120C A differential equation is said to have constant coefficients if the coefficients of all the terms which involve the
dependent variable or its derivatives are constants. If, after cleared of any common factors, any of the terms with the
dependent variable or its derivatives involve the independent variable as a coefficient, that equation is said to have variable
coefficients The equation y ′′ − 4 x 2 y = 0 has variable coefficients whereas the equation y ′′ − 4 y = 0 has constant
coefficients.
2-121C A linear differential equation that involves a single term with the derivatives can be solved by direct integration.
2-122C The general solution of a 3rd order linear and homogeneous differential equation will involve 3 arbitrary constants.
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2-66
Review Problems
2-123 In a quenching process, steel ball bearings at a given instant have a rate of temperature decrease of 50 K/s. The rate of
heat loss is to be determined.
Assumptions 1 Heat conduction is one-dimensional. 2 There is no heat generation. 3 Thermal properties are constant.
Properties The properties of the steel ball bearings are given to be c = 500 J/kg · K, k = 60 W/m · K, and ρ = 7900 kg/m3.
Analysis The thermal diffusivity on the steel ball bearing is
α=
k
ρc
=
60 W/m ⋅ K
3
(7900 kg/m )(500 J/kg ⋅ K )
= 15.19 × 10 −6 m 2 /s
The given rate of temperature decrease can be expressed as
dT (ro )
= −50 K/s
dt
For one-dimensional transient heat conduction in a sphere with no heat generation, the differential equation is
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎜r
⎟=
r 2 ∂r ⎝ ∂r ⎠ α ∂t
Substituting the thermal diffusivity and the rate of temperature decrease, the differential equation can be written as
− 50 K/s
1 d ⎛ 2 dT ⎞
⎟=
⎜r
2 dr
dr
r
⎠ 15.19 × 10 −6 m 2 /s
⎝
Multiply both sides of the differential equation by r
2
and rearranging gives
− 50 K/s
d ⎛ 2 dT ⎞
r2
⎜r
⎟=
dr ⎝ dr ⎠ 15.19 × 10 −6 m 2 /s
Integrating with respect to r gives
r2
⎛ r3 ⎞
− 50 K/s
dT
⎜ ⎟ + C1
=
dr 15.19 × 10 − 6 m 2 /s ⎜⎝ 3 ⎟⎠
(a)
Applying the boundary condition at the midpoint (thermal symmetry about the midpoint),
r = 0:
0×
− 50 K/s
dT (0)
⎛0⎞
=
⎜ ⎟ + C1
6
2
−
dr
15.19 × 10 m /s ⎝ 3 ⎠
→
C1 = 0
Dividing both sides of Eq. (a) by r 2 gives
dT
− 50 K/s
⎛r⎞
=
⎜ ⎟
6
2
−
dr 15.19 × 10 m /s ⎝ 3 ⎠
The rate of heat loss through the steel ball bearing surface can be determined from Fourier’s law to be
dT
Q& loss = −kA
dr
= −k (4π ro2 )
dT (ro )
50 K/s
⎛ ro ⎞
= k (4π ro2 )
⎜ ⎟
−
6
2
dr
15.19 × 10 m /s ⎝ 3 ⎠
= (60 W/m ⋅ K )(4π )(0.025 m) 2
50 K/s
⎛ 0.025 m ⎞
⎟
⎜
3
15.19 × 10 −6 m 2 /s ⎝
⎠
= 12.9 kW
Discussion The rate of heat loss through the steel ball bearing surface determined here is for the given instant when the rate
of temperature decrease is 50 K/s.
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2-124 A spherical reactor of 5-cm diameter operating at steady condition has its heat generation suddenly set to 9 MW/m3.
The time rate of temperature change in the reactor is to be determined.
Assumptions 1 Heat conduction is one-dimensional. 2 Heat generation is uniform. 3 Thermal properties are constant.
Properties The properties of the reactor are given to be c = 200 J/kg·°C, k = 40 W/m·°C, and ρ = 9000 kg/m3.
Analysis The thermal diffusivity of the reactor is
α=
40 W/m ⋅ °C
k
=
= 22.22 × 10 −6 m 2 /s
ρc (9000 kg/m 3 )(200 J/kg ⋅ °C)
For one-dimensional transient heat conduction in a sphere with heat generation, the differential equation is
1 ∂ ⎛ 2 ∂T ⎞ e&gen
1 ∂T
=
⎟+
⎜r
2 ∂r
k
α ∂t
r
⎝ ∂r ⎠
or
⎡ 1 ∂ ⎛ 2 ∂T ⎞ e&gen ⎤
∂T
= α⎢ 2
⎟+
⎜r
⎥
∂t
k ⎦
⎣ r ∂r ⎝ ∂r ⎠
At the instant when the heat generation of reactor is suddenly set to 90 MW/m3 (t = 0), the temperature variation can be
expressed by the given T(r) = a – br2, hence
e&gen ⎫
⎧1 ∂ 2
⎧1 ∂ ⎡ 2 ∂
∂T
⎤ e&gen ⎫
( a − br 2 ) ⎥ +
= α⎨ 2
r (−2br ) +
r
⎬
⎬ = α⎨ 2
⎢
∂t
k ⎭
k ⎭
⎦
⎩ r ∂r
⎩ r ∂r ⎣ ∂r
[
]
e&gen ⎞
e&gen ⎤
⎛
⎡1
⎟
= α ⎢ 2 (−6br 2 ) +
⎥ = α ⎜⎜ − 6b +
k ⎟⎠
k ⎦
⎝
⎣r
The time rate of temperature change in the reactor when the heat generation suddenly set to 9 MW/m3 is determined to be
e&gen ⎞
⎡
⎛
∂T
9 × 10 6 W/m 3 ⎤
⎟ = (22.22 × 10 −6 m 2 /s) ⎢− 6(5 × 10 5 °C/m 2 ) +
= α ⎜⎜ − 6b +
⎥
∂t
k ⎟⎠
40 W/m ⋅ °C ⎦⎥
⎝
⎣⎢
= −61.7 °C/s
Discussion Since the time rate of temperature change is a negative value, this indicates that the heat generation of reactor is
suddenly decreased to 9 MW/m3.
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2-68
2-125 A small hot metal object is allowed to cool in an environment by convection. The differential equation that describes
the variation of temperature of the ball with time is to be derived.
Assumptions 1 The temperature of the metal object changes uniformly with time during cooling so that T = T(t). 2 The
density, specific heat, and thermal conductivity of the body are constant. 3 There is no heat generation.
Analysis Consider a body of arbitrary shape of mass m, volume V, surface area A, density ρ, and specific heat cp initially at a
uniform temperature Ti. At time t = 0, the body is placed into a medium at temperature T∞ , and heat transfer takes place
between the body and its environment with a heat transfer coefficient h.
A
During a differential time interval dt, the temperature of the body rises by a
differential amount dT. Noting that the temperature changes with time only, an energy
balance of the solid for the time interval dt can be expressed as
m, c, Ti
T=T(t)
⎛ Heat transfer from the body ⎞ ⎛ The decrease in the energy ⎞
⎟⎟ = ⎜⎜
⎟⎟
⎜⎜
during dt
⎠ ⎝ of the body during dt ⎠
⎝
or
h
T∞
hAs (T − T∞ )dt = mc p ( −dT )
Noting that m = ρV and dT = d (T − T∞ ) since T∞ = constant, the equation above can be rearranged as
hAs
d (T − T∞ )
=−
dt
T − T∞
ρVc p
which is the desired differential equation.
2-126 A long rectangular bar is initially at a uniform temperature of Ti. The surfaces of the bar at x = 0 and y = 0 are
insulated while heat is lost from the other two surfaces by convection. The mathematical formulation of this heat conduction
problem is to be expressed for transient two-dimensional heat transfer with no heat generation.
Assumptions 1 Heat transfer is transient and two-dimensional. 2 Thermal conductivity is constant. 3 There is no heat
generation.
Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as
∂ 2T
∂x
2
+
∂ 2T
∂y
2
=
1 ∂T
α ∂t
h, T∞
b
∂T ( x,0, t )
=0
∂x
∂T (0, y , t )
=0
∂y
h, T∞
∂T ( a, y , t )
= h[T ( a, y , t ) − T∞ ]
∂y
∂T ( x, b, t )
−k
= h[T ( x, b, t ) − T∞ ]
∂x
−k
a
Insulated
T ( x, y,0) = Ti
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2-69
2-127 Heat is generated at a constant rate in a short cylinder. Heat is lost from the cylindrical surface at r = ro by convection
to the surrounding medium at temperature T∞ with a heat transfer coefficient of h. The bottom surface of the cylinder at r =
0 is insulated, the top surface at z = H is subjected to uniform heat flux q& h , and the cylindrical surface at r = ro is subjected
to convection. The mathematical formulation of this problem is to be expressed for steady two-dimensional heat transfer.
Assumptions 1 Heat transfer is given to be steady and two-dimensional. 2 Thermal conductivity is constant. 3 Heat is
generated uniformly.
Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as
1 ∂ ⎛ ∂T ⎞ ∂ 2 T e& gen
+
=0
⎜r
⎟+
k
r ∂r ⎝ ∂r ⎠ ∂z 2
∂T (r ,0)
=0
∂z
∂T (r , H )
k
= q& H
∂z
qH
egen
h
T∞
z
ro
∂T (0, z )
=0
∂r
∂T (ro , z )
−k
= h[T (ro , z ) − T∞ ]
∂r
2-128 The concrete slab roof of a house is subjected to specified temperature at the bottom surface and convection and
radiation at the top surface. The temperature of the top surface of the roof and the rate of heat transfer are to be determined
when steady operating conditions are reached.
Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the roof area is large
relative to its thickness, and the thermal conditions on both sides of the roof are uniform. 3 Thermal properties are constant.
4 There is no heat generation in the wall.
Properties The thermal conductivity and emissivity are given to be k = 1.9 W/m⋅°C and ε = 0.8.
Analysis In steady operation, heat conduction through the roof must be
equal to net heat transfer from the outer surface. Therefore, taking the
x
outer surface temperature of the roof to be T2 (in °C),
T −T
4
kA 1 2 = hA(T2 − T∞ ) + εAσ [(T2 + 273) 4 − Tsky
]
L
L
T∞
h
Tsky
Canceling the area A and substituting the known quantities,
(16 − T2 )°C
T1
0.25 m
= (18 W/m 2 ⋅ °C)(T2 − 10)°C + 0.8(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(T2 + 273) 4 − 170 4 ]K 4
(1.9 W/m ⋅ °C)
Using an equation solver (or the trial and error method), the outer surface temperature is determined to be
T2 = 3.0°C
Then the rate of heat transfer through the roof becomes
T −T
(16 − 3)°C
Q& = kA 1 2 = (1.9 W/m ⋅ °C)(8 × 10 m 2 )
= 7904 W
0.25 m
L
Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside. Therefore, the
house is losing heat as expected.
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2-129 A steam pipe is subjected to convection on both the inner and outer surfaces. The mathematical formulation of the
problem and expressions for the variation of temperature in the pipe and on the outer surface temperature are to be obtained
for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is
thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe.
Analysis (a) Noting that heat transfer is steady and one-dimensional in the radial
r direction, the mathematical formulation of this problem can be expressed as
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
−k
and
−k
dT ( r1 )
= hi [Ti − T ( r1 )]
dr
Ti
hi
dT (r2 )
= ho [T (r2 ) − To ]
dr
r1
r2
r
To
ho
(b) Integrating the differential equation once with respect to r gives
r
dT
= C1
dr
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,
dT C1
=
dr
r
T (r ) = C1 ln r + C 2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
r = r1:
−k
C1
= hi [Ti − (C1 ln r1 + C2 )]
r1
r = r2:
−k
C1
= ho [(C1 ln r2 + C2 ) − To ]
r2
Solving for C1 and C2 simultaneously gives
C1 =
T0 − Ti
r2
k
k
ln +
+
r1 hi r1 ho r2
and
⎛
⎛
T0 − Ti
k ⎞
k ⎞
⎟ = Ti −
⎜ ln r1 −
⎟
C2 = Ti − C1⎜⎜ ln r1 −
⎟
⎜
⎟
r
k
k
h
r
hi r1 ⎠
2
i
1
⎝
⎝
⎠
ln +
+
r1 hi r1 ho r2
Substituting C1 and C 2 into the general solution and simplifying, we get the variation of temperature to be
r
k
+
k
r1 hi r1
) = Ti +
T (r ) = C1 ln r + Ti − C1 (ln r1 −
r2
k
k
hi r1
ln +
+
r1 hi r1 ho r2
ln
(c) The outer surface temperature is determined by simply replacing r in the relation above by r2. We get
r
k
ln 2 +
r1 hi r1
T ( r2 ) = Ti +
r
k
k
+
ln 2 +
r1 hi r1 ho r2
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2-71
2-130 A spherical liquid nitrogen container is subjected to specified temperature on the inner surface and convection on the
outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of nitrogen are to be
determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal
symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation.
Properties The thermal conductivity of the tank is given to be k = 12 W/m⋅°C. Also, hfg = 198 kJ/kg for nitrogen.
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this
problem can be expressed as
d ⎛ 2 dT ⎞
⎜r
⎟=0
dr ⎠
dr ⎝
and
h
T∞
T ( r1 ) = T1 = −196°C
N2
dT (r2 )
= h[T ( r2 ) − T∞ ]
dr
−k
r1
-196°C
r2
r
(b) Integrating the differential equation once with respect to r gives
r2
dT
= C1
dr
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,
dT C1
=
dr r 2
→
T (r ) = −
C1
+ C2
r
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
r = r1:
T ( r1 ) = −
r = r2:
−k
C1
+ C 2 = T1
r1
⎞
⎛ C
C1
= h⎜⎜ − 1 + C2 − T∞ ⎟⎟
2
r2
⎠
⎝ r2
Solving for C1 and C2 simultaneously gives
C1 =
r2 (T1 − T∞ )
r
k
1− 2 −
r1 hr2
and
C 2 = T1 +
C1
= T1 +
r1
T1 − T∞ r2
r
k r1
1− 2 −
r1 hr2
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T (r ) = −
=
⎛ 1 1⎞
C1
C
+ T1 + 1 = C1 ⎜⎜ − ⎟⎟ + T1 =
r
r1
⎝ r1 r ⎠
T1 − T∞ ⎛ r2 r2 ⎞
⎜ − ⎟⎟ + T1
r
k ⎜⎝ r1
r ⎠
1− 2 −
r1 hr2
(−196 − 20)°C
⎛ 2.1 2.1 ⎞
−
⎜
⎟ + ( −196)°C = 1013(1.05 − 2.1 / r ) − 196
2.1
12 W/m ⋅ °C
r ⎠
2
⎝
−
1−
2
(35 W/m 2 ⋅ °C)(2.1 m)
(c) The rate of heat transfer through the wall and the rate of evaporation of nitrogen are determined from
C
r (T − T∞ )
dT
= −k (4πr 2 ) 21 = −4πkC1 = −4πk 2 1
Q& = −kA
r
k
dx
r
1− 2 −
r1 hr2
= −4π (12 W/m ⋅ °C)
m& =
( 2.1 m)(−196 − 20)°C
= −320,710 W (to the tank since negative)
2.1
12 W/m ⋅ °C
1−
−
2
(35 W/m 2 ⋅ °C)(2.1 m)
Q&
320,700 J/s
=
= 1.62 kg/s
h fg 198,000 J/kg
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2-131 A spherical liquid oxygen container is subjected to specified temperature on the inner surface and convection on the
outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of oxygen are to be
determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal
symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation.
Properties The thermal conductivity of the tank is given to be k = 12 W/m⋅°C. Also, hfg = 213 kJ/kg for oxygen.
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this
problem can be expressed as
d ⎛ 2 dT ⎞
⎜r
⎟=0
dr ⎠
dr ⎝
and
h
T∞
T ( r1 ) = T1 = −183°C
O2
dT (r2 )
= h[T ( r2 ) − T∞ ]
dr
−k
r1
-183°C
r2
r
(b) Integrating the differential equation once with respect to r gives
r2
dT
= C1
dr
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,
dT C1
=
dr r 2
→
T (r ) = −
C1
+ C2
r
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
r = r1:
T ( r1 ) = −
r = r2:
−k
C1
+ C 2 = T1
r1
⎞
⎛ C
C1
= h⎜⎜ − 1 + C2 − T∞ ⎟⎟
2
r2
⎠
⎝ r2
Solving for C1 and C2 simultaneously gives
C1 =
r2 (T1 − T∞ )
r
k
1− 2 −
r1 hr2
and
C 2 = T1 +
C1
= T1 +
r1
T1 − T∞ r2
r
k r1
1− 2 −
r1 hr2
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T (r ) = −
=
⎛ 1 1⎞
C1
C
+ T1 + 1 = C1 ⎜⎜ − ⎟⎟ + T1 =
r
r1
⎝ r1 r ⎠
T1 − T∞ ⎛ r2 r2 ⎞
⎜ − ⎟⎟ + T1
r
k ⎜⎝ r1
r ⎠
1− 2 −
r1 hr2
(−183 − 20)°C
⎛ 2.1 2.1 ⎞
−
⎜
⎟ + ( −183)°C = 951.9(1.05 − 2.1 / r ) − 183
2.1
12 W/m ⋅ °C
r ⎠
2
⎝
−
1−
2
(35 W/m 2 ⋅ °C)(2.1 m)
(c) The rate of heat transfer through the wall and the rate of evaporation of oxygen are determined from
C
r (T − T∞ )
dT
= −k (4πr 2 ) 21 = −4πkC1 = −4πk 2 1
Q& = −kA
r
k
dx
r
1− 2 −
r1 hr2
= −4π (12 W/m ⋅ °C)
m& =
(2.1 m)(−183 − 20)°C
= −301,400 W (to the tank since negative)
2.1
12 W/m ⋅ °C
1−
−
2
(35 W/m 2 ⋅ °C)(2.1 m)
Q&
301,400 J/s
=
= 1.42 kg/s
h fg 213,000 J/kg
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2-132 A large plane wall is subjected to convection, radiation, and specified temperature on the right surface and no
conditions on the left surface. The mathematical formulation, the variation of temperature in the wall, and the left surface
temperature are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal
conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the
wall.
Properties The thermal conductivity and emissivity are given to be k = 8.4 W/m⋅°C and ε = 0.7.
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, and the
mathematical formulation of this problem can be expressed as
d 2T
=0
dx 2
and
−k
dT ( L)
4
4
] = h[T2 − T∞ ] + εσ [(T2 + 273) 4 − Tsurr
]
= h[T ( L) − T∞ ] + εσ [T ( L) 4 − Tsurr
dx
T ( L) = T2 = 45°C
Tsurr
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
45°C
ε
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
Convection at x = L
Temperature at x = L:
4
]
− kC1 = h[T2 − T∞ ] + εσ[(T2 + 273) 4 − Tsurr
4
]} / k
→ C1 = −{h[T2 − T∞ ] + εσ[(T2 + 273) 4 − Tsurr
h
T∞
L
x
T ( L) = C1 × L + C 2 = T2 → C 2 = T2 − C1 L
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
4
]
h[T2 − T∞ ] + εσ[(T2 + 273)4 − Tsurr
(L − x)
k
(14 W/m2 ⋅ °C)(45 − 25)°C + 0.7(5.67×10−8 W/m2 ⋅ K4 )[(318K)4 − (290 K)4 ]
(0.4 − x) m
= 45°C +
8.4 W/m⋅ °C
= 45 + 48.23(0.4 − x)
T ( x) = C1x + (T2 − C1L) = T2 − (L − x)C1 = T2 +
(c) The temperature at x = 0 (the left surface of the wall) is
T (0) = 45 + 48.23(0.4 − 0) = 64.3°C
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2-74
2-133 The base plate of an iron is subjected to specified heat flux on the left surface and convection and radiation on the
right surface. The mathematical formulation, and an expression for the outer surface temperature and its value are to be
determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat
generation. 4 Heat loss through the upper part of the iron is negligible.
Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7.
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire
heat generated in the resistance wires is transferred to the base plate, the heat flux
through the inner surface is determined to be
Q&
1200 W
= 80,0000 W/m 2
q& 0 = 0 =
Abase 150 × 10 − 4 m 2
Tsurr
q
ε
h
T∞
Taking the direction normal to the surface of the wall to be the x direction with x = 0 at
the left surface, the mathematical formulation of this problem can be expressed as
d 2T
dx 2
and
−k
=0
L
x
dT (0)
= q& 0 = 80,000 W/m 2
dx
dT ( L)
4
4
= h[T ( L) − T∞ ] + εσ [T ( L) 4 − Tsurr
] = h[T2 − T∞ ] + εσ [(T2 + 273) 4 − Tsurr
]
dx
(b) Integrating the differential equation twice with respect to x yields
−k
dT
= C1
dx
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
q& 0
k
x = 0:
− kC1 = q& 0 → C1 = −
x = L:
4
− kC1 = h[T2 − T∞ ] + εσ [(T2 + 273) 4 − Tsurr
]
Eliminating the constant C1 from the two relations above gives the following expression for the outer surface temperature T2,
4
h(T2 − T∞ ) + εσ [(T2 + 273) 4 − Tsurr
] = q&0
(c) Substituting the known quantities into the implicit relation above gives
(30 W/m 2 ⋅ °C)(T2 − 26) + 0.7(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(T2 + 273) 4 − 295 4 ] = 80,000 W/m 2
Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above
to be
T2 = 819°C
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2-134 The base plate of an iron is subjected to specified heat flux on the left surface and convection and radiation on the
right surface. The mathematical formulation, and an expression for the outer surface temperature and its value are to be
determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat
generation. 4 Heat loss through the upper part of the iron is negligible.
Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7.
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire
heat generated in the resistance wires is transferred to the base plate, the heat flux
through the inner surface is determined to be
Q&
1500 W
q& 0 = 0 =
= 100,000 W/m 2
Abase 150 × 10 − 4 m 2
Tsurr
q
ε
h
T∞
Taking the direction normal to the surface of the wall to be the x direction with x = 0 at
the left surface, the mathematical formulation of this problem can be expressed as
d 2T
dx 2
and
−k
=0
L
x
dT (0)
= q& 0 = 100,000 W/m 2
dx
dT ( L)
4
4
] = h[T2 − T∞ ] + εσ [(T2 + 273) 4 − Tsurr
]
= h[T ( L) − T∞ ] + εσ [T ( L) 4 − Tsurr
dx
(b) Integrating the differential equation twice with respect to x yields
−k
dT
= C1
dx
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
q& 0
k
x = 0:
− kC1 = q& 0 → C1 = −
x = L:
4
− kC1 = h[T2 − T∞ ] + εσ [(T2 + 273) 4 − Tsurr
]
Eliminating the constant C1 from the two relations above gives the following expression for the outer surface temperature T2,
4
h(T2 − T∞ ) + εσ [(T2 + 273) 4 − Tsurr
] = q&0
(c) Substituting the known quantities into the implicit relation above gives
(30 W/m 2 ⋅ °C)(T2 − 26) + 0.7(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(T2 + 273) 4 − 295 4 ] = 100,000 W/m 2
Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above
to be
T2 = 896°C
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2-76
2-135 The surface and interface temperatures of a resistance wire covered with a plastic layer are to be determined.
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since this twolayer heat transfer problem possesses symmetry about the center line and involves no change in the axial direction, and thus
T = T(r) . 3 Thermal conductivities are constant. 4 Heat generation in the wire is uniform.
Properties It is given that k wire = 18 W/m ⋅ °C and k plastic = 1.8 W/m ⋅ °C .
Analysis Letting TI denote the unknown interface temperature, the mathematical formulation of the heat transfer problem in
the wire can be expressed as
1 d ⎛ dT ⎞ e& gen
=0
⎜r
⎟+
r dr ⎝ dr ⎠
k
with
T (r1 ) = TI
and
T∞
h
dT (0)
=0
dr
Multiplying both sides of the differential equation by r,
rearranging, and integrating give
r1
egen
e& gen r 2
dT
=−
+ C1
→ r
dr
k 2
e& gen
d ⎛ dT ⎞
r
⎜r
⎟=−
dr ⎝ dr ⎠
k
r2
r
(a)
Applying the boundary condition at the center (r = 0) gives
0×
B.C. at r = 0:
e& gen
dT (0)
=−
× 0 + C1
2k
dr
→ C1 = 0
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating,
e& gen
dT
r
=−
dr
2k
T (r ) = −
→
e& gen
4k
r 2 + C2
(b)
Applying the other boundary condition at r = r1 ,
TI = −
B. C. at r = r1 :
e& gen
4k
r12 + C 2
→ C 2 = TI +
e& gen
4k
r12
Substituting this C 2 relation into Eq. (b) and rearranging give
Twire ( r ) = T I +
e& gen
4k wire
( r12 − r 2 )
(c)
Plastic layer The mathematical formulation of heat transfer problem in the plastic can be expressed as
d ⎛ dT ⎞
⎟=0
⎜r
dr ⎝ dr ⎠
with
T (r1 ) = TI
and
−k
dT (r2 )
= h[T ( r2 ) − T∞ ]
dr
The solution of the differential equation is determined by integration to be
r
dT
= C1
dr
→
dT C1
=
dr
r
→
T (r ) = C1 ln r + C 2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
C1 ln r1 + C 2 = T I
r = r1:
r = r2:
−k
→ C 2 = T I − C1 ln r1
C1
= h[(C1 ln r2 + C2 ) − T∞ ]
r2
→
C1 =
T∞ − TI
r
k
ln 2 +
r1 hr2
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2-77
Substituting C1 and C2 into the general solution, the variation of temperature in plastic is determined to be
Tplastic (r ) = C1 ln r + TI − C1 ln r1 = T I +
T∞ − T I
r
ln
r1
r2 k plastic
ln +
r1
hr2
We have already utilized the first interface condition by setting the wire and plastic layer temperatures equal to TI at the
interface r = r1 . The interface temperature TI is determined from the second interface condition that the heat flux in the wire
and the plastic layer at r = r1 must be the same:
− k wire
dTplastic (r1 )
e& gen r1
dTwire (r1 )
= −k plastic
→
= −k plastic
2
dr
dr
T∞ − T I
1
k
r
r
plastic 1
ln 2 +
r1
hr2
Solving for TI and substituting the given values, the interface temperature is determined to be
TI =
=
e&gen r12 ⎛ r2 k plastic ⎞
⎟ + T∞
⎜ ln +
hr2 ⎟⎠
2k plastic ⎜⎝ r1
⎞
1.8 W/m ⋅ °C
(4.8 × 10 6 W/m 3 )(0.003 m) 2 ⎛⎜ 0.007 m
⎟ + 25°C = 255.6°C
+
ln
2
⎟
⎜
2(1.8 W/m ⋅ °C)
⎝ 0.003 m (14 W/m ⋅ °C)(0.007 m) ⎠
Knowing the interface temperature, the temperature at the center line (r = 0) is obtained by substituting the known quantities
into Eq. (c),
Twire (0) = TI +
e&gen r12
4k wire
= 255.6°C +
( 4.8 × 10 6 W/m 3 )(0.003 m) 2
= 256.2°C
4 × (18 W/m ⋅ °C)
Thus the temperature of the centerline will be slightly above the interface temperature.
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2-78
2-136 A cylindrical shell with variable conductivity is subjected to
specified temperatures on both sides. The rate of heat transfer through
the shell is to be determined.
k(T)
Assumptions 1 Heat transfer is given to be steady and onedimensional. 2 Thermal conductivity varies quadratically. 3 There is
no heat generation.
T1
T2
Properties The thermal conductivity is given to be
k (T ) = k 0 (1 + β T 2 ) .
r1
r2
r
Analysis When the variation of thermal conductivity with temperature
k(T) is known, the average value of the thermal conductivity in the
temperature range between T1 and T2 is determined from
T
T2
k avg
T2
∫ k (T )dT = ∫ k (1 + βT )dT =
=
T1
T1
T2 − T1
2
0
β
⎛
⎞ 2
k 0 ⎜T + T 3 ⎟
3
⎝
⎠ T1
T2 − T1
β
⎤
⎡
k 0 ⎢(T2 − T1 ) + T23 − T13 ⎥
3
⎦
⎣
=
T2 − T1
(
T2 − T1
)
⎡ β
⎤
= k 0 ⎢1 + T22 + T1T2 + T12 ⎥
⎣ 3
⎦
(
)
This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal
conductivity k avg equals the rate of heat transfer through the same medium with variable conductivity k(T).
Then the rate of heat conduction through the cylindrical shell can be determined from Eq. 2-77 to be
T − T2
⎤ T − T2
⎡ β
= 2πk 0 ⎢1 + T22 + T1T2 + T12 ⎥ L 1
Q& cylinder = 2πk avg L 1
ln(r2 / r1 )
3
⎦ ln(r2 / r1 )
⎣
(
)
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-77, and
performed the indicated integration.
2-137 Heat is generated uniformly in a cylindrical uranium fuel rod. The temperature difference between the center and the
surface of the fuel rod is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal
conductivity is constant. 4 Heat generation is uniform.
Properties The thermal conductivity of uranium at room temperature
is k = 27.6 W/m⋅°C (Table A-3).
Analysis The temperature difference between the center and the
surface of the fuel rods is determined from
To − Ts =
e&gen ro2
4k
=
Ts
e
D
(4 × 10 7 W/m 3 )(0.005 m) 2
= 9.1°C
4( 27.6 W/m.°C)
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2-79
2-138 A large plane wall is subjected to convection on the inner and outer surfaces. The mathematical formulation, the
variation of temperature, and the temperatures at the inner and outer surfaces to be determined for steady one-dimensional
heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat
generation.
Properties The thermal conductivity is given to be k = 0.77 W/m⋅°C.
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the inner surface, the
mathematical formulation of this problem can be expressed as
d 2T
=0
dx 2
and
k
dT (0)
h1 [T∞1 − T (0)] = − k
dx
−k
h2
T∞2
h1
T∞1
dT ( L)
= h 2 [T ( L) − T∞ 2 ]
dx
L
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
T ( x) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
x = 0:
h1 [T∞1 − (C1 × 0 + C 2 )] = −kC1
x = L:
−kC1 = h2 [(C1 L + C 2 ) − T∞ 2 ]
Substituting the given values, these equations can be written as
8( 22 − C 2 ) = −0.77C1
−0.77C1 = (12)(0.2C1 + C 2 − 8)
Solving these equations simultaneously give
C1 = −38.84
C 2 = 18.26
Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be
T ( x) = 18.26 − 38.84 x
(c) The temperatures at the inner and outer surfaces are
T (0) = 18.26 − 38.84 × 0 = 18.3°C
T ( L) = 18.26 − 38.84 × 0.2 = 10.5°C
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2-80
2-139 A hollow pipe is subjected to specified temperatures at the inner and outer surfaces. There is also heat generation in
the pipe. The variation of temperature in the pipe and the center surface temperature of the pipe are to be determined for
steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is
thermal symmetry about the centerline. 2 Thermal conductivity is constant.
Properties The thermal conductivity is given to be k = 14 W/m⋅°C.
Analysis The rate of heat generation is determined from
W&
W&
25,000 W
e& gen =
=
=
= 26,750 W/m 3
2
V π ( D 2 − D1 2 ) L / 4 π (0.4 m) 2 − (0.3 m) 2 (17 m) / 4
[
]
Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be
expressed as
1 d ⎛ dT ⎞ e& gen
=0
⎟+
⎜r
k
r dr ⎝ dr ⎠
egen
and
T ( r1 ) = T1 = 60°C
T1
T (r2 ) = T2 = 80°C
T2
Rearranging the differential equation
d ⎛ dT ⎞ −e& gen r
r1
r2
=0
⎟=
⎜r
r
k
dr ⎝ dr ⎠
and then integrating once with respect to r,
2
dT − e& gen r
=
+ C1
dr
2k
Rearranging the differential equation again
dT −e& gen r C1
=
+
2k
dr
r
and finally integrating again with respect to r, we obtain
r
T (r ) =
− e& gen r 2
+ C1 ln r + C 2
4k
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
T ( r1 ) =
r = r1:
− e& gen r1 2
T (r2 ) =
4k
+ C1 ln r1 + C 2
− e& gen r2 2
+ C1 ln r2 + C 2
4k
Substituting the given values, these equations can be written as
r = r2:
60 =
− (26,750)(0.15) 2
+ C1 ln(0.15) + C 2
4(14)
80 =
− (26,750)(0.20) 2
+ C1 ln(0.20) + C 2
4(14)
Solving for C1 and C 2 simultaneously gives
C1 = 98.58
C 2 = 257.8
Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be
T (r ) =
− 26,750r 2
+ 98.58 ln r + 257.8 = 257.8 − 477.7r 2 + 98.58 ln r
4(14)
The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is the average of the
inner radius and outer radius.
T (r ) = 257.8 − 477.7(0.175) 2 + 98.58 ln(0.175) = 71.3°C
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2-81
2-140 Heat is generated in a plane wall. Heat is supplied from one side which is insulated while the other side is subjected to
convection with water. The convection coefficient, the variation of temperature in the wall, and the location and the value of
the maximum temperature in the wall are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform.
Analysis (a) Noting that the heat flux and the heat generated will be
transferred to the water, the heat transfer coefficient is determined from
Ts
the Newton’s law of cooling to be
k
Heater
q& s + e& gen L
q& s
h=
Ts − T∞
T∞ , h
(16,000 W/m 2 ) + (10 5 W/m 3 )(0.04 m)
Insulation
= 400 W/m 2 ⋅ °C
=
(90 − 40)°C
e& gen x
(b) The variation of temperature in the wall is in the form of
T(x) = ax2+bx+c. First, the coefficient a is determined as
follows
L
e& gen
d 2T
d 2T
&
k
+ e gen = 0 → k
=−
k
dx 2
dT 2
e& gen 2
e& gen
e& gen
dT
10 5 W/m 3
x + bx + c → a = −
x + b and T = −
=−
=
= −2500°C/m 2
2k
dx
k
2k
2(20 W/m ⋅ °C)
Applying the first boundary condition:
x = 0, T(0) = Ts → c = Ts = 90ºC
As the second boundary condition, we can use either
or
−k
⎞
⎛ e& gen L
1
1
dT
16000 + 10 5 × 0.04 = 1000°C/m
+ b ⎟⎟ = q s → b = q s + e& gen L =
= − q s → k ⎜⎜ −
k
20
k
dx x = L
⎠
⎝
−k
dT
= − h(Ts − T∞ )
dx x =0
(
(
)
)
400
(90 − 40) = 1000°C/m
20
Substituting the coefficients, the variation of temperature becomes
k(a×0+b) = h(Ts -T∞) → b =
T ( x) = −2500 x 2 + 1000 x + 90
(c) The x-coordinate of Tmax is xvertex= -b/(2a) = 1000/(2×2500) = 0.2 m = 20 cm. This is outside of the wall boundary, to the
left, so Tmax is at the left surface of the wall. Its value is determined to be
Tmax = T ( L) = −2500 L2 + 1000 L + 90 = −2500(0.04) 2 + 1000(0.04) + 90 = 126 °C
The direction of qs(L) (in the negative x direction) indicates that at x = L the temperature increases in the positive x direction.
If a is negative, the T plot is like in Fig. 1, which shows Tmax at x=L. If a is positive, the T plot could only be like in Fig. 2,
which is incompatible with the direction of heat transfer at the surface in contact with the water. So, temperature distribution
can only be like in Fig. 1, where Tmax is at x=L, and this was determined without using numerical values for a, b, or c.
Slope
qs(L)
Fig. 1
Slope
qs(L)
Fig. 2
qs(0)
qs(0)
Here, heat transfer
and slope are
incompatible
This part could also be answered to without any information about the nature of the T(x) function, using qualitative
arguments only. At steady state, heat cannot go from right to left at any location. There is no way out through the left surface
because of the adiabatic insulation, so it would accumulate somewhere, contradicting the steady state assumption. Therefore,
the temperature must continually decrease from left to right, and Tmax is at x = L.
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2-82
2-141 Heat is generated in a plane wall. The temperature distribution in the wall is given. The surface temperature, the heat
generation rate, the surface heat fluxes, and the relationship between these heat fluxes, the heat generation rate, and the
geometry of the wall are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform.
Analysis (a) The variation of temperature is symmetric about x = 0. The surface temperature is
Ts = T ( L) = T (− L) = a − bL2 = 80°C − (2 × 10 4 °C/m 2 )(0.025 m) 2 = 67.5°C
The plot of temperatures across the wall thickness is given below.
82
T [C]
80
72ºC
78
k
76
e&gen
72ºC
T∞
h
74
72
70
68
-L
66
-0.025
-0.015
-0.005
0.005
0.015
x
L
0.025
x [m]
(b) The volumetric rate of heat generation is
k
d 2T
dx 2
+ e&gen = 0 ⎯
⎯→ e&gen = −k (−2b) = 2(8 W/m ⋅ °C)(2 × 10 4 °C/m 2 ) = 3.2 × 10 5 W/m 3
(c) The heat fluxes at the two surfaces are
q& s ( L) = − k
dT
= − k (−2bL) = 2(8 W/m ⋅ °C)(2 × 10 4 °C/m 2 )(0.025 m) = 8000 W/m 2
dx L
q& s (− L) = − k
dT
= − k [(−2b(− L)] = −2(8 W/m ⋅ °C)(2 × 10 4 °C/m 2 )(0.025 m) = −8000 W/m 2
dx L
(d) The relationship between these fluxes, the heat generation rate and the geometry of the wall is
E& out = E& gen
[q& s ( L) + q& s (− L)]A = e&genV
[q& s ( L) + q& s (− L)]WH = e&gen (2 LWH )
q& s ( L) + q& s (− L) = 2e& gen L
Discussion Note that in this relation the absolute values of heat fluxes should be used. Substituting numerical values gives
1000 W/m2 on both sides of the equation, and thus verifying the relationship.
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2-83
2-142 Steady one-dimensional heat conduction takes place in a long slab. It is to be shown that the heat flux in steady
*
k * ⎛⎜ T + T0 ⎞⎟
operation is given by q& =
ln *
. Also, the heat flux is to be calculated for a given set of parameters.
W ⎜⎝ T + Tw ⎟⎠
Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness.
Analysis The derivation is given as follows
q& = − k
or
Tw
dT
T0
*
∫ T +T
ln(T * + T )
Tw
T0
=−
=−
dT
− k * dT
= *
dx (T + T ) dx
q&
W
∫ dx
k* 0
q&
k*
(W − 0)
⎛T * +T ⎞
W
ln⎜ * w ⎟ = − q& *
⎜ T +T ⎟
k
0 ⎠
⎝
*
k * ⎛⎜ T + T0 ⎞⎟
q& =
ln *
W ⎜⎝ T + Tw ⎟⎠
The heat flux for the given values is
q& =
*
k * ⎛⎜ T + T0 ⎞⎟ 7 × 10 4 W/m ⎛ (1000 − 600)K ⎞
⎟⎟ = −1.42 × 10 5 W/m 2
=
ln *
ln⎜⎜
−
(
1000
400
)
K
W ⎜⎝ T + Tw ⎟⎠
0.2 m
⎝
⎠
2-143 A spherical ball in which heat is generated uniformly is exposed to iced-water. The temperatures at the center and at
the surface of the ball are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional, and there is thermal symmetry about the center point. 3 Thermal conductivity is constant. 4 Heat generation is
uniform.
Properties The thermal conductivity is given to be k = 45 W/m⋅°C.
Analysis The temperatures at the center and at the surface of the ball
are determined directly from
Ts = T∞ +
T0 = Ts +
e&gen ro
3h
e&gen ro2
6k
6
= 0°C +
(4.2 × 10 W/m )(0.12 m)
= 140°C +
D
3
3(1200 W/m 2 .°C)
= 140°C
h
T∞
e&gen
(4.2 × 10 6 W/m 3 )(0.12 m) 2
= 364°C
6(45 W/m.°C)
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2-84
2-144 A 10-m tall exhaust stack discharging exhaust gases at a rate of 1.2 kg/s is subjected to solar radiation and convection
at the outer surface. The variation of temperature in the exhaust stack and the inner surface temperature of the exhaust stack
are to be determined.
Assumptions 1 Heat conduction is steady and one-dimensional and there is thermal symmetry about the centerline. 2
Thermal properties are constant. 3 There is no heat generation in the pipe.
Properties The constant pressure specific heat of exhaust gases is given to be 1600 J/kg · °C and the pipe thermal
conductivity is 40 W/m · K. Both the emissivity and solar absorptivity of the exhaust stack outer surface are 0.9.
Analysis The outer and inner radii of the pipe are
r2 = 1 m / 2 = 0.5 m
r1 = 0.5 m − 0.1 m = 0.4 m
The outer surface area of the exhaust stack is
As , 2 = 2π r2 L = 2π (0.5 m)(10 m) = 31.42 m 2
The rate of heat loss from the exhaust gases in the exhaust stack can be determined from
Q& loss = m& c p (Tin − Tout ) = (1.2 kg/s)(1600 J/kg ⋅ °C)(30) °C = 57600 W
The heat loss on the outer surface of the exhaust stack by radiation and convection can be expressed as
Q& loss
4
] − α s q& solar
= h [T (r2 ) − T∞ ] + εσ [T (r2 ) 4 − Tsurr
As , 2
57600 W
31.42 m 2
= (8 W/m 2 ⋅ K )[T (r2 ) − (27 + 273)] K
+ (0.9)(5.67 × 10 −8 W/m 2 ⋅ K 4 )[T (r2 ) 4 − (27 + 273) 4 ] K 4 − (0.9)(150 W/m 2 )
Copy the following line and paste on a blank EES screen to solve the above equation:
57600/31.42=8*(T_r2-(27+273))+0.9*5.67e-8*(T_r2^4-(27+273)^4)-0.9*150
Solving by EES software, the outside surface temperature of the furnace front is
T (r2 ) = 412.7 K
(a) For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
and
−k
Q&
dT ( r1 ) Q& loss
=
= loss
2π r1 L
dr
As ,1
T (r2 ) = 412.7 K
(heat flux at the inner exhaust stack surface)
(outer exhaust stack surface temperature)
Integrating the differential equation once with respect to r gives
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2-85
dT C1
=
r
dr
Integrating with respect to r again gives
T (r ) = C1 ln r + C 2
where C1 and C 2 are arbitrary constants. Applying the boundary conditions gives
r = r1 :
dT (r1 )
C
1 Q& loss
=−
= 1
dr
k 2π r1 L r1
r = r2 :
T (r2 ) = −
1 Q& loss
ln r2 + C 2
2π kL
→
→
C1 = −
C2 =
1 Q& loss
2π kL
1 Q& loss
ln r2 + T (r2 )
2π kL
Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be
1 Q& loss
1 Q& loss
ln r +
ln r2 + T (r2 )
2π kL
2π kL
1 Q& loss
=−
ln(r / r2 ) + T (r2 )
2π kL
T (r ) = −
(b) The inner surface temperature of the exhaust stack is
1 Q& loss
ln(r1 / r2 ) + T ( r2 )
2π kL
1
57600 W
⎛ 0.4 ⎞
=−
ln⎜
⎟ + 412.7 K
2π (40 W/m ⋅ K )(10 m) ⎝ 0.5 ⎠
= 417.7 K = 418 K
T (r1 ) = −
Discussion There is a temperature drop of 5 °C from the inner to the outer surface of the exhaust stack.
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2-86
Fundamentals of Engineering (FE) Exam Problems
2-145 The heat conduction equation in a medium is given in its simplest form as
1 d ⎛ dT ⎞
⎟ + e& gen = 0 . Select the wrong
⎜ rk
r dr ⎝ dr ⎠
statement below.
(a) the medium is of cylindrical shape.
(b) the thermal conductivity of the medium is constant.
(c) heat transfer through the medium is steady.
(d) there is heat generation within the medium.
(e) heat conduction through the medium is one-dimensional.
Answer (b) thermal conductivity of the medium is constant
2-146 Heat is generated in a long 0.3-cm-diameter cylindrical electric heater at a rate of 180 W/cm3. The heat flux at the
surface of the heater in steady operation is
(a) 12.7 W/cm2
(b) 13.5 W/cm2 (c) 64.7 W/cm2
(d) 180 W/cm2
(e) 191 W/cm2
Answer (b) 13.5 W/cm2
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
"Consider a 1-cm long heater:"
L=1 [cm]
e=180 [W/cm^3]
D=0.3 [cm]
V=pi*(D^2/4)*L
A=pi*D*L "[cm^2]”
Egen=e*V "[W]"
Qflux=Egen/A "[W/cm^2]"
“Some Wrong Solutions with Common Mistakes:”
W1=Egen "Ignoring area effect and using the total"
W2=e/A "Threating g as total generation rate"
W3=e “ignoring volume and area effects”
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2-87
2-147 Heat is generated in a 10-cm-diameter spherical radioactive material whose thermal conductivity is 25 W/m.°C
uniformly at a rate of 15 W/cm3. If the surface temperature of the material is measured to be 120°C, the center temperature of
the material during steady operation is
(a) 160°C
(b) 205°C
(c) 280°C
(d) 370°C
(e) 495°C
Answer (d) 370°C
D=0.10
Ts=120
k=25
e_gen=15E+6
T=Ts+e_gen*(D/2)^2/(6*k)
“Some Wrong Solutions with Common Mistakes:”
W1_T= e_gen*(D/2)^2/(6*k) "Not using Ts"
W2_T= Ts+e_gen*(D/2)^2/(4*k) "Using the relation for cylinder"
W3_T= Ts+e_gen*(D/2)^2/(2*k) "Using the relation for slab"
2-148 Consider a medium in which the heat conduction equation is given in its simplest form as
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎜r
⎟=
∂r ⎠ α ∂t
r 2 ∂r ⎝
(a) Is heat transfer steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or variable?
(e) Is the medium a plane wall, a cylinder, or a sphere?
(f) Is this differential equation for heat conduction linear or nonlinear?
Answers: (a) transient, (b) one-dimensional, (c) no, (d) constant, (e) sphere, (f) linear
2-149 An apple of radius R is losing heat steadily and uniformly from its outer surface to the ambient air at temperature T∞
with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are absolute
temperatures). Also, heat is generated within the apple uniformly at a rate of e& gen per unit volume. If Ts denotes the outer
surface temperature, the boundary condition at the outer surface of the apple can be expressed as
(a) − k
(c) k
dT
dT
4
4
) + e& gen
= h(Ts − T∞ ) + εσ (Ts4 − Tsurr
= h(Ts − T∞ ) + εσ (Ts4 − Tsurr
) (b) − k
dr r = R
dr r = R
dT
4
)
= h(Ts − T∞ ) + εσ (Ts4 − Tsurr
dr r = R
(d) k
4πR 3 / 3
dT
4
)+
= h(Ts − T∞ ) + εσ (Ts4 − Tsurr
e& gen
dr r = R
4πR 2
(e) None of them
Answer: (a) − k
dT
4
= h(Ts − T∞ ) + εσ (Ts4 − Tsurr
)
dr r = R
Note: Heat generation in the medium has no effect on boundary conditions.
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2-150 A furnace of spherical shape is losing heat steadily and uniformly from its outer surface of radius R to the ambient air
at temperature T∞ with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are
absolute temperatures). If To denotes the outer surface temperature, the boundary condition at the outer surface of the furnace
can be expressed as
(a) − k
(c) k
dT
dT
4
4
= h(To − T∞ ) + εσ (To4 − Tsurr
) (b) − k
= h(To − T∞ ) − εσ (To4 − Tsurr
)
dr r = R
dr r = R
dT
4
= h(To − T∞ ) + εσ (To4 − Tsurr
)
dr r = R
(e) k (4πR 2 )
(d) k
dT
4
= h(To − T∞ ) − εσ (To4 − Tsurr
)
dr r = R
dT
4
= h(To − T∞ ) + εσ (To4 − Tsurr
)
dr r = R
Answer (a) − k
dT
4
)
= h(To − T∞ ) + εσ (To4 − Tsurr
dr r = R
2-151 A plane wall of thickness L is subjected to convection at both surfaces with ambient temperature T∞1 and heat transfer
coefficient h1 at inner surface, and corresponding T∞2 and h2 values at the outer surface. Taking the positive direction of x to
be from the inner surface to the outer surface, the correct expression for the convection boundary condition is
dT (0)
= h1 [T (0) − T∞1 )]
dx
dT (0)
= h1 [T∞1 − T∞ 2 )]
(c) − k
dx
(a) k
Answer (a) k
dT ( L)
= h2 [T ( L) − T∞ 2 )]
dx
dT ( L)
= h2 [T∞1 − T∞ 2 )]
(d) − k
dx
(b) k
(e) None of them
dT (0)
= h1 [T (0) − T∞1 )]
dx
2-152 Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of
uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the
variation of temperature in the direction of heat transfer be linear is
(a) plane wall
(b) cylindrical shell (c) spherical shell
(d) all of them
(e) none of them
Answer (a) plane wall
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2-89
2-153 Consider a large plane wall of thickness L, thermal conductivity k, and surface area A. The left surface of the wall is
exposed to the ambient air at T∞ with a heat transfer coefficient of h while the right surface is insulated. The variation of
temperature in the wall for steady one-dimensional heat conduction with no heat generation is
(a) T ( x) =
h( L − x )
T∞
k
(b) T ( x) =
⎛ xh ⎞
(c) T ( x) = ⎜1 − ⎟T∞
k ⎠
⎝
k
T∞
h( x + 0.5 L)
(d) T ( x) = ( L − x)T∞
(e) T ( x) = T∞
Answer (e) T ( x) = T∞
2-154 The variation of temperature in a plane wall is determined to be T(x)=52x+25 where x is in m and T is in °C. If the
temperature at one surface is 38ºC, the thickness of the wall is
(a) 0.10 m
(b) 0.20 m
(c) 0.25 m
(d) 0.40 m
(e) 0.50 m
Answer (c) 0.25 m
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
38=52*L+25
2-155 The variation of temperature in a plane wall is determined to be T(x)=110 - 60x where x is in m and T is in °C. If the
thickness of the wall is 0.75 m, the temperature difference between the inner and outer surfaces of the wall is
(a) 30ºC
(b) 45ºC
(c) 60ºC
(d) 75ºC
(e) 84ºC
Answer (b) 45ºC
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
T1=110 [C]
L=0.75
T2=110-60*L
DELTAT=T1-T2
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2-90
2-156 The temperatures at the inner and outer surfaces of a 15-cm-thick plane wall are measured to be 40ºC and 28ºC,
respectively. The expression for steady, one-dimensional variation of temperature in the wall is
(a) T ( x) = 28 x + 40
(b) T ( x) = −40 x + 28
(d) T ( x) = −80 x + 40
(e) T ( x) = 40 x − 80
(c) T ( x) = 40 x + 28
Answer (d) T ( x) = −80 x + 40
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
T1=40 [C]
T2=28 [C]
L=0.15 [m]
"T(x)=C1x+C2"
C2=T1
T2=C1*L+T1
2-157 Heat is generated in a 3-cm-diameter spherical radioactive material uniformly at a rate of 15 W/cm3. Heat is dissipated
to the surrounding medium at 25°C with a heat transfer coefficient of 120 W/m2⋅°C. The surface temperature of the material
in steady operation is
(a) 56°C
(b) 84°C
(c) 494°C
(d) 650°C
(e) 108°C
Answer (d) 650°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
h=120 [W/m^2-C]
e=15 [W/cm^3]
Tinf=25 [C]
D=3 [cm]
V=pi*D^3/6 "[cm^3]"
A=pi*D^2/10000 "[m^2]"
Egen=e*V "[W]"
Qgen=h*A*(Ts-Tinf)
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2-91
2-158 Which one of the followings is the correct expression for one-dimensional, steady-state, constant thermal conductivity
heat conduction equation for a cylinder with heat generation?
(a)
∂T
1 ∂ ⎛ ∂T ⎞
⎜ rk
⎟ + e& gen = ρc
r ∂r ⎝ ∂r ⎠
∂t
(b)
1 ∂ ⎛ ∂T ⎞ e& gen
1 ∂T
=
⎜r
⎟+
k
α ∂t
r ∂r ⎝ ∂r ⎠
(d)
1 d ⎛ dT ⎞ e& gen
=0
⎜r
⎟+
r dr ⎝ dr ⎠
k
(e)
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
Answer (d)
(c)
1 ∂ ⎛ ∂T ⎞ 1 ∂T
⎜r
⎟=
r ∂r ⎝ ∂r ⎠ α ∂t
1 d ⎛ dT ⎞ e& gen
=0
⎜r
⎟+
r dr ⎝ dr ⎠
k
2-159 A solar heat flux q& s is incident on a sidewalk whose thermal conductivity is k, solar absorptivity is αs and convective
heat transfer coefficient is h. Taking the positive x direction to be towards the sky and disregarding radiation exchange with
the surroundings surfaces, the correct boundary condition for this sidewalk surface is
dT
= α s q& s
dx
(d) h(T − T∞ ) = α s q& s
(a) − k
Answer (c) − k
dT
= h(T − T∞ )
dx
(e) None of them
(b) − k
(c) − k
dT
= h(T − T∞ ) − α s q& s
dx
dT
= h(T − T∞ ) − α s q& s
dx
2-160 Hot water flows through a PVC (k = 0.092 W/m⋅K) pipe whose inner diameter is 2 cm and outer diameter is 2.5 cm.
The temperature of the interior surface of this pipe is 50oC and the temperature of the exterior surface is 20oC. The rate of
heat transfer per unit of pipe length is
(a) 77.7 W/m
(b) 89.5 W/m
(c) 98.0 W/m
(d) 112 W/m
(e) 168 W/m
Answer (a) 77.7 W/m
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
do=2.5 [cm]
di=2.0 [cm]
k=0.092 [W/m-C]
T2=50 [C]
T1=20 [C]
Q=2*pi*k*(T2-T1)/LN(do/di)
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2-92
2-161 The thermal conductivity of a solid depends upon the solid’s temperature as k = aT + b where a and b are constants.
The temperature in a planar layer of this solid as it conducts heat is given by
(a) aT + b = x + C2
(b) aT + b = C1x2 + C2
(c) aT2 + bT = C1x + C2
2
2
(d) aT + bT = C1x + C2 (e) None of them
Answer (c) aT2 + bT = C1x + C2
2-162 Harvested grains, like wheat, undergo a volumetric exothermic reaction while they are being stored. This heat
generation causes these grains to spoil or even start fires if not controlled properly. Wheat (k = 0.5 W/m⋅K) is stored on the
ground (effectively an adiabatic surface) in 5-m thick layers. Air at 22°C contacts the upper surface of this layer of wheat
with h = 3 W/m2⋅K. The temperature distribution inside this layer is given by
T − Ts
⎛x⎞
= 1− ⎜ ⎟
T0 − T s
⎝L⎠
2
where Ts is the upper surface temperature, T0 is the lower surface temperature, x is measured upwards from the ground, and L
is the thickness of the layer. When the temperature of the upper surface is 24oC, what is the temperature of the wheat next to
the ground?
(a) 42oC
(b) 54oC
(c) 58oC
(d) 63oC
(e) 76°C
o
Answer (b) 54 C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
k=0.5 [W/m-K]
h=3 [W/m2-K]
L=5[m]
Ts=24 [C]
Ta=22 [C]
To=(h*L/(2*k))*(Ts-Ta)+Ts
2-163 The conduction equation boundary condition for an adiabatic surface with direction n being normal to the surface is
(a) T = 0
(b) dT/dn = 0
(c) d2T/dn2 = 0
(d) d3T/dn3 = 0
(e) -kdT/dn = 1
Answer (b) dT/dn = 0
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2-93
2-164 Heat is generated uniformly in a 4-cm-diameter, 12-cm-long solid bar (k = 2.4 W/m⋅ºC). The temperatures at the
center and at the surface of the bar are measured to be 210ºC and 45ºC, respectively. The rate of heat generation within the
bar is
(a) 597 W
(b) 760 W
b) 826 W
(c) 928 W
(d) 1020 W
Answer (a) 597 W
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
D=0.04 [m]
L=0.12 [m]
k=2.4 [W/m-C]
T0=210 [C]
T_s=45 [C]
T0-T_s=(e*(D/2)^2)/(4*k)
V=pi*D^2/4*L
E_dot_gen=e*V
"Some Wrong Solutions with Common Mistakes"
W1_V=pi*D*L "Using surface area equation for volume"
W1_E_dot_gen=e*W1_V
T0=(W2_e*(D/2)^2)/(4*k) "Using center temperature instead of temperature difference"
W2_Q_dot_gen=W2_e*V
W3_Q_dot_gen=e "Using heat generation per unit volume instead of total heat generation as the result"
2-165 .... 2-167 Design and Essay Problems

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