ch02
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ch02


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PROBLEM 2.1
KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape.
FIND: Sketch temperature distribution and explain shape of curve.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) No
internal heat generation.
ANALYSIS: Performing an energy balance on the object according to Eq. 1.11a, \ufffd \ufffd ,E Ein out\u2212 = 0 it
follows that
\ufffd \ufffdE E qin out x\u2212 =
and that q q xx x\u2260 \ufffd \ufffd. That is, the heat rate within the object is everywhere constant. From Fourier\u2019s
law,
q kA dT
dxx x
= \u2212 ,
and since qx and k are both constants, it follows that
A dT
dx
Constant.x =
That is, the product of the cross-sectional area normal to the heat rate and temperature gradient
remains a constant and independent of distance x. It follows that since Ax increases with x, then
dT/dx must decrease with increasing x. Hence, the temperature distribution appears as shown above.
COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution. (2)
What would the distribution be when T2 > T1? (3) How does the heat flux, \u2032\u2032qx , vary with distance?
PROBLEM 2.2
KNOWN: Hot water pipe covered with thick layer of insulation.
FIND: Sketch temperature distribution and give brief explanation to justify shape.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No
internal heat generation, (4) Insulation has uniform properties independent of temperature and
position.
ANALYSIS: Fourier\u2019s law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the form
q kA dT
dr
k 2 r dT
drr r
= \u2212 = \u2212 \u3c0 \ufffd\ufffd \ufffd
where A r and r = 2\u3c0 \ufffd \ufffd is the axial length of the pipe-insulation system. Recognize that for steady-
state conditions with no internal heat generation, an energy balance on the system requires
\ufffd \ufffd \ufffd \ufffd
.E E since E Ein out g st= = = 0 Hence
qr = Constant.
That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows that
r
dT
dr
Constant.\ufffd
\ufffd\ufffd
\ufffd
\ufffd\ufffd
=
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r,
remains constant throughout the insulation. For our situation, the temperature distribution must
appear as shown in the sketch.
COMMENTS: (1) Note that, while qr is a constant and independent of r, \u2032\u2032qr is not a constant. How
does \u2032\u2032q rr \ufffd \ufffd vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with
increasing radius.
PROBLEM 2.3
KNOWN: A spherical shell with prescribed geometry and surface temperatures.
FIND: Sketch temperature distribution and explain shape of the curve.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial (spherical
coordinates) direction, (3) No internal generation, (4) Constant properties.
ANALYSIS: Fourier\u2019s law, Eq. 2.1, for this one-dimensional, radial (spherical coordinate) system
has the form
( )2r r dT dTq k A k 4 rdr dr\u3c0= \u2212 = \u2212
where Ar is the surface area of a sphere. For steady-state conditions, an energy balance on the system
yields \ufffd \ufffd ,E Ein out= since \ufffd \ufffd .E Eg st= = 0 Hence,
( )in out r rq q q q r .= = \u2260
That is, qr is a constant, independent of the radial coordinate. Since the thermal conductivity is
constant, it follows that
2 dTr Constant.
dr
\uf8ee \uf8f9
=\uf8ef \uf8fa\uf8f0 \uf8fb
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius
squared, r2, remains constant throughout the shell. Hence, the temperature distribution appears as
shown in the sketch.
COMMENTS: Note that, for the above conditions, ( )r rq q r ;\u2260 that is, qr is everywhere constant.
How does \u2032\u2032qr vary as a function of radius?
PROBLEM 2.4
KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature
distribution and heat rate.
FIND: Expression for the thermal conductivity, k.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3)
No internal heat generation.
ANALYSIS: Applying the energy balance, Eq. 1.11a, to the system, it follows that, since
\ufffd \ufffd
,E Ein out=
( )xq Constant f x .= \u2260
Using Fourier\u2019s law, Eq. 2.1, with appropriate expressions for Ax and T, yields
( ) ( )
x x
2 3
dTq k A
dx
d K6000W=-k 1-x m 300 1 2x-x .
dx m
= \u2212
\uf8ee \uf8f9
\u22c5 \u22c5 \u2212\uf8ef \uf8fa\uf8f0 \uf8fb
Solving for k and recognizing its units are W/m\u22c5K,
( ) ( ) ( )( )22
-6000 20k= .
1 x 2 3x1-x 300 2 3x
=\uf8ee \uf8f9
\u2212 +
\u2212 \u2212\uf8ef \uf8fa\uf8f0 \uf8fb
<
COMMENTS: (1) At x = 0, k = 10W/m\u22c5K and k \u2192 \u221e as x \u2192 1. (2) Recognize that the 1-D
assumption is an approximation which becomes more inappropriate as the area change with x, and
hence two-dimensional effects, become more pronounced.
PROBLEM 2.5
KNOWN: End-face temperatures and temperature dependence of k for a truncated cone.
FIND: Variation with axial distance along the cone of q q k, and dT / dx.x x, ,\u2032\u2032
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in x (negligible temperature gradients along y),
(2) Steady-state conditions, (3) Adiabatic sides, (4) No internal heat generation.
ANALYSIS: For the prescribed conditions, it follows from conservation of energy, Eq. 1.11a, that
for a differential control volume, \ufffd \ufffd .E E or q qin out x x+dx= = Hence
qx is independent of x.
Since A(x) increases with increasing x, it follows that \u2032\u2032 =q q A xx x / \ufffd \ufffd decreases with increasing x.
Since T decreases with increasing x, k increases with increasing x. Hence, from Fourier\u2019s law, Eq.
2.2,
\u2032\u2032 = \u2212q k dT
dxx
,
it follows that | dT/dx | decreases with increasing x.
PROBLEM 2.6
KNOWN: Temperature dependence of the thermal conductivity, k(T), for heat transfer through a
plane wall.
FIND: Effect of k(T) on temperature distribution, T(x).
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) No internal heat
generation.
ANALYSIS: From Fourier\u2019s law and the form of k(T),
\u2032\u2032 = \u2212 = \u2212 +q k dT
dx
k aT dT
dxx o
\ufffd \ufffd . (1)
The shape of the temperature distribution may be inferred from knowledge of d2T/dx2 = d(dT/dx)/dx.
Since \u2032\u2032qx is independent of x for the prescribed conditions,
dq
dx
-
d
dx
k aT dT
dx
k aT d T
dx
a
dT
dx
x
o
o
2
2
\u2032\u2032
= +
\ufffd
\ufffd\ufffd
\ufffd
\ufffd\ufffd
=
\u2212 + \u2212
\ufffd
\ufffd\ufffd
\ufffd
\ufffd\ufffd
=
\ufffd \ufffd
\ufffd \ufffd
0
0
2
.
Hence,
d T
dx
-a
k aT
dT
dx
 where 
k aT = k > 0
dT
dx
2
2
o
o
=
+
\ufffd
\ufffd\ufffd
\ufffd
\ufffd\ufffd
+
\ufffd
\ufffd\ufffd
\ufffd
\ufffd\ufffd
>
	
\ufffd
\ufffd
\ufffd
2
2
0
from which it follows that for
a > 0: d T / dx < 02 2
a = 0: d T / dx 02 2 =
a < 0: d T / dx > 0.2 2
COMMENTS: The shape of the distribution could also be inferred from Eq. (1). Since T decreases
with increasing x,
a > 0: k decreases with increasing x = > | dT/dx | increases with increasing x
a = 0: k = ko = > dT/dx is constant
a < 0: k increases with increasing x = > | dT/dx | decreases with increasing x.
PROBLEM 2.7
KNOWN: Thermal conductivity and thickness of a one-dimensional system with no internal heat
generation and steady-state conditions.
FIND: Unknown surface temperatures, temperature gradient or heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat flow, (2) No internal heat generation, (3) Steady-state
conditions, (4) Constant properties.
ANALYSIS: The rate equation and temperature gradient for this system are
1 2
x
dT dT T T
q k and .
dx dx L
\u2212
\u2032\u2032 = \u2212 = (1,2)
Using Eqs. (1) and (2), the unknown quantities can be determined.
(a) ( )400 300 KdT 200 K/m
dx 0.5m
\u2212
= =
2
x
W K
q 25 200 5000 W/m .
m K m
\u2032\u2032 = \u2212 × = \u2212
\u22c5
<
(b) 2x
W K
q 25 250 6250 W/m
m K m
\u2032\u2032 = \u2212 × \u2212 =
\u22c5
\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb
2 1
dT K
T T L 1000 C-0.5m -250
dx m
= \u2212 =
\uf8ee \uf8f9 \uf8ee